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  StyleBox["\n\nRen\[EAcute] Schoof\[CloseCurlyQuote]s Algorithm\n for \
Determining the Order of the Group of Points\n  on an Elliptic Curve over a \
Finite Field\n\n",
    FontSize->18],
  "\nJohn J. McGee\n\n\nThesis submitted to the faculty of the Virginia \
Polytechnic Institute and State University \nin partial fulfillment of the \
requirements for the degree of\n\nMaster of Science\nin\nMathematics\n\n\n\n\n\
\nDr. Ezra Brown, Chair\nDr. Charles Parry\nDr. Michael Williams\n\n\nApril \
25, 2006\nBlacksburg, Virginia\n\nKeywords: Elliptic Curve, Schoof, \
Cryptography\n"
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Cell[TextData[{
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  StyleBox["\n\n",
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  "\nJohn McGee\n\nABSTRACT\n\n\nElliptic curves have a rich mathematical \
history dating back to Diophantus (c. 250 C.E.), who used a form of these \
cubic equations to find right triangles of integer area with rational sides.  \
 In more recent times the deep mathematics of elliptic curves was used by \
Andrew Wiles et. al., to construct a proof of Fermat's last theorem, a \
problem which challenged mathematicians for more than 300 years.  In \
addition, elliptic curves over finite fields find practical application in \
the areas of cryptography and coding theory.  For such problems, knowing the \
order of the group of points satisfying the elliptic curve equation is \
important to the security of these applications.  In 1985 Ren\[EAcute] Schoof \
published a paper [5] describing a polynomial time algorithm for solving this \
problem.  In this thesis we explain some of the key mathematical principles \
that provide the basis for Schoof's method.  We also present an \
implementation of Schoof's algorithm as a collection of ",
  StyleBox["Mathematica",
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  " functions.  The operation of each algorithm is illustrated by way of \
numerical examples.                                                        "
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  "Chapter 1 - Introduction ",
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..............\t  ", "Leader"],
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  "\n\t1.1 Background \
..............................................................................\
..\t  1\n\t1.2 When is f(x,y) an Elliptic Curve? \
..............................................",
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....................................\t",
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",
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",
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.............................................\t  8\n\t2.2 The Euclidean \
Algorithm ...........................................................\t",
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.........................................",
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  "\n\t2.4 Finding the modular inverse \
.....................................................\t",
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  "\n\t\tExample 3 - Multiplicative Inverse (mod ",
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  "\n\t2.5 Modular Exponentiation \
..........................................................",
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..................................",
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  "\n\t2.6 Square roots modulo ",
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  "\n\t2.7 Shanks-Tonelli Modular Square Root Algorithm \
......................",
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  "\n\t\tExample 5 - Computing Square Roots Modulo ",
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  " ................",
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............................................",
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  "\nChapter 3 - Arithmetic of Elliptic Curves over ",
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  "\n\t\tExample 7 - Arithmetic in ",
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  "\nChapter 4 - Computing the Order of the Group ",
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  "\n\t4.1 A direct method of computing ",
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  "\n\t4.2 Overview of Schoof's Algorithm \
...............................................",
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  "\n\t4.3 Hasse's Theorem \
.........................................................................",
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  "\n\t4.5 Baby Step, Giant Step Method \
................................................",
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  "\n\t\tExample 8 - Determining Group Order using Hasse's\n\t\t\tTheorem \
.................................................................",
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  "\nChapter 5 - Schoof's Algorithm Implementation \
........................................ ",
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  "\n\t5.3 The Division Polynomials \
.........................................................",
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.............................................",
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  "\n\t\tExample 10 - Computation of the Division Polynomials .... ",
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  " with the Division Polynomials ...........................",
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  "\n\t5.6 The Frobenius Endomorphism \
................................................",
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  "\n\t5.7 The Characteristic Equation of the Frobenius \
............................",
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  "\n\t5.8 Schoof's Algorithm: Case One \
.................................................",
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  "\n\t5.9 Schoof Equation (17) \
................................................................",
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  "\n\t5.10 Schoof Equation (18) \
..............................................................",
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  "\n\t5.11 Schoof's Algorithm: Case Two \
..............................................",
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  "\n\t5.12 Schoof Equation (19x) \
..........................................................",
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  "\n\t5.13 Schoof Equation (19y) \
..........................................................",
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  "\n\t5.14 Schoof's Algorithm Summary \
...............................................",
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  "\nChapter 6 - Results of Running Schoof's Algorithm \
...................................",
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  "\n\t6.1 A Detailed Example \
.................................................................",
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  "\n\t6.2 Other Experiments \
...................................................................",
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...............................................................",
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  "\nChapter 7 - Applications \
..............................................................................\
",
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  "\n\t7.1 The Elliptic Curve Discrete Log Problem \
..................................",
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  "\n\t7.2 Anomalous Curves and the MOV attack \
...................................",
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  "\nReferences \
..............................................................................\
.......................\t47\nAppendix A - Dictionary of ",
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  " Functions for Elliptic Curves ...",
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  "\nAppendix B - ",
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  " Code for Our Elliptic Curves Functions ......",
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......................................................... ",
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..............................................",
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...................",
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  "\n\tThe Functions that Comprise Schoof's Algorithm \
...........................",
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  "Figure 1",
  " - Ren",
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  " Schoof ......",
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  "\nFigure 2",
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....................................................\t45"
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\"In re mathematica ars propendi pluris facienda est quam solvendi\"
 - Georg Cantor.
\
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  "Consider the following cubic polynomial in ",
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      \(TraditionalForm\`x, y\)]],
  " over the field of real numbers ",
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      \(TraditionalForm\`\[DoubleStruckCapitalR]\)]],
  ":"
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  "Suppose further that the right hand side of equation (1) has distinct \
roots.  Then the graph of this curve is called an elliptic curve.  Elliptic \
curves have a rich mathematical history dating back to Diophantus (c. 200 \
C.E.), who used a form of these cubic equations to find right triangles of \
integer area with rational sides.   In more recent times some deep \
mathematical properties of elliptic curves were used by Andrew Wiles et. al., \
to construct a proof of Fermat's last theorem, a problem that had challenged \
mathematicians for more than 300 years.  The Birch-Swinnerton-Dyer \
conjecture, one of the Clay Math Institute's million dollar problems, is also \
a question about certain mathematical properties of elliptic curves.\n\nIn \
addition, elliptic curves over the finite field ",
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      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " for some large integer ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(q\)\(,\)\)\)]],
  " find practical application in the areas of cryptography and coding \
theory.  One example of this is the Massey-Omura encryption method which \
relies on the difficulty of solving the elliptic curve discrete logarithm \
problem for security.  For such methods, knowing the order of the group of \
points satisfying (1) with coefficients and coordinates in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  ", written as ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  ", is very important because a poor choice of curve parameters can lead to \
a situation that gives a potential eavesdropper the ability to break the code \
in reasonable time.   \n\nIn 1985 Ren\[EAcute] Schoof ( Figure 1) published a \
paper entitled \"Elliptic curves over finite fields and the computation of \
square roots mod ",
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      \(TraditionalForm\`p\)]],
  "\" [5].  His paper describes a polynomial time algorithm for determining \
",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  ".   Refinements to his method by Elkies and Atkin have resulted in \
computer algorithms capable of finding results for elliptic curves over \
fields with orders greater than ",
  Cell[BoxData[
      \(TraditionalForm\`10\^100\)]],
  "[3,6]. \n"
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Cell[TextData[{
  "The purpose of this thesis is to explain the mathematical basis for \
Schoof's algorithm and to provide a ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " reference implementation of it.  In order to achieve this goal we first \
present some background on elliptic curves in the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-",
  StyleBox["y",
    FontSlant->"Italic"],
  " plane.  In particular we will see that the set points on the curve have \
the structure of an algebraic group.  In chapter 2 we review the finite field \
arithmetic needed to work with elliptic curves over finite fields.  In \
chapter 3 we present some basic algorithms for arithmetic in the group of \
points on an elliptic curve over a finite field.  Chapter 4 describes some \
methods for computing the elliptic curve group order, and includes an \
introduction to Schoof's algorithm.  We present the details of Schoof's \
algorithm in chapter 5.   For each algorithm we first give a mathematical \
justification for the method or provide a reference to such.  Next we present \
numerical examples that illustrates the operation of the algorithm. In \
chapter 6 we present results of running Schoof's algorithm against various \
curves.  We conclude in chapter 7 with some applications that motivate \
efficient solutions to the elliptic curve group order problem.  Appendix A \
contains a listing of the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " functions that implement these algorithms, and Appendix B presents the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " code for their implementation."
}], "Text",
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Cell["1.2 When is f(x,y) an Elliptic Curve?", "Subsection",
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Cell[TextData[{
  "Much of the following discussion is based on material presented in \
Lawrence Washington's book \"Elliptic Curves - Number Theory and Cryptography\
\" [7].   An",
  StyleBox[" ",
    FontWeight->"Bold",
    FontSlant->"Italic"],
  StyleBox["elliptic curve",
    FontSlant->"Italic"],
  " is the set of points satisfying a nonsingular cubic polynomial in two \
variables.  If ",
  Cell[BoxData[
      \(TraditionalForm\`K\)]],
  " is a field, then an elliptic curve can be specified as"
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    \(TraditionalForm\`\(\(E : {\((x, y)\) \[Element] K\[Cross]K | 
          f(x, y) \[Element] F[x, y], 
        f(x, y) = 0}\)\(,\)\)\)], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "where ",
  Cell[BoxData[
      \(TraditionalForm\`f(x, y)\)]],
  " is a particular nonsingular polynomial in ",
  Cell[BoxData[
      \(TraditionalForm\`x, y\)]],
  " of degree 3.  A polynomial is nonsingular if it has distinct roots.  If \
the field ",
  Cell[BoxData[
      \(TraditionalForm\`K\)]],
  " has characteristic other than ",
  Cell[BoxData[
      \(TraditionalForm\`2, 3\)]],
  ", then by a transformation of variables it can be shown that ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " has the same behavior as an elliptic curve of the form:"
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    \(TraditionalForm\`E : y\^2 = 
      x\^3 + A\ x + \(\(B\)\(.\)\)\)], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "Equation (3) is called Weierstrass equation of an elliptic curve.\n\n\
Suppose ",
  Cell[BoxData[
      \(TraditionalForm\`a, b, c\)]],
  " are the roots of the right hand side of (3).  Then\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = \(\((x - a)\) \((x - b)\) \((x - c)\) = 
          x\^3 - \((a + b + c)\) x\^2 + \((a\ b + a\ c + b\ c)\)\ x - 
            a\ b\ c\)\)]],
  ".\n\t\nSince the coefficient of ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2\)]],
  " is zero we must have ",
  Cell[BoxData[
      \(TraditionalForm\`a + b + c = 0\)]],
  ".  Suppose that (3) has a double root.  Without loss of generality let the \
double root be ",
  Cell[BoxData[
      \(TraditionalForm\`a = b\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`2  a + c = 0\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`a = \(-c\)/2\)]],
  ", and we have\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = 
        x\^3 - \(3\/4\) \(c\^2\) x - \(1\/4\) c\^3\)]],
  ".\n\t\nPut ",
  Cell[BoxData[
      \(TraditionalForm\`A = \(\(-3\)\/4\) c\^2, B = \(\(-1\)\/4\) c\^3\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + A\ x + B\)]],
  " then\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`4 
           A\^3 = \(\(-\(27\/16\)\) c\^6 = \(-27\) B\^2\)\)]],
  ".\n\t\nHence if (3) has a double root then ",
  Cell[BoxData[
      \(TraditionalForm\`4\ A\^2 + 27\ B\^2 = 0\)]],
  ". The contrapositive gives that (3) has distinct roots only if"
}], "Text",
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Cell[TextData[{
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      \(TraditionalForm\`4\ A\^2 + 27\ B\^3 \[NotEqual] 0\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "The negative of the left hand side of (4) is called the ",
  StyleBox["discriminant",
    FontSlant->"Italic"],
  " of the elliptic curve."
}], "Text",
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Cell[CellGroupData[{

Cell["1.3 Addition of Points on an Elliptic Curve", "Subsection",
  CellTags->"c:4"],

Cell[TextData[{
  "For an elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ", take any two points ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(P\)\(,\)\(Q\)\(\ \ \)\)\)]],
  "that lie on ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ", then by Bezout's Theorem, the line between ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`Q\)]],
  " will intersect the curve ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " at a third point ",
  Cell[BoxData[
      \(TraditionalForm\`R\)]],
  ".  This is illustrated in Figure 2 for the elliptic curve given by ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 - 5\ x - 2\)]],
  ".  Notice that the line between two points would be vertical if the two \
points had the same ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate.  Such a line does not appear to intersect the curve.  In this \
case we define the third point of intersection to be a special point at \
infinity, \[ScriptCapitalO].  This definition is justified based on a \
consideration of the elliptic curve in projective coordinates, where \
\[ScriptCapitalO] is a well-defined point."
}], "Text",
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Cell[TextData[{
  StyleBox["Figure 2 - Plot of the Elliptic Curve ",
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    FontWeight->"Bold"],
  "\n\n",
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Cell[TextData[{
  "Using this idea, we can define an addition operation on the elements of ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ".  We define ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q\)]],
  " as follows.  Let ",
  Cell[BoxData[
      \(TraditionalForm\`R\)]],
  " be the third point of intersection of the line ",
  Cell[BoxData[
      \(TraditionalForm\`P - Q\)]],
  " with the elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q\)]],
  " is the reflection of ",
  Cell[BoxData[
      \(TraditionalForm\`R\)]],
  " about the line of symmetry of ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ", which is the ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  "-axis for curves in Weierstrass form.  If the line ",
  Cell[BoxData[
      \(TraditionalForm\`P - Q\)]],
  " is vertical we define ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q = \[ScriptCapitalO]\)]],
  ".  To add ",
  Cell[BoxData[
      \(TraditionalForm\`P + P\)]],
  ", we find the line tangent to the curve at ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " and intersect it with the curve to arrive at ",
  Cell[BoxData[
      \(TraditionalForm\`R\)]],
  ".\n\nWith these definitions it can be shown that ",
  Cell[BoxData[
      \(TraditionalForm\`\((E, +)\)\)]],
  " actually forms an abelian group with identity \[ScriptCapitalO].  That \
is, the operation is commutative, associative, and every point ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " has an inverse ",
  Cell[BoxData[
      \(TraditionalForm\`\(-P\)\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`P + \((\(-P\))\) = \[ScriptCapitalO]\)]],
  ".   In fact, for an elliptic curve in Weierstrass form if ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((x, y)\)\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`\(-P\) = \((x, \(-y\))\)\)]],
  ", the reflection of ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " about the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-axis.  We have demonstrated that every element has an inverse and the \
proof of commutativity follows directly from the fact that the line through \
",
  Cell[BoxData[
      \(TraditionalForm\`P, Q\)]],
  " is the same as a line through ",
  Cell[BoxData[
      \(TraditionalForm\`Q, P\)]],
  ".   The proof of associativity is nontrivial.  One method of proof is \
given in Washington [7] \[Section] 2.4\n\nWe can also define this addition \
operation using analytic geometry.  Suppose that ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((x\_1, y\_1)\), Q = \((x\_2, y\_2)\)\)]],
  " are two points on the elliptic curve  ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x + B\)]],
  " with distinct ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinates.  Then the slope of the line between the two points is given \
by"
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Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \((y\_2 - y\_1)\)/\((x\_2 - x\_1)\)\)]],
  "."
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  FontSize->14],

Cell[TextData[{
  "The point-slope formula of the line passing through ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q\)]],
  " is given by"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`y - y\_1 = \[Lambda](x - x\_1)\)]],
  "."
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  FontSize->14],

Cell[TextData[{
  "Substituting (6) into the elliptic curve equation (3) yields \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Lambda](x - x\_1) + y\_1)\)\^2 = 
        x\^3 + A\ x + B\)]],
  ".\n\nExpanding and collecting terms in ",
  StyleBox["x",
    FontSlant->"Italic"],
  " gives the following monic polynomial in \[DoubleStruckCapitalF][x]"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[Cell[BoxData[{
    \(TraditionalForm\`x\^3 - \(\[Lambda]\^2\) 
        x\^2 + \((2 \( \[Lambda]\^2\) x\_1 - 2\ \[Lambda]\ y\_1 + A)\) 
        x\), "\[IndentingNewLine]", 
    \(TraditionalForm\`\ \ \ \ \(+\ \((B - \(\[Lambda]\^2\) x\_1\^2 - 
            y\_1\^2 + 2\ \[Lambda]\ \(x\_1\) y\_1)\)\) = 
      0. \)}]]], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "We know that ",
  Cell[BoxData[
      \(TraditionalForm\`x\_1, x\_2\)]],
  " satisfy (7) because ",
  Cell[BoxData[
      \(TraditionalForm\`P\ and\ Q\)]],
  " satisfy both the line and the elliptic curve equations.  So we can factor \
the cubic (7) as\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((x - x\_1)\) \((x - x\_2)\) \((x - x\_3)\) = 
        0\)]],
  ",\n\nwhere ",
  Cell[BoxData[
      \(TraditionalForm\`x\_3\)]],
  " must be the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate of the third point of intersection.  Expanding and collecting \
terms in ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " we obtain"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`x\^3 - \((x\_1 + x\_2 + x\_3)\) 
          x\^2 + \((x\_1\ x\_2 + x\_1\ x\_3 + 
              x\_2\ x\_3)\)\ x - \(x\_1\) \(x\_2\) x\_3\)]],
  "."
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  FontSize->14],

Cell[TextData[{
  "Because (7) and (8)  represent the same polynomial, the coefficients of ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2\)]],
  " must be equal, giving\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(-\[Lambda]\^2\) = \(-\((x\_1 + x\_2 + 
              x\_3)\)\)\)]],
  ".\nHence we can compute the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate of the third point of intersection as"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`x\_3 = \[Lambda]\^2 - x\_1 - x\_2\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "We can compute the corresponding  ",
  StyleBox["y",
    FontSlant->"Italic"],
  "-coordinate using the equation of the line (6) and then negate the result \
to obtain the ",
  StyleBox["y",
    FontSlant->"Italic"],
  "-coordinate of ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q\)]],
  ","
}], "Text",
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  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`y\_3 = \(-\((\[Lambda](x\_3 - x\_1) + y\_1)\)\)\)]],
  " = ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda](x\_1 - x\_3) - y\_1\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "On the other hand, if ",
  Cell[BoxData[
      \(TraditionalForm\`x\_1 = x\_2\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(y\_1\%2\)\(=\)\(y\_2\%2\)\(\ \)\)\)]],
  ", so either ",
  Cell[BoxData[
      \(TraditionalForm\`y\_2 = y\_1\)]],
  " or ",
  Cell[BoxData[
      \(TraditionalForm\`y\_2 = \(-y\_1\)\)]],
  ".  If ",
  Cell[BoxData[
      \(TraditionalForm\`y\_2 = \(-y\_1\)\)]],
  " then the line between ",
  Cell[BoxData[
      \(TraditionalForm\`P\ and\ Q\_2\)]],
  " is vertical, and we define ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q = \[ScriptCapitalO]\)]],
  ", the identity.   Otherwise ",
  Cell[BoxData[
      \(TraditionalForm\`P = Q\)]],
  ", so we want to compute ",
  Cell[BoxData[
      \(TraditionalForm\`P + P = 2\ P\)]],
  ".  To accomplish this we define ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda]\)]],
  " as the slope of the tangent to the curve at ",
  Cell[BoxData[
      \(TraditionalForm\`\((x\_1, y\_1)\)\)]],
  ".   We can compute this slope by implicit differentiation of (3) giving\n\n\
\t",
  Cell[BoxData[
      \(TraditionalForm\`2  y\ d\ y = \((3\ x\^2 + A)\) d\ x\)]],
  "\nso that"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      FormBox[
        RowBox[{"\[Lambda]", "=", 
          RowBox[{
            StyleBox[\(\(d\ y\)\/\(d\ x\)\),
              FontSize->18], "=", 
            StyleBox[\(\(3\ x\_1\^2 + A\)\/\(2\ y\_1\)\),
              FontSize->18]}]}], TraditionalForm]]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "Using this \[Lambda] with equations (9) and (10), along with the fact that \
",
  Cell[BoxData[
      \(TraditionalForm\`x\_1 = x\_2\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`y\_1 = y\_2\)]],
  " gives"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`\((x\_1, y\_1)\) + \((x\_1, 
            y\_1)\) = \((\[Lambda]\^2 - 2\ x\_1, \[Lambda](x\_1 - x\_3) - 
            y\_1)\)\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "Equations (5) through (12) are incorporated into the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function ",
  StyleBox["EcAdd",
    FontSize->12,
    FontWeight->"Bold"],
  ", which adds two points on an elliptic curve over \[DoubleStruckCapitalR] \
or \[DoubleStruckCapitalQ]."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Example 1 - Elliptic Curve Point Addition", "Subsubsection",
  CellTags->"c:5"],

Cell[TextData[{
  "Let an elliptic curve be given by ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 2\ x + 1\)]],
  " over the rational numbers \[DoubleStruckCapitalQ].  Then the discriminant \
is  ",
  Cell[BoxData[
      \(TraditionalForm\`d = \(\(-4\) A\^2 - 
            27  B\^3 = \(\(-4\)*4 - 27 = \(-43\) \[NotEqual] \ 0\)\)\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalQ])\)]],
  " is a valid elliptic curve.  It is easy to check, by substituting the \
point coordinates into the equation for ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ", that the points ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((0, 1)\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`Q = \((1, \(-2\))\)\)]],
  " are elements of ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalQ])\)]],
  ".   Also, the ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " coordinates of ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q\)]],
  " are distinct so we can compute the slope of the line passing through ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q\)]],
  " using (5) so that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \(\(\(-2\) - 1\)\/\(1 - 0\) = \
\(-3\)\)\)]],
  ".\n\t\nThen equations (9) and (10) give\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`x\_3 = \(\((\(-3\))\)\^2 - 0 - 1 = 8\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`y\_3 = \(\(-3\) \((0 - 8)\) - 1 = 23\)\)]],
  ".\n\t\nSo that ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q = \((8, 23)\)\)]],
  ", which is also a point in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalQ])\)]],
  " since ",
  Cell[BoxData[
      \(TraditionalForm\`8\^3 + 2*8 + 1 = \(529 = 23\^2\)\)]],
  ".  We obtain the same result using the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function EcAdd[{2,1},{0,1},{1,-2}] which returns ",
  Cell[BoxData[
      \(TraditionalForm\`{8, 23}\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "Chapter 2 - Arithmetic in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]]
}], "Section",
  CellTags->"c:6"],

Cell[CellGroupData[{

Cell["2.1 Elliptic Curves over Finite Fields", "Subsection",
  CellTags->"c:7"],

Cell[TextData[{
  "Suppose ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " is a finite field of order ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`q = p\^k\)]],
  " for some integer ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " (see, for example Dummit and Foote [2] \[Section]14.3).  Suppose also \
that ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x + B\)]],
  " is an element of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q[x, y]\)]],
  ", which means that ",
  Cell[BoxData[
      \(TraditionalForm\`A, B \[Element] \[DoubleStruckCapitalF]\_q\)]],
  ".  Then, if ",
  Cell[BoxData[
      \(TraditionalForm\`4\ a\^3 + 27\ b\^2 \[NotEqual] 0\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " it can be shown that\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q) = {\((x, 
                  y)\) \[Element] \[DoubleStruckCapitalF]\_q\[Cross]\
\[DoubleStruckCapitalF]\_q | y\^2 = x\^3 + A\ x\  + B}\)]],
  "\n\t \nis an elliptic curve.  Further, equations (5) through (12) of the \
previous section still obey the group axioms when arithmetic is done in the \
finite field, so that ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " is a finite abelian group.  Since we have at most ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  " choices for ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  ", and for each of these at most 2 choices for ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " contains at most ",
  Cell[BoxData[
      \(TraditionalForm\`2\ q\)]],
  " points.  It turns out that the actual bound is closer to ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  ".\n\nBefore we consider this question in detail, we introduce some of the \
number-theoretic functions which are required for computation on elliptic \
curves over a finite field.   For the purpose of our goal, which is to \
determine the group order ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  ", it turns out that it will be sufficient to work with fields of prime \
order, ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".  This allows us to perform arithmetic in the field using modular \
arithmetic in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalZ]/
          p\ \[DoubleStruckCapitalZ]\)]],
  ".  That is, we can perform addition, subtraction and multiplication using \
ordinary integer arithmetic and then reduce each result mod ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " by dividing it by ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " and keeping the remainder.\n\nConsider the following example in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_7\)]],
  ":\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`5*5 = \(25 = \(21 + 4 = 
            3*7 + 4 \[Congruent] 4\ \((mod\ 7)\)\)\)\)]],
  ",\n\t\nso that in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_7\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`5*5 = 4\)]],
  ".\n\nPerforming division in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " is a little bit more complicated.  To compute ",
  Cell[BoxData[
      \(TraditionalForm\`a/b\)]],
  " we need to multiply ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " by the modular inverse of ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  ".   For example, to compute ",
  Cell[BoxData[
      \(TraditionalForm\`4/5 \((mod\ 7)\)\)]],
  " we must first find ",
  Cell[BoxData[
      \(TraditionalForm\`c\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`5\ c \[Congruent] 1\ \((mod\ 7)\)\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`4/5 = 4*c\)]],
  ".   By trial and error we can find that ",
  Cell[BoxData[
      \(TraditionalForm\`5*3 = 15 \[Congruent] 1 \((mod\ 7)\)\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`4/5 \[Congruent] 4*3 \[Congruent] 
        5 \((mod\ 7)\)\)]],
  ".  We will later see how to use the Euclidean algorithm to accomplish this \
for problems involving larger integers.\n\nWe will also have occasion to \
compute ",
  Cell[BoxData[
      \(TraditionalForm\`\(a\^k\)(mod\ p)\)]],
  ".  This can be done by direct computation, for example ",
  Cell[BoxData[
      \(TraditionalForm\`3\^4 = 81 \[Congruent] 4 \((mod\ 7)\)\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`3\^4 = 4\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[DoubleStruckCapitalF]\_7\)\(.\)\)\)]],
  "  Fortunately, for large ",
  Cell[BoxData[
      \(TraditionalForm\`k\ and\ p\)]],
  ", there exists a much more efficient method based on doubling and reducing \
the result mod ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " at each step.  The following sections describe these algorithms."
}], "Text",
  TextJustification->1,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["2.2 The Euclidean Algorithm", "Subsection",
  CellTags->"c:7"],

Cell[TextData[{
  "The Euclidean Algorithm computes the greatest common divisor ",
  Cell[BoxData[
      \(TraditionalForm\`d\)]],
  " of the integers ",
  Cell[BoxData[
      \(TraditionalForm\`a, b\)]],
  ".  It dates back to at least 300 B.C.E., where it appeared in geometric \
form in Euclid's ",
  StyleBox["Elements",
    FontSlant->"Italic"],
  ". The most obvious way to find the greatest common divisor is to \
completely factor both ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " into powers of primes such as\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`a = \(\(p\_1\%\(r\_1\)\) p\_2\%\(r\_2\) ... \) 
          p\_n\%\(r\_n\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`b = \(\(p\_1\%\(s\_1\)\) p\_2\%\(\_\(s\_2\)\) ... \) 
          p\_n\%\(\_\(s\_n\)\)\)]],
  ",\n\t\nwhere ",
  Cell[BoxData[
      \(TraditionalForm\`r\_i\ and\ s\_i\)]],
  " are integers greater than or equal to zero and ",
  Cell[BoxData[
      \(TraditionalForm\`r\_i = 0\)]],
  " if ",
  Cell[BoxData[
      \(TraditionalForm\`p\_i\)]],
  " does not divide ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`s\_i = 0\)]],
  " of ",
  Cell[BoxData[
      \(TraditionalForm\`p\_i\)]],
  " does not divide ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  ".  Then the greatest common divisor ",
  Cell[BoxData[
      \(TraditionalForm\`d\)]],
  " is given by\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`d = \(\(p\_1\%\(t\_1\)\) p\_2\%\(\_\(t\_2\)\) ... \) 
          p\_n\%\(\_\(t\_n\)\)\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`t\_i = min(r\_i, s\_i)\)]],
  ".\n\t\nFor example, if ",
  Cell[BoxData[
      \(TraditionalForm\`a = \(7960 = 2\^3*\ 5\ *199\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(b = \(6580 = 2\^4*3\^4*5\)\)\)\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, b) = \(2\^3*\ 5 = 40\)\)]],
  ".\n\nHowever the integer factoring problem is currently understood to be \
difficult when the integer has no small prime divisors.  The Euclidean \
algorithm is far superior because it can compute the greatest common divisor \
of two integers without factoring them.  Suppose we wish to find the greatest \
common divisor of ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  ", called ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, b)\)]],
  ".   Let ",
  Cell[BoxData[
      \(TraditionalForm\`a > b\)]],
  " and divide ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " by ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " producing\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`a = \(s\_0\) b + r\_0\)]],
  "\twhere ",
  Cell[BoxData[
      \(TraditionalForm\`0 \[LessEqual] r\_0 < b\)]],
  "    (by the division algorithm).\n\t\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`r\_0 = 0\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " divides ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " so ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, b) = b\)]],
  ".  Otherwise, divide ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " by ",
  Cell[BoxData[
      \(TraditionalForm\`r\_0\)]],
  " giving\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`b = \(s\_1\) r\_0 + r\_1\)]],
  "\twhere ",
  Cell[BoxData[
      \(TraditionalForm\`0 \[LessEqual] r\_1 < r\_0\)]],
  ".\n\t\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`r\_1 = 0\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`r\_0\)]],
  " divides ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " and since ",
  Cell[BoxData[
      \(TraditionalForm\`a = \(s\_0\) b + r\_0\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`r\_0\)]],
  " also divides ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " giving ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, b) = r\_0\)]],
  ".  If ",
  Cell[BoxData[
      \(TraditionalForm\`r\_i \[NotEqual] 0\)]],
  ", we can continue the process dividing ",
  Cell[BoxData[
      \(TraditionalForm\`r\_\(i - 1\)\)]],
  " by ",
  Cell[BoxData[
      \(TraditionalForm\`r\_i\)]],
  " giving\n\t",
  Cell[BoxData[{
      \(TraditionalForm\`r\_0 = \(s\_2\) r\_1 + r\_2\), 
    "\[IndentingNewLine]", 
      \(TraditionalForm\` ... \), "\[IndentingNewLine]", 
      \(TraditionalForm\`r\_\(i - 1\) = \(s\_\(i + 1\)\) 
            r\_i + \(\(r\_\(i + 1\)\)\(.\)\)\)}]],
  "\n\t\nEventually we must find ",
  Cell[BoxData[
      \(TraditionalForm\`r\_\(n + 1\) = 0\)]],
  " because at each step ",
  Cell[BoxData[
      \(TraditionalForm\`0 \[LessEqual] r\_i < r\_\(i - 1\)\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`r\_\(n - 1\) = \(s\_\(n + 1\)\) r\_n\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(r\_\(n - 1\), r\_n) = \(\(r\_n\)\(.\)\)\)]],
  "\n\nNote, however, that if ",
  Cell[BoxData[
      \(TraditionalForm\`x = q\ y + r\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x, y) = gcd(y, r)\)]],
  ".  This is true because ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(y, r)\)]],
  " divides both ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`r\)]],
  ", so it divides ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " also, hence ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(y, r)\)]],
  " divides ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x, y)\)]],
  ".  But we can write ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(y, r)\)]],
  " as a linear combination of ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`r\)]],
  " so that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(y, 
          r) = \(u\ y + 
            v\ r = \(u\ y + v(x - q\ y) = \((u - v\ q)\) y + v\ x\)\)\)]],
  ",\n\t\nso ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x, y)\)]],
  " divides ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(y, r)\)]],
  ", hence ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x, y) = gcd(y, r)\)]],
  ".  Applying this to the chain of divisions above gives ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(r\_\(i - 1\), r\_i) = 
        gcd(r\_\(i - 2\), r\_\(i - 1\))\)]],
  ", so in particular ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(r\_0, r\_1) = \(gcd(b, r\_0) = gcd(a, b)\)\)]],
  ".  Therefore the last nonzero divisor ",
  Cell[BoxData[
      \(TraditionalForm\`r\_n = gcd(a, b)\)]],
  ".   Euclid's algorithm for computing the greatest common divisor is \
implemented in the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function ",
  StyleBox["EuclideanAlgorithm",
    FontSize->12,
    FontWeight->"Bold"],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["2.3 The Extended Euclidean Algorithm", "Subsection",
  CellTags->"c:8"],

Cell[TextData[{
  "The Extended Euclidean Algorithm computes the greatest common divisor ",
  Cell[BoxData[
      \(TraditionalForm\`d\)]],
  " of the integers ",
  Cell[BoxData[
      \(TraditionalForm\`a, b\)]],
  " and also computes two integers ",
  Cell[BoxData[
      \(TraditionalForm\`r, s\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`d = r\ a\  + \ s\ b\)]],
  ".  This method provides a fast way to compute multiplicative inverses in \
",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".  The algorithm proceeds as follows.\n\nStarting with ",
  Cell[BoxData[
      \(TraditionalForm\`r\_0 = 1, \ s\_1 = 0\)]],
  " we take ",
  Cell[BoxData[
      \(TraditionalForm\`d\_0 = \(a = \(r\_0\) a + \(s\_0\) b\)\)]],
  ".  For step 1 we take ",
  Cell[BoxData[
      \(TraditionalForm\`r\_1 = 0, \ s\_1 = 1\)]],
  " so we can write ",
  Cell[BoxData[
      \(TraditionalForm\`d\_1 = \(b = \(r\_1\) a + \(s\_1\) b\)\)]],
  ".  At each succeeding step we compute the smallest positive ",
  Cell[BoxData[
      \(TraditionalForm\`d\_i \[Congruent] \(d\_\(\(i\)\(-\)\(2\)\(\ \)\)\)(
          mod\ d\_\(i - 1\))\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`d\_i = d\_\(i - 2\) - k\ d\_\(i - 1\)\)]],
  " for some positive integer ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  ".  The algorithm maintains ",
  Cell[BoxData[
      \(TraditionalForm\`d\_i = r\_i\ a + \(s\_i\) b\)]],
  " at each step so that\n\n\t",
  Cell[BoxData[{
      \(TraditionalForm\`d\_i = \((\(r\_\(i - 2\)\) a + s\_\(i - 2\)\ b)\) - 
          k(\(r\_\(i - 1\)\) a + s\_\(i - 1\)\ b)\), "\[IndentingNewLine]", 
      \(TraditionalForm\`\ \ \ \ \(\(=\)\(\((r\_\(i - 2\) - 
                k\ r\_\(i - 1\))\) 
            a + \((s\_\(i - 2\) - k\ s\_\(i - 1\))\) \(\(b\)\(.\)\)\)\)\)}]],
  "\n\t\nHence, we must have ",
  Cell[BoxData[
      \(TraditionalForm\`r\_i = \(r\_\(i - 2\) - 
            k\ r\_\(i - 1\)\ and\ \ s\_i = 
          s\_\(i - 2\) - k\ s\_\(i - 1\)\)\)]],
  ", which completes the formulation of the recursion definition.  Our ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " implementation is based on  Rosen [4] \[Section]3.3."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Example 2 - The Extended Euclidean Algorithm", "Subsubsection",
  CellTags->"c:9"],

Cell[TextData[{
  "If we can find the prime factorization of two numbers then we can write \
down the greatest common divisor directly.  It is the product of the largest \
prime powers that divide both numbers. For example, let ",
  Cell[BoxData[
      \(TraditionalForm\`a = \(7960 = 2\^3*\ 5\ *199\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \ \)\(b = \(6580 = 2\^4*3\^4*5\)\)\)\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, b) = \(2\^3*\ 5 = 40. \)\)]],
  "  Even for such easy problems, however, the determination of ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, b)\)]],
  " as a linear combination of ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " is best accomplished using the Extended Euclidean algorithm.  Using \
ExtendedEucideanAlgorithm[ ",
  Cell[BoxData[
      \(TraditionalForm\`a, b\)]],
  " ] we find\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(7960, 
          6480) = \(40 = \(-35\)*7960 + 43*6480\)\)]],
  "."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["2.4 Finding the modular inverse", "Subsection",
  CellTags->"c:10"],

Cell[TextData[{
  "We can also use the Euclidean algorithm to find the modular inverse.  The \
extended Euclidean algorithm finds the greatest common divisor of two numbers \
",
  Cell[BoxData[
      \(TraditionalForm\`d = gcd(a, b)\)]],
  ".  It  also computes two numbers ",
  Cell[BoxData[
      \(TraditionalForm\`r, s\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`d = r\ a + s\ b\)]],
  ".  If ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " is relatively prime to ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ", which is always true if ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is a prime and ",
  Cell[BoxData[
      \(TraditionalForm\`1 < a < p\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, p) = 1\)]],
  ".  So to find the modular inverse of ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ", we use the Euclidean algorithm to compute\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(a, p) = \(1 = 
          r\ a + s\ p \[Congruent] r\ a\ \((mod\ p)\)\)\)]],
  ", hence ",
  Cell[BoxData[
      \(TraditionalForm\`a\^\(-1\) \[Congruent] r\ \((mod\ p)\)\)]],
  ".\n\t\nWe need to compute modular inverses in order to perform the \
divisions in the elliptic curve point addition formulas."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell[TextData[{
  "Example 3 - Multiplicative Inverse (mod ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ")"
}], "Subsubsection",
  CellTags->"c:11"],

Cell[TextData[{
  "As an example, lets work over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_19\)]],
  ", the field with 19 elements ",
  Cell[BoxData[
      \(TraditionalForm\`{0, 1,  ... , 18}\)]],
  " with arithmetic modulo 19, a prime.  In a field, every nonzero element \
has a multiplicative inverse, so lets find the inverse of 7.  The Euclidean \
algorithm gives \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(7, 
          19) = \(1 = \(-8\)*7 + 
              3*19 \[Congruent] \(-8\)*7 \((mod\ 19)\)\)\)]],
  ". \n\t\nBut ",
  Cell[BoxData[
      \(TraditionalForm\`\(-8\) \[Congruent] 11 \((mod\ 19)\)\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`11*7 \[Congruent] 1\ \((mod\ 19)\)\)]],
  ".  Hence ",
  Cell[BoxData[
      \(TraditionalForm\`7\^\(-1\) \[Congruent] 11\ \((mod\ 19)\)\)]],
  ".  This is verified by the fact that ",
  Cell[BoxData[
      \(TraditionalForm\`11*7 = \(77 = 
          4*19 + 1 \[Congruent] 1\ \((mod\ 19)\)\)\)]],
  "."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["2.5 Modular Exponentiation", "Subsection",
  CellTags->"c:12"],

Cell[TextData[{
  "This method uses the binary representation of ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " to construct the result.\nStarts with ",
  Cell[BoxData[
      \(TraditionalForm\`a\^1 \[Congruent] a(mod\ p), \ x = 1\)]],
  " then at each iteration ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " we compute\n\n\tIf the ",
  Cell[BoxData[
      \(TraditionalForm\`k\^th\)]],
  " bit of ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " is 1 then  ",
  Cell[BoxData[
      \(TraditionalForm\`x \[Congruent] x*\(\(a\^k\)(mod\ p)\)\)]],
  ".\n\tThen ",
  Cell[BoxData[
      \(TraditionalForm\`a\^\(2  k\) \[Congruent] 
        a\^k*\(\(a\^k\)(mod\ p)\)\)]],
  " for the next iteration.\n\t\nIn this way the arithmetic is done with \
relatively small integers, even though ",
  Cell[BoxData[
      \(TraditionalForm\`a\^n\)]],
  " may have hundreds or thousands of digits.  In fact, at each step of the \
algorithm we multiply two numbers which are less than ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ", so the largest product we ever compute is less than ",
  Cell[BoxData[
      \(TraditionalForm\`p\^2\)]],
  ".  Hence, if the binary representing of ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " requires ",
  Cell[BoxData[
      \(TraditionalForm\`m\)]],
  " bits, then we need no more than ",
  Cell[BoxData[
      \(TraditionalForm\`2  m\)]],
  " bits to store the intermediate results.  On the other hand, if we compute \
",
  Cell[BoxData[
      \(TraditionalForm\`a\^n\)]],
  " directly, and ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " has ",
  Cell[BoxData[
      \(TraditionalForm\`m\)]],
  " bits then we would need  ",
  Cell[BoxData[
      \(TraditionalForm\`n\ m\)]],
  " bits to hold the intermediate result."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Example 4 - Modular Exponentiation", "Subsubsection",
  CellTags->"c:13"],

Cell[TextData[{
  "As an example, let's compute ",
  Cell[BoxData[
      \(TraditionalForm\`\(11\^37\) \((mod\ 97)\)\)]],
  ".   The direct method would first compute ",
  Cell[BoxData[
      \(TraditionalForm\`11\^37 = 
        340039485861577398992406882305761986971\ in\ \
\[DoubleStruckCapitalZ]\)]],
  " and then find ",
  Cell[BoxData[
      \(TraditionalForm\`11\^37 = 
        3505561709913169061777390539234659659*97 + 48\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`11\^37 \[Congruent] 48 \((mod\ \ 97)\)\)]],
  ".   However, in binary ",
  Cell[BoxData[
      FormBox[
        RowBox[{"37", "=", 
          TagBox[
            InterpretationBox[\("100101"\_"2"\),
              37,
              Editable->False],
            (BaseForm[ #, 2]&)]}], TraditionalForm]]],
  ", so we can compute ",
  Cell[BoxData[
      \(TraditionalForm\`\(11\^37\) \((mod\ 97)\)\)]],
  " as follows.\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`11\^37 = \(11\^\(32 + 4 + 1\) = \(11\^\(2\^5 + 2\^2 \
+ 1\) = \(11\^\(\((2\^3 + 1)\) 2\^2 + 1\) = \((\((\((\((11\^2)\)\^2)\)\^2*11)\
\)\^2)\)\^2*11\)\)\)\)]],
  ".\n\t\nBut ",
  Cell[BoxData[
      \(TraditionalForm\`11\^2 \[Congruent] 24 \((mod\ 97)\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`24\^2 \[Congruent] 91 \((mod\ 97)\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`91\^2 \[Congruent] 36 \((mod\ 97)\)\)]],
  " so that\n",
  Cell[BoxData[
      \(TraditionalForm\`11\^37 \[Congruent] \((\((36*11)\)\^2)\)\^2*11 \
\((mod\ 97)\)\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`396 \[Congruent] 8 \((mod\ 97)\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`\((8\^2)\)\^2 \[Congruent] 22 \((mod97)\)\)]],
  ",\nhence ",
  Cell[BoxData[
      \(TraditionalForm\`11\^37 \[Congruent] 22*11 \[Congruent] 
        48 \((mod\ 97)\)\)]],
  ".\n\nNotice that we performed these computations without using any number \
larger than ",
  Cell[BoxData[
      \(TraditionalForm\`97\^2\)]],
  ".   We will see later how this same idea, applied to polynomial arithmetic \
will allow us to compute ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(f(x)\)\^k\) \((mod\ \(g(x)\))\)\)]],
  " in an efficient manner."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "2.6 Square roots modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]]
}], "Subsection",
  CellTags->"c:14"],

Cell[TextData[{
  "One of the steps in Schoof's algorithm requires the solution of the \
congruence ",
  Cell[BoxData[
      \(TraditionalForm\`w\^2 \[Congruent] p\ \((mod\ l)\)\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  ", with ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  " a prime number less than ",
  Cell[BoxData[
      \(TraditionalForm\`\@p\)]],
  ".   In other words we need to find the square root of ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " modulo ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  ".  Unlike the multiplicative inverse problem, this problem does not always \
have a solution.  When it does, we say that ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is a quadratic residue modulo ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  ", else ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is called quadratic nonresidue.  If there is an ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2 \[Congruent] a\ \((mod\ p)\)\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " is a quadratic residue mod ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " and \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`a\^\(\((p - 1)\)/2\) \[Congruent] \((x\^2)\)\^\(\((p \
- 1)\)/2\) \[Congruent] x\^\(p - 1\) \[Congruent] 1 \((mod\ p)\)\)]],
  " by Fermat's little theorem.\n\t\nOtherwise, for all ",
  Cell[BoxData[
      \(TraditionalForm\`i < p\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(i\ , p) = 1\)]],
  ". Then for each ",
  Cell[BoxData[
      \(TraditionalForm\`i\)]],
  " less than ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ",  ",
  Cell[BoxData[
      \(TraditionalForm\`i\ j \[Congruent] a\ \((mod\ p)\)\)]],
  " has a unique solution, which can not be ",
  Cell[BoxData[
      \(TraditionalForm\`j = i\)]],
  ", else ",
  Cell[BoxData[
      \(TraditionalForm\`i\^2 \[Congruent] a(mod\ p)\)]],
  ".  Hence we can group the solutions into ",
  Cell[BoxData[
      \(TraditionalForm\`\((p - 1)\)/2\)]],
  " pairs, each with product congruent to ",
  Cell[BoxData[
      \(TraditionalForm\`a\ \((mod\ p)\)\)]],
  ".  Taking the product of all of the solutions gives:\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`a\^\(\((p - 1)\)/2\) \[Congruent] \(\((p - 
                1)\)!\)\ \((mod\ p)\)\)]],
  ",\n\t\nsince each number less than ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is included exactly once in the product.  Then Wilson's Theorem gives ",
  Cell[BoxData[
      \(TraditionalForm\`\(\((p - 
              1)\)!\) \[Congruent] \(-1\) \((mod\ p)\)\)]],
  ", so that \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`a\^\(\((p - 1)\)/2\) \[Congruent] \(-1\)\ \((mod\ p)\
\)\)]],
  ".\n\t\nNote that a more efficient algorithm exists which makes use of the \
Quadratic Reciprocity Theorem of Gauss, but we do not need the complexity of \
this method because we will be testing for quadratic residues for only modest \
sized integers."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["2.7 Shanks-Tonelli Modular Square Root Algorithm", "Subsection",
  CellTags->"c:15"],

Cell[TextData[{
  "Once we have determined that integer ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " is a quadratic residue modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ", we need a method to find the square root.  One method to accomplish this \
is called the Shanks-Tonelli modular square root algorithm.  The details of \
the algorithm, which has performance  logarithmic in the number of digits of \
",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ",  are described in the paper \"Square Roots from 1; 24, 51, 10 to Dan \
Shanks\" by  Ezra Brown [1]."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Example 5 - Computing Square Roots Modulo p", "Subsubsection",
  CellTags->"c:16"],

Cell[TextData[{
  "We consider the following nontrivial example.  Let \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`p = 360027784083079948259017962255826129\)]],
  ".\n\t\nWe want to  find ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2 \[Congruent] 2865\ \((mod\ p)\)\)]],
  ".  The Shanks-Tonelli algorithm gives\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`x = 203744876602447660339212047901408164\)]],
  ".\n\t \nWe could verify that this is the correct solution using the \
modular exponentiation method described above, but since we are only \
computing ",
  Cell[BoxData[
      \(TraditionalForm\`\(x\^2\)(mod\ p)\)]],
  " we can compute this directly,  showing that ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2 \[Congruent] 2865 \((mod\ p)\)\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["2.8 The Chinese Remainder Theorem", "Subsection",
  CellTags->"c:17"],

Cell[TextData[{
  "The Chinese Remainder Theorem provides a method to compute the smallest \
positive integer satisfying a set of congruences. It first appeared as a \
method of solution to a particular modular congruence problem in a \
third-century book by Chinese mathematician Sun Tzu [4] \[Section] 4.5.  In \
Schoof's algorithm, we compute ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau]\_i \[Congruent] t\ \((mod\ l\_i)\)\)]],
  " for a set of small primes ",
  Cell[BoxData[
      \(TraditionalForm\`l\_i\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " satisfies ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ".  The Chinese Remainder Theorem allows us to recover ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " from this set of congruences, thus determining the order of the group.\n\n\
We are given the following information, for the unknown ",
  Cell[BoxData[
      \(TraditionalForm\`z < N\)]],
  "\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`r\_i \[Congruent] z\ \((mod\ n\_i)\)\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`i = 1, 2,  ... , k\)]],
  "\n\t\nwhere ",
  Cell[BoxData[
      \(TraditionalForm\`N = \(n\_1\ n\_2 ... \) n\_k\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(n\_i, n\_j) = 1  when\ i \[NotEqual] j\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`N\)]],
  " is also the least common multiple of  the ",
  Cell[BoxData[
      \(TraditionalForm\`{n\_i}\)]],
  ".\n\nLet ",
  Cell[BoxData[
      \(TraditionalForm\`a\_i = N\/n\_i\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`b\_i = \(a\_i\%\(-1\)\)(mod\ n\_i)\)]],
  ".  The modular inverses\t",
  Cell[BoxData[
      \(TraditionalForm\`b\_i\)]],
  " exist because  ",
  Cell[BoxData[
      \(TraditionalForm\`n\_i\)]],
  " does not divide ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(a\_i\)\(\ \)\)\)]],
  "because ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(n\_i, n\_j) = 1  when\ i \[NotEqual] j\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`\(a\_i\) b\_i \[Congruent] 1 \((mod\ n\_i)\)\)]],
  " and since ",
  Cell[BoxData[
      \(TraditionalForm\`n\_j\)]],
  " divides ",
  Cell[BoxData[
      \(TraditionalForm\`a\_i\)]],
  " when ",
  Cell[BoxData[
      \(TraditionalForm\`j \[NotEqual] i\)]],
  ", hence ",
  Cell[BoxData[
      \(TraditionalForm\`\(a\_i\) b\_i \[Congruent] 0 \((mod\ n\_j)\)\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`j \[NotEqual] i\)]],
  ".  So we can compute\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`z = \((\[Sum]\+\(i = 1\)\%k\( a\_i\) \(b\_i\) 
                r\_i)\)\ \((mod\ N)\)\)]],
  ", \n\t \nwhich is the unique ",
  Cell[BoxData[
      \(TraditionalForm\`0 < z < N\)]],
  " for which all the congruences hold."
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Example 6 - Determining the Chinese Remainder", "Subsubsection",
  CellTags->"c:18"],

Cell[TextData[{
  "For example, suppose we have ",
  Cell[BoxData[
      \(TraditionalForm\`z \[Congruent] 1 \((mod\ 2)\), 
      z \[Congruent] 0\ \((mod\ 5)\), z \[Congruent] 6\ \((mod\ 7)\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`z \[Congruent] 7 \((mod\ 11)\)\)]],
  ".  Then  ",
  Cell[BoxData[
      \(TraditionalForm\`N = 770\)]],
  " and the Chinese remainder algorithm gives\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`a\_i = {385, 154, 110, 70}\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`b\_i = \(\(a\_i\%\(-1\)\)(mod\ n\_i) = {1, 4, 3, 
            3}\)\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`e\_i = \(\(a\_i\) b\_i = {385, 616, 330, 210}\)\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`z = \(e\[CenterDot]r = 
          3835 \[Congruent] 755 \((mod\ N)\)\)\)]],
  "\n\t\nSo ",
  Cell[BoxData[
      \(TraditionalForm\`755\)]],
  " is the smallest positive integer satisfying this set of congruences."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "Chapter 3 - Arithmetic of Elliptic Curves over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]]
}], "Section",
  PageBreakAbove->True,
  CellTags->"c:19"],

Cell[TextData[{
  "Now that we have a set of methods for performing arithmetic in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ", we can apply these to performing arithmetic in the group ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  ".  We provide a set of ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " functions to implement algorithms for testing if a cubic equation is an \
elliptic curve over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ", for adding points on the curve, and for efficiently computing ",
  Cell[BoxData[
      \(TraditionalForm\`k\ P\)]],
  ", the sum of ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " copies of the point ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  ".  In each case it is assumed, and for efficiency sake not verified, that \
",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is a prime. "
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell[TextData[{
  "Example 7 - Arithmetic in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]]
}], "Subsubsection",
  CellTags->"c:20"],

Cell[TextData[{
  "We will now review several of the mathematical ideas and methods related \
to elliptic curves over a prime field by way of an example.  Consider the \
elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46\ x + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  ".  It has discriminant\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`d \[Congruent] \(-\((4\ a\^3 + 
              27\ b\^2)\)\) \[Congruent] \(-\((4*\ 46\^3 + 
              27\ *74\^2)\)\) \[Congruent] 537196 \[Congruent] 
        87 \((mod\ p)\)\)]],
  "\n\t \nSince the discriminant is nonzero modulo",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " is a valid elliptic curve."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "At ",
  Cell[BoxData[
      \(TraditionalForm\`x = 1\)]],
  " we have\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + 46\ x + 74 = \(1 + 46 + 74 = 
          121 \[Congruent] 24 \((mod\ 97)\)\)\)]],
  "\n\t \nso ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " has a point with ",
  Cell[BoxData[
      \(TraditionalForm\`x = 1\)]],
  " if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`24\)]],
  " is a quadratic residue modulo ",
  Cell[BoxData[
      \(TraditionalForm\`97\)]],
  ".  Using modular exponentiation we find with ",
  Cell[BoxData[
      \(TraditionalForm\`p = 97\)]],
  " that ",
  Cell[BoxData[
      \(TraditionalForm\`24\^\(\((p - 1)\)/2\) \[Congruent] 
        1 \((mod\ p)\)\)]],
  ", so that 24 has a square root modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ".  We then employ the Shanks-Tonelli algorithm to find this square root \
giving ",
  Cell[BoxData[
      \(TraditionalForm\`11\^2 \[Congruent] 24 \((mod\ p)\)\)]],
  " so that the point ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((1, 11)\)\)]],
  " is an element of ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "Next let us compute ",
  Cell[BoxData[
      \(TraditionalForm\`P + P = 2  P\)]],
  " using Equation (12) modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " as\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \(3\ x\^3 + A\)\/\(2\ y\) \[Congruent] \
\((3 + 46)\) \((22\^\(-1\))\) \((mod\ 97)\) \[Congruent] 
          49*75 \((mod\ 97)\) \[Congruent] 86 \((mod\ 97)\)\)]],
  ".\n\t\nThen\n\t",
  Cell[BoxData[
      \(TraditionalForm\`x\_3 = \[Lambda]\^2 - 2\ x\_1 \[Congruent] 
          86\^2 - 2 \((mod\ 97)\) \[Congruent] 22 \((mod\ 97)\)\)]],
  ",  also\n\t",
  Cell[BoxData[
      \(TraditionalForm\`y\_3 = \(\[Lambda](x\_1 - x\_3) - y\_1 = 
          86 \((1 - 22)\) - 11 \((mod\ 97)\) \[Congruent] 
            26 \((mod\ 97)\)\)\)]],
  ".\n\t\nHence ",
  Cell[BoxData[
      \(TraditionalForm\`2*\((1, 11)\) = \((22, 26)\)\)]],
  " on ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_97)\)]],
  ".  We can also compute ",
  Cell[BoxData[
      \(TraditionalForm\`4*\((1, 11)\)\)]],
  " by adding ",
  Cell[BoxData[
      \(TraditionalForm\`\((22, 26)\) + \((22, 26)\)\)]],
  " giving ",
  Cell[BoxData[
      \(TraditionalForm\`\((4, 15)\)\)]],
  " or obtain the same result using the function ",
  Cell[BoxData[
      \(TraditionalForm\`EcPowerMod[{46, 74}, {1, 11}, 4, 97]\)]],
  "."
}], "Text",
  TextJustification->1,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "The order of a point ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E(\[DoubleStruckCapitalF]\_p)\)]],
  " is defined as the smallest ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`k\ P = \[ScriptCapitalO]\)]],
  ".  One way to determine ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " is to compute ",
  Cell[BoxData[
      \(TraditionalForm\`n\ P\)]],
  " for each successive ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " starting at 1 until the result is the identity.  For the previous example \
with  ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46\ x + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((1, 11)\)\)]],
  " we find, using repeated application of ",
  StyleBox["EcAddMod",
    FontSize->12,
    FontWeight->"Bold"],
  ", that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`2  P = \((2, 26)\), 3  P = \((27, 12)\), 
      4  P = \((4, 15)\),  ... , 15  P = \((1, 86)\)\)]],
  ".\n\t\nSince ",
  Cell[BoxData[
      \(TraditionalForm\`86 + 11 \[Congruent] 0 \((mod\ 97)\)\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`15  P = \(-P\)\)]],
  ", so we know that ",
  Cell[BoxData[
      \(TraditionalForm\`16  P = \[ScriptCapitalO]\)]],
  ".  Since 16 is the smallest multiple of ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " for which this occurs we have the order of ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " is 16 and we write ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(P\)\(|\)\) = 16\)]],
  ".  This gives a hint as to one way to determine ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  ".  By Lagrange's theorem the order of the element of a finite group must \
divide the order of the group.  So we know that 16 divides ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_97)\)\)]],
  " for our sample curve."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "Chapter 4 - Computing the Order of the Group ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]]
}], "Section",
  PageBreakAbove->True,
  CellTags->"c:21"],

Cell[CellGroupData[{

Cell[TextData[{
  "4.1 A direct method of computing ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]]
}], "Subsection",
  CellTags->"c:22"],

Cell[TextData[{
  "A direct approach to determining ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  " is to compute ",
  Cell[BoxData[
      \(TraditionalForm\`z = x\^3 + A\ x + B\)]],
  " for each ",
  Cell[BoxData[
      \(TraditionalForm\`x \[Element] \[DoubleStruckCapitalF]\_q\)]],
  ", and then to test if ",
  Cell[BoxData[
      \(TraditionalForm\`z\)]],
  " has a square root in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  ".  If ",
  Cell[BoxData[
      \(TraditionalForm\`z = 0\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, 0)\) \[Element] 
        E(\[DoubleStruckCapitalF]\_q)\)]],
  ".  If there exists ",
  Cell[BoxData[
      \(TraditionalForm\`y \[Element] \[DoubleStruckCapitalF]\_q\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = z\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y)\), \((x, \(-y\))\) \[Element] 
        E(\[DoubleStruckCapitalF]\_q)\)]],
  ", else there is no point in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " with ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(x\)\(.\)\)\)]],
  "  This means there are at most ",
  Cell[BoxData[
      \(TraditionalForm\`2\ q + 1\)]],
  "elements in the group.  However, a theorem of finite fields states that \
exactly 1/2 of the non-zero elements of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " are quadratic residues.  This means that, on average, there will be \
approximately ",
  Cell[BoxData[
      \(TraditionalForm\`q + 1\)]],
  " elements in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  ".  As we shall see next, specific bounds on ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  " can be established, and this is one key to more efficient methods of \
determining the group order. "
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "Before proceeding, we want to fully characterize all of the points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " for our example curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = 
        x\^3 + 46\ x + 74\ over\ \ \[DoubleStruckCapitalF]\_97\)]],
  ".   We can do this using the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function ",
  StyleBox["FindEcPointSet",
    FontSize->12,
    FontWeight->"Bold"],
  " which encodes the technique above to find every point on the curve.  Then \
for each point we can determine its order using the technique outlined at the \
end of chapter 3.   This method is encoded in the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function ",
  StyleBox["EcPointOrderMod",
    FontSize->12,
    FontWeight->"Bold"],
  ".  The results of applying these methods to our example are shown in the \
table of Figure 3.   Note that for each ",
  StyleBox["x",
    FontSlant->"Italic"],
  " there are two values of ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  ", which are distinct unless ",
  Cell[BoxData[
      \(TraditionalForm\`y = 0\)]],
  ".  This is so because each solution has ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 \[Congruent] z\ \((mod\ p)\)\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`z = x\^3 + A\ x + B\)]],
  " for a particular value of ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  ", so if ",
  Cell[BoxData[
      \(TraditionalForm\`y\_1\^2 = z\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(-y\_1\))\)\^2 = z\)]],
  ".  There can be no other distinct solutions because the quadratic equation \
",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 - z = 0\)]],
  " can have only two solutions.  Observe that for each pair of points in \
Figure 3 with the same value of ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " we have ",
  Cell[BoxData[
      \(TraditionalForm\`y\_2 \[Congruent] \(-y\_1\)\ \((mod\ p)\)\)]],
  ", so that the points ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y\_1)\), \((x, y\_2)\)\)]],
  " are indeed inverses in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  ".\n\nIf we count these points we find 39 pairs of points ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y\_1)\), \((x, y\_2)\)\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`y\_1 \[NotEqual] y\_2\)]],
  ".  We also have one point, ",
  Cell[BoxData[
      \(TraditionalForm\`\((57, 0)\)\)]],
  ", with ",
  Cell[BoxData[
      \(TraditionalForm\`y = 0\)]],
  ".   Including the identity \[ScriptCapitalO], there are a total of ",
  Cell[BoxData[
      \(TraditionalForm\`2*39 + 1 + 1 = 80\)]],
  " points so that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 80\)]],
  ". The last column in the table gives the order of each point.  Notice also \
that each point order divides 80, the order of the group, as must be so by \
Lagrange's theorem."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  StyleBox["Table 1 - Points for ",
    FontWeight->"Bold"],
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = \(\(x\^3\)\(+\)\(46  
            x\)\(\ \)\(+\)\(74\)\(\ \)\)\)],
    FontWeight->"Bold"],
  StyleBox[" over ",
    FontWeight->"Bold"],
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)],
    FontWeight->"Bold"],
  "\n\n",
  Cell[BoxData[
      TagBox[
        FrameBox[GridBox[{
              {"x", \(y\_1\), \(y\_2\), \(Ord \((P)\)\)},
              {"1", "11", "86", "16"},
              {"4", "15", "82", "4"},
              {"6", "9", "88", "80"},
              {"8", "9", "88", "40"},
              {"9", "21", "76", "10"},
              {"10", "46", "51", "8"},
              {"15", "29", "68", "80"},
              {"19", "12", "85", "80"},
              {"20", "19", "78", "80"},
              {"22", "26", "71", "8"},
              {"24", "8", "89", "40"},
              {"27", "12", "85", "16"},
              {"30", "18", "79", "40"},
              {"32", "48", "49", "40"},
              {"34", "28", "69", "80"},
              {"35", "6", "91", "20"},
              {"37", "7", "90", "80"},
              {"43", "46", "51", "80"},
              {"44", "46", "51", "80"},
              {"46", "2", "95", "20"},
              {"49", "45", "52", "5"},
              {"51", "12", "85", "10"},
              {"52", "22", "75", "80"},
              {"57", "0", "0", "2"},
              {"60", "14", "83", "40"},
              {"63", "25", "72", "40"},
              {"64", "35", "62", "16"},
              {"65", "47", "50", "40"},
              {"66", "24", "73", "20"},
              {"67", "42", "55", "80"},
              {"70", "2", "95", "80"},
              {"75", "32", "65", "20"},
              {"76", "41", "56", "80"},
              {"78", "2", "95", "80"},
              {"83", "9", "88", "16"},
              {"85", "5", "92", "80"},
              {"88", "17", "80", "40"},
              {"90", "31", "66", "5"},
              {"94", "43", "54", "80"},
              {"96", "30", "67", "80"}
              },
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            ColumnLines->True]],
        DisplayForm]]]
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Cell[CellGroupData[{

Cell["4.2 Overview of Schoof's Algorithm", "Subsection",
  CellTags->"c:23"],

Cell[TextData[{
  "The method for determining ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " outlined above is feasible only for small ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ".  In modern cryptographic applications of elliptic curves the cardinality \
of the field is typically a number with at least 50 decimal digits.  Thus \
there is a need for an efficient means of computing ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " for large primes ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ".  Ren\[EAcute] Schoof's 1985 paper entitled \"Elliptic curves over finite \
fields and the computation of square roots mod ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  "\", details a polynomial time algorithm for determining ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  ".  Versions of this algorithm, enhanced by Elkies and Atkins, have been \
used successfully for ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  " with hundreds of decimal digits [3,6].  The following steps sketch the \
outline of Schoof's method.\n\nLet ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " be an elliptic curve over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " given by"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x\  + B\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`A, B \[Element] \[DoubleStruckCapitalF]\_q\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell["\<\
Hasse's Theorem tells us that the cardinality of the group of points is\
\>", "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) = 
        q + 1 - t\)]],
  ", with ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(t\)\(|\)\(\(\[LessEqual]\)\(2 \@ \
q\)\)\)\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "Let ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_q\) : 
            E(\[DoubleStruckCapitalF]\&_\_q) \[Rule] \(E(\
\[DoubleStruckCapitalF]\&_\_q)\)\ such\ that\ \(\(\[Phi]\_q\)(\((x, 
                y)\))\) = \((x\^q, y\^q)\)\)]],
  ".  Note that this is map of points with coordinates in the algebraic \
closure of  ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\)]],
  " is an endomorphism called the Frobenius map.  It has the following \
property, crucial to Schoof's algorithm."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\^2 - t\ \[Phi]\_q + q = 0\)]],
  "    ",
  Cell[BoxData[
      \(TraditionalForm\`\[ForAll] P \[Element] 
          E(\[DoubleStruckCapitalF]\&_\_q)\)]]
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "We can use (15) to compute ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ p\_i)\)\)]],
  " for a set of ",
  Cell[BoxData[
      \(TraditionalForm\`L\)]],
  " primes ",
  Cell[BoxData[
      \(TraditionalForm\`p\_1, p\_2,  ... , p\_L\)]],
  " such that  \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`K = \[Product]\+\(i = 1\)\%L p\_i > 4 \@ q\)]],
  ",\n\t\nThe Chinese Remainder Theorem is then applied to compute the unique\
\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`t\ \ mod\ K\)]],
  " such that  ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(t\)\(|\)\(\(\[LessEqual]\)\(2 \@ \
q\)\)\)\)]],
  ".\n\t\nOnce we know ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " we can compute the order of the group as ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) = 
        q + 1 - t\)]],
  ".  Schoof showed that this algorithm will run time proportional to ",
  Cell[BoxData[
      \(TraditionalForm\`\(log\^9\) q\)]],
  ", based on analysis of the number of elementary operations required.  The \
following sections will explain the details of this algorithm, along with \
some important observations that permit its efficient implementation."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["4.3 Hasse's Theorem", "Subsection",
  CellTags->"c:24"],

Cell[TextData[{
  "The following theorem, first proved by Helmut Hasse in 1933, places \
specific bounds on ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  ".  Let ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " be an elliptic curve over the finite field ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`q = p\^k, 
      k \[Element] \(\[DoubleStruckCapitalZ]\^+\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " a prime. Then there exists a unique ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Element] \[DoubleStruckCapitalZ]\)]],
  " such that"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) = 
        q + 1 - t\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(t\)\(|\)\(\(<\)\(2 \@ q\)\)\)\)]],
  "\n"
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  StyleBox["Sketch of the Proof",
    FontWeight->"Bold"],
  "\nDefine  the map ",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Phi]\_q - 1)\) : 
          E(\[DoubleStruckCapitalF]\&_\_q) \[Rule] 
        E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  ", then the set of points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " which are sent to the identity by  this map is called the kernel.  Then \
",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{"ker", "(", 
            FormBox[\(\[Phi]\_q - 1\),
              "TraditionalForm"], ")"}], 
          "=", \(E(\[DoubleStruckCapitalF]\_q)\)}], TraditionalForm]]],
  ", since ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\)]],
  " is the identity on ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  ".  Further, since ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q - 1\)]],
  " is a separable polynomial then\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`# \( 
            E(\[DoubleStruckCapitalF]\_q)\) = \(# \( ker(\[Phi]\_q - 1)\) = 
          deg(\[Phi]\_q - 1)\)\)]],
  ".\n\t \nNow let ",
  Cell[BoxData[
      \(TraditionalForm\`t = 
        q + 1 - # \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  ".  Then by Washington [7] Proposition 3.16 for ",
  Cell[BoxData[
      \(TraditionalForm\`r, 
      s \[Element] \[DoubleStruckCapitalZ]\ and\ \(gcd(s, q)\) = 1\)]],
  " we have ",
  Cell[BoxData[
      \(TraditionalForm\`deg(r\ \[Phi]\_q - s)\)]],
  "\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(\(=\)\(\(r\^2\)(
            deg\ \[Phi]\_q) + \(s\^2\) \(deg(\(-1\))\) + 
          r\ \(s(deg(\ \[Phi]\_q - 1) - deg(\[Phi]\_q) - 
                deg(\(-1\)))\)\)\)\)]],
  "\n\t",
  Cell[BoxData[{
      \(TraditionalForm\`\(\(=\)\(\(r\^2\) q + s\^2 + 
          r\ s\ \((\ # \( E(\[DoubleStruckCapitalF]\_q)\) - q - 1)\)\)\)\), 
    "\[IndentingNewLine]", 
      \(TraditionalForm\`\(\(=\)\(\(r\^q\) q + s\^2 + 
          r\ \(\(s(\ q + 1 - t - q - 1)\)\(.\)\)\)\)\)}]],
  "\n\t\nSo we can conclude that"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`deg(r\ \[Phi]\_q - s) = \(r\^2\) q + s\^2 - 
        r\ s\ \(\(t\)\(.\)\)\)]]], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "Since ",
  Cell[BoxData[
      \(TraditionalForm\`deg(r\ \[Phi]\_q - s) \[GreaterEqual] 0\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`s \[NotEqual] 0\)]],
  " then dividing through by ",
  Cell[BoxData[
      \(TraditionalForm\`s\^2\)]],
  " gives"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`q\ \((r\/s)\)\^2 - \[Tau]\ \((r\/s)\) + 
        1 \[GreaterEqual] 0. \)]]], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "Having that the set of rational numbers ",
  Cell[BoxData[
      \(TraditionalForm\`r\/s\ with\ \(gcd(s, q)\) = 1\)]],
  " is dense in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalR]\)]],
  " implies that for all ",
  Cell[BoxData[
      \(TraditionalForm\`x \[Element] \[DoubleStruckCapitalR]\)]],
  " we have"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`q\ x\^2 - \[Tau]\ x + 1 \[GreaterEqual] 
      0. \)]]], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "So quadratic equation (19) has no real roots, hence its discriminant is \
less than zero. Thus\n\t\t",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[Tau]\^2 - 4\ q < 
            0\)\(\[Implies]\)\) | \[Tau] | \(\(<\)\(2 \@ q\)\)\)]],
  ",\ncompleting the proof of (16)."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell[TextData[{
  "4.4 Reducing the problem to that for ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]]
}], "Subsection",
  CellTags->"c:25"],

Cell[TextData[{
  "A beautiful result due to Andre Weil, and explained in ",
  "Washington [7]",
  " Theorem 4.12,  shows that if we can compute ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  ", then we can compute ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_\(p\^n\))\)\)]],
  " in a direct manner.\n\nLet ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ".  Write ",
  Cell[BoxData[
      \(TraditionalForm\`X\^2 - t\ X + 
          p = \((X - \[Alpha])\) \((X - \[Beta])\)\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha]\^n + \[Beta]\^n \[Element] \
\[DoubleStruckCapitalZ]\)]],
  " and"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_\(p\^n\))\) = 
        p\^n + 1 - \((\[Alpha]\^n + \[Beta]\^n)\)\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "So we only need to use Schoof's Algorithm to solve for ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  ".  Then we can compute ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) = # \( 
            E(\[DoubleStruckCapitalF]\_\(p\^n\))\)\)]],
  " via (20).  Of course, if ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is a small prime, then it is easy to determine ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " by direct counting or other simple methods, so the complexity of Schoof's \
method would not be justified.  Assuming ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is large enough to warrant the use of Schoof's method we may employ \
another useful result allowing us to determine the integer ",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Alpha]\^n + \[Beta]\^n)\)\)]],
  " without explicitly computing \[Alpha] and \[Beta].  The following \
recursion relation computes ",
  Cell[BoxData[
      FormBox[
        RowBox[{\(s\_n\), "=", " ", 
          FormBox[\((\[Alpha]\^n + \[Beta]\^n)\),
            "TraditionalForm"]}], TraditionalForm]]],
  " where"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`s\_0 = 2, \ \ \ \ s\_1 = t, \ \ \ \ \ s\_\(n + 1\) = 
        t\ s\_n - p\ s\_\(n - 1\)\)]],
  "."
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "The ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function ",
  Cell[BoxData[
      \(TraditionalForm\`ComputeOrderEFq[\ t, p, n\ ]\)]],
  " implements equations (21) and (22).\n\nFor our example elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46\ x + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  " we determined that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_97)\) = 80\)]],
  ".  By Hasse's theorem ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_97)\) = 
        p + 1 - t\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`80 = 97 + 1 - t\)]],
  ", hence ",
  Cell[BoxData[
      \(TraditionalForm\`t = 18\)]],
  ".   Then we can determine ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_\(97\^4\))\)\)]],
  " using ",
  Cell[BoxData[
      FormBox[
        StyleBox["ComputeOrderEFq",
          FontSize->12,
          FontWeight->"Bold"], TraditionalForm]]],
  ", giving ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_\(p\^4\))\) = 
        88531200\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["4.5 Baby Step, Giant Step Method", "Subsection",
  CellTags->"c:26"],

Cell[TextData[{
  "One way to use Hasse's bound to compute ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " is based on Lagrange's theorem which states that the order of any element \
of a finite group must divide the order of the group.   Hence if we can find \
the order ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " of a point ",
  Cell[BoxData[
      \(TraditionalForm\`Q \[Element] E(\[DoubleStruckCapitalF]\_p)\)]],
  " then the group order must be a multiple of ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " falling inside of Hasse's bounds.   If we compute the order for several \
different points then some common multiple of these orders must fall inside \
of Hasse's bounds.  Let ",
  Cell[BoxData[
      \(TraditionalForm\`{k\_i}\)]],
  " be the set of orders of ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  ". By Hasse's theorem we have for some integer ",
  Cell[BoxData[
      \(TraditionalForm\`r\)]],
  " that ",
  Cell[BoxData[
      \(TraditionalForm\`p + 1 - 2 \@ p < r*\(lcm({k\_i}}\) < 
        p + 1 + 2 \@ p\)]],
  ".  If there is only one ",
  Cell[BoxData[
      \(TraditionalForm\`r\)]],
  " for which this is true, then we must have ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        r*\(lcm({k\_i}}\)\)]],
  ".\n\nIn order for this method to be efficient we need a high performance \
method to compute point orders, that is, a method far better than exhaustive \
search to find the smallest ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`k\ P = \[ScriptCapitalO]\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  ".  The Baby Step-Giant Step method outlined in Washington [7] ",
  "\[Section]",
  " 4.3 provides such a method with runtime proportional to ",
  Cell[BoxData[
      \(TraditionalForm\`\@p\%4\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Example 8 - Determining Group Order using Hasse's Theorem", \
"Subsubsection",
  CellTags->"c:27"],

Cell[TextData[{
  "For this example we again take ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x + B\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  ".  Since ",
  Cell[BoxData[
      \(TraditionalForm\`9 < \@97 < 10\)]],
  ", Hasse's theorem gives\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`97 - 
          2*9 = \(79 \[LessEqual] \ # \( 
                E(\[DoubleStruckCapitalF]\_q)\)\  \[LessEqual] 117 = 
          97 + 2*9\)\)]],
  ".\n\t\nWe randomly found that ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((64, 35)\)\)]],
  " is a point on the curve, and that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(P\)\(|\)\) = 16\)]],
  ".  Since there are three multiples of 16 between 79 and 117, we need to \
choose another point.  We randomly found a second point ",
  Cell[BoxData[
      \(TraditionalForm\`Q = \((46, 95)\)\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(Q\)\(|\)\) = 20\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`lcm(16, 20) = 80\)]],
  ".  Since ",
  Cell[BoxData[
      \(TraditionalForm\`79 \[LessEqual] 80 \[LessEqual] 117\)]],
  " we can conclude that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) = 80\)]],
  ", in agreement with the direct counting method."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Chapter 5 - Schoof's Algorithm Implementation", "Section",
  PageBreakAbove->True,
  CellTags->"c:28"],

Cell[TextData[{
  "\"A four-year-old child could understand that.  Run out and find me a \
four-year-old child, I can't make head or tail out of it.\" - Groucho Marx \
(",
  StyleBox["Duck Soup-1933",
    FontSlant->"Italic"],
  ")"
}], "Text",
  TextAlignment->Center,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "In this chapter we present the algorithms that embody the key ideas of \
Schoof's method.  For the curve given by ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x\  + B\)]],
  ", over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ",  Hasse's theorem tells us that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ".  The main objective of Schoof's algorithm is to determine ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ l)\)\)]],
  " for a set of small primes ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  ".  For the case of  ",
  Cell[BoxData[
      \(TraditionalForm\`l = 2\)]],
  " we have a special method, so we outline this first.  For ",
  Cell[BoxData[
      \(TraditionalForm\`l > 2\)]],
  " we must employ more sophisticated mathematics including the Frobenius \
endomorphism and the so-called division polynomials.  We will examine these \
in more detail after describing the method for computing ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ 2)\)\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell[TextData[{
  "5.1 Computing ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ 2)\)\)]]
}], "Subsection",
  CellTags->"c:29"],

Cell[TextData[{
  "As before, let ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " be an elliptic curve over the finite field ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " given by ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x\  + B\)]],
  ".  A point ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E(\[DoubleStruckCapitalF]\_p)\)]],
  " has order 2 if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`P + P = \[ScriptCapitalO]\)]],
  " which means that ",
  Cell[BoxData[
      \(TraditionalForm\`P = \(-P\)\)]],
  ".  As we have seen, this is true only if the ",
  StyleBox["y",
    FontSlant->"Italic"],
  "-coordinate of ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " is zero.  Now ",
  Cell[BoxData[
      \(TraditionalForm\`y = 0\)]],
  " if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + A\ x\  + B\  = 0\)]],
  ".\n\nSuppose that there exists some ",
  Cell[BoxData[
      \(TraditionalForm\`e \[Element] \[DoubleStruckCapitalF]\_p\ such\ that\ \
e\^3 + A\ e + B = 0\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(\((e, 0)\) \[Element] 
        E(\[DoubleStruckCapitalF]\_p)\)\)\)]],
  ".  Also, by the definition of elliptic curve addition, ",
  Cell[BoxData[
      \(TraditionalForm\`2 \((e, 0)\) = \[Infinity]\)]],
  " , so that ",
  Cell[BoxData[
      \(TraditionalForm\`\((e, 0)\) \[Element] E[2]\)]],
  ".  Then ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " has a point of order 2, so that, by Langrange's Theorem, ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  " is even.  Since ",
  Cell[BoxData[
      \(TraditionalForm\`p + 1\)]],
  " is even then ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " also is even, therefore ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ 2)\)\)]],
  ".\n\nAlternatively, suppose that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  " is even.  Then by Theorem 4.1 of Washington [7] either\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p) \[TildeFullEqual] \
\[DoubleStruckCapitalZ]\_n\)]],
  " or ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p) \[TildeFullEqual] \
\[DoubleStruckCapitalZ]\_\(n\_1\)\[CirclePlus]\[DoubleStruckCapitalZ]\_\(n\_2\
\)\ with\ \ n, n\_1, 
      n\_2 \[Element] \[DoubleStruckCapitalZ]\ and\ n\_1 | \
\(\(n\_2\)\(.\)\)\)]],
  "\n\t\nIf two groups are isomorphic, there is a 1-1 mapping between the \
elements of the two groups which preserves the group operation.  This means, \
in particular, that if one has a nonzero point of order ",
  Cell[BoxData[
      \(TraditionalForm\`2\)]],
  ", then the other has a nonzero point of order ",
  Cell[BoxData[
      \(TraditionalForm\`2\)]],
  ".\n\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p) \[TildeFullEqual] \
\[DoubleStruckCapitalZ]\_n\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " is even, because ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " is even.  We know that ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]],
  " is cyclic with generator 1, and  ",
  Cell[BoxData[
      \(TraditionalForm\`2*\((n\/2*1)\) = n \[Congruent] 0 \((mod\ n)\)\)]],
  ", so ",
  Cell[BoxData[
      \(TraditionalForm\`n\/2\)]],
  " is an element of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]],
  " of order 2, therefore ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " is cyclic with some generator ",
  Cell[BoxData[
      \(TraditionalForm\`P\_g\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`2*\((n\/2*P\_g)\) = \(n\ P\_g = \[Infinity]\)\)]],
  ",  so that ",
  Cell[BoxData[
      \(TraditionalForm\`P\_2 = \(n\/2\) P\_g\)]],
  " has order 2,  therefore ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(P\_2\)\(=\)\((e, 0)\)\(\ \)\)\)]],
  " for some ",
  Cell[BoxData[
      \(TraditionalForm\`e \[Element] \[DoubleStruckCapitalF]\_p\)]],
  ".\n\nOtherwise, ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p) \[TildeFullEqual] \
\[DoubleStruckCapitalZ]\_\(n\_1\)\[CirclePlus]\[DoubleStruckCapitalZ]\_\(n\_2\
\)\ and\ \ n\_1*n\_2\)]],
  " is even, so that either ",
  Cell[BoxData[
      \(TraditionalForm\`n\_2\)]],
  " is even or ",
  Cell[BoxData[
      \(TraditionalForm\`n\_1\)]],
  " is even, which implies that ",
  Cell[BoxData[
      \(TraditionalForm\`n\_2\)]],
  " is even, because ",
  Cell[BoxData[
      \(TraditionalForm\`n\_1 | n\_2\)]],
  ".  So we have that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((0, n\_2\/2)\)\)]],
  " is a point of order ",
  Cell[BoxData[
      \(TraditionalForm\`2\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalZ]\_\(n\_1\)\[CirclePlus]\
\[DoubleStruckCapitalZ]\_\(n\_2\)\)]],
  ",\n\t\ntherefore ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " has a point of order 2, call it ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(P\_2\)\(.\)\)\)]],
  "  So we conclude that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) \[Congruent] 
            0 \((mod\ 2)\) \[Implies] \[Exists] P\_2 \[Element] \(E(\
\[DoubleStruckCapitalF]\_p)\)\ with\ 2\ P\_2 = \[Infinity]\)]],
  ".\n\nThe contrapositive gives that if ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " does not have a point of order 2, then ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) \[Congruent] 
          1 \((mod\ 2)\) \[Implies] 
        t \[Congruent] 1 \(\((mod\ 2)\)\(.\)\)\)]],
  "  Hence to compute ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ 2)\)\)]],
  " it suffices to determine if ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + A\ x + B\)]],
  " has a root in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "5.2 Determining if ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + A\ x + B\)]],
  " has a root in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]]
}], "Subsection",
  CellTags->"c:30"],

Cell[TextData[{
  "A basic theorem of algebra tells us if ",
  Cell[BoxData[
      \(TraditionalForm\`g(x)\)]],
  " is a polynomial of degree ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " with coefficients in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`g(x)\)]],
  " has ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " roots in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\&_\_p\)]],
  ", the algebraic closure of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".  In addition if ",
  Cell[BoxData[
      \(TraditionalForm\`g(x)\)]],
  " has no roots in common with ",
  Cell[BoxData[
      \(TraditionalForm\`g' \((x)\)\)]],
  ", then the roots of ",
  Cell[BoxData[
      \(TraditionalForm\`g(x)\)]],
  " are distinct.\n\nTake ",
  Cell[BoxData[
      \(TraditionalForm\`g(x) = x\^p - x\)]],
  ".  Then",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(g' \((x)\) = \(p\ x\^\(p - 1\) - 
            1 = \(-1\)\), \ since\ p \[Congruent] 0 \((mod\ p)\)\)\)\)]],
  ", so ",
  Cell[BoxData[
      \(TraditionalForm\`g' \((x)\)\)]],
  " has no roots in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\&_\_p\)]],
  ".  Therefore, the ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " roots of ",
  Cell[BoxData[
      \(TraditionalForm\`g(X)\)]],
  " are distinct, and these are just the set of elements \[Alpha] of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\&_\_p\)]],
  " satisfying ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha]\^p - \[Alpha] = \(0 \[Implies] \
\[Alpha]\^\(p - 1\) = 1. \)\)]],
  "  But the ",
  Cell[BoxData[
      \(TraditionalForm\`p - 1\)]],
  " non-zero elements of ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[DoubleStruckCapitalF]\_p\^\[Times]\)\)]],
  " all have order ",
  Cell[BoxData[
      \(TraditionalForm\`p - 1\)]],
  ", so the roots of ",
  Cell[BoxData[
      \(TraditionalForm\`g(X)\)]],
  " are precisely the elements of ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[DoubleStruckCapitalF]\_p\)\(.\)\)\)]],
  "\n\nThen ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + A\ x + B\)]],
  " has a root in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " if and only if it has a root in common with ",
  Cell[BoxData[
      \(TraditionalForm\`g(x)\)]],
  ".  So if ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x\^3 + A\ x + B, \ x\^p - x) = 1\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + A\ x + B\)]],
  " has no root in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ", else it has at least one such root.   From a practical standpoint we can \
compute ",
  Cell[BoxData[
      \(TraditionalForm\`x\_p \[Congruent] \(x\^p\)(
          mod\ x\^3 + A\ x + B)\)]],
  " using an efficient algorithm for modular polynomial exponentiation, and \
then compute\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`g = \(gcd(x\^3 + A\ x + B, \ x\_p - x) = 
          gcd(x\^3 + A\ x + B, \ x\^p - x)\)\)]],
  ",\n\t\nusing the Euclidean algorithm for polynomials.\n\nGiven ",
  Cell[BoxData[
      \(TraditionalForm\`g\)]],
  ", we determine ",
  Cell[BoxData[
      \(TraditionalForm\`t(mod\ 2)\)]],
  " as\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 1 \((mod\ 2)\)\ if\ g = 1, \ 
      else\ t \[Congruent] 0 \((mod\ 2)\)\)]],
  ".\n\t\nThis method in encoded in the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " function ",
  StyleBox["ComputeTModTwo",
    FontSize->10,
    FontWeight->"Bold"],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell[TextData[{
  "Example 9 - Computation of ",
  Cell[BoxData[
      \(TraditionalForm\`t(mod\ 2)\)]]
}], "Subsubsection",
  CellTags->"c:31"],

Cell[TextData[{
  "We know from previous examples that for ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46\ x\  + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  " that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 80\)]],
  ".  Hasse's theorem tells us that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ", hence ",
  Cell[BoxData[
      \(TraditionalForm\`t = 18\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ 2)\)\)]],
  ".   By the previous discussion ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " has a point of order two if and only if\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x\^p - x, x\^3 + 46\ x\  + 74) \[NotEqual] 
        1\)]],
  "\n\t\nwhere we can compute ",
  Cell[BoxData[
      \(TraditionalForm\`x\^p - x\)]],
  " modulo ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + 46\ x\  + 74\)]],
  ".   We find, using modular polynomial arithmetic,  that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`x\^p\ \((mod\ x\^3 + 46\ x\  + 
              74)\) = \(\(30\ x\^2\)\(+\)\(60\ x\)\(+\)\(47\)\(\ \)\)\)]],
  ".\n\t\nThen using a modular polynomial version of the Euclidean algorithm \
we compute\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(30\ x\^2 + 59\ x + 47, x\^3 + 46\ x\  + 74) = 
        x + 40 \[NotEqual] 1\)]],
  ".\n\t\nHence ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " has at least one point of order two.  In fact, the table in Figure 3 \
shows it has exactly one such point, namely ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((57, 0)\)\)]],
  ",  thus ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " is even.  Since ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ", and ",
  Cell[BoxData[
      \(TraditionalForm\`p + 1 = 98\)]],
  " is even, then ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " is even.  Hence ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ 2)\)\)]],
  ". "
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.3 The Division Polynomials", "Subsection",
  CellTags->"c:32"],

Cell[TextData[{
  "In order to determine ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ l\_i)\)\)]],
  " for primes ",
  Cell[BoxData[
      \(TraditionalForm\`l\_i > 2\)]],
  ", we need to make use of what are called the ",
  StyleBox["division polynomials",
    FontSlant->"Italic"],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + b\)]],
  ".  These are polynomials which go to zero on points of a particular order. \
 We define ",
  Cell[BoxData[
      \(TraditionalForm\`E[n]\)]],
  " as the set of ",
  StyleBox["n",
    FontSlant->"Italic"],
  "-torsion points of an elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + b\)]],
  ", that is, the set of points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_p)\)]],
  " with order dividing ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`E[
          n] = {P \[Element] E(\[DoubleStruckCapitalF]\&_\_p) | 
              n\ P = \[ScriptCapitalO]}\)]],
  ".  Note that this set includes points with coordinates in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\&_\_p\)]],
  ", the algebraic closure of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".\n\nWith this definition the division polynomials ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_n\)]],
  " of an elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  " are elements of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p[x, y]\)]],
  " with the property that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Psi]\_n\)(x, y) = 0\)]],
  " if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y)\) \[Element] E[n]\)]],
  ".   These polynomials are defined recursively as follows.\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_0 = 0, \[Psi]\_1 = 1, \[Psi]\_2 = 2\ y\)]],
  ",\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_3 = 
        3\ x\^4 + 6\ a\ x\^2 + 12\ b\ x - a\^2\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_4 = 
        4\ \(y(x\^6 + 5\ a\ x\^4 + 20\ b\ x\^3 - 5\ a\^2\ x\^2 - 4\ a\ b\ x - 
              8\ b\^2 - a\^3)\)\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(2  n\) = \(\[Psi]\_n\)(\(\[Psi]\_\(n + 
                    2\)\) \[Psi]\_\(n - 1\)\%2 - \(\[Psi]\_\(n - 
                    2\)\) \[Psi]\_\(n + 1\)\%2)\)]],
  "\t",
  Cell[BoxData[
      \(TraditionalForm\`n \[Element] \[DoubleStruckCapitalZ], n > 2\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(2  n + 1\) = \(\[Psi]\_\(n + 
                  2\)\) \[Psi]\_n\%3 - \(\[Psi]\_\(n + 
                  1\)\%3\) \[Psi]\_\(n - 1\)\)]],
  "\t\t",
  Cell[BoxData[
      \(TraditionalForm\`n \[Element] \[DoubleStruckCapitalZ], n > 1\)]],
  "\n\t\nLets see why ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_3\)]],
  " is the correct polynomial. First, if ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((x, y)\) \[Element] E[3]\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`3  P = 0\)]],
  " which means that ",
  Cell[BoxData[
      \(TraditionalForm\`2  P = \(-P\)\)]],
  ", hence the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinates of ",
  Cell[BoxData[
      \(TraditionalForm\`2  P\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\(-P\)\)]],
  " must be the same.   Using Equations (11,12) to compute ",
  Cell[BoxData[
      \(TraditionalForm\`2  P\)]],
  " we find\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`x = \(\[Lambda]\^2 - 
            2\ x = \((3\ x\^2 + A)\)\^2\/\(4\ y\^2\) - \ 2\ x\)\)]],
  " \nso that\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(-3\)\ x)\) \((4\ y\^2)\) = 
        9\ x\^4 + 6\ A\ x\^2 + A\^2\)]],
  ".\n\t \nBut ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + A\ x + B\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`\(-12\)\ \((x\^4 + A\ x\^2 + B\ x)\) = 
        9\ x\^4 + 6\ A\ x\^2 + A\^2\)]],
  ".   Collecting terms and multiplying through by -1 gives\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`3\ x\^4 + 6\ A\ x\^2 + 12  B\ x - 
          A\^2 = \[Psi]\_3\)]],
  ".\n\t \nSo if ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_3 = 0\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`2  P = \(\[PlusMinus]P\)\)]],
  ", meaning that ",
  Cell[BoxData[
      \(TraditionalForm\`P = \[ScriptCapitalO]\)]],
  " or ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " is a point of order 3.  In either case ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[3]\)]],
  ".\n\nThe division polynomials are polynomials in ",
  Cell[BoxData[
      \(TraditionalForm\`x, y\)]],
  ".  Using the elliptic curve equation we can replace ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + A\ x + B\)]],
  ".  More generally we can replace ",
  Cell[BoxData[
      \(TraditionalForm\`y\^\(2  k\)\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`\((x\^3 + A\ x + B)\)\^k\)]],
  ".  This allows us to express the division polynomials as elements of ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p[x]\)]],
  " or  ",
  Cell[BoxData[
      \(TraditionalForm\`y\ \[DoubleStruckCapitalF]\_p[x]\)]],
  ",  so that no power of ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " greater than 1 will appear.   It can be further proved that we can \
produce polynomials in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p[x]\)]],
  " with the following replacements.\n\n\t",
  Cell[BoxData[
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              RowBox[{"{", 
                StyleBox[GridBox[{
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                            y)\)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ n\ is\
\ odd\)},
                      {\(\(\ \)\(\(\(\[Psi]\_n\)(x, y)\)/
                            y\ \ \ \ \ \ \ \ \ \ \ \ \ if\ n\ is\ even\)\)}
                      }],
                  ShowAutoStyles->True]}],
              ShowAutoStyles->False],
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  "\t\n\t\t\t\nThese polynomials, by definition, also have the property that \
",
  Cell[BoxData[
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  " if and only if ",
  Cell[BoxData[
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    FontSlant->"Italic"],
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      \(TraditionalForm\`n\)]],
  "."
}], "Text",
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  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.4 How many division polynomials?", "Subsection",
  CellTags->"c:33"],

Cell[TextData[{
  "How many division polynomials will we need for the execution of Schoof's \
algorithm?  As noted in the outline of Schoof's algorithm in chapter 4, we \
need to test Equation (15) for a set of primes ",
  Cell[BoxData[
      \(TraditionalForm\`l\_i\)]],
  " such the product of these primes is greater than 4",
  Cell[BoxData[
      \(TraditionalForm\`\@p\)]],
  ".   The function ",
  StyleBox["ComputePrimeSet",
    FontSize->12,
    FontWeight->"Bold"],
  " determines this set.  If the cardinality of ",
  Cell[BoxData[
      \(TraditionalForm\`{l\_i} = k\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(k + 2\)\)]],
  " is the highest order division polynomial required."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "What is the relationship between ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  "?  Figure 4 contains a plot of ",
  Cell[BoxData[
      \(TraditionalForm\`Log\_10[p]\)]],
  " vs. ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  ", which indicates that ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " grows approximately logarithmically with ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ".   The horizontal axis is the number of primes ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  ", the vertical axis is the number of decimal digits in ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ", the size ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".  A statistical fit of this data gives the approximate relationship for \
",
  Cell[BoxData[
      \(TraditionalForm\`k > 10. \)]],
  "\n\n\t",
  Cell[BoxData[
      \(Log\_10[p] = 0.012\ k\^2 + 3.34\ k - 15.98\)],
    FontFamily->"Times New Roman"],
  "\n\t\nGiven that we wish to apply Schoof's algorithm to an elliptic curve \
over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " we could use this graph to estimate the number of small primes ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " that would be required."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  StyleBox["Figure 3 -  Number of digits in ",
    FontWeight->"Bold"],
  Cell[BoxData[
      \(TraditionalForm\`p\)],
    FontWeight->"Bold"],
  StyleBox[" vs. number of small primes.",
    FontWeight->"Bold"],
  " \n\n",
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Cell[CellGroupData[{

Cell["Example 10 - Computation of the Division Polynomials", "Subsubsection",
  CellTags->"c:34"],

Cell[TextData[{
  "For our example ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46\ x\  + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  ", we first compute the set of small primes whose product is greater than \
",
  Cell[BoxData[
      \(TraditionalForm\`4 \@ 97\)]],
  ", such that ",
  Cell[BoxData[
      \(TraditionalForm\`p \[NotCongruent] 1 \((mod\ l\_i)\)\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`l\_i > 2\)]],
  ".  The necessary primes are ",
  Cell[BoxData[
      \(TraditionalForm\`2, 5, 7\)]],
  " whose product is ",
  Cell[BoxData[
      \(TraditionalForm\`70 > 4 \(\(\@ 97\)\(.\)\)\)]],
  "  Then ",
  Cell[BoxData[
      \(TraditionalForm\`p \[Congruent] 2\ \((mod\ 5)\)\)]],
  ", and ",
  Cell[BoxData[
      \(TraditionalForm\`p \[Congruent] 6 \((mod\ 7)\)\)]],
  ".  Therefore we will need division polynomials up to and including ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_9\)]],
  ", so we compute these at this time.   Note that the odd numbered \
polynomials, such as ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_1, \[Psi]\_3,  ... \)]],
  " are polynomials in ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " only, while the even numbered polynomials are polynomials in ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " multiplied by ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  ".  More precisely ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(2\ n + 1\) \[Element] \
\[DoubleStruckCapitalF]\_p[x]\)]],
  " and  ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(2\ n\) \[Element] 
        y\ \[DoubleStruckCapitalF]\_p[x]\)]],
  ".\n\nFor our sample curve we find that the first five division polynomials \
are\n\n",
  StyleBox["\t",
    FontFamily->"Times New Roman"],
  Cell[BoxData[{
      \(\(\[Psi]\_1\) \((x, y)\) = 
        1, \[IndentingNewLine]\(\[Psi]\_2\) \((x, y)\) = 
        2\ y, \[IndentingNewLine]\(\[Psi]\_3\) \((x, y)\) = 
        18 + 15\ x + 82\ x\^2 + 
          3\ x\^4, \(\[Psi]\_4\) \((x, 
            y)\) = \((61 + 50\ x + 69\ x\^2 + 3\ x\^3 + 47\ x\^4 + 
              4\ x\^6)\)\ y, \[IndentingNewLine]\(\[Psi]\_5\) \((x, y)\) = 
        23 + 67\ x + 11\ x\^2 + 38\ x\^3 + 77\ x\^4 + 43\ x\^5 + 93\ x\^6\), 
    "\[IndentingNewLine]", 
      \(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(+26\)\ x\^7 + 47\ x\^8 + 
        87\ x\^9 + 39\ x\^10 + 5\ \(\(x\^12\)\(.\)\)\)}],
    FontFamily->"Times New Roman"]
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "With ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46\ x + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  " we have that ",
  Cell[BoxData[
      \(TraditionalForm\`\((4, 15)\)\)]],
  " is a point of order 4.  Then we must have ",
  Cell[BoxData[
      \(TraditionalForm\`f\_n[4] = 0\)]],
  " if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`4 | n\)]],
  ".  To check the function ",
  StyleBox["ComputeDivisionPolynomials",
    FontSize->12,
    FontWeight->"Bold"],
  " we calculate ",
  Cell[BoxData[
      \(TraditionalForm\`f\_n[4]\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`2 \[LessEqual] n \[LessEqual] 8\)]],
  " giving\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`f\_2[4] = 2, f\_3[4] = 24, f\_4[4] = 0, 
      f\_5[4] = 47, \[IndentingNewLine]\ \ f\_6[4] = 25, f\_7[4] = 22, 
      f\_8[4] = 0, \)]],
  "\n\t\nas expected, since ",
  Cell[BoxData[
      \(TraditionalForm\`\((4, 15)\) \[Element] E[4]\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`E[4] \[SubsetEqual] E[8]\)]],
  ".  Similarly the point ",
  Cell[BoxData[
      \(TraditionalForm\`\((90, 31)\)\)]],
  " is of order 5 and we find\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`f\_2[90] = 2, f\_3[90] = 76, f\_4[90] = 14, 
      f\_5[90] = 0, f\_6[90] = 21, f\_7[90] = 23\)]],
  ".\n\t\nSo the division polynomials are correct, at least for this \
particular case."
}], "Text",
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  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell[TextData[{
  "5.5 Computing ",
  Cell[BoxData[
      \(TraditionalForm\`n\ P\)]],
  " with the Division Polynomials"
}], "Subsection",
  CellTags->"c:35"],

Cell[TextData[{
  "If ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((x, y)\)\)]],
  " is a point in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_p)\)]],
  " then"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`n\ P = \((x - \(\(\[Psi]\_\(n - 1\)\) \[Psi]\_\(n + \
1\)\)\/\[Psi]\_n\%2, \(\(\[Psi]\_\(n + 2\)\) \[Psi]\_\(n - 1\)\%2 - \
\(\[Psi]\_\(n - 2\)\) \[Psi]\_\(n + 1\)\%2\)\/\(4\ y\ \[Psi]\_n\%3\))\)\)]]], \
"NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "It can be shown that multiplication by ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " is an endomorphism ",
  Cell[BoxData[
      \(TraditionalForm\`\[Mu]\_n\)]],
  " of ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_p)\)]],
  ".  This follows from the fact that ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_p)\)]],
  " is an abelian group so that ",
  Cell[BoxData[
      \(TraditionalForm\`n(P + Q) = n\ P + n\ Q\)]],
  ".   Since ",
  Cell[BoxData[
      \(TraditionalForm\`n\ P = \[ScriptCapitalO]\)]],
  " if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[n]\)]],
  " we have that the kernel of ",
  Cell[BoxData[
      \(TraditionalForm\`\[Mu]\_n\)]],
  " is ",
  Cell[BoxData[
      \(TraditionalForm\`E[n]\)]],
  ".  Further, because ",
  Cell[BoxData[
      \(TraditionalForm\`\[Mu]\_n\)]],
  " is expressed as a separable rational polynomial of degree ",
  Cell[BoxData[
      \(TraditionalForm\`n\^2\)]],
  ", we have that ",
  Cell[BoxData[
      \(TraditionalForm\`#  E[n] = \(deg(\[Mu]\_n) = n\^2\)\)]],
  ".   For a proof of (21) see Washington [7] ",
  "\[Section]",
  " 9.5.\n\nIt should be noted that Equation (21) does not provide an \
efficient way to compute ",
  Cell[BoxData[
      \(TraditionalForm\`n\ P\)]],
  " for specific points ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  ".  Rather, it provides the basis of proof for the characteristic equation \
of the Frobenius (15), one of the key equations used in Schoof's method."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.6 The Frobenius Endomorphism", "Subsection",
  CellTags->"c:36"],

Cell[TextData[{
  "Let ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_q\) : 
          E(\[DoubleStruckCapitalF]\&_\_q) \[Rule] 
        E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_q\)(x, y) = \((x\^q, y\^q)\)\)]],
  ", called the Frobenius endomorphism.  Since ",
  Cell[BoxData[
      \(TraditionalForm\`a\^q = a\)]],
  " for all ",
  Cell[BoxData[
      \(TraditionalForm\`a \[Element] \[DoubleStruckCapitalF]\_q\)]],
  " this map is the identity for points with coordinates in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  ".  Let ",
  Cell[BoxData[
      \(TraditionalForm\`\((x\_1, y\_1)\) \[Element] 
        E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " then \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`y\_1\^2 = x\_1\%3 + A\ x\_1 + B\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\&_\_q\)]],
  ".\n\t\nNow ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_q\)(x\_1, y\_1) = \((x\_1\%q, 
          y\_1\%q)\)\)]],
  ".  Substituting this into the elliptic curve equation gives\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((y\_1\%q)\)\^2 = \(\((y\_1\%2)\)\^q = \
\(\(\((x\_1\%3 + A\ x\_1 + B)\)\^q\)\(.\)\)\)\)]],
  "\n\t\nHowever, for all ",
  Cell[BoxData[
      \(TraditionalForm\`a, b \[Element] \[DoubleStruckCapitalF]\&_\_q\)]],
  " we have ",
  Cell[BoxData[
      \(TraditionalForm\`\((a + b)\)\^q = a\^q + b\^q\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  " so that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((y\_1\%q)\)\^2 = \(\((x\_1\%3)\)\^q + 
            A\^q\ x\_1\%q + \((B)\)\^q = \((x\_1\%q)\)\^3 + A\ x\_1\%q + 
            B\)\)]],
  ", since ",
  Cell[BoxData[
      \(TraditionalForm\`A, B \[Element] \[DoubleStruckCapitalF]\_q\)]],
  ".\n\t\nHence ",
  Cell[BoxData[
      \(TraditionalForm\`\((x\_1\%q, 
          y\_1\%q)\) = \(\[Phi]\_q\)(x\_1, y\_1) \[Element] 
          E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\)]],
  " maps a point on the curve to another point on the curve.\n\nLet ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((x\_1, y\_1)\), Q = \((x\_2, y\_2)\)\)]],
  " be two points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`x\_1 \[NotEqual] x\_2\)]],
  ", then with ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \((y\_2 - y\_1)\)/\((x\_2 - x\_1)\)\)]],
  " we have\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`P + Q = \((x\_3, y\_3)\)\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`x\_3 = \[Lambda]\^2 - x\_1 - x\_2, \ 
      y\_3 = \[Lambda](x\_1 - x\_3) - y\_1\)]],
  ".\n\t\nUsing the same properties of ",
  Cell[BoxData[
      \(TraditionalForm\`q\^th\)]],
  " powers in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\&_\_q\)]],
  " we have\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_q\)(
          P + Q) = \((\[Lambda]\^\(2\ q\) - x\_1\%q - 
            x\_2\%q, \(\[Lambda]\^q\)(x\_1\^q - x\_3\%q) - y\_1\%q)\)\)]],
  " ,\n\t \n\twith ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda]\^q = \((y\_2\%q - y\_1\%q)\)/\((x\_2\%q - 
              x\_1\%q)\)\)]],
  ".\n\t \nTherefore ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_q\)(
          P + Q) = \(\[Phi]\_q\)(P) + \(\[Phi]\_q\)(Q)\)]],
  ".  It can also be shown that this holds also for ",
  Cell[BoxData[
      \(TraditionalForm\`Q = \(P\ and\ Q = \(-P\)\)\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\)]],
  " is a homomorphism from ",
  Cell[BoxData[
      \(TraditionalForm\`\(E(\[DoubleStruckCapitalF]\&_\_q)\) 
        to\ \(E(\[DoubleStruckCapitalF]\&_\_q)\)\)]],
  ", hence an endomorphism. "
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.7 The Characteristic Equation of the Frobenius", "Subsection",
  CellTags->"c:37"],

Cell[TextData[{
  "We now proceed to the equation that provides the foundation for Schoof's \
algorithm.  Remember that ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " is the set of points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " who's order divides ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  ".   First we show that ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " is a subgroup of ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  ". Clearly ",
  Cell[BoxData[
      \(TraditionalForm\`\[ScriptCapitalO] \[Element] E[l]\)]],
  ".  If ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q \[Element] E[l]\)]],
  " then, because ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " is abelian,  ",
  Cell[BoxData[
      \(TraditionalForm\`\[ScriptCapitalO] = \(l\ P + l\ Q = l(P + Q)\)\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`P + Q \[Element] E[l]\)]],
  ".  Further,\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[ScriptCapitalO] = \(l\ \[ScriptCapitalO] = \(l(
              P + \((\(-P\))\)) = \(l\ P + 
                l(\(-P\)) = \(\[ScriptCapitalO] + l(\(-P\)) = 
                l(\(-P\))\)\)\)\)\)]],
  ",\n\t \ntherefore ",
  Cell[BoxData[
      \(TraditionalForm\`\(-P\) \[Element] E[l]\)]],
  ".  Hence ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " is a subgroup of ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  ".\n\nSince ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " is an abelian group of order ",
  Cell[BoxData[
      \(TraditionalForm\`l\^2\)]],
  ", it follows from the structure theorem for finite abelian groups that ",
  Cell[BoxData[
      \(TraditionalForm\`E[
          l] \[TildeFullEqual] \[DoubleStruckCapitalZ]\_l\[CirclePlus]\
\[DoubleStruckCapitalZ]\_l\)]],
  ".  The integer ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  " is a prime so that ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalZ]\_l\)]],
  " is a cyclic group generated by any number ",
  Cell[BoxData[
      \(TraditionalForm\`1 \[LessEqual] \ a\  < l\)]],
  ", and there exists points ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1, \[Beta]\_2 \[Element] E[l]\)]],
  " such that any point in ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " can be written as a ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalZ]\_l\)]],
  "-linear combination ",
  Cell[BoxData[
      \(TraditionalForm\`P = \(m\_1\) \[Beta]\_1 + \(m\_2\) \[Beta]\_2\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`m\_1, 
      m\_2 \[Element] \[DoubleStruckCapitalZ]\_l\)]],
  ".  Suppose ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha]\)]],
  " is any homomorphism of ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha](P) \[Element] E[l]\)]],
  " for all ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " because ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(\[Alpha](P)\)\(|\)\)\)]],
  " divides ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(P\)\(|\)\)\)]],
  ".  Then, in particular, ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha](\[Beta]\_1) = 
        s\ \[Beta]\_1 + t\ \[Beta]\_2\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha](\[Beta]\_2) = 
        u\ \[Beta]\_1 + v\ \[Beta]\_2\)]],
  ", so that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha](
          P) = \(\[Alpha](\(m\_1\) \[Beta]\_1 + \(m\_2\) \[Beta]\_2) = \(m\_1\
\) \(\[Alpha](\[Beta]\_1)\) + \(m\_2\) \(\[Alpha](\[Beta]\_2)\)\)\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(\(=\)\(\(m\_1\) s\ \[Beta]\_1 + \(m\_1\) 
            t\ \[Beta]\_2 + 
          m\_2\ u\ \[Beta]\_1 + \(m\_2\) v\ \(\(\[Beta]\_2\)\(.\)\)\)\)\)]],
  "\n\t\nWe can express this in matrix form as"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  Cell[BoxData[
      FormBox[
        RowBox[{\(\[Alpha](P)\), "=", 
          RowBox[{
            RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                  {"s", "t"},
                  {"u", "v"}
                  }], "\[NegativeThinSpace]", ")"}], 
            RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                  {\(\(m\_1\) \[Beta]\_1\)},
                  {\(\(m\_2\) \[Beta]\_2\)}
                  }], "\[NegativeThinSpace]", ")"}]}]}], TraditionalForm]]],
  ".\n"
}], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "In particular, the action of the Frobenius on ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  ", denoted ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_\(q, l\)\)]],
  " can be described by such a ",
  Cell[BoxData[
      \(TraditionalForm\`2\[Times]2\)]],
  " matrix.  Applied to ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " we have\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`deg(\[Phi]\_q - 
            1) \[Congruent] \(det(\[Phi]\_\(q, l\) - I)\)\ \((mod\ l)\)\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(\(=\)\(\((s - 1)\) \((v - 1)\) - t\ u = 
          s\ v - t\ u - \((s + v)\) + 1\)\)\)]],
  ".\n\t\nBut ",
  Cell[BoxData[
      \(TraditionalForm\`s\ v - u\ t = 
        det(\[Phi]\_\(q, l\)) \[Congruent] q\ \((mod\ l)\)\)]],
  " (by Washington [7] Proposition 3.15).\n\nThen by Hasse's theorem, using \
",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " instead of ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " to avoid conflicting variable names, we have\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`#\ \(ker(\[Phi]\_q - 1)\) = 
        q + 1 - a\  \[Congruent] q - \((s + v)\) + 1 \((mod\ l)\)\)]],
  ".\n\t\nHence we have the following congruences.   First, ",
  Cell[BoxData[
      \(TraditionalForm\`a \[Congruent] \((s + v)\) \((mod\ l)\)\)]],
  ", the trace of ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_\(q, l\)\)]],
  " . Also ",
  Cell[BoxData[
      \(TraditionalForm\`t\ u \[Congruent] 
        s\ v - q\ \(\((mod\ l)\)\(.\)\)\)]],
  "\n\nWe can now compute ",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Phi]\_\(q, l\))\)\^2 - 
        a\ \((\[Phi]\_\(q, l\))\) + q\)]],
  " using *, so that\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            SuperscriptBox[
              RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                    {"s", "t"},
                    {"u", "v"}
                    }], "\[NegativeThinSpace]", ")"}], "2"], "-", 
            RowBox[{"a", " ", 
              RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                    {"s", "t"},
                    {"u", "v"}
                    }], "\[NegativeThinSpace]", ")"}]}], "+", 
            RowBox[{"q", "(", "\[NegativeThinSpace]", GridBox[{
                  {"1", "0"},
                  {"0", "1"}
                  }], "\[NegativeThinSpace]", ")"}]}], "=", 
          RowBox[{
            RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                  {\(s\^2 + t\ u\), \(s\ t\  + t\ v\)},
                  {\(s\ u + u\ v\), \(t\ u\  + v\^2\)}
                  }], "\[NegativeThinSpace]", ")"}], "-", 
            RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                  {\(a\ s - q\), \(a\ t\)},
                  {\(a\ y\), \(a\ v - q\)}
                  }], "\[NegativeThinSpace]", ")"}]}]}], TraditionalForm]]],
  "\n\t\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"=", 
          RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                {\(s\^2 + t\ u - \ a\ s + q\), \(s\ t + t\ v - a\ t\)},
                {\(s\ u + u\ v - a\ u\), \(v\^2 + t\ u\  - a\ v + q\)}
                }], "\[NegativeThinSpace]", ")"}]}], TraditionalForm]]],
  ".\n\t\nApplying the congruences yeild\n\t\n\t",
  Cell[BoxData[{
      FormBox[
        RowBox[{"\[Congruent]", 
          RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                {\(s\^2 + s\ v - q - \((s + v)\) s + 
                    q\), \(\((s + v)\) t - \((s + v)\) t\)},
                {\(\((s + v)\) u - \((s + v)\) u\), \(v\^2 + s\ v - 
                    q - \((s + v)\) v + q\)}
                }], "\[NegativeThinSpace]", ")"}]}], TraditionalForm], 
    "\[IndentingNewLine]", 
      FormBox[
        RowBox[{"\[Congruent]", 
          RowBox[{
            RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
                  {"0", "0"},
                  {"0", "0"}
                  }], "\[NegativeThinSpace]", ")"}], 
            " ", \(\((mod\ l)\)\(.\)\)}]}], TraditionalForm]}]],
  ".\n\t\nTherefore  ",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Phi]\_\(q, l\))\)\^2 - 
          a\ \((\[Phi]\_\(q, l\))\) + q \[Congruent] 0 \((mod\ l)\)\)]],
  " for all ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(l, q) = 1\)]],
  ".   Since there are an infinite number of choices for such ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  ", the kernel of ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\%2 + q - a\ \[Phi]\_q\)]],
  " is not finite, hence,  as stated in Equation (15),\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\%2 + q - a\ \[Phi]\_q = 0\)]],
  " for all ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  ".\n\t \nIn particular, if ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " then,\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_q\%2 + 
          k \[Congruent] \[Tau]\ \[Phi]\_q\ \ \((mod\ l)\)\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`k \[Congruent] q\ \((mod\ l)\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] \[Congruent] a\ \((mod\ l)\)\)]],
  ".\n\t \nSo for a particular point ",
  Cell[BoxData[
      \(TraditionalForm\`P = \((x, y)\) \[Element] E[l]\)]],
  " it must be true that"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    \(TraditionalForm\`\(\(\((x\^\(q\^2\), y\^\(q\^2\))\) + 
        k(x, y) \[Congruent] \(\[Tau](x\^q, 
          y\^q)\)\ \((mod\ l)\)\)\(,\)\)\)], "NumberedEquation",
  FontSize->14],

Cell[TextData[{
  "where addition is performed in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\&_\_q)\)]],
  " using (9) through (12), and scalar point multiplication is performed \
using the division polynomials as in Equation (22).  We can simplify the \
computation of (24) further by noting that if  ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y)\) \[Element] E[l]\)]],
  " then the division polynomial ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Psi]\_l\)(x, y) = 0\)]],
  ", so we can reduce (24) mod ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_l\)]],
  " without changing the set of points which satisfy the equation."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.8 Schoof's Algorithm: Case One", "Subsection",
  CellTags->"c:38"],

Cell[TextData[{
  "In order to use Equation (24) we must first test if ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(\[PlusMinus]k\)\ P\)]],
  " for some ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  ".   We can determine this by computing the test condition for the ",
  StyleBox["x",
    FontSlant->"Italic"],
  " coordinate using ",
  StyleBox["Schoof (16)",
    FontWeight->"Bold"],
  ":\n",
  StyleBox["\n\t",
    FontWeight->"Bold"],
  Cell[BoxData[
      \(TraditionalForm\`x\^\(p\^2\) \[Congruent] 
        x - \(\(\(\[Psi]\_\(k - 
                        1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\) \((mod\ \
f\_l, p)\)\)]],
  ".\n\t\nThis is true if and only if\n",
  StyleBox["\n\t",
    FontWeight->"Bold"],
  Cell[BoxData[
      \(TraditionalForm\`\(p\_16\)(x, 
          y) = \((x\^\(q\^2\) - 
                  x)\) \[Psi]\_k\%2 - \(\[Psi]\_\(k - 
                    1\)\) \[Psi]\_\(k + 1\) \[Congruent] 
          0 \((mod\ f\_l, p)\)\)]],
  ".\n\t\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " even, ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_k = y\ f\_k\)]],
  " and since ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + a\ x\  + b\)]],
  " we obtain\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_16\)(
          x) = \((x\^\(p\^2\) - x)\) \(\(f\_k\%2\)(
              x)\) \((x\^3 + a\ x + b)\) + \(\(f\_\(k - 1\)\)(
              x)\) \(\(f\_\(k + 1\)\)(x)\)\)]],
  ".\n\t\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " odd, ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(k - 1\) = y\ f\_\(k - 1\)\)]],
  " and  ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_\(k + 1\) = y\ f\_\(k + 1\)\)]],
  " so that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_16\)(
          x) = \((x\^\(p\^2\) - x)\) \(\(f\_k\%2\)(x)\) + \(\(f\_\(k - 1\)\)(
              x)\) \(\(f\_\(k + 1\)\)(x)\) \((x\^3 + a\ x + b)\)\)]],
  ".\n\t\nNotice that ",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_16\)(x, y)\)]],
  " is a polynomial in ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " only. Hence if ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(\(p\_16\)(x), \(f\_l\)(x)) \[NotEqual] 1\)]],
  " then some point ",
  Cell[BoxData[
      \(TraditionalForm\`P\)]],
  " exists in ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  "  which satisfies ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(\[PlusMinus]k\)\ P\)]],
  ", so we are in case 1.  Otherwise we must proceed to case 2 where we test \
equation (24) for various values of \[Tau]."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.9 Schoof Equation (17)", "Subsection",
  CellTags->"c:39"],

Cell[TextData[{
  "Given that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(\[PlusMinus]k\)\ P\)]],
  " for some ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Element] {0, \(-2\) w, 2  w}\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`w\^2 \[Congruent] k\ \((mod\ l)\)\)]],
  ".  This is shown as follows.  Suppose ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = k\ P\)]],
  ", then, by equation (24) we have ",
  Cell[BoxData[
      \(TraditionalForm\`2  k \[Congruent] \[Tau]\ \[Phi]\_l\)]],
  ".  Squaring both sides gives ",
  Cell[BoxData[
      \(TraditionalForm\`4 
           k\^2 = \(\(t\^2\) \[Phi]\_l\%2 = \(t\^2\) k\)\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`4  k = t\^2\)]],
  ", then we must have that ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " is a quadratic residue.  If so, find ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`w\^2 \[Congruent] k\ \((mod\ l)\)\)]],
  ",  then ",
  Cell[BoxData[
      \(TraditionalForm\`4\ w\^2 = t\^2\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`t = \(\[PlusMinus]2\)\ w\)]],
  ".  Now we can compute\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Phi]\_l - w)\) \((\[Phi]\_l + 
              w)\) = \(\[Phi]\_l\%2 - k = 0\)\)]],
  ", so ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = \(\[PlusMinus]w\)\ P\)]],
  ".\n\t\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " is not a quadratic residue, we can not be in this case, hence  ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(-k\)\ P\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P + 
          k\ P = \(0 = \(\[Tau]\[Phi]\_l\) P\)\)]],
  " for all ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] \[Congruent] 0 \((mod\ l)\)\)]],
  ". \n\nWe can test if ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = \(\[PlusMinus]\ w\)\ P\)]],
  " using the point multiplication formula (22) again yielding for the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate the test\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`x\^p \[Congruent] 
        x - \(\(\(\[Psi]\_\(w - 
                        1\)\) \[Psi]\_\(w + 1\)\)\/\[Psi]\_w\%2\) \((mod\ \
f\_l, p)\)\)]],
  ".\n\t\nMultiplying through by ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_w\%2\)]],
  " produces Schoof equation (17)\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_17\)(x, 
          y) = \((x\^p - 
                x)\) \[Psi]\_w\%2 - \(\[Psi]\_\(w - 
                  1\)\) \[Psi]\_\(w + 1\)\)]],
  ".\n\t\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " even or odd this can be reduced to a polynomial in ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " only, as in\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_17\)(
          x) = \((x\^p - x)\) \(\(f\_w\%2\)(
              x)\) \((x\^3 + a\ x + b)\) + \(\(f\_\(w - 1\)\)(
              x)\) \(\(f\_\(w + 1\)\)(x)\)\)]],
  "\t",
  StyleBox["w",
    FontSlant->"Italic"],
  " even,\n\t\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_17\)(
          x) = \((x\^p - x)\) \(\(f\_w\%2\)(x)\) + \(\(f\_\(w - 1\)\)(
              x)\) \(\(f\_\(w + 1\)\)(x)\) \((x\^3 + a\ x + b)\)\)]],
  "\t",
  StyleBox["w",
    FontSlant->"Italic"],
  " odd.\n\t\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(\(p\_17\)(x), \(f\_l\)(x)) = 1\)]],
  " then we must have ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(-k\)\ P\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0\ \((mod\ l)\)\)]],
  ".  Otherwise ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(\[PlusMinus]\(w(mod\ l)\)\)\)]],
  " and we test the ",
  StyleBox["y",
    FontSlant->"Italic"],
  "-coordinate of ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = \(\[PlusMinus]\ w\)\ P\)]],
  " to determine the sign."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.10 Schoof Equation (18)", "Subsection",
  CellTags->"c:40"],

Cell[TextData[{
  "After we know that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = \(\[PlusMinus]w\)\ P\)]],
  " we need test the ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  "-coordinate of ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = w\ P\)]],
  ".  Equation (22) gives for the ",
  StyleBox["y",
    FontSlant->"Italic"],
  "-coordinate,\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{\(y\^p\), "\[Congruent]", 
          RowBox[{
            
            StyleBox[\(\(\(\[Psi]\_\(w + 
                            2\)\) \[Psi]\_\(w - 1\)\%2 - \ \(\[Psi]\_\(w - 
                            2\)\) \[Psi]\_\(w + 1\)\%2\)\/\(4\ y\ \
\[Psi]\_w\%3\)\),
              FontSize->16], \((mod\ \[Psi]\_l, p)\)}]}], TraditionalForm]]],
  ".\n\t\nMultiplying through by the denominator of the right hand side and \
collecting terms gives\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x, y) = 
        4\ \ \(\[Psi]\_w\%3\) 
            y\^\(p + 1\) - \(\[Psi]\_\(w + 
                  2\)\) \[Psi]\_\(w - 1\)\%2 - \ \(\[Psi]\_\(w - 
                  2\)\) \[Psi]\_\(w + 1\)\%2\)]],
  ".\n\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " even:\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x) = 
        4 \(\((y\^2)\)\^\(\((p + 3)\)/2\)\) \(\(f\_w\%3\)(
              x)\) - \(\(f\_\(w + 2\)\)(x)\) \(\(f\_\(w - 1\)\%2\)(
              x)\) + \(\(f\_\(w - 2\)\)(x)\) \(\(f\_\(w + 1\)\%2\)(x)\)\)]],
  ".\n\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " odd:\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x) = 
        4 \(\((y\^2)\)\^\(\((p - 1)\)/2\)\) \(\(f\_w\%3\)(
              x)\) - \(\(f\_\(w + 2\)\)(x)\) \(\(f\_\(w - 1\)\%2\)(
              x)\) + \(\(f\_\(w - 2\)\)(x)\) \(\(f\_\(w + 1\)\%2\)(x)\)\)]],
  ".\n\t\nNotice that ",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x)\)]],
  " is also a polynomial in ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " only since all exponents of ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " are even.\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(\(p\_18\)(x), \(f\_l\)(x)) = 1\)]],
  " then there is no ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " for which ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = w\ P\)]],
  ", so ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(-2\) w\ \((mod\ l)\)\)]],
  ", else such a point exists and ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 2\ w\ \((mod\ l)\)\)]],
  ".  This completes the equations required to test for case 1."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.11 Schoof's Algorithm: Case Two", "Subsection",
  CellTags->"c:41"],

Cell[TextData[{
  "If there is no ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(\[PlusMinus]k\)\ P\)]],
  "  then we are in case 2 so we need to test for each ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] \[Element] \[DoubleStruckCapitalZ]/
            l\ \[DoubleStruckCapitalZ]\^x\)]],
  " if there exist ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " such that (24) holds.  In order to perform this test we apply addition \
formulas ",
  Cell[BoxData[
      \(TraditionalForm\`\((5)\), \((9)\)\ and\ \((10)\)\)]],
  " to compute polynomials representing\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((x\^\(p\^2\), y\^\(p\^2\))\) + k(x, y)\)]],
  ",\n\t\nwhere ",
  Cell[BoxData[
      \(TraditionalForm\`k(x, y)\)]],
  " is computed using the division polynomials and equation (22).  Since we \
are in case 2 we know that ",
  Cell[BoxData[
      \(TraditionalForm\`x\_1 \[NotEqual] x\_2\)]],
  " so we can compute ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \((y\_2 - y\_1)\)/\((x\_2 - x\_1)\)\)]],
  ".  We find\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{\(y\_2 - y\_1\), "=", 
          RowBox[{
            
            StyleBox[\(\(\(\[Psi]\_\(k + 
                            2\)\) \[Psi]\_\(k - 1\)\%2 - \ \(\[Psi]\_\(k - 
                            2\)\) \[Psi]\_\(k + 1\)\%2\)\/\(4\ y\ \
\[Psi]\_k\%3\)\),
              FontSize->16], "-", \(y\^\(p\^2\)\)}]}], TraditionalForm]]],
  ",\n\t\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{\(x\_2 - x\_1\), "=", 
          RowBox[{"x", "-", 
            
            StyleBox[\(\(\(\[Psi]\_\(k - 
                          1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\),
              FontSize->16], "-", \(x\^\(p\^2\)\)}]}], TraditionalForm]]],
  "\nSo then\n\t ",
  Cell[BoxData[
      FormBox[
        RowBox[{"\[Lambda]", "=", 
          RowBox[{
            StyleBox[\(\(y\_2 - y\_1\)\/\(x\_2 - x\_1\)\),
              FontSize->16], "=", 
            
            StyleBox[\(\((\(\[Psi]\_\(k + 
                              2\)\) \[Psi]\_\(k - 1\)\%2 - \ \(\[Psi]\_\(k - 
                              2\)\) \[Psi]\_\(k + 1\)\%2 - 
                      4 \( \[Psi]\_k\%3\) y\^\(p\^2 + 1\))\)\/\(4  
                    y\ \(\(\[Psi]\_k\)(\(-\[Psi]\_\(k - 1\)\) \[Psi]\_\(k + 1\
\) - \(\[Psi]\_k\%2\)(x\^\(p\^2\) - x))\)\)\),
              FontSize->18]}]}], TraditionalForm]]],
  ".\n\nPut ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \[Alpha]/\[Beta]\)]],
  " then\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = \(\[Psi]\_\(k + 
                  2\)\) \[Psi]\_\(k - 1\)\%2 - \(\[Psi]\_\(k - 
                  2\)\) \[Psi]\_\(k + 1\)\%2 - 
          4 \( \[Psi]\_k\%3\) y\^\(p\^2 + 1\)\)]],
  "\nand\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = 
        4\ y\ \(\(\[Psi]\_k\)(\(\[Psi]\_k\%2\)(
                x - x\^\(p\^2\)) - \(\[Psi]\_\(k - 
                      1\)\) \[Psi]\_\(k + 1\))\)\)]],
  ".\n\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " even we have\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = 
        y(\(f\_\(k + 2\)\) f\_\(k - 1\)\%2 - \(f\_\(k - 2\)\) 
              f\_\(k + 1\)\%2 - 4 \( f\_k\%3\) y\^\(p\^2 + 3\))\)]],
  "\nand\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = 
        4\ y\^2\ \(\(f\_k\)(\(y\^2\) \(\(f\_k\%2\)(
                  x - x\^\(p\^2\))\) - \(f\_\(k - 1\)\) f\_\(k + 1\))\)\)]],
  "\t.\n\nOtherwise, for ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " odd\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = \(y\^2\)(\(f\_\(k + 2\)\) 
                f\_\(k - 1\)\%2 - \(f\_\(k - 2\)\) f\_\(k + 1\)\%2) - 
          4 \( f\_k\%3\) y\^\(p\^2 + 1\)\)]],
  "\nand\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = 
        y(4\ \(\(f\_k\)(\(f\_k\%2\)(
                  x - x\^\(p\^2\)) - \(y\^2\) \(f\_\(k - 1\)\) 
                  f\_\(k + 1\))\))\)]],
  ".\n\t\nWe use these equations for \[Alpha] and \[Beta] to formulate the \
tests of Schoof equations (19)."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.12 Schoof Equation (19x)", "Subsection",
  CellTags->"c:42"],

Cell[TextData[{
  "Using the equations we just derived for ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \[Alpha]/\[Beta]\)]],
  " we can now compute the addition of points using equation (22) so that if\n\
\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\((x\_3, y\_3)\) = \((x\^\(p\^2\), y\^\(p\^2\))\) + 
          k(x, y)\)]],
  "\n\t\nthen ",
  Cell[BoxData[
      \(TraditionalForm\`x\_3\)]],
  " is given by\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{\(x\_3\), "=", 
          RowBox[{\(\[Lambda]\^2 - x\_1 - x\_2\), "=", 
            
            RowBox[{\(\[Alpha]\^2\/\[Beta]\^2\), "-", \((x\^\(p\^2\) + x)\), 
              "+", 
              
              StyleBox[\(\(\(\[Psi]\_\(k - 
                            1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\),
                FontSize->16]}]}]}], TraditionalForm]]],
  ".\nAlso ",
  Cell[BoxData[
      \(TraditionalForm\`y\_3\)]],
  " is given by\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{\(y\_3\), "=", 
          RowBox[{\(\[Lambda](2  x\_1 + x\_2 - \[Lambda]\^2) - y\_1\), "=", 
            RowBox[{
              RowBox[{\(\[Alpha]\/\[Beta]\), 
                RowBox[{"(", 
                  RowBox[{\(2\ x\^\(p\^2\)\), "+", "x", "-", 
                    
                    StyleBox[\(\(\(\[Psi]\_\(k - 
                                  1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\),
                      FontSize->16], 
                    StyleBox["-",
                      FontSize->16], \(\[Alpha]\^2\/\[Beta]\^2\)}], ")"}]}], 
              "-", \(y\^\(p\^2\)\)}]}]}], TraditionalForm]]],
  ".\n\t\nFurther, since  ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Psi]\_n\)(x\^p, 
          y\^p) = \(\(\[Psi]\_n\)(x, y)\)\^p\)]],
  " we have\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau](x\^p, 
          y\^p) = \((\((x - \(\(\[Psi]\_\(\[Tau] - 1\)\) \[Psi]\_\(\[Tau] + 1\
\)\)\/\[Psi]\_\[Tau]\%2)\)\^p, \((\(\(\[Psi]\_\(\[Tau] + 2\)\) \[Psi]\_\(\
\[Tau] - 1\)\%2 - \ \(\[Psi]\_\(t - 2\)\) \[Psi]\_\(\[Tau] + 1\)\%2\)\/\(4\ y\
\ \[Psi]\_\[Tau]\%3\))\)\^p)\)\)]],
  ".\n\nThen (23) holds if and only if (for the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate)\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{
          
          RowBox[{\(\[Alpha]\^2\/\[Beta]\^2\), "-", \((x\^\(p\^2\) + x)\), 
            "+", 
            
            StyleBox[\(\(\(\[Psi]\_\(k - 
                          1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\),
              FontSize->16]}], 
          StyleBox["=",
            FontSize->16], 
          SuperscriptBox[
            RowBox[{
              StyleBox["(",
                
                FontSize->
                  16], \(x - \(\(\[Psi]\_\(\[Tau] - 1\)\) \[Psi]\_\(\[Tau] + \
1\)\)\/\[Psi]\_\[Tau]\%2\), ")"}], "p"]}], TraditionalForm]]],
  ".\n\t\nExpanding and clearing the (nonzero) denominators gives\n\n\t",
  Cell[BoxData[{
      \(TraditionalForm\`\(\[Psi]\_\[Tau]\%\(2  
              p\)\)(\[Alpha]\ \[Psi]\_k\%2 - \(\[Beta]\^2\) \
\(\(\[Psi]\_k\%2\)(
              x\^\(p\^2\) + 
                x)\) + \(\[Beta]\^2\)(\(\[Psi]\_\(k - 
                    1\)\) \[Psi]\_\(k + 1\)))\), "\[IndentingNewLine]", 
      \(TraditionalForm\`\ \ \(\(=\)\(\(\[Psi]\_k\%2\) \(\(\[Beta]\^2\)(\(\
\[Psi]\_\[Tau]\%\(2  p\)\) 
                x\^p - \((\(\[Psi]\_\(\[Tau] - 1\)\) \[Psi]\_\(\[Tau] - \
1\))\)\^p)\)\)\)\)}]],
  ".\n\t\nBringing everything to the left hand side we have\n\n\t",
  Cell[BoxData[{
      \(TraditionalForm\`\(p\_\(19\_x\)\)(x, 
          y) = \(\[Psi]\_\[Tau]\%\(2  
                p\)\)(\(\[Beta]\^2\)(\(\[Psi]\_\(k - 
                      1\)\) \[Psi]\_\(k + 1\) - \(\[Psi]\_k\%2\)(
                x\^\(p\^2\) + x\^p + x) + \[Alpha]\ \[Psi]\_k\%2))\), 
    "\[IndentingNewLine]", 
      \(TraditionalForm\`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(+\ \[Psi]\_k\
\%2\) \(\(\[Beta]\^2\)(\(\[Psi]\_\(\[Tau] - 1\)\) \[Psi]\_\(\[Tau] - \
1\))\)\^p = 0\)}]],
  ",\n\t\nwhich is Schoof equation (19) for the ",
  StyleBox["x",
    FontSlant->"Italic"],
  "-coordinate.  Since we are testing this equation for points in ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " we perform all of the polynomial arithmetic modulo ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_l\)]],
  ".\n\nNow there exists a point in ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " satisfying ",
  Cell[BoxData[
      \(TraditionalForm\`p\_\(19\_x\)\)]],
  " if and only if ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_\(19\_x\), f\_l) \[NotEqual] 1\)]],
  ".  If such a point exist, then ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(\[PlusMinus]\(\[Tau](
            mod\ l)\)\)\)]],
  ".  We use Schoof equation (19y), explained in the next section, to \
determine the sign of \[Tau]."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell["5.13 Schoof Equation (19y)", "Subsection",
  CellTags->"c:43"],

Cell[TextData[{
  "Once we know that there exists ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y)\) \[Element] E[l]\)]],
  " such that\n\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\((x\^\(p\^2\), y\^\(p\^2\))\) + 
          k(x, y) = \(\[PlusMinus]\(\[Tau](x\^p, y\^p)\)\)\)\(,\)\)\)],
    FontSize->14],
  "\n\nwe must test the",
  StyleBox[" y",
    FontSlant->"Italic"],
  "-coordinate to determine the sign of ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau]\)]],
  ".  We have from the previous section,\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{\(y\_3\), "=", 
          RowBox[{
            RowBox[{\(\[Alpha]\/\[Beta]\), 
              RowBox[{"(", 
                RowBox[{\(2\ x\^\(p\^2\)\), "+", "x", "-", 
                  
                  StyleBox[\(\(\(\[Psi]\_\(k - 
                                1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\),
                    FontSize->16], 
                  StyleBox["-",
                    FontSize->16], \(\[Alpha]\^2\/\[Beta]\^2\)}], ")"}]}], 
            "-", \(y\^\(p\^2\)\)}]}], TraditionalForm]]],
  ".\n\t\nThen if (24) holds for the ",
  StyleBox["y",
    FontSlant->"Italic"],
  "-coordinate we have\n\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            RowBox[{\(\[Alpha]\/\[Beta]\), 
              RowBox[{"(", 
                RowBox[{\(2\ x\^\(p\^2\)\), "+", "x", "-", 
                  
                  StyleBox[\(\(\(\[Psi]\_\(k - 
                                1\)\) \[Psi]\_\(k + 1\)\)\/\[Psi]\_k\%2\),
                    FontSize->16], 
                  StyleBox["-",
                    FontSize->16], \(\[Alpha]\^2\/\[Beta]\^2\)}], ")"}]}], 
            "-", \(y\^\(p\^2\)\)}], 
          "=", \(\((\(\(\[Psi]\_\(t + 2\)\) \[Psi]\_\(t - 1\)\%2 - \ \(\[Psi]\
\_\(t - 2\)\) \[Psi]\_\(t + 1\)\%2\)\/\(4\ y\ \[Psi]\_t\%3\))\)\^p\)}], 
        TraditionalForm]]],
  ".\n\t\nMultiplying through by ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Beta]\^3\) \(\[Psi]\_k\%2\) 
        4 \( y\^p\) \[Psi]\_t\%\(3  p\)\)]],
  " to clear the denominators we have\n\n\t",
  Cell[BoxData[{
      FormBox[
        RowBox[{"4", " ", \(y\^p\), " ", \(\[Psi]\_t\%\(3  p\)\), 
          RowBox[{"(", 
            RowBox[{
              RowBox[{"\[Alpha]", " ", 
                RowBox[{\(\[Beta]\^2\), "(", 
                  RowBox[{\(\(\[Psi]\_k\%2\)(2\ x\^\(p\^2\) + x)\), "-", 
                    StyleBox[\(\(\[Psi]\_\(k - 1\)\) \[Psi]\_\(k + 1\)\),
                      FontSize->16]}], 
                  StyleBox[")",
                    FontSize->16]}]}], 
              StyleBox["-",
                
                FontSize->
                  16], \(\(\[Psi]\_k\%2\)(\[Alpha]\^3 + \(\[Beta]\^3\) 
                    y\^\(p\^2\))\)}], ")"}]}], TraditionalForm], 
    "\[IndentingNewLine]", 
      FormBox[
        RowBox[{
          "        ", \(\(=\)\(\(\[Beta]\^3\) \(\(\(\(\[Psi]\_k\%2\) \((\(\
\[Psi]\_\(t + 2\)\) \[Psi]\_\(t - 1\)\%2 - \ \(\[Psi]\_\(t - 
                                  2\)\) \[Psi]\_\(t + 1\)\%2)\)\)\^p\)\(.\)\)\
\)\)}], TraditionalForm]}]],
  "\n\t\nRearranging gives for Schoof (19y)  (corrected)\n\n\t",
  Cell[BoxData[{
      \(TraditionalForm\`\(p\_\(19  y\)\)(x, 
          y) = \(\(4\) \( 
              f\_\[Tau]\%\(3\ p\)\) \(y\^p\)\(\ \)\((\((\((2\ x\^\(p\^2\) + 
                          x)\)\ \[Alpha]\ \[Beta]\^2 - \[Beta]\^3\ \
y\^\(p\^2\) - \[Alpha]\^3)\)\ f\_k\%2 - \[Alpha]\ \(\[Beta]\^2\) 
                f\_\(k - 1\)\ f\_\(k + 1\)\ )\)\(\ \)\)\), 
    "\[IndentingNewLine]", 
      \(TraditionalForm\`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\(-\[Beta]\^3\)\ f\_k\%2\ \(\(\((f\_\(\[Tau] - 1\)\%2\ f\_\(\[Tau] + 2\) - 
                  f\_\(\[Tau] - 2\)\ f\_\(\[Tau] + \
1\)\%2)\)\^p\)\(.\)\)\)}]],
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  StyleBox["For ",
    FontSize->16],
  Cell[BoxData[
      \(TraditionalForm\`k\)],
    FontSize->14],
  StyleBox[" even, \[Tau] even, we must take ",
    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] \[Rule] y\ \[Alpha]\)],
    FontSize->14],
  " so that,\n\t\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_\(19  y\)\)(x) = 
        4 \( f\_\[Tau]\%\(3\ p\)\) 
            y\^\(3\ p - 1\)\ \((\((\((2\ x\^\(p\^2\) + 
                            x)\)\ \ \[Alpha]\ \[Beta]\^2 - \[Beta]\^3\ y\^\(p\
\^2 - 1\) - \(y\^2\) \[Alpha]\^3)\)\ \(y\^2\) 
                  f\_k\%2\[IndentingNewLine] - \[Alpha]\ \(\[Beta]\^2\) 
                  f\_\(k - 1\)\ f\_\(k + 1\)\ )\)\  - \(\[Beta]\^3\) 
            f\_k\%2\ \(\(\((f\_\(\[Tau] - 1\)\%2\ f\_\(\[Tau] + 2\) - 
                      f\_\(\[Tau] - 2\)\ f\_\(\[Tau] + \
1\)\%2)\)\^p\)\(.\)\)\)]],
  "\n   \n",
  StyleBox["For ",
    FontSize->16],
  Cell[BoxData[
      \(TraditionalForm\`k\)],
    FontSize->14],
  StyleBox[" even, \[Tau] odd,\n\n",
    FontSize->14],
  "\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_\(19  y\)\)(x) = 
        4  f\_\[Tau]\%\(3\ p\)\ \((\((\((2\ x\^\(p\^2\) + 
                            x)\)\ \[Alpha]\ \ \[Beta]\^2 - \[Beta]\^3\ y\^\(p\
\^2 - 1\) - \[Alpha]\^3\ y\^2)\)\ \(y\^2\) 
                  f\_k\%2\[IndentingNewLine] - \[Alpha]\ \ \(\[Beta]\^2\) 
                  f\_\(k - 1\)\ f\_\(k + 1\)\ )\)\  - \(\[Beta]\^3\) 
            f\_k\%2\ \(\(\(\(y\^\(p + 1\)\)(
                    f\_\(\[Tau] - 1\)\%2\ f\_\(\[Tau] + 2\) - 
                      f\_\(\[Tau] - 2\)\ f\_\(\[Tau] + \
1\)\%2)\)\^p\)\(.\)\)\)]],
  "\n   \n",
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    FontSize->14],
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      \(TraditionalForm\`k\)],
    FontSize->14],
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    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] \[Rule] y\ \[Beta]\)],
    FontSize->14],
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  Cell[BoxData[
      \(TraditionalForm\`\(p\_\(19  y\)\)(x) = 
        4 \( f\_\[Tau]\%\(3\ p\)\) 
            y\^\(3  p - 3\)\ \((\((\((2\ x\^\(p\^2\) + 
                            x)\)\ \[Alpha]\ y\^2\ \[Beta]\^2 - \[Beta]\^3\ \
y\^\(p\^2 + 3\) - \[Alpha]\^3)\)\ f\_k\%2\[IndentingNewLine] - \[Alpha]\ y\^2\
\ \(\[Beta]\^2\) f\_\(k - 1\)\ f\_\(k + 1\)\ )\)\  - \(\[Beta]\^3\) 
            f\_k\%2\ \(\(\((f\_\(\[Tau] - 1\)\%2\ f\_\(\[Tau] + 2\) - 
                      f\_\(\[Tau] - 2\)\ f\_\(\[Tau] + \
1\)\%2)\)\^p\)\(.\)\)\)]],
  "\n       \n",
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    FontSize->14],
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      \(TraditionalForm\`k\)],
    FontSize->14],
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    FontSize->14],
  "\n      ",
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        4  f\_\[Tau]\%\(3\ p\)\ \((\((\((2\ x\^\(p\^2\) + 
                            x)\)\ \[Alpha]\ \(y\^2\) \[Beta]\^2 - \[Beta]\^3\ \
y\^\(p\^2 + 3\) - \[Alpha]\^3)\)\ f\_k\%2\[IndentingNewLine] - \[Alpha]\ y\^2\
\ \(\[Beta]\^2\) 
                  f\_\(k - 1\)\ f\_\(k + 1\)\ )\)\  - \[Beta]\^3\ f\_k\%2\ \(\
\(\(\(y\^\(p + 3\)\)(
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                      f\_\(\[Tau] - 2\)\ f\_\(\[Tau] + \
1\)\%2)\)\^p\)\(.\)\)\)]],
  "\n      \nNow if ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_\(19  y\), f\_l) \[NotEqual] 1\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " has a point satisfying ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P + k\ P = \[Tau]\[Phi]\ P\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \[Tau](mod\ l)\)]],
  ", else ",
  Cell[BoxData[
      \(TraditionalForm\`t = \(-\(\[Tau](mod\ l)\)\)\)]],
  "."
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  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell["5.14 Schoof's Algorithm Summary", "Subsection",
  PageBreakAbove->True,
  CellTags->"c:44"],

Cell[TextData[{
  "We can now summarize Schoof's algorithm for ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x\  + b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " as follows.  By Hasse's theorem we have ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ".\n\n\t1. If ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x\^3 + a\ x + b, \ x\^p - x) = 1\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0\ \((mod\ 2)\)\)]],
  ", else ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 1\ \((mod\ 2)\)\)]],
  "\n\t2. Create a set of small primes ",
  Cell[BoxData[
      \(TraditionalForm\`S = {l\_i}\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Product]\+\(i = 1\)\%L l\_i > 4\ \@p\)]],
  ".\n\t3. Compute the first ",
  Cell[BoxData[
      \(TraditionalForm\`l\_L + 2\)]],
  " division polynomials ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_k\)]],
  ".\n\t4. For each ",
  Cell[BoxData[
      \(TraditionalForm\`l \[Element] S\)]],
  ", compute ",
  Cell[BoxData[
      \(TraditionalForm\`k \[Congruent] p(\ mod\ l)\)]],
  "\n\t5. \tIf ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_16, f\_l) \[NotEqual] 1\)]],
  " then there exists ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " such that  ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(\[PlusMinus]k\)\ P\)]],
  ".\n\t6.\t\tIf ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " is not a quadratic residue mod ",
  StyleBox["l",
    FontSlant->"Italic"],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ l)\)\)]],
  " else\n\t7.\t\tCompute ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`w\^2 \[Congruent] k\ \((mod\ l)\)\)]],
  "\n\t8.\t\tIf ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_17, f\_l) = 1\)]],
  " then  ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ l)\)\)]],
  ", else\n\t9.\t\tIf ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_18, f\_l) \[NotEqual] 1\)]],
  " then  ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 2 \( w(mod\ l)\)\)]],
  ", else  ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(-2\) \(w(mod\ l)\)\)]],
  ". \n\t10.\telse we are in case two\n\t11.\t    For each ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] \[LessEqual] \((l + 1)\)/2\)]],
  "\n\t12.\t\tIf ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(gcd(p\_19, f\_l)\),
            "TraditionalForm"], "\[NotEqual]", "1"}], TraditionalForm]]],
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  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_p\^2 + 
          k \[Congruent] \(\[PlusMinus]\[Tau]\)\ \[Phi]\_p\ \((mod\ l)\)\)]],
  "\n\t13.\t\t    for some point in ",
  Cell[BoxData[
      \(TraditionalForm\`E[l]\)]],
  " so we test\n\t14.\t\t\tIf ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_19, f\_l) \[NotEqual] 1\)]],
  " then  ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \[Tau](mod\ l)\)]],
  " else ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(-\[Tau]\)\ \((mod\ l)\)\)]],
  " \n\t15.\t    Next \[Tau]\n\t16. Next ",
  Cell[BoxData[
      \(TraditionalForm\`l\)]],
  "\n\t17. At this point we have computed ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ l\_i)\)\)]],
  " for all ",
  Cell[BoxData[
      \(TraditionalForm\`l\_i \[Element] S\)]],
  ", \n\t18.\tso we use the Chinese Remainder Theorem to compute\n\t19.\t     \
",
  Cell[BoxData[
      \(TraditionalForm\`T \[Congruent] t\ \((mod\ N)\)\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`N = \[Product]\+\(i = 1\)\%L l\_i\)]],
  ".\n\t20. If ",
  Cell[BoxData[
      \(TraditionalForm\`T\)]],
  " is within Hasse's bounds then ",
  Cell[BoxData[
      \(TraditionalForm\`t = T\)]],
  ", else ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(-\(T(mod\ N)\)\)\)]],
  " and\n\t21.\t",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ".\n\nThis completes the description of Schoof's algorithm."
}], "Text",
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  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell["Chapter 6 - Results of Running Schoof's Algorithm", "Section",
  PageBreakAbove->True,
  CellTags->"c:45"],

Cell["\<\
This chapter contains the results of running our implementation of Schoof's \
algorithm for several different elliptic curves.   We present detailed \
results for one particular curve and then summarize the results for other \
curves in table 1.  We conclude this section with a discussion of lessons \
learned from these experiments.\
\>", "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["6.1 A Detailed Example", "Subsection",
  PageBreakAbove->False,
  CellTags->"c:54"],

Cell[TextData[{
  "For our example curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 46  x + 74\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_97\)]],
  ", Schoof's algorithm produces the following results.  First, since ",
  Cell[BoxData[
      \(TraditionalForm\`\[LeftCeiling]4 \@ p\[RightCeiling] = 40\)]],
  ", we need a product of small primes at least this large so the algorithm \
selects the primes ",
  Cell[BoxData[
      \(TraditionalForm\`{2, 5, 7}\)]],
  ", with ",
  Cell[BoxData[
      \(TraditionalForm\`\[Product]l\_i = 70\)]],
  ".\nAlso ",
  Cell[BoxData[
      \(TraditionalForm\`9\  < \@97 < 10\)]],
  ", so Hasse's theorem gives ",
  Cell[BoxData[
      \(TraditionalForm\`79 \[LessEqual] \ # \( 
            E(\[DoubleStruckCapitalF]\_q)\)\  \[LessEqual] 117\)]],
  ".\n\nThe next step is to compute ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ 2)\)\)]],
  ".  For this step we find\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(x\^p\)(\ mod\ x\^3 + a\ x\  + b) = 
        47 + 60\ x + 30\ x\^2\)]],
  " and\n\t",
  Cell[BoxData[
      \(TraditionalForm\`gcd(x\^p - x, \ x\^3 + a\ x + b) = 40\  + \ x\)]],
  "\n\t\nSince the gcd is not equal to ",
  Cell[BoxData[
      \(TraditionalForm\`1\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`E[2]\)]],
  " is not empty so ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((\ mod\ 2)\)\)]],
  ".\n\nNext we test ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_p\^2\) P = \(\[PlusMinus]k\)\ P\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`k \[Congruent] p(mod\ 5)\)]],
  " and we find  \n",
  StyleBox["\n\t",
    FontFamily->"Times New Roman"],
  Cell[BoxData[{
      RowBox[{"\<\"f[\"\>", 
        StyleBox["\[InvisibleSpace]",
          "Text"], "5", 
        StyleBox["\[InvisibleSpace]",
          "Text"], "\<\"] =\"\>", 
        StyleBox["\[InvisibleSpace]",
          "Text"], \(23 + 67\ x + 11\ x\^2 + 38\ x\^3 + 77\ x\^4 + 
          43\ x\^5\)}], "\[IndentingNewLine]", 
      RowBox[{
        " ", \(\(+\ 93\)\ x\^6 + 26\ x\^7 + 47\ x\^8 + 87\ x\^9 + 39\ x\^10 + 
          5\ x\^12\)}]}],
    FontFamily->"Times New Roman"],
  StyleBox["\n\t\n\t",
    FontFamily->"Times New Roman"],
  Cell[BoxData[{
      RowBox[{"\<\"\\!\\(p\\_16\\) = \"\>", 
        StyleBox["\[InvisibleSpace]",
          "Text"], \(7 + 91\ x + 40\ x\^2 + 24\ x\^3 + 81\ x\^4 + 
          69\ x\^5\)}], "\[IndentingNewLine]", 
      RowBox[{
        " ", \(\(+\ 43\)\ x\^6 + 45\ x\^7 + 39\ x\^8 + 14\ x\^9 + 30\ x\^10 + 
          79\ x\^11\)}]}],
    FontFamily->"Times New Roman"],
  StyleBox["\n",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  StyleBox["\n\t",
    FontFamily->"Times New Roman"],
  Cell[BoxData[
      \(TraditionalForm\`gcd(\(p\_16\) f\_l) = 1. \)]],
  StyleBox["\n",
    FontFamily->"Times New Roman"],
  "\t\nHence, there is no point in ",
  Cell[BoxData[
      \(TraditionalForm\`E[5]\)]],
  " satisfying  ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_p\^2\) P = \(\[PlusMinus]k\)\ P\)]],
  " so we proceed to case two.\n\nNext we test ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_p\^2 + k = \[Tau]\ \[Phi]\_p\)]],
  " until we find for ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] = 2\)]],
  " that\n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`p\_\(19  x\) \[Congruent] 0 \((mod\ f\_l, p)\)\)]],
  " and",
  StyleBox["\n\n\t",
    FontFamily->"Times New Roman"],
  Cell[BoxData[{
      RowBox[{
        RowBox[{"\<\"gcd(\\!\\(p\\_\\(19  x\\)\\),\\!\\(f\\_l\\)) = \"\>", 
          StyleBox["\[InvisibleSpace]",
            "Text"], "23"}], "+", \(67\ x\), "+", \(11\ x\^2\), 
        "+", \(38\ x\^3\), "+", \(77\ x\^4\), "+", \(43\ x\^5\)}], 
    "\[IndentingNewLine]", 
      RowBox[{
        "  ", \(\(+\ 93\)\ x\^6 + 26\ x\^7 + 47\ x\^8 + 87\ x\^9 + 
          39\ x\^10 + 5\ \(\(x\^12\)\(.\)\)\)}]}],
    FontFamily->"Times New Roman"],
  StyleBox["\n",
    FontFamily->"Times New Roman"],
  "\t\nSince the gcd is not equal to 1, we know that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] = \(\[PlusMinus]2\)\)]],
  " so we compute\n\n\t",
  Cell[BoxData[{
      RowBox[{"\<\"p19y =\"\>", 
        StyleBox["\[InvisibleSpace]",
          "Text"], \(39 + 52\ x + 48\ x\^2 + 33\ x\^3 + 91\ x\^4 + 
          3\ x\^5\)}], "\[IndentingNewLine]", 
      RowBox[{
        " ", \(\(+\ 23\)\ x\^6 + 59\ x\^7 + 16\ x\^8 + 37\ x\^9 + 33\ x\^10 + 
          74\ x\^11\)}]}],
    FontFamily->"Times New Roman"],
  StyleBox["\n\t\n\t",
    FontFamily->"Times New Roman"],
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_\(19  y\), f\_l) = 1. \)]],
  StyleBox["\n",
    FontFamily->"Times New Roman"],
  "\t\nSince this gcd is 1, there is no point in ",
  Cell[BoxData[
      \(TraditionalForm\`E[5]\)]],
  " satisfying  ",
  Cell[BoxData[
      \(TraditionalForm\`\[Phi]\_p\^2 + k = \[Tau]\ \[Phi]\_p\)]],
  ", so we must have ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(-2\) \((mod\ 5)\) \[Congruent] 
        3 \((mod\ 5)\)\)]],
  ".  Similarly, for ",
  Cell[BoxData[
      \(TraditionalForm\`l = 7\)]],
  " we find at ",
  Cell[BoxData[
      \(TraditionalForm\`\[Tau] = 3\)]],
  " that ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_\(19  x\), f\_7) \[NotEqual] 1\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`gcd(p\_\(19  y\), f\_7) = 1\)]],
  " so that ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] \(-3\) \((mod\ 7)\) \[Congruent] 
        4 \((mod\ 7)\)\)]],
  ".  Thus we have the following set of simultaneous conguences. \n\n\t",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ 2)\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 3 \((mod\ 5)\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 4 \((mod\ 7)\)\)]],
  ".\n\t\nUsing the Chinese Remainder theorem we find that the smallest \
positive integer satisfying this set of congruences is ",
  Cell[BoxData[
      \(TraditionalForm\`t = 18\)]],
  ".  Since ",
  Cell[BoxData[
      \(TraditionalForm\`p + 1 - t = 80\)]],
  " and 80 is within Hasse's bounds we can conclude ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 80. \)]]
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Cell[CellGroupData[{

Cell["6.2 Other Experiments", "Subsection",
  CellTags->"c:54"],

Cell[TextData[{
  "For many other curves our implementation of Schoof's method also produces \
the correct results, as verified by point counting.  For the curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + 1333  x + 1129\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_3571\)]],
  ", however, the computed point order is incorrect.  The correct order, as \
determined by direct numerical tests is 3559, giving ",
  Cell[BoxData[
      \(TraditionalForm\`t = 13\)]],
  ", so that ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 1 \((mod\ 2)\), \ 
      t \[Congruent] 2\ \((mod\ 11)\)\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Congruent] 0 \((mod\ 13)\)\)]],
  ".  Our implementation of Schoof's algorithm wrongly determines that ",
  Cell[BoxData[
      \(TraditionalForm\`t \[Equal] \(-2\) \((mod\ 11)\)\)]],
  ".  Analysis of this problem indicates that the most likely source of the \
error  is an incorrect computation for the ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  "-coordinate part of Schoof equation (19).  We are continuing efforts to \
resolve this issue."
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Cell[TextData[{
  StyleBox["Table 1 summarizes the results of running Schoof's algorithms on \
several curves.  The first column describes the curve ",
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      \(TraditionalForm\`y\^2 = x\^3 + A\ x + B\)],
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  StyleBox[" over ",
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)],
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
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    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
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      \(TraditionalForm\`A, B, p\)],
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  StyleBox[".  The second column show the results of computing ",
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    FontColor->RGBColor[0, 0, 1]],
  Cell[BoxData[
      \(TraditionalForm\`\[Tau]\_i \[Congruent] t\ \((mod\ l\_i)\)\)],
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    FontColor->RGBColor[0, 0, 1]],
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    FontColor->RGBColor[0, 0, 1]],
  Cell[BoxData[
      \(TraditionalForm\`{\[Tau]\_i, l\_i}\)],
    FontSize->14,
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      \(TraditionalForm\`t\)],
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  StyleBox[" within Hasses's bounds that satisfies the congruences of column \
two.   ",
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    FontColor->RGBColor[0, 0, 1]],
  StyleBox[" is then computed as ",
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  Cell[BoxData[
      \(TraditionalForm\`p + 1 - t\)],
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  StyleBox[". ",
    FontSize->14,
    FontColor->RGBColor[0, 0, 1]],
  StyleBox["\n ",
    FontSize->14],
  "\n                                                ",
  StyleBox["Table 2 - Results from Schoof's Algorithm",
    FontWeight->"Bold"],
  "\n   ",
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          {\(A, B, p\), \(t \((mod\ l\_i)\)\), 
            "t", \(#  E \((\[DoubleStruckCapitalF]\_p)\)\), "Correct"},
          {\(13, 215, 229\), \({0, 2}, {0, 5}, {4, 7}\), \(-10\), "240", 
            "yes"},
          {\(106, 166, 197\), \({0, 2}, {2, 3}, {0, 5}, {1, 11}\), \(-10\), 
            "208", "yes"},
          {\(31, 16, 137\), \({1, 2}, {0, 3}, {4, 5}, {2, 7}\), "9", "129", 
            "yes"},
          {\(503, 367, 523\), \({1, 2}, {0, 5}, {6, 7}, {7, 11}\), \(-15\), 
            "539", "yes"},
          {\(1333, 1129, 3571\), \({1, 2}, {9, 11}, {0, 13}\), \(-13\), 
            "3585", "no"}
          },
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Cell[CellGroupData[{

Cell["6.3 Discussion of Results", "Subsection",
  CellTags->"c:54"],

Cell[TextData[{
  "Schoof's method, as here implemented, has only educational value.  This is \
so for several reasons.  First, as noted in references [3,6], the \
practicality of the algorithm is greatly limited by the quadratic growth of \
the degree of the division polynomials.  For example, for an elliptic curve \
over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " has 200 digits, we must perform modular polynomial arithmetic using the \
",
  Cell[BoxData[
      \(TraditionalForm\`55\^th\)]],
  " division polynomial, which produces intermediate products of degree \
greater than ",
  Cell[BoxData[
      \(TraditionalForm\`9\[Cross]10\^6\)]],
  ".  The improvements due to Elkies and Atkin, often called the SEA \
algorithm, reduce to nearly linear growth the degree of the division \
polynomials, making the method applicable to elliptic curves of with  \
cryptographic utility.    Another limiting factor is the peformance of \
modular polynomial arithmetic in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ".   Even for the small problems presented in this section, the run-time of \
our implemention exceeded 10 minutes.   Finally, the current implementation \
is limited by an apparent software bug in the implementation of Schoof \
equation (19y).\n\nOn the other hand, the major advantage of our \
implementation is as an exploration tool.  All of the algorithms we \
implemented are well-documented, and rely only on low-level functions with ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", making the operation of the algorithms transparent and open to \
experimentation.   The author plans to correct the programming difficiencies \
just cited and to make this suite of elliptic curve functions  available as a \
",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " package."
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Cell[CellGroupData[{

Cell["Chapter 7 - Applications", "Section",
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  CellTags->"c:52"],

Cell[CellGroupData[{

Cell["7.1 The Elliptic Curve Discrete Log Problem", "Subsection",
  CellTags->"c:53"],

Cell[TextData[{
  "One way to apply elliptic curve methods to cryptography is via the \
discrete log problem, which can be described as follows.   Suppose ",
  Cell[BoxData[
      \(TraditionalForm\`G\)]],
  " is a group of finite order and ",
  Cell[BoxData[
      \(TraditionalForm\`g, q\)]],
  " are elements of G.  Then suppose there exists an integer ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`q = g\^k\)]],
  ".  Given the elements ",
  Cell[BoxData[
      \(TraditionalForm\`g, q\)]],
  "  we are asked to find the corresponding exponent ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  ".  Since this looks similar to taking logarithms, we call this the \
discrete log problem or the DLP.  For example, we could choose ",
  Cell[BoxData[
      \(TraditionalForm\`G = \[DoubleStruckCapitalF]\_q\%x\)]],
  ", the multiplicative group of the finite field ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\)]],
  ". \n\nIf ",
  Cell[BoxData[
      \(TraditionalForm\`G\)]],
  " is a finite abelian group and ",
  Cell[BoxData[
      \(TraditionalForm\`g\)]],
  " is an element of the group, then for any integer ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " we define ",
  Cell[BoxData[
      \(TraditionalForm\`k\[CenterDot]g = \(g + g +  ... \) + g\)]],
  " (a total of ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " times).  If ",
  Cell[BoxData[
      \(TraditionalForm\`q = k\[CenterDot]g\)]],
  " and we give someone the two elements ",
  Cell[BoxData[
      \(TraditionalForm\`g, q\)]],
  " and asks them to find the integer ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  ", then we have given them the discrete log problem in additive notation.  \
Since the elements of ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " form an abelian group we can state the discrete log problem for elliptic \
curves in the following way.  Given two points ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q \[Element] E(\[DoubleStruckCapitalF]\_q)\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`Q = k\ P\)]],
  ", find ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  ".  For a given field order ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  ", this problem is understood to be more difficult than the equivalent \
problem in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_q\%x\)]],
  "."
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  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell["7.2 Anomalous Curves and the MOV attack", "Subsection",
  CellTags->"c:54"],

Cell[TextData[{
  "An elliptic curve  is called ",
  StyleBox["anomalous",
    FontWeight->"Bold"],
  " if ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\) = q\)]],
  ".  When this occurs, the discrete log problem for the group ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " can be solved in linear time using an algorithm by Satoh & Araki, \
outlined in Washington [7] \[Section] 5.4.\n\nThe MOV attack, first presented \
by Menezes, Okamoto and Vanstone, uses the Weil pairing to convert the \
discrete log problem for ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_q)\)]],
  " into a discrete log problem for ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_\(q\^s\)\)]],
  ".  When ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  ", the characteristic of the field, divides the trace of the Frobenius ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  ", then ",
  Cell[BoxData[
      \(TraditionalForm\`s\)]],
  " will be small.  In this case the index calculus can be used to attack the \
DLP in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_\(q\^s\)\)]],
  "  (Washington [7] \[Section]5.3).  Hence, when selecting an elliptic curve \
for cryptographic applications, it is important to determine the order of the \
resulting group to avoid cases which simplify the discrete log problem for \
potential eavesdroppers.  Ren\[EAcute] Schoof's method, and in particular its \
extensions, provide a way of efficiently determining ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_q)\)\)]],
  " for large integers ",
  Cell[BoxData[
      \(TraditionalForm\`q\)]],
  ", thus assisting in the creation of secure codes."
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Cell[CellGroupData[{

Cell["References", "Section",
  PageBreakAbove->True,
  CellTags->"c:55"],

Cell[TextData[{
  "[1] Brown, Ezra - ",
  StyleBox["Square Roots from 1; 24, 51, 10 to Dan Shanks",
    FontSlant->"Italic"],
  ", College Math Journal (1999).\n\n[2] Dummit, D., Foote, R. ",
  StyleBox["Abstract Algebra",
    FontSlant->"Italic"],
  ", John Wiley and Sons (2004)\n\n[3] Lercier, R. and Morain, F., ",
  StyleBox["Counting the number of points on elliptic curves over finite \
fields: strategies and performances",
    FontSlant->"Italic"],
  ", Advances in Cryptology, Proc. Eurocrypt'95, LNCS 921, L.C. Guillou and \
J.J. Quisquater, Eds., Springer-Verlag, 1995, pp. 79--94.\n\n[4] Rosen, \
Kenneth H. -  ",
  StyleBox["Elementary Number Theory and its Applications, ",
    FontSlant->"Italic"],
  "Addison and Wesley",
  " (2000)\n\n[5] Schoof, Ren\[EAcute] - ",
  StyleBox["Elliptic curves over finite fields and the computation of square \
roots mod p,",
    FontSlant->"Italic"],
  " Mathematics of Computation, Vol. 44, No 170 (1985), 482-494\n\n[6] \
Schoof, Ren\[EAcute] - ",
  StyleBox["Counting points on elliptic curves over finite fields",
    FontSlant->"Italic"],
  ", Journal de T\[EAcute]orie des Nombres de Bordeaux 7 (1995), 219-254\n\n\
[7] Washington, Lawrence - ",
  StyleBox["Elliptic Curves - Number Theory and Cryptography,",
    FontSlant->"Italic"],
  " Chapman & Hall (2003)"
}], "Text",
  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "Appendix A - Dictionary of Our ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " Functions for Elliptic Curves"
}], "Section",
  PageBreakAbove->True,
  CellTags->"c:56"],

Cell[TextData[{
  "This section gives a summary of the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " functions we developed in order to implement, test and explain Schoof's \
algorithm.  Many of these functions have value independent of Schoof's method \
and may be useful for other elliptic curve computation projects."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[CellGroupData[{

Cell["Number Theoretic Functions", "Subsubsection",
  CellTags->"c:57"],

Cell[TextData[{
  "EuclideanAlgorithm[ ",
  Cell[BoxData[
      \(TraditionalForm\`a, b\)]],
  " ]\n\tCompute the greatest common divisor ",
  Cell[BoxData[
      \(TraditionalForm\`d\)]],
  " of the two integers ",
  Cell[BoxData[
      \(TraditionalForm\`a, b\)]],
  ".\n\t\n",
  Cell[BoxData[
      \(TraditionalForm\`ExtendedEuclideanAlgorithm[a, b]\)]],
  "\n\tCompute ",
  Cell[BoxData[
      \(TraditionalForm\`d = gcd(a, b)\)]],
  " and return ",
  Cell[BoxData[
      \(TraditionalForm\`{d, r, s}\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`d = r\ a + s\ b\)]],
  "\n\t\n",
  Cell[BoxData[
      \(TraditionalForm\`MultiplicativeInverse[a, p]\)]],
  "\n\tCompute ",
  Cell[BoxData[
      \(TraditionalForm\`b\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`a\ b \[Congruent] 1 \((mod\ p)\)\)]],
  "\n\t\nModularExponentiation[ ",
  Cell[BoxData[
      \(TraditionalForm\`a, n, p\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`a\^n\ \((mod\ p)\)\)]],
  "\n\t\nSquareRootModPShanksTonelli[ ",
  Cell[BoxData[
      \(TraditionalForm\`a, p\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`c\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`c\^2 \[Congruent] a\ \((mod\ p)\)\)]],
  " if ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " is a quadratic residue mod ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  ".\n\t\n",
  Cell[BoxData[
      \(TraditionalForm\`EcAddMod[\ {a, b}, P\_1, P\_2, p\ ]\)]],
  "\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`P\_1 + P\)]],
  " using elliptic curve addition in\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x\  + b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".\n\t\nEcPowerMod[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, Q, k, p\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`k\ Q\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x\  + b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".\n\t\nDetermineChineseRemainder[ ",
  Cell[BoxData[
      \(TraditionalForm\`t\)]],
  " ]\n\tUses the Chinese Remainder Theorem to solve the set of congruences\n\
\t",
  Cell[BoxData[
      \(TraditionalForm\`z \[Congruent] r\_i\ \((mod\ n\_i)\)\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`i = 1, 2,  ... , k\)]],
  " for the unknown ",
  Cell[BoxData[
      \(TraditionalForm\`z\)]],
  ",\n\twhere these congruences are represented by the set of ordered pairs\n\
\t",
  Cell[BoxData[
      \(TraditionalForm\`t = {{\(r\_\(\(1\)\(,\)\)\) n\_1}, {r\_2, 
            n\_2},  ... , {r\_k, n\_k}}\)]]
}], "Text",
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  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell["Elliptic Curve Functions", "Subsubsection",
  CellTags->"c:58"],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`EcAdd[{\ a, b}, P, Q]\)]],
  "\n\tAdd the points ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q\)]],
  " on the elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x\  + b\)]],
  "\n\t\n",
  Cell[BoxData[
      \(TraditionalForm\`IsEcQ[\ {a, b}, p\ ]\)]],
  "\n\tReturns true if ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + b\)]],
  " is an elliptic curve, that is,\n\tif ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(E\)\(\ \)\)\)]],
  "has nonzero discriminant modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  "\n\t\nEcPointQ[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, {x, y}, p\)]],
  " ]\n\tReturn true if ",
  Cell[BoxData[
      \(TraditionalForm\`\((x, y)\)\)]],
  " satistfies  ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + a\ x + b\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".\n\t\nEcAddMod[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, P, Q, p\)]],
  " ]\n\tAdd the points ",
  Cell[BoxData[
      \(TraditionalForm\`P, Q\)]],
  " on the elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x\  + b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  "\n\t\nEcPowerMod[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, Q, k, p\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`k\ Q\)]],
  " on the elliptic curve ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x\  + b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  "\n\t\nEcPointOrderMod[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, Q, p, m\)]],
  " ]\n\tDetermines the value of ",
  Cell[BoxData[
      \(TraditionalForm\`k \[LessEqual] m\)]],
  " for which ",
  Cell[BoxData[
      \(TraditionalForm\`k\ Q = \[ScriptCapitalO]\)]],
  ", if any.\n\t\nFindEcPointSet[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, p\)]],
  " ]\n\tDetermines all points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " for the curve ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + a\ x + \ b\)]],
  "\n\tby exhaustive search.\n\n",
  Cell[BoxData[
      \(TraditionalForm\`ComputeOrderEFq[\ t, p, n\ ]\)]],
  "\n\tGiven that ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ", this method computes ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_\(p\^n\))\)\)]],
  " as\n\t\t",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_\(p\^n\))\) = 
        p\^n + 1 - \((\[Alpha]\^n + \[Beta]\^n)\)\)]],
  ", \n\t\twhere ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\(X\^2 + t\ X + p\)\(=\)\),
            "TraditionalForm"], \((X - \[Alpha])\), \((X - \[Beta])\)}], 
        TraditionalForm]]],
  "\n\t\tand ",
  Cell[BoxData[
      \(TraditionalForm\`\((\[Alpha]\^n + \[Beta]\^n)\)\)]],
  " is computed using a recursion relation\n\t\t\nEcPointOrderBabyGiant[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, Q, p\)]],
  " ]\n\tComputes the order of the point ",
  Cell[BoxData[
      \(TraditionalForm\`Q\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + \ b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  "\n\tusing the Baby Step, Giant Step method.\n\t\nGenerateRandomPointEC[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, p\)]],
  " ]\n\tFinds a random point on ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + \ b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  "\n\t\nEcGroupOrder[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, p\)]],
  " ]\n\tDetermines ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + \ b\)]],
  ".\n\tIt uses EcPointOrderBabyGiant on random points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  "\n\tuntil the least common multiple of their orders implies a\n\tunique \
value of the group order within Hasse's bounds."
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  TextJustification->0,
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]]
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Cell[CellGroupData[{

Cell["Functions that Comprise Schoof's Algorithm", "Subsubsection",
  CellTags->"c:59"],

Cell[TextData[{
  "ComputeTModTwo[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, p\)]],
  " ]\n\tDetermines ",
  Cell[BoxData[
      \(TraditionalForm\`t(mod\ 2)\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\) = 
        p + 1 - t\)]],
  ".\n\t\nComputePrimeSet[ ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " ]\n\tDetermines the set of small primes ",
  Cell[BoxData[
      \(TraditionalForm\`{l\_i}\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`\[Sum]\+\(i = 1\)\%L l\_i > 4 \@ p\)]],
  ".\n\t\nComputeDivisionPolynomials[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, p, k\)]],
  " ]\n\tComputes the first ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " division polynomials for\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + a\ x + \ b\)]],
  " over ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  ".\n\t\nComputeEquation16[ ",
  Cell[BoxData[
      \(TraditionalForm\`p, l, {a, b}\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_16\)(x, 
          y) = \((x\^\(q\^2\) - 
                x)\) \[Psi]\_k\%2 - \(\[Psi]\_\(k - 
                  1\)\) \[Psi]\_\(k + 1\)\ \((mod\ f\_l, p)\)\)]],
  "..\n\t\nComputeEquation17[ ",
  Cell[BoxData[
      \(TraditionalForm\`p, l, w, {a, b}\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_17\)(x, 
          y) = \((x\^p - 
                x)\) \[Psi]\_w\%2 - \(\[Psi]\_\(w - 
                  1\)\) \[Psi]\_\(w + 1\)\ \((mod\ f\_l, p)\)\)]],
  ".\n\t\nComputeEquation18[ ",
  Cell[BoxData[
      \(TraditionalForm\`p, l, w, {a, b}\)]],
  " ]\n\tComputes ",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x, y) = 
        4\ \ \(\[Psi]\_w\%3\) 
            y\^\(p + 1\) - \(\[Psi]\_\(w + 
                  2\)\) \[Psi]\_\(w - 1\)\%2 - \ \(\[Psi]\_\(w - 
                  2\)\) \[Psi]\_\(w + 1\)\%2\ \((mod\ f\_l, p)\)\)]],
  ".\n\t\nComputeTmodLCaseOne[ ",
  Cell[BoxData[
      \(TraditionalForm\`p, l, {a, b}\)]],
  " ]\n\tUses Schoof equations 16, 17, 18 to compute ",
  Cell[BoxData[
      \(TraditionalForm\`t(mod\ l)\)]],
  ".\n\t\nComputeAlpha[ ",
  Cell[BoxData[
      \(TraditionalForm\`p, l, {a, b}\)]],
  " ]\n\tGiven ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \[Alpha]/\[Beta]\)]],
  " for the elliptic curve addition ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_p\%2\) P + k\ P\)]],
  ",\n\tcompute the polynomial representing \[Alpha].\n\t\nComputeBeta[ ",
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  " Code for Our Elliptic Curve Functions"
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  FontColor->RGBColor[0, 0, 1]],

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Cell[TextData[{
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Cell[TextData[{
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Cell[CellGroupData[{

Cell[TextData[{
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Cell["Elliptic Curve Arithmetic Algorithms", "Subsection",
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                  "\[IndentingNewLine]", \(m = \((y2 - y1)\)/\((x2 - 
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              "\[IndentingNewLine]", \(x3 = m\^2 - x1 - x2\), ";", 
              "\[IndentingNewLine]", \(y3 = m \((x1 - x3)\) - y1\), ";", 
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Cell[TextData[Cell[BoxData[
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is  ",
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  " over ",
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Cell[TextData[Cell[BoxData[
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Cell[TextData[{
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  " if and only if the right hand side has distinct roots.  This is true if \
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  "."
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Cell[TextData[Cell[BoxData[
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  " is the characteristic of the field, which must be prime.  The function \
accepts and returns ",
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              "\[IndentingNewLine]", \(x3 = Mod[m\^2 - x1 - x2, p]\), ";", 
              "\[IndentingNewLine]", \(y3 = Mod[m \((x1 - x3)\) - y1, p]\), 
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Cell[TextData[{
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Cell[TextData[{
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  Cell[BoxData[
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  ".  This algorithm is similar to the modular exponentiation method, in that \
it uses the binary representation of ",
  Cell[BoxData[
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  " to convert the problem into a series of doublings and additions in ",
  Cell[BoxData[
      \(TraditionalForm\`E\)]],
  ".  The algorithm is based on Washington ",
  "\[Section]",
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          s = q; \[IndentingNewLine]While[
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                    p];\[IndentingNewLine], \[IndentingNewLine]i = 
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                  EcAddMod[{a, b}, r, s, 
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\[IndentingNewLine]Return[r];\[IndentingNewLine]];\)\)], "Input"]
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Cell[TextData[Cell[BoxData[
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Cell[TextData[{
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  Cell[BoxData[
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              RowBox[{"(*", " ", 
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              RowBox[{"(*", " ", 
                
                StyleBox[\(d\^\(-1\)\ \((mod\ p)\)\ exists\ because\ \
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              "\[IndentingNewLine]", \(j = Mod[1728*4*a3*di, p]\), ";", 
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Cell[TextData[{
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Cell[TextData[{
  "This function computes the order of a point ",
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  " on the elliptic curve ",
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      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " as the smallest\n\t ",
  Cell[BoxData[
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  " on ",
  Cell[BoxData[
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  ".\nIt works by direct search so is suitable only if the order is small.  \
It tries orders up to ",
  Cell[BoxData[
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  Cell[BoxData[
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            "\[IndentingNewLine]", \(k = 3\), ";", "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(compute\  - P\ \((mod\ p)\)\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(pn3 = Mod[p1*{1, \(-1\)}, p]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(while\ \((k - 1)\) P\  \[NotEqual] \ \(-P\)\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", 
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              RowBox[{\(p2 \[NotEqual] pn3 && k < m\), ",", 
                "\[IndentingNewLine]", 
                RowBox[{"(*", " ", 
                  StyleBox[\(kP\  = \ P\  + \ \((k - 1)\) P\),
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                "\[IndentingNewLine]", \(p2 = 
                  EcAddMod[ec, p2, p1, p]; \[IndentingNewLine]\(++k\);\)}], 
              "\[IndentingNewLine]", "]"}], ";", "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(did\ not\ find\ a\ solution\  < \ m, \ return\ 0\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(If[k \[GreaterEqual] m, Return[0]]\), 
            ";", "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(return\ the\ order\ of\ P\  = \ k\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(Return[k]\), ";"}]}], 
        "\[IndentingNewLine]", "]"}]}]], "Input"]
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Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`FindEcPointSet[\ {a, b}, p\ ]\)]]], "Subsubsection",
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  CellTags->"c:80"],

Cell[TextData[{
  "Find all points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  " by testing every value of ",
  Cell[BoxData[
      \(TraditionalForm\`x \[Element] \[DoubleStruckCapitalF]\_p\)]],
  " to determine if there are values of ",
  Cell[BoxData[
      \(TraditionalForm\`y \[Element] \[DoubleStruckCapitalF]\_p\)]],
  " satisfying ",
  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x\  + B\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

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          RowBox[{\({s, x, e1, x1, y1, y2}\), ",", "\[IndentingNewLine]", 
            
            RowBox[{\(e1 = x\^3 + a\ x\  + \ b\), ";", "\[IndentingNewLine]", 
              
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                      RowBox[{\(x1 = 0\), ",", \(x1 < p\), ",", \(++x1\), ",",
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                        RowBox[{\(y2 = Mod[e1 /. x \[Rule] x1, p]\), ";", 
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                          RowBox[{"If", "[", 
                            
                            RowBox[{\(y2 \[Equal] 0\), ",", 
                              "\[IndentingNewLine]", \(Sow[{x1, 0, 0}];\), 
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                              "\[IndentingNewLine]", 
                              RowBox[{"(*", " ", 
                                
                                StyleBox[\(If\ y2\ has\ a\ square\ root\ mod\ \
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                                  FontSlant->"Italic",
                                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
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                              RowBox[{
                                RowBox[{"If", "[", 
                                  
                                  RowBox[{\(QuadraticResidueQ[y2, p]\), ",", 
                                    "\[IndentingNewLine]", 
                                    RowBox[{"(*", " ", 
                                      
                                      StyleBox[\(find\ the\ smallest\ y1\  \
\[SuchThat] \ y1\^2\  \[Congruent] \ y2 \((mod\ p)\)\),
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                                      "*)"}], 
                                    "\[IndentingNewLine]", \(y1 = 
                                      SqrtModPShanksTonelli[y2, 
                                        p]; \[IndentingNewLine]Sow[{x1, y1, 
                                        Mod[\(-y1\), p]}];\)}], 
                                  "\[IndentingNewLine]", "]"}], ";"}]}], 
                            "\[IndentingNewLine]", "]"}], ";"}]}], 
                      "\[IndentingNewLine]", "]"}], ";"}], 
                  "\[IndentingNewLine]", "]"}]}], ";", 
              "\[IndentingNewLine]", \(Return[First[Last[s]]]\), ";"}]}], 
          "\[IndentingNewLine]", "]"}]}], ";"}]], "Input"]
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Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`ProduceEcPointTable[\ {a, b}, 
      p\ ]\)]]], "Subsubsection",
  CellTags->"c:81"],

Cell[TextData[{
  "This function produces a table of all points in ",
  Cell[BoxData[
      \(TraditionalForm\`E(\[DoubleStruckCapitalF]\_p)\)]],
  ", along with the order of each point.  It returns the order of the group \
",
  Cell[BoxData[
      \(TraditionalForm\`# \( E(\[DoubleStruckCapitalF]\_p)\)\)]],
  " by counting the points."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
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        RowBox[{\({s, c, t}\), ",", "\[IndentingNewLine]", 
          RowBox[{"(*", " ", 
            
            StyleBox[\(Find\ the\ set\ of\ points\ on\ y\^\(\(2\)\(\ \)\) = \ 
                x\^3 + a\ x\  + \ b\ over\ \[DoubleStruckF]\_P\),
              FontSlant->"Italic",
              FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
          "\[IndentingNewLine]", 
          
          RowBox[{\(s = FindEcPointSet[{a, b}, p]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              
              StyleBox[\(Compute\ #  
                  E \((\[DoubleStruckCapitalF]\_p)\)\ by\ counting\ the\ \
points, \ including\ \[ScriptCapitalO]\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(c = 
              2*Length[s] + 1 - Length[Select[s, Last[#] \[Equal] 0 &]]\), 
            ";", "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              
              StyleBox[\(Determine\ the\ order\ of\ each\ point\ an\ put\ it\ \
in\ the\ last\ column\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(t = 
              Map[Append[#, EcPointOrderMod[{a, b}, Take[#, 2], p, 2*p]] &, 
                s]\), ";", "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(Produce\ a\ nice\ table\ with\ headings\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(t2 = 
              Prepend[t, {"\<x\>", "\<\!\(y\_1\)\>", "\<\!\(y\_2\)\>", \
"\<Ord(P)\>"}]\), ";", 
            "\[IndentingNewLine]", \(Print[
              FrameBox[
                  GridBox[t2, RowLines \[Rule] True, 
                    ColumnLines \[Rule] True]] // DisplayForm]\), ";", 
            "\[IndentingNewLine]", \(Return[c]\), ";"}]}], 
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Cell[CellGroupData[{

Cell["Methods to Determine the Elliptic Curve Group Order", "Subsection",
  PageBreakAbove->True,
  CellTags->"c:82"],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`ComputeOrderEFq[\ t, p, n\ ]\)]]], "Subsubsection",
  CellTags->"c:83"],

Cell[TextData[{
  "Given that  ",
  Cell[BoxData[
      \(TraditionalForm\`#  E[\[DoubleStruckCapitalF]\_p] = p + 1 - t\)]],
  ", this method computes and returns ",
  Cell[BoxData[
      \(TraditionalForm\`#  E[\[DoubleStruckCapitalF]\_\(p\^n\)]\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    \(\(ComputeOrderEFq[t_, p_, n_] := 
        Module[{sn0, sn1, sn2, k}, \[IndentingNewLine]sn0 = 
            2; \[IndentingNewLine]sn1 = t; \[IndentingNewLine]For[k = 2, 
            k \[LessEqual] n, \(++k\), \[IndentingNewLine]sn2 = 
              t*\ sn1 - p*\ sn0; \[IndentingNewLine]sn0 = 
              sn1; \[IndentingNewLine]sn1 = 
              sn2;\[IndentingNewLine]]; \[IndentingNewLine]Return[
            p\^n + 1 - sn2];\[IndentingNewLine]];\)\)], "Input"]
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Cell[CellGroupData[{

Cell[TextData[{
  "EcPointOrderBabyGiant[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, Q, p\)]],
  " ]"
}], "Subsubsection",
  CellTags->"c:84"],

Cell[TextData[{
  "Use the BabyStep-GiantStep method to find order of the point ",
  Cell[BoxData[
      \(TraditionalForm\`Q\)]],
  ", that is\n\tfind ",
  Cell[BoxData[
      \(TraditionalForm\`m \[Element] \[DoubleStruckCapitalZ]\_q \[SuchThat] 
          m\ Q = \(\[Infinity]\ \ on\ the\ elliptic\ curve\ \(E : y\^2\) = 
          x\^3 + a\ x + b\)\)]],
  ".\n\tThis method is described in Washington[7] ",
  "\[Section]",
  " 4.3."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

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          RowBox[{\({Q1, i, j, k, m, b, t1, t2, p2, pm3, pm4, m1, f1, x1, q2, 
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            RowBox[{"(*", " ", 
              StyleBox[\(Check\ for\ small\ order\ directly\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", 
            
            RowBox[{\(k = EcPointOrderMod[ec, P1, p, Min[100, p]]\), ";", 
              "\[IndentingNewLine]", \(If[k > 0, \ Return[k]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[\(Compute\ Q\  = \ \((p + 1)\) P\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(Q1 = EcPowerMod[ec, P1, p + 1, p]\), 
              ";", "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Table\ size\ m\  = \ \[LeftCeiling]\@p\%4\
\[RightCeiling]\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(m = Ceiling[Power[p, 1/4]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[\(Create\ table\ of\ iP\ for\ i = 1, m\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(t1 = {P1}\), ";", 
              "\[IndentingNewLine]", \(p2 = P1\), ";", 
              "\[IndentingNewLine]", \(For[i = 2, 
                i \[LessEqual] m, \(++i\), \[IndentingNewLine]p2 = 
                  EcAddMod[ec, p2, P1, p]; \[IndentingNewLine]AppendTo[t1, 
                  p2];\[IndentingNewLine]]\), ";", 
              "\[IndentingNewLine]", \(t2 = First[Transpose[t1]]\), ";", 
              "\[IndentingNewLine]", \(q2 = Q1\), ";", 
              "\[IndentingNewLine]", \(q3 = Q1\), ";", "\[IndentingNewLine]", 
              
              RowBox[{"(*", " ", 
                
                StyleBox[\(Create\ table\ of\ Q\  + \ 
                      2  kmP\ for\ k = \(-m\),  ... , m\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], "*)"}], 
              "\[IndentingNewLine]", \(pm3 = EcPowerMod[ec, P1, 2*m, p]\), 
              ";", "\[IndentingNewLine]", \(pm4 = pm3\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[\(Compute\  - P \((mod\ p)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(pn5 = Mod[P1*{1, \(-1\)}, p]\), ";", 
              "\[IndentingNewLine]", \(pn6 = EcPowerMod[ec, pn5, 2*m, p]\), 
              ";", "\[IndentingNewLine]", \(pn7 = pn6\), ";", 
              "\[IndentingNewLine]", \(k = 0\), ";", "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Compute\ for\ k = \(\[PlusMinus]1\), \
\(\[PlusMinus]2\),  ... \),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", 
              RowBox[{"While", "[", 
                RowBox[{"True", ",", "\[IndentingNewLine]", 
                  RowBox[{"(*", " ", 
                    
                    StyleBox[\(Until\ Q\  + \ 
                          2  kmP\  = \ \(\[PlusMinus]jP\)\ or\  ... \),
                      FontSlant->"Italic",
                      FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                  "\[IndentingNewLine]", 
                  
                  RowBox[{\(If[
                      MemberQ[t2, 
                        First[q2]], \[IndentingNewLine]\(Break[];\)\
\[IndentingNewLine]]\), ";", "\[IndentingNewLine]", 
                    RowBox[{"(*", " ", 
                      
                      StyleBox[\(Until\ Q\  - \ 
                            2  kmP\  = \ \(\[PlusMinus]jP\)\ or\  ... \),
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                    "\[IndentingNewLine]", \(If[
                      MemberQ[t2, 
                        First[q3]], \[IndentingNewLine]k = \(-k\); \
\[IndentingNewLine]q2 = q3; \[IndentingNewLine]Break[];\[IndentingNewLine]]\),
                     ";", "\[IndentingNewLine]", 
                    RowBox[{"(*", " ", 
                      
                      StyleBox[\(Compute\ next\ Q\  \[PlusMinus] \ 
                          2  kmP\ for\ next\ k\),
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                    "\[IndentingNewLine]", \(q2 = EcAddMod[ec, Q1, pm4, p]\), 
                    ";", "\[IndentingNewLine]", \(q3 = 
                      EcAddMod[ec, Q1, pn7, p]\), ";", 
                    "\[IndentingNewLine]", \(pm4 = 
                      EcAddMod[ec, pm4, pm3, p]\), ";", 
                    "\[IndentingNewLine]", \(pn7 = 
                      EcAddMod[ec, pn7, pn6, p]\), ";", 
                    "\[IndentingNewLine]", \(++k\), ";"}]}], 
                "\[IndentingNewLine]", "]"}], ";", "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Here\ we\ have\ Q\  \[PlusMinus] \ 
                      2  kmP\  = \ \(\[PlusMinus]jP\), \ figure\ out\ j\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(j = 
                First[Flatten[Position[t2, First[q2]]]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[\(Determine\ which\  \[PlusMinus] j\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p2 = t1[\([j]\)]\), ";", 
              "\[IndentingNewLine]", \(If[
                Last[q2] \[NotEqual] 
                  Last[p2]\ , \[IndentingNewLine]\(If[
                    Last[q2] \[Equal] 
                      Mod[\(-Last[p2]\), 
                        p], \[IndentingNewLine]\(j = \(-j\);\)\
\[IndentingNewLine], \[IndentingNewLine]Print["\<ERROR: Q + kmP \[NotEqual] \
\[PlusMinus]jP\>"]; \[IndentingNewLine]Return[
                      0];\[IndentingNewLine]];\)\[IndentingNewLine]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Compute\ M\  \[SuchThat] \ Order \((P)\)\  | \ M\),
                  
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(m1 = p + 1 + 2*m*k - j\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(TBD\  - \ 
                      This\ fails\ if\ m1\  = \ \(0\  = \(\(\(>\)\(\ \
\)\(j\)\)\  = \ p\  + \ 1\  + \ 2*m*k\)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[1, 0, 0]], " ", "*)"}], 
              "\[IndentingNewLine]", \(If[m1\  \[Equal] \ 0, \ 
                Print["\<ERROR - M == 0\>"]; Return[0]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(If\ M\ is\ prime, \ 
                  it\ must\ be\ the\ order\ of\ P\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(If[PrimeQ[m1], \ Return[m1]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Factor\ can\ be\ VERY\ hard\ if\ no\ small\ \
factors\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(f1 = FactorInteger[m1]\), ";", 
              "\[IndentingNewLine]", \(b = True\), ";", "\[IndentingNewLine]",
               
              RowBox[{"While", "[", 
                RowBox[{"b", ",", "\[IndentingNewLine]", 
                  RowBox[{\(b = False\), ";", "\[IndentingNewLine]", 
                    RowBox[{"For", "[", 
                      
                      RowBox[{\(i = 1\), ",", \(i \[LessEqual] Length[f1]\), 
                        ",", \(++i\), ",", "\[IndentingNewLine]", 
                        RowBox[{"(*", " ", 
                          StyleBox[\(Divide\ out\ a\ prime\ factor\),
                            FontSlant->"Italic",
                            FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                        "\[IndentingNewLine]", 
                        
                        RowBox[{\(x1 = m1/First[f1[\([i]\)]]\), ";", 
                          "\[IndentingNewLine]", 
                          RowBox[{"(*", " ", 
                            RowBox[{
                              RowBox[{
                                RowBox[{
                                  StyleBox["If",
                                    FontSlant->"Italic",
                                    FontColor->RGBColor[0, 0, 1]], 
                                  StyleBox[" ",
                                    FontSlant->"Italic",
                                    FontColor->RGBColor[0, 0, 1]], 
                                  StyleBox["x",
                                    FontSlant->"Italic",
                                    FontColor->RGBColor[0, 0, 1]], 
                                  StyleBox[" ",
                                    FontSlant->"Italic",
                                    FontColor->RGBColor[0, 0, 1]], 
                                  StyleBox["P",
                                    FontSlant->"Italic",
                                    FontColor->RGBColor[0, 0, 1]], 
                                  StyleBox[\((mod\ p)\),
                                    FontSlant->"Plain",
                                    FontColor->RGBColor[0, 0, 1]]}], 
                                StyleBox[" ",
                                  FontSlant->"Italic",
                                  FontColor->RGBColor[0, 0, 1]], 
                                StyleBox["=",
                                  FontSlant->"Italic",
                                  FontColor->RGBColor[0, 0, 1]], 
                                StyleBox[" ",
                                  FontSlant->"Italic",
                                  FontColor->RGBColor[0, 0, 1]], 
                                StyleBox["P",
                                  FontSlant->"Italic",
                                  FontColor->RGBColor[0, 0, 1]]}], 
                              StyleBox[",",
                                FontSlant->"Italic",
                                FontColor->RGBColor[0, 0, 1]], 
                              StyleBox[" ",
                                FontSlant->"Italic",
                                FontColor->RGBColor[0, 0, 1]], 
                              StyleBox[\(M\  = \ x\),
                                FontSlant->"Italic",
                                FontColor->RGBColor[0, 0, 1]]}], " ", "*)"}], 
                          "\[IndentingNewLine]", \(pm3 = 
                            EcPowerMod[ec, P1, x1 - 1, p]\), ";", 
                          "\[IndentingNewLine]", \(If[
                            pm3 \[Equal] pn5, \[IndentingNewLine]m1 = 
                              x1; \[IndentingNewLine]b = 
                              True; \[IndentingNewLine]If[\(--f1[\([i, 
                                      2]\)]\)\  \[LessEqual] \ 0, \ 
                              f1 = Delete[f1, 
                                  i]]; \[IndentingNewLine]Break[];\
\[IndentingNewLine]]\), ";"}]}], "\[IndentingNewLine]", "]"}], ";"}]}], 
                "\[IndentingNewLine]", "]"}], ";", 
              "\[IndentingNewLine]", \(Return[m1]\), ";"}]}], 
          "\[IndentingNewLine]", "]"}]}], ";"}]], "Input"]
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Cell[CellGroupData[{

Cell[TextData[{
  "GenerateRandomPointEC[ ",
  Cell[BoxData[
      \(TraditionalForm\`{a, b}, p\)]],
  " ]"
}], "Subsubsection",
  CellTags->"c:85"],

Cell[TextData[{
  "Generate a random point ",
  Cell[BoxData[
      \(TraditionalForm\`{x, y}\)]],
  " on the elliptic curve given by ",
  Cell[BoxData[
      \(TraditionalForm\`ec = {A, B}\)]],
  " such that\n\t",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + A\ x\  + B\)]],
  " with all arithmetic modulo ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  "."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    RowBox[{
      RowBox[{\(GenerateRandomPointEC[{a_, b_}, p_]\), ":=", 
        RowBox[{"Module", "[", 
          RowBox[{\({x, e1, x1, y1, y2}\), ",", "\[IndentingNewLine]", 
            
            RowBox[{\(e1 = x\^3 + a\ x\  + \ b\), ";", 
              "\[IndentingNewLine]", \(y2 = 1\), ";", "\[IndentingNewLine]", 
              RowBox[{"While", "[", 
                RowBox[{"True", ",", "\[IndentingNewLine]", 
                  
                  RowBox[{\(x1 = Random[Integer, {1, p - 1}]\), ";", 
                    "\[IndentingNewLine]", \(y2 = 
                      Mod[e1 /. x \[Rule] x1, p]\), ";", 
                    "\[IndentingNewLine]", 
                    RowBox[{"(*", " ", 
                      StyleBox[\(If\ y2\ has\ a\ square\ root\ mod\ p\),
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                    "\[IndentingNewLine]", 
                    RowBox[{"If", "[", 
                      
                      RowBox[{\(QuadraticResidueQ[y2, p]\), ",", 
                        "\[IndentingNewLine]", 
                        RowBox[{"(*", " ", 
                          
                          StyleBox[\(find\ the\ smallest\ y1\  \[SuchThat] \ 
                              y1\^2\  \[Congruent] \ y2 \((mod\ p)\)\),
                            FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                        "\[IndentingNewLine]", \(y1 = 
                          SqrtModPShanksTonelli[y2, 
                            p]; \[IndentingNewLine]Break[];\)}], 
                      "\[IndentingNewLine]", "]"}], ";"}]}], 
                "\[IndentingNewLine]", "]"}], ";", 
              "\[IndentingNewLine]", \(Return[{x1, y1}]\), ";"}]}], 
          "\[IndentingNewLine]", "]"}]}], ";"}]], "Input"]
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Cell[CellGroupData[{

Cell["GenerateRandomEC[ p ]", "Subsubsection",
  CellTags->"c:86"],

Cell[TextData[{
  "Generate a random elliptic curve over the field ",
  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
  " where ",
  Cell[BoxData[
      \(TraditionalForm\`p\)]],
  " is a prime greater than 3.\n\tThe curve is of the form ",
  Cell[BoxData[
      \(TraditionalForm\`y\^2 = x\^3 + A\ x\  + B\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`4  A\^2 + 27  B\^3 \[NotEqual] 0\)]],
  "\n\tReturns {A,B}"
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              "\[IndentingNewLine]", \(p1 = 
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                  "\[IndentingNewLine]", \(p17 = p2 + p3*y2;\)}], 
                "\[IndentingNewLine]", "]"}], ";", 
              "\[IndentingNewLine]", \(p17 = 
                PolynomialPowerMod[p17, 1, {fl, p}]\), ";", 
              "\[IndentingNewLine]", \(Return[p17]\), ";"}]}], 
          "\[IndentingNewLine]", "]"}]}], ";"}]], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["ComputeEquation18[ p, l, w, {a, b} ]", "Subsubsection",
  PageBreakAbove->True,
  CellTags->"c:96"],

Cell[TextData[{
  "After we know from using ",
  StyleBox["Schoof(17)",
    FontWeight->"Bold"],
  " that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = \(\[PlusMinus]w\)\ P\)]],
  " we use ",
  StyleBox["Schoof (18)",
    FontWeight->"Bold"],
  " to test the ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " coordinate of ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\) P = w\ P\)]],
  ".\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " even:\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x) = 
        4 \(\((y\^2)\)\^\(\((q + 3)\)/2\)\) \(\(f\_w\%3\)(
              x)\) - \(\(f\_\(w + 2\)\)(x)\) \(\(f\_\(w - 1\)\%2\)(
              x)\) + \(\(f\_\(w - 2\)\)(x)\) \(\(f\_\(w + 1\)\%2\)(x)\)\)]],
  "\nFor ",
  Cell[BoxData[
      \(TraditionalForm\`w\)]],
  " odd:\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\(p\_18\)(x) = 
        4 \(\((y\^2)\)\^\(\((q - 1)\)/2\)\) \(\(f\_w\%3\)(
              x)\) - \(\(f\_\(w + 2\)\)(x)\) \(\(f\_\(w - 1\)\%2\)(
              x)\) + \(\(f\_\(w - 2\)\)(x)\) \(\(f\_\(w + 1\)\%2\)(x)\)\)]]
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

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          RowBox[{\({y2, fl, k, p1, p2, p3, p4, p18y}\), ",", 
            "\[IndentingNewLine]", 
            
            RowBox[{\(y2 = x\^3 + a\ x\  + \ b\), ";", 
              "\[IndentingNewLine]", \(fl = f[l]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(k\  = \ \((p + 3)\)/2\ \((w\ even)\), \ 
                  k\  = \ \((p - 1)\)/2\ \((w\ odd)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(If[EvenQ[w], \(k = \((p + 3)\)/2;\), 
                k = \((p - 1)\)/2]\), ";", "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[
                  RowBox[{"p1", " ", "=", " ", 
                    RowBox[{\(\((x\^3\  + \ a\ x\  + \ b)\)\^k\), "=", " ", 
                      SuperscriptBox[
                        RowBox[{
                          StyleBox["(",
                            FontSlant->"Italic",
                            FontColor->RGBColor[0, 0, 1]], 
                          SuperscriptBox[
                            StyleBox["y",
                              FontSlant->"Italic",
                              FontColor->RGBColor[0, 0, 1]], "2"], ")"}], 
                        "k"]}]}],
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p1 = 
                PolynomialPowerMod[y2, k, {fl, p}]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[\(p2\  = \ \(f\_w\%3\) \((x)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p2 = 
                PolynomialPowerMod[f[w], 3, {fl, p}]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                StyleBox[\(p3\  = \ \(f\_\(w + 2\)\) \((x)\) \(f\_\(w - 
                            1\)\%2\) \((x)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p3 = 
                PolynomialPowerMod[f[w - 1], 2, {fl, p}]\), ";", 
              "\[IndentingNewLine]", \(p3\  = 
                PolynomialPowerMod[\ f[w + 2]*p3, 1, {fl, p}]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(p4\  = \ \(f\_\(w - 2\)\) \((x)\)\ \(f\_\(w + 
                            1\)\%2\) \((x)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p4 = 
                PolynomialPowerMod[f[w + 1], 2, {fl, p}]\), ";", 
              "\[IndentingNewLine]", \(p4 = 
                PolynomialPowerMod[f[w - 2]*p4, 1, {fl, p}]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(p18y\  = \ 
                    4 \( y\^\(2  k\)\) \(f\_w\%3\) \((x)\) - \(f\_\(w + 
                              2\)\) \((x)\) \(f\_\(w - 
                              1\)\%2\) \((x)\) + \(f\_\(w - 
                              2\)\) \((x)\) \(f\_\(w + 1\)\%2\) \((x)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p18y = 4*p1*p2 - p3 + p4\), ";", 
              "\[IndentingNewLine]", \(p18y = 
                PolynomialPowerMod[p18, 1, {fl, p}]\), ";", 
              "\[IndentingNewLine]", \(Return[p18y]\), ";"}]}], 
          "\[IndentingNewLine]", "]"}]}], ";"}]], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["ComputeTmodLCaseOne[ p, l, {a, b} ]", "Subsubsection",
  PageBreakAbove->True,
  CellTags->"c:97"],

Cell[TextData[{
  "Compute ",
  Cell[BoxData[
      \(TraditionalForm\`t\ \((mod\ l)\)\)]],
  " for the case when there exists ",
  Cell[BoxData[
      \(TraditionalForm\`P \[Element] E[l]\)]],
  " such that ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P = \(\[PlusMinus]k\)\ P\)]],
  ".  This method uses Schoof equations (17) and (18) to perform these \
tests."
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

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        RowBox[{"Module", "[", 
          RowBox[{\({j, tl, w, fl, p17, p18}\), ",", "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              
              StyleBox[\(If\ p\ is\ a\ quadratic\ nonresidue\ of\ l\
\[IndentingNewLine]
                \ then\ t\  \[Congruent] \ 0\ \((mod\ l)\)\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", 
            
            RowBox[{\(If[\(! QuadraticResidueQ[p, l]\), Return[0]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Else\ find\ w\ such\ that\ w\^2\  \[Congruent] \ 
                    p\ \((mod\ l)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(w = SqrtModPShanksTonelli[p, l]\), ";",
               "\[IndentingNewLine]", 
              StyleBox[\(Print["\<Sqrt of q mod l = \>", w]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
                FontColor->RGBColor[1, 0, 0]], 
              "\[IndentingNewLine]", \(fl = f[l]\), ";", 
              "\[IndentingNewLine]", 
              StyleBox[\(Print["\<f[\>", l, "\<] = \>", fl]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
                FontColor->RGBColor[1, 0, 0]], "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Use\ Schoof\ 17\ to\ test\ \(\[Phi]\_l\) 
                      P\  = \ \(\[PlusMinus]w\)*P\ \((x\ coord)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p17 = 
                ComputeEquation17[p, l, w, {a, b}]\), ";", 
              "\[IndentingNewLine]", 
              StyleBox[\(Print[\*"\"\<\!\(p\_17\) = \>\"", p17]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox["\[IndentingNewLine]",
                FontColor->RGBColor[1, 0, 0]], \(g = 
                PolynomialGCD[p17, fl, Modulus \[Rule] p]\), ";", 
              "\[IndentingNewLine]", 
              
              StyleBox[\(Print[\*"\"\<gcd(\!\(p\_17\),\!\(f\_l\)) = \>\"", 
                  g]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
                FontColor->RGBColor[1, 0, 0]], "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(g\  = \ 1\ means\ no\ such\ P, \ 
                  so\ t\  \[Congruent] \ 0\ \((mod\ l\ )\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(If[SameQ[g, 1], Return[0]]\), ";", 
              "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(Use\ Schoof\ 18\ to\ test\ \(\[Phi]\_l\) P\  = \ 
                    w*P\ \((y\ coord)\)\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p18 = 
                ComputeEquation18[p, l, w, {a, b}]\), ";", 
              "\[IndentingNewLine]", 
              StyleBox[\(Print[\*"\"\<\!\(p\_18\) = \>\"", p18]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox["\[IndentingNewLine]",
                FontColor->RGBColor[1, 0, 0]], \(g = 
                PolynomialGCD[p18, fl, Modulus \[Rule] p]\), ";", 
              "\[IndentingNewLine]", 
              
              StyleBox[\(Print[\*"\"\<gcd(\!\(p\_18\),\!\(f\_l\)) = \>\"", 
                  g]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
                FontColor->RGBColor[1, 0, 0]], "\[IndentingNewLine]", 
              RowBox[{"(*", " ", 
                
                StyleBox[\(g\  = \ 1\ means\ no\ such\ P, \ 
                  so\ t\  = \ \(-2\) w, \ else\ t\  = \ 2  w\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], 
                StyleBox[" ",
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], "*)"}], 
              "\[IndentingNewLine]", \(If[SameQ[g, 1], \ 
                tl = Mod[\(-2\) w, l], tl = Mod[2  w, l]]\), ";", 
              "\[IndentingNewLine]", \(Return[tl]\), ";"}]}], 
          "\[IndentingNewLine]", "]"}]}], ";"}]], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["ComputeAlpha[ p, l, {a, b} ]", "Subsubsection",
  PageBreakAbove->True,
  CellTags->"c:98"],

Cell[TextData[{
  "ComputeAlpha",
  Cell[BoxData[
      \(TraditionalForm\`\([p, l, {a, b}]\)\)]],
  "\n\tCompute \[Alpha], where ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \[Alpha]\/\[Beta]\)]],
  " is the slope of the line between ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`k\ P\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = \ \(\[Psi]\_\(k + 
                  2\)\) \[Psi]\_\(k - 1\)\%2 - \(\[Psi]\_\(k - 
                  2\)\) \[Psi]\_\(k + 1\)\%2 - 
          4 \( y\^\(p\^2 + 1\)\) \[Psi]\_k\%3\)]],
  "\nfor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " even\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = 
        y\ \((\(f\_\(k + 2\)\) f\_\(k - 1\)\%2 - \(f\_\(k - 2\)\) 
                f\_\(k + 1\)\%2 - 
              4 \(\((y\^2)\)\^\(\((p\^2 + 3)\)/2\)\) f\_k\%3)\)\)]],
  " \n\thowever we return ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha]/y\)]],
  " in this case and compensate at a higher level\nfor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " odd\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = \(y\^2\)(\(f\_\(k + 2\)\) 
                f\_\(k - 1\)\%2 - \(f\_\(k - 2\)\) f\_\(k + 1\)\%2) - 
          4 \(\((y\^2)\)\^\(\((p\^2 + 1)\)/2\)\) f\_k\%3\)]]
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    RowBox[{\(ComputeAlpha[p_, l_, {a_, b_}]\), ":=", 
      RowBox[{"Module", "[", 
        RowBox[{\({k, fl, y2, y4, p1, p2, p3, p4, p5, \[Alpha]}\), ",", 
          "\[IndentingNewLine]", 
          
          RowBox[{\(k = Mod[p, l]\), ";", 
            "\[IndentingNewLine]", \(fl = f[l]\), ";", "\[IndentingNewLine]", 
            
            RowBox[{"(*", " ", 
              RowBox[{
                StyleBox[\(y\^2\),
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], 
                StyleBox[",",
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], 
                StyleBox[" ",
                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], 
                RowBox[{
                  StyleBox[
                    SuperscriptBox[
                      StyleBox["y",
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]], "4"],
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox["are",
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox[" ",
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox["polynomials",
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox[" ",
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox["in",
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox[" ",
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], 
                  StyleBox["x",
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]]}]}], " ", "*)"}], 
            "\[IndentingNewLine]", \(y2 = x\^3 + a\ x + b\), ";", 
            "\[IndentingNewLine]", \(y4 = 
              PolynomialPowerMod[y2, 2, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(p\_1\  = \ \(f\_\(k + 2\)\) f\_\(k - 1\)\%2\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(p1 = 
              PolynomialPowerMod[f[k - 1], 2, {fl, p}]\), ";", 
            "\[IndentingNewLine]", \(p1 = 
              PolynomialPowerMod[f[k + 2]*p1, 1, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(p\_2\  = \ \(f\_\(k - 2\)\) f\_\(k + 1\)\%2\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], "*)"}], 
            "\[IndentingNewLine]", \(p2 = 
              PolynomialPowerMod[f[k + 1], 2, {fl, p}]\), ";", 
            "\[IndentingNewLine]", \(p2 = 
              PolynomialPowerMod[f[k - 2]*p2, 1, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              
              StyleBox[\(p\_3\  = \ \(\((y\^2)\)\^\(\((p\^2 - 1)\)/2\) = \ \
\((y\^2)\)\^\(\((p - 1)\) \((p + 1)\)/2\)\)\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(p3 = 
              PolynomialPowerMod[y2, \((p - 1)\), {fl, p}]\), ";", 
            "\[IndentingNewLine]", \(p3 = 
              PolynomialPowerMod[p3, \((p + 1)\)/2, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(p\_4\  = \ f\_k\%3\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(p4 = 
              PolynomialPowerMod[f[k], 3, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"If", "[", 
              RowBox[{\(EvenQ[k]\), ",", "\[IndentingNewLine]", 
                RowBox[{"(*", 
                  RowBox[{
                    RowBox[{" ", 
                      StyleBox[
                        
                        FormBox[\(\[Alpha]\  = \ \(\(y\)\((\)\(\(f\_\(k + 
                                        2\)\) 
                                  f\_\(k - 1\)\%2\  - \ \(f\_\(k - 2\)\) 
                                  f\_\(k + 1\)\%2\  - \ 
                                4 \(\((y\^2)\)\^\(\((p\^2 + 3)\)/2\)\) 
                                  f\_k\%3\)\)\),
                          "TraditionalForm"],
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]], 
                      StyleBox[")",
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]]}], 
                    StyleBox["/",
                      FontSlant->"Italic",
                      FontColor->RGBColor[0, 0, 1]], 
                    StyleBox["y",
                      FontSlant->"Italic",
                      FontColor->RGBColor[0, 0, 1]]}], " ", "*)"}], 
                "\[IndentingNewLine]", \(p5 = 
                  PolynomialPowerMod[4*y4*p3*p4, 
                    1, {fl, p}]; \[IndentingNewLine]\[Alpha] = 
                  PolynomialPowerMod[p1 - p2 - p5, 1, {fl, p}];\), 
                "\[IndentingNewLine]", ",", "\[IndentingNewLine]", 
                RowBox[{"(*", " ", 
                  StyleBox[
                    
                    FormBox[\(\[Alpha]\  = \ \(y\^2\)(\(f\_\(k + 2\)\) 
                                f\_\(k - 1\)\%2\  - \ \(f\_\(k - 2\)\) 
                                f\_\(k + 1\)\%2)\  - \ 
                          4 \(\((y\^2)\)\^\(\((q\^2 + 1)\)/2\)\) f\_k\%3\),
                      "TraditionalForm"],
                    FontSlant->"Italic",
                    FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                "\[IndentingNewLine]", \(p5 = 
                  PolynomialPowerMod[4*y2*p3*p4, 
                    1, {fl, p}]; \[IndentingNewLine]\[Alpha] = 
                  PolynomialPowerMod[y2*\((p1 - p2)\) - p5, 1, {fl, p}];\)}], 
              "\[IndentingNewLine]", "]"}], ";", 
            "\[IndentingNewLine]", \(Return[\[Alpha]]\), ";"}]}], 
        "\[IndentingNewLine]", "]"}]}]], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["ComputeBeta[ p, l, {a, b} ]", "Subsubsection",
  CellTags->"c:99"],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`ComputeBeta[p, l, {a, b}]\)]],
  "\n\tCompute \[Beta], where ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] = \[Alpha]\/\[Beta]\)]],
  " is the slope of the line between ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[Phi]\_l\%2\) P\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`k\ P\)]],
  "\n\t",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = 
        4  y\ \[Psi]\_k\ \((\((x - 
                    x\^\(p\^2\))\) \[Psi]\_k\%2 - \(\[Psi]\_\(k - 
                      1\)\) \[Psi]\_\(k + 1\))\)\)]],
  "\nfor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " even\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = 
        4\ \(y\^2\) \(\(f\_k\)(\((\ x - x\^\(p\^2\))\) \(y\^2\) 
                f\_k\%2 - \(f\_\(k - 1\)\) f\_\(k + 1\))\)\)]],
  "\nfor ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " odd\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = 
        4\ y\ \(\(f\_k\)(\((\ x - x\^\(p\^2\))\) 
                f\_k\%2 - \(y\^2\) \(f\_\(k - 1\)\) f\_\(k + 1\))\)\)]],
  "\n\t however we return ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]/y\)]],
  " in this case and compensate at a higher level"
}], "Text",
  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

Cell[BoxData[
    RowBox[{\(ComputeBeta[p_, l_, {a_, b_}]\), ":=", 
      RowBox[{"Module", "[", 
        RowBox[{\({k, fl, y2, p1, p2, p3, p4, \[Beta]}\), ",", 
          "\[IndentingNewLine]", 
          
          RowBox[{\(k = Mod[p, l]\), ";", 
            "\[IndentingNewLine]", \(fl = f[l]\), ";", "\[IndentingNewLine]", 
            
            RowBox[{"(*", " ", 
              StyleBox[\(y\^2\ is\ a\ polynomial\ in\ x\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(y2 = x\^3 + a\ x + b\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(p1\  = \ x\  - \ x\^\(p\^2\)\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], "*)"}], 
            "\[IndentingNewLine]", \(p1 = PolynomialPowerMod[x, p, {fl, p}]\),
             ";", "\[IndentingNewLine]", \(p1 = 
              PolynomialPowerMod[p1, p, {fl, p}]\), ";", 
            "\[IndentingNewLine]", \(p1 = x - p1\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(p\_2\  = \ \(f\_\(k - 1\)\) f\_\(k + 1\)\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
            "\[IndentingNewLine]", \(p2 = 
              PolynomialPowerMod[f[k - 1]*f[k + 1], 1, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"(*", " ", 
              StyleBox[\(p\_3\  = \ f\_k\%2\),
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], 
              StyleBox[" ",
                FontSlant->"Italic",
                FontColor->RGBColor[0, 0, 1]], "*)"}], 
            "\[IndentingNewLine]", \(p3 = 
              PolynomialPowerMod[f[k], 2, {fl, p}]\), ";", 
            "\[IndentingNewLine]", 
            RowBox[{"If", "[", 
              RowBox[{\(EvenQ[k]\), ",", "\[IndentingNewLine]", 
                RowBox[{"(*", 
                  StyleBox[
                    RowBox[{" ", 
                      StyleBox[" ",
                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]]}]], 
                  StyleBox[
                    
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                "\[IndentingNewLine]", \(p4 = 
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Cell["\<\
ComputeEquation19X[ p, l, \[Tau], {a, b} ]\
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  "Compute the polynomial, modulo ",
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  " representing Schoof equation (19x) as\n\t",
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      \(TraditionalForm\`\(p\_\(19\_x\)\)(x, 
          y) = \(\[Psi]\_\[Tau]\%\(2  
                p\)\)(\(\[Beta]\^2\)(\(\[Psi]\_\(k - 
                      1\)\) \[Psi]\_\(k + 1\) - \(\[Psi]\_k\%2\)(
                x\^\(p\^2\) + x\^p + x) + \[Alpha]\ \[Psi]\_k\%2))\), 
    "\[IndentingNewLine]", 
      \(TraditionalForm\`\(+\[Psi]\_k\%2\) \
\(\(\[Beta]\^2\)(\(\[Psi]\_\(\[Tau] - 1\)\) \[Psi]\_\(\[Tau] - \
1\))\)\^p\)}]],
  "."
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              "\[IndentingNewLine]", 
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                StyleBox[\(\[Lambda]\  = \ \[Alpha]/\[Beta], \ 
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                  Note\ \[Alpha], \[Beta]\ are\ globals\ so\ we\ can\ share\ \
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              "\[IndentingNewLine]", \(p1 = f[k - 1]*f[k + 1]\), ";", 
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              RowBox[{"(*", " ", 
                StyleBox[\(p2\  = \ x\^p\),
                  FontSlant->"Italic",
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                StyleBox[\(p3\  = \ x\^\(p\^2\)\),
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                PolynomialPowerMod[p2, p, {fl, p}]\), ";", 
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              RowBox[{"(*", " ", 
                StyleBox[\(p4\  = \ \((x\^\(p\^2\)\  + \ x\^p\  + \ x)\)\),
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              "\[IndentingNewLine]", \(p4 = p3 + p2 + x\), ";", 
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              RowBox[{"(*", " ", 
                StyleBox[\(p5\  = \ f\_k\%2\),
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                StyleBox[\(p6\  = \ \((\(f\_\(\[Tau] - 1\)\) f\_\(\[Tau] + \
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              RowBox[{"(*", " ", 
                
                StyleBox[\(Compute\ \[Alpha]\^2, \ \[Beta]\^2\ compensating\ \
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              RowBox[{"If", "[", 
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                  "\[IndentingNewLine]", 
                  
                  RowBox[{\(\[Alpha]2 = 
                      y2*PolynomialPowerMod[\[Alpha], 2, {fl, p}]\), ";", 
                    "\[IndentingNewLine]", \(\[Beta]2 = 
                      PolynomialPowerMod[\[Beta], 2, {fl, p}]\), ";", 
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                          "TraditionalForm"],
                        FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                    "\[IndentingNewLine]", \(p7 = 
                      PolynomialPowerMod[p1 - y2*p5*p4, 1, {fl, p}]\), ";", 
                    "\[IndentingNewLine]", \(p7 = 
                      PolynomialPowerMod[p7*\[Beta]2 + y2*p5*\[Alpha]2, 
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                    StyleBox[
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                  "\[IndentingNewLine]", 
                  
                  RowBox[{\(\[Alpha]2 = 
                      PolynomialPowerMod[\[Alpha], 2, {fl, p}]\), ";", 
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                      y2*PolynomialPowerMod[\[Beta], 2, {fl, p}]\), ";", 
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                          "TraditionalForm"],
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                    "\[IndentingNewLine]", \(p7 = 
                      PolynomialPowerMod[y2*p1 - p5*p4, 1, {fl, p}]\), ";", 
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                      PolynomialPowerMod[p7*\[Beta]2 + p5*\[Alpha]2, 
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                RowBox[{\(EvenQ[k] && EvenQ[\[Tau]]\), ",", 
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                          FormBox[\(\((\((\(f\_\(k - 1\)\) 
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                                        x)\))\) \[Beta]\^2\  + \ \(y\^2\) \(f\
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                                f\_\[Tau]\%\(2  p\)\  + \ \(f\_\(\[Tau] - 
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                  "\[IndentingNewLine]", \(p8 = 
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                      2*p, {fl, p}]; \[IndentingNewLine]p9 = 
                    PolynomialPowerMod[y2, 
                      p, {fl, p}]; \[IndentingNewLine]p19 = 
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                RowBox[{\(EvenQ[k] && OddQ[\[Tau]]\), ",", 
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                                        x)\))\) \[Beta]\^2\  + \ \(y\^2\) \(f\
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                                  2\)\) \(f\_\(\[Tau] - 
                                  1\)\%p\) \(f\_\(\[Tau] + 
                                  1\)\%p\) \(\[Beta]\^2\) f\_k\%2\),
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                      FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                  "\[IndentingNewLine]", \(p8 = 
                    PolynomialPowerMod[f[\[Tau]], 
                      2*p, {fl, p}]; \[IndentingNewLine]p9 = 
                    PolynomialPowerMod[y2, 
                      p + 1, {fl, p}]; \[IndentingNewLine]p19 = 
                    PolynomialPowerMod[p7*p8 + p9*p6*\[Beta]2*p5, 
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                            f\_\[Tau]\%\(2  p\)\  + \ \(f\_\(\[Tau] - 
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                      FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
                  "\[IndentingNewLine]", \(p8 = 
                    PolynomialPowerMod[f[\[Tau]], 
                      2*p, {fl, p}]; \[IndentingNewLine]p9 = 
                    PolynomialPowerMod[y2, 
                      p, {fl, p}]; \[IndentingNewLine]p19 = 
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                      FontColor->RGBColor[0, 0, 1]], 
                    StyleBox[" ",
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                      2*p, {fl, p}]; \[IndentingNewLine]p9 = 
                    PolynomialPowerMod[y2, 
                      p, {fl, p}]; \[IndentingNewLine]p19 = 
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Cell["\<\
ComputeEquation19Y[ p, l, \[Tau], {a, b} ]\
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  CellTags->"c:101"],

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      \(TraditionalForm\`\(p\_\(19  y\)\)(x, y) = 
        4 \( f\_\[Tau]\%\(3\ p\)\) 
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                            x)\)\ \[Alpha]\ \[Beta]\^2 - \[Beta]\^3\ \
y\^\(p\^2\) - \[Alpha]\^3)\)\ f\_k\%2\[IndentingNewLine] - \[Alpha]\ \
\(\[Beta]\^2\) 
                  f\_\(k - 1\)\ f\_\(k + 1\)\ )\)\  - \[Beta]\^3\ f\_k\%2\ \
\((f\_\(\[Tau] - 1\)\%2\ f\_\(\[Tau] + 2\) - f\_\(\[Tau] - 2\)\ f\_\(\[Tau] + \
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              "\[IndentingNewLine]", \(k = Mod[q, l]\), ";", 
              "\[IndentingNewLine]", 
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                StyleBox[\(Make\ sure\ we\ have\ no\ junk\ floating\ around\),
                  
                  FontSlant->"Italic",
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              "\[IndentingNewLine]", \(Print[{q, l, \[Tau], x}]\), ";", 
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              RowBox[{"(*", " ", 
                
                StyleBox[\(\[Lambda]\  = \ \[Alpha]/\[Beta], \ 
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                StyleBox[" ",
                  FontSlant->"Italic",
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              StyleBox[\(\[Beta]\  = \ ComputeBeta[q, l, {a, b}]\),
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                  StyleBox[" ",
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                  StyleBox[\(\[Beta]\^2\),
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                PolynomialPowerMod[\[Beta], 2, {fl, q}]\), ";", 
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                  FontSlant->"Italic",
                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p1 = \[Alpha]*b2*f[k - 1]*f[k + 1]\), 
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                PolynomialPowerMod[p1, 1, {fl, q}]\), ";", 
              "\[IndentingNewLine]", 
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                StyleBox[\(p\_2\  = \ f\_k\%2\),
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              RowBox[{"(*", " ", 
                StyleBox[\(p\_3\  = \ 2\ x\^\(q\^2\) + \ x\),
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              "\[IndentingNewLine]", \(p3 = 
                PolynomialPowerMod[x, q, {fl, q}]\), ";", 
              "\[IndentingNewLine]", \(p3 = 
                PolynomialPowerMod[p3, q, {fl, q}]\), ";", 
              "\[IndentingNewLine]", \(p3 = 
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                  StyleBox["=",
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                    StyleBox[\(4  f\_\[Tau]\%\(3  q\)\),
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                    StyleBox[" ",
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              RowBox[{"(*", " ", 
                
                StyleBox[\(p\_5\  = \ \((\(f\_\(\[Tau] - 1\)\%2\) f\_\(\[Tau] \
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                  FontColor->RGBColor[0, 0, 1]], " ", "*)"}], 
              "\[IndentingNewLine]", \(p5 = 
                f[\[Tau] + 2]*PolynomialPowerMod[f[\[Tau] - 1], 2, {fl, q}]\),
               ";", "\[IndentingNewLine]", \(p5 = 
                p5 - f[\[Tau] - 2]*
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Cell[CellGroupData[{

Cell["\<\
ComputeTmodLCaseTwo[ q, l, \[Tau], {a, b} ]\
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  CellTags->"c:102"],

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  "Given that ",
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          l\ \(\[DoubleStruckCapitalZ]\^\[Cross]\)\)]],
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  Cell[BoxData[
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          q\ P = \(\[PlusMinus]\[Tau]\)\ \(\[Phi]\_l\) P\)]],
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  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

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                  StyleBox[\(Print[\*"\"\<\!\(p\_\(19\_x\)\) = \>\"", p19x]\),
                    
                    FontColor->RGBColor[1, 0, 0]], 
                  StyleBox[";",
                    FontColor->RGBColor[1, 0, 0]], 
                  "\[IndentingNewLine]", \(g = 
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                  "\[IndentingNewLine]", 
                  
                  StyleBox[\(Print[\*"\"\<gcd(\!\(p\_\(19\_x\)\),\!\(f\_l\)) \
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                    StyleBox[\(g\  \[NotEqual] \ 1\ means\ such\ P\ exists, \ 
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                        "\[IndentingNewLine]", 
                        
                        StyleBox[\(Print[\*"\"\<gcd(\!\(p\_\(19\_y\)\),\!\(f\_\
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                        "\[IndentingNewLine]", \(If[SameQ[g, 1], 
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        "\[IndentingNewLine]", "]"}]}]], "Input"]
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Cell[CellGroupData[{

Cell[TextData[{
  "ComputeGroupOrderSchoof[ ",
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  CellTags->"c:103"],

Cell[TextData[{
  "Computes the order of the group of points ",
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  Cell[BoxData[
      \(TraditionalForm\`E : y\^2 = x\^3 + A\ x + B\)]],
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  Cell[BoxData[
      \(TraditionalForm\`\[DoubleStruckCapitalF]\_p\)]],
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  FontSize->14,
  FontColor->RGBColor[0, 0, 1]],

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            RowBox[{\({prmprod, s} = ComputePrimeSet[p]\), ";", 
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                StyleBox[\(Compute\ t\ \((mod\ 2)\)\),
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              "\[IndentingNewLine]", \(tl = ComputeTModTwo[{a, b}, p]\), ";", 
              "\[IndentingNewLine]", 
              StyleBox[\(Print["\<\[Tau] (mod 2 ) = \>", tl\ ]\),
                FontColor->RGBColor[1, 0, 0]], 
              StyleBox[";",
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              "\[IndentingNewLine]", \(t = \ {{2, tl}}\), ";", 
              "\[IndentingNewLine]", 
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                StyleBox[\(Create\ the\ required\ division\ polynomials\ f\_k \
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              "\[IndentingNewLine]", \(m = Last[s]\), ";", 
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              RowBox[{"(*", " ", 
                
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              RowBox[{"For", "[", 
                
                RowBox[{\(i = 2\), ",", \(i \[LessEqual] Length[s]\), 
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                      StyleBox[\(If\ \(\[Exists] \ 
                                P\  \[Element] \ 
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                        FontColor->RGBColor[0, 0, 1]], 
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                        FontSlant->"Italic",
                        FontColor->RGBColor[0, 0, 1]], "*)"}], 
                    "\[IndentingNewLine]", \(p16 = 
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                      PolynomialGCD[p16, fl, Modulus \[Rule] p]\), ";", 
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                        "\[IndentingNewLine]", \(tl = 
                          ComputeTmodLCaseOne[p, 
                            l, {a, b}]; \[IndentingNewLine]AppendTo[
                          t, {l, tl}];\), "\[IndentingNewLine]", ",", 
                        "\[IndentingNewLine]", 
                        RowBox[{"(*", " ", 
                          
                          StyleBox[\(Else, \ 
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                          Determine\ if\ \(\[Exists] \ \[Tau]\  < \ 
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                          RowBox[{"For", "[", 
                            
                            RowBox[{\(\[Tau] = 1\), 
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                                StyleBox[\(Print[\*"\"\<\!\(\[Tau]\_l\) = \
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                                  "*)"}], 
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                                  ComputeEquation19X[p, l, \[Tau], {a, b}]\), 
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                                  FontColor->RGBColor[1, 0, 0]], 
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                                StyleBox[\(Print[\*"\"\<gcd(\!\(p\_\(19  \
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                                        P\  + \ 
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                                      StyleBox[\(Print["\<p19y = \>", p19y]\),
                                        
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                                        Modulus \[Rule] p]\), ";", 
                                      "\[IndentingNewLine]", 
                                      
                                      StyleBox[\(Print[\*"\"\<gcd(\!\(p\_\(19 \
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                                      StyleBox[";",
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                                      "\[IndentingNewLine]", \(If[\(! SameQ[
                                        gy, 1]\), tl = \[Tau], 
                                        tl = l - \[Tau]]\), ";", 
                                      "\[IndentingNewLine]", \(AppendTo[
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