On Recognizing Congruent Primes Brett Hemenway B.Sc., Brown University 2004 A THESIS SUBMITTED I N PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTEROF SCIENCE I N THE DEPARTMENT OF MATHEMATICS @ Brett Hemenway 2006 SIMON FRASER UNIVERSITY Fall 2006 All rights reserved. This work may not be reproduced in whole or in part, by photocopy or other means, without the permission of the author APPROVAL Name: Brett Hemenway Degree: Master of Science Title of thesis: On Recognizing Congruent Primes Examining Committee: Dr. Jason Bell Chair Dr. Nils Bruin Senior Supervisor Dr. Peter Borwein Supervisor Dr. Stephen Choi Supervisor Dr. Imin Chen Internal/External Examiner Date Approved: September 12th, 2006 DECLARATION OF PARTIAL COPYRIGHT LICENCE The author, whose copyright is declared on the title page of this work, has gra to Simon Fraser University the right to lend this thesis, project or extended essay to users of the Simon Fraser University Library, and to make partial or single copies only for such users or in response to a request from the library of any other university, or other educational institution, on its own behalf or for one of its users. The author has further granted permission to Simon Fraser University to keep or make a digital copy for use in its circulating collection, and, without changing the content, to translate the thesislproject or extended essays, if technically possible, to any medium or format for the purpose of preservation of the digital work. The author has further agreed that permission for multiple copying of this work for scholarly purposes may be granted by either the author or the Dean of Graduate Studies. It is understood that copying or publication of this work for financial gain shall not be allowed without the author's written permission. Permission for public performance, or limited permission for private scholarly use, of any multimedia materials forming part of this work, may have been granted by the author. This information may be found on the separately catalogued multimedia material and in the signed Partial Copyright Licence. The original Partial Copyright Licence attesting to these terms, and signed by this author, may be found in the original bound copy of this work, retained in the Simon Fraser University Archive. Simon Fraser University Library Bumaby, BC, Canada Abstract An integer n is called "congruent" if it corresponds to the area of a right triangle with three rational sides. The problem of classifying congruent numbers has an extensive history, and is as yet unresolved. The most promising approach to this problem utilizes elliptic curves. In this thesis we explicitly lay out the correspondence between the congruence of a number n and the rank of the elliptic curve y2 = x3 - n 2 x . By performing two-descents on this curve and isogenous curves for n = p a prime, we - are able to obtain a simple and unified proof of the majority of the known results concerning the congruence of primes. Finally, by calculating the equations for homogeneous spaces associated to the curve when p for future analysis. 1 mod 8, we position the problem Acknowledgments I would like to thank my entire thesis committee. In particular, thanks to Peter Borwein for welcon~ingme into t,he SFU mathematical community, for making my time in Canada possible, and for his continued guidance throughout my time at SFU. Thanks also to Nils Bruin for introducing me to the study of elliptic curves, for helping me navigate this project, and for reading many many draftasof this thesis. His constant encouragement, feedback and support has made him a wonderful advisor. Dedication To Emily. Contents .. Approval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Congruent Numbers . . . . . . . . . . . . . . . . . . . . . . . Known Results . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 1.1 1.2 Genocchi's 1855 Argument . . . . . . . . . . . . . . . . . . . Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 2 2.1 2.2 2.3 2.4 2.5 2.6 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Group Law . . . . . . . . . . . . . . . . . . . . . . . . . . The Structure of the Group . . . . . . . . . . . . . . . . . . . Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... 2 4 9 9 9 10 11 14 14 15 2.8 Torsion in the Congruent Number Curve . . . . . . . . . . . . Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 2-Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 The Curve y2 = x3 - n2x . . . . . . . . . . . . . . . . . . . . . TwoDescent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.2 The map p . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.7 2.10 3 11 18 24 The Two-descent . . . . . . . . . . . . . . . . . . . . . . . . . Twedescent applied t o congruent number curves . . . . . . . . . . . 4.1 General Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 4 4.2 4.3 5 Specific Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . =1 30 32 33 35 mod 8 . . . . . . . . . . . . . . . . . . . . . . 36 36 5.2 A Specific Two-Isogeny . . . . . . . . . . . . . . . . . . . . . . Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 5.4 General Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . p-adics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 38 39 5.5 The Quadratic Character of 1 + i . . . . . . . . . . . 2-adics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Case When p 5.1 5.4.1 6 28 Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 6.3 Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Follow up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography vii 37 40 43 43 45 48 Chapter 1 Introduction 1.1 Congruent Numbers A positive rational number n E Q is said to be a Congruent Number if it is the area of a right triangle with rational sides, i.e. if there are rational numbers a, b, c such that If n is a congruent number, then nr2 is also congruent for any r E Q , since nr2 is the area of the triangle with sides ra, rb, re. Thus whether a number is congruent is determined solely by its residue class in the group Q*/(Q*)2.With this in mind, when searching for congruent numbers we restrict our attention to square free integers. The question of determining which numbers are congruent is known as the Congruent Number Problem, and while the statement is completely elementary, this problem has been investigated by mathematicians for over one thousand years (see [lo]) without leading to a complete solution. By performing elementary transforma- tions on the equations (1.1),the problem can be formulated in many different ways. The earliest known reference the Congruent Number Problem is from A.D. 972 (see [lo]), and states the problem in another form. A rational number n is said to be congruent if there exists a rational number x such that x2 + n and x2 - n are both CHAPTER 1. INTRODUCTION 2 squares. This is also the formulation that Leonardo Pisano (Fibonacci) used to show that 5 is a congruent number in 1220 [lo]. This formulation gives some insight into + the name, since the three squares, x2 - n, x 2 ,x2 n are congruent modulo n. The two definitions of a Congruent Number are easily seen to be equivalent by the maps For most of this thesis we will be using a third characterization of congruent numbers, one relating t o rational solutions to a cubic equation. A square-free natural number n is a congruent number if we can simultaneously solve two equations over the rationals which means ( a i b)2 4 Then setting x = = (f) 2 i n . (i)2 and y = (a2- b 2 ) i gives a solution t o the cubic equation We will c,arefully examine the curve give by equation 1.2 in the case when n is prime to determine whether n is congruent. 1.2 Known Results We now give a brief review of what is currently known about the Congruent Number Problem. Assuming the Birch and Swinnerton-Dyer Conjecture [4], the problem was essentially solved by Jerold B. Tunnel1 in 1983. C H A P T E R 1. INTRODUCTION Tunnell's Theorem. Define Suppose n is congruent, if n is even then A, = Bn and if n is odd, then 2Cn = D,. If the Birch and Swinnerton-Dyer Conjecture holds for curves of the form y2 = x3 - n2x then, conversely, these equalities imply n is a congruent number. Proof. See [34], or [20] Chapter IV section 4. 0 Currently this provides a method for showing certain numbers are not congruent, and assuming the Birch and Swinnerton-Dyer Conjecture this would provide a fairly efficient method for determining whether a given number is congruent, since counting solutions to these equations can be done easily. We will not go into details about the Birch and Swinnerton-Dyer conjecture, but it should be noted that it is widely believed to be true, and is one of the Clay Mathematics Institute's Millenium Prize Problems. Much study has gone into Congruent Number Problem, not assuming the Birch and Swinnerton-Dyer Conjecture, and we will list the known results here. Let p, and q, denote distinct primes with p, = q, = i mod 8 Then the following results are known p3 is not a congruent number Genocchi, 1855 [13] and Tunnell, 1983 [34] p3q3,2p5,p5q5 are not congruent numbers Genocchi [13] and Tunnel1 [34] p5,p7 are congruent numbers Monsky, 1990 [24] 1 2p7,2p3_p3q7,2p3q5,2p5q7 are congruent numbers Monsky, 1990 1241 Overviews of progress made on the Congruent Number Problem can be found in [lo], [141, PI, [81, [281 and [271. In this thesis, we obtain upper bounds for the rank of the elliptic curve y2 = x3-n2x using the method of 2-descent. In this way, we show that pa is not a congruent, C H A P T E R 1. INTRODUCTION 4 number, p5:p7 are congruent numbers assuming the conjecture that UT(E/K)[2] is + + finite, and for pl, if we decompose pl = a2 b2, then pl is not congruent if (a b)2 $ 1 mod 16. This last result was stated by Bastien in 1915 [2], but a proof does not seem to have appeared in the literature until Tunnell's proof in [34]. Tunnell's proof comes - as a consequence of Tunnell's Theorem, and hence his method is very different from the one presented here. The case of when p 3 mod 8 was the first to be resolved. In 1855 Angelo Genocchi showed that p3 is not a congruent number. His paper predates much of the general machinery of elliptic curves, and his argument is fairly elementary. Nevertheless, Genocchi's method bears many similarities to the method of descent used in this thesis. To show that p3 is not congruent, Genocchi shows that if p3 were congruent, this would lead to an integral point on a quartic, then he shows that because 2 are not squares in Z/p& - 1 and these quartics have no rational points. To illustrate his technique, we give an overview of his original argument in 51.3. 1.3 Genocchi's 1855 Argument We now make a brief digression to give Genocchi's argument [13] that if p is a prime with p EE 3 mod 8 then p is not a congruent number. Genocchi's paper is often cited, as it is one of the earliest demonstrations that an entire class of numbers is not congruent. Unfortunately, his paper has become very difficult to obtain. For its historical significance, as well as its ingenuity, we give a detailed account of his argument. The terminology has been updated, but the content of the proof remains the same. Fibonacci gave t,he following characterization of congruent numbers. A number is congruent if it is one of the four numbers a , b, a + b, a - b and the remaining three numbers are square. This is equivalent to the statement that n is congruent if there is a rational number x such that x2 f n are both square. Genocchi begins with Fibonacci's characterization, and considers four cases separately. Throughout the proof, we will make use of the fact that all Pythagorean Triples CHAPTER 1. INTRODUCTION can be paramatrized as ( r 2- s2,~ T Sr2 , + s2). Suppose n is a congruent number then one of the following four cases holds. Case 1: ~ = n f ~ , 2 ,ba + = b~= h 2 , a - b = k 2 Thus we have a2 - b2 = h2k2,a2 = b2 + ( f ~ k ) ~ So b, h k , a is a Pythagorean Triple. Then parametrizing we have a = r2 + s2 and b = r2 - s2 or b = 2 ~ s This . gives In the first case, since r , s are relatively prime, we must have r r - s = P2. +s = a2, This gives In the second case, we get Case 2: a = f 2 , b = g 2 , a + b = h2 , a - b = n k .2 Thus we have f 2 +g2 = h2. Thus ( f , g, h ) is a pythagorean triple, so ( f , g) = ( r 2 - s2,2rs) or ( f , g) (2rs,r2 - s 2 ) . From the equation f 2 Expanding gives - g2 = n k 2 , we then obtain = C H A P T E R 1. INTRODUCTION Case 3: a = f 2 , b = g 2 , a + b = n h 2 , a - b = k .2 Thus we have k2 + g2 = f2. So parametrizing this pythagorean triple, we have f = r2 + s2 and g = 2rs or g = r2 - s2, plugging into the equation gives + + ( r 2 s2)2 ( 2 ~ s=) nh2 ~ or Thus Case 4: a = f 2,b=ng 2: a + b = h 2 , a - b = k 2 . Here, we are forced to use a different method since and f 2 - ng2 = k2 tell us only that n is a congruent number. Going back t o the definition of a congruent number, if n is congruent, we can find x l , x2, X Q E Q such that, CHAPTER 1. INTRODUCTION 7 Clearing denominators we can find q such that xi = where pi, q E Z. Here, we assume that q is the smallest integer such that pl,p2,p3 E Z. Thus we have the equations - nq2 =p:,pi + nq 2 = p 32 . Since 2 p2 3 - pl we conclude that pl = p3 = 2nq 2 , mod 2, thus we can define integers This gives pi = rf + r,2, which is a Pythagorean Triple, so we can parametrize this as Now, we also have ng2 = 2rlr3, so substituting our parametrization for r l , r3 we get ng2 = 4ab(a - b) (a + b). Since n is congruent, we also know that we can find integers a , b such that three of the four integers a, b, a + b, a - b are square and n divides fourth. Assuming we are not in one of the three previous cases, if n is prime we must have nlb, so Then, these equations give So ng2 < ng2, but this contradicts the minimality of q. Thus we conclude that this case is impossible for n prime. CHAPTER 1. INTRODUCTION 8 It remains to show that the first three cases are impossible. Genocchi does this by examining congruence conditions mod p. Gathering equations (1.3) - (1.7), and unifying notation, we have If p =3 mod 8 and pin, then taking remainders modulo p shows that Equations (1.8),(1.9), (1.12) are impossible since - 1 is not a square mod p. To deal with equation (1.l o ) , we notice that and 2 is not a square mod p. Similarly for equation (1.11),we observe and it suffices to notice that either -1 or 2 is not a square mod p. Thus if p =3 mod 8, then p is not a congruent number. Chapter 2 Elliptic Curves 2.1 Introduction We will be looking for rational solutions (x,y ) t o the equation E : y2 = x3 - n2x. This is an example of an elliptic curve. The theory of elliptic curves is well-developed, and before we begin analyzing the curve E,we review some of the general properties of elliptic curves that we will use. 2.2 Definition Let P2 denote the projective plane. An elliptic curve is the locus of points in P2 E : y 2 Z + a l X Y Z + a 3 Y Z 2= x3+ a 2 x 2 2+ a 4 x Z 2+ a 6 z 3 . It is usually convenient t o de-homogenize, i.e. change variables and let x = X / Z and y = Y / Z . Then we have the equation When the characteristic of the base field is not equal t o 2 or 3, then we can make a change of variables resulting in the simpler form CHAPTER 2. ELLIPTIC CURVES 10 We denote by 0 the point [0 : 1 : 0] on E. Since this is the only point with Z = 0 on E , for convenience, we will often denote a point P on E with P # 0 simply as P = (x, y), where this is shorthand for the point P = (x : y : 1). Recall that a function f (x) = x3+ax+b has a double root if and only if 4a3+27b2 = 0. A curve given by the equation y2 = f (x) is called singular if f (x) has a double root. 2.3 Finite Fields An elliptic curve E : y2 = x3 + ax + b, is said to be defined over Q if a , b E Q. A point on E is called rational if its coordinates are rational numbers. When studying the rational points, it can be useful to examine the points in IF, for q = p f , where IF, denotes the finite field with q elements. For the curve to remain nonsingular in IF,, we need -16(4a3 + 27b" # 0 E IF, which means p f 2 and p f 4a3 + 27b2. Assuming this is the case, we denote the natural map for any rational point ( X : Y : 2) on E, we can choose X: IT Z such , that X , Y, Z E Z and gcd(X, Y, Z ) = 1. Then we have a map 2) = ,1, we cannot have )? = Y = 2 = 0, thus P is a point in P2(IF,). Since gcd(X, IT We can sometimes (as in the proof of Theorem 1) use the finiteness of P"IF,) to great advantage. CHAPTER 2. ELLIPTIC CURVES The Group Law The set of rational points on an elliptic curve can be made into an abelian group, with 0 acting as the identity, and it is towards this group that we direct our further attentions. For a more in depth discussion of the group law, see [29] 111.2, [5] Chapter 7, or [18] Chapter 3. We describe the group law geometrically. First, we need a lemma. Lemma 1. If P = (XI : Yl : Z1) and Q = (X2 : Y2 : Z2) are two rational points on the curve E , then the line L through P and Q intersects E in a third rational point, R. Proof. This is a direct consequence of Bkzout's Theorem ([I51 Theorem 18.3 or [16] Corollary 1.7.8). 0 This means that the line at infinity intersects E with multiplicity 3 at 0. The line through the point 0 and R intersects the curve E in a third point by lemma 1, and we call this third point of intersection P + Q (see figure 2.1). This operation makes the rational points on E into an abelian group. It is straightforward to check that) Associativity can be checked directly, but the calculations are long. A verification using MAGMA can be found at [30]. We give a short argument for the associativity given in [5] Chapter 7. First, we note that three points P, Q, R are collinear if and only if there exists a linear form L1 such that L1 has zeros at P, Q, R. Thus if 0, R, S are collinear, then there is a linear form La with zeros at 0, R, S. CHAPTER 2. ELLIPTIC CURVES Figure 2.1: The addition law on the curve y2 = 2 -412x. Suppose the points P, Q, R are collinear and 0, R, S are collinear, i.e. P + Q = S. Since P, Q, R are collinear there exists a linear form L1, such that L1 has zeros at P, Q, R. Similarly there exists a linear form L2 with zeros at 0, R, S. Thus the rational function is defined on the curve E and has zeros at P, Q and poles at 0, S. In fact, the converse holds as well: if there exists a rational function f with zeros at P, Q and poles at 0, S, then P + Q = S. With this characterization 2 of the addition law, it becomes straightforward to see that associativity holds. The equation X = (P+Q)+T, is thus equivalent to the existence of a function with simple poles at P, Q, T a double CHAPTER 2. ELLIPTIC CURVES 13 zero at 0 , and a simple zero at X. But this is exactly the same as the function corresponding to the equation so we conclude that ( P + Q) + T = P + ( Q + T). This characterization of will also be useful in our study of isogenies. The group law can be given explicitly as functions on the coordinates of the points. As we will not have occasion to use the group la,w in its full generality, we will calculate only a few special cases here. We will be interested in curves of the form Let P = (xo,yo) be a point on E. This curve E is now symmetric about the x-axis. The line through P and 0 is vertical, so it intersects the curve again a t the point (50, -yo). Thus ( s o ,-yo) + (xo,yo) = 0 , because the line through 0 and 0 intersects at 0 with multiplicity 3. Thus we have - As is the case with any abelian group, we can consider the group E as a Zmodule under the action [ n ] P =P + . . . + P n times Let us calculate 2P for P = (so,yo) on the curve E : tangent at the point (xo,yo) has slope So the equation of the tangent line becomes This intersects the curve at the point (xl, yl) where y2 = x" ax + b. The CHAPTER 2. ELLIPTIC CURVES 2.5 The Structure of the Group The group of rational points on an elliptic curve is clearly abelian, but we can say much more. Mordell-Weil Theorem. The group of rational points on an elliptic curve is finitely generated. Proof. See [29]Theorem 4.1, [5]Theorem 13.1 and [18]Theorem 7.4. 0 Fundamental Theorem of Finitely Generated Abelian Groups. Every finitely generated abelian group is the direct product of a finite torsion group and a number of copies of infinite cyclic groups (i.e. Z ) . Proof. See [26]Theorem 10.20. 0 We now know that the group E ( Q ) can be decomposed as where the integer r is known as the rank of the curve. Thus to fully describe the group E ( Q ) ,we only need to calculate E(Q)torsand r. 2.6 Torsion The torsion subgroup of an elliptic curve is well-understood. Mazur's Theorem. For an elliptic curve E over Q the torsion subgroup E ( Q t o r sis one of the following 15 groups Z / N Z for 1 5 N 5 10 or N = 12 Z / 2 Z x Z / 2 N Z for 1 <N 54 This was originally proven by Barry Mazur in [21]and [22].It is also stated without proof as [29] Theorem VIII.7.5, [18] Chapter 1 Theorem 5.3, and [27] Theorem 2.2. CHAPTER 2. ELLIPTIC CURVES 15 For our purposes we will only consider the two-torsion of a curve, i.e. the points of order dividing two. Since the group law gives us - ( x , y) = ( x ,- y ) , it is easy to see that a nonzero point is a two-torsion point if and only if its second coordinate is zero. We will not be overly concerned with the computation of the torsion group of E ( Q ) . It should be noted, however, that the problem of calculating torsion points over the rationals has been solved, and a general algorithm for calculating the torsion group can be found in [9], Section 3.3. Torsion in the Congruent Number Curve On the curve, E : y2 = x3 - n 2 x , we can see that there are four two-torsion points ( 0 , (0,0 ) ,(f n , 0)). Since 0, ztn are the only roots of x3 - n2x,we conclude that these are the only two-torsion points on E. We now show that these are the only torsion points on E following the method in [20]. We begin by calculating the number of points on E over the finite field IFp. Lemma 2. Let p be a prime, with p { n and p p + 1 points on E over IFp. =3 mod 4. Then there are exactly Proof. The curve E always has the four points ( 0 , (0;0 ) ,( f n , 0)). Notice that these remain distinct over IF, since p 1 n, and p odd. If p = 3, then these are the only four points because 0, fn are the only three possible values of the x-coordinate of a point on E. Let us now examine the points where x # 0, fn. There are p - 3 such values for x. We can group them in to pairs {fx ) . Now, we have a point on our curve E exactly when "x not a square in IF,. n2x is a square in IFp. Since p =3 mod 4, we know that -1 is Since the squares form a subgroup of index 2 in (IF,)*, we can see that every pair { f x ) in (IFp)* contains exactly one square. But we also have that x3 - n2x = - ( ( - x ) ~- n 2 ( - x ) ) ,SO for every pair {fx ) , exactly one will lead to a point on our curve in IFp. So we have (p - 3)/2 distinct x values. Since each x value leads to exactly two solutions f y , we have p -3 additional points on E over Fp. Adding in the four two-torsion points, we find there are exactly p + 1 points on E over Fp. CHAPTER 2. ELLIPTIC CURVES If PI, P2 are points on the curve condition that 16 E(Q),we now give a necessary and sufficient PI = P2. Lemma 3. Let PI, P2 E P2(FP), i.e. gcd(Xi, Y,, Zi) = 1. Then Pl = Pi fi ifl p = (Xi : Y , : Zi) with Xi, Y,, Zi E 7,and divides Y1Z2 - Y2Z1,X2Z1 - X1Z2 and X1Y2 - X2Yl. Proof. Notice that these are the con~ponentsof the cross-product of Pl and P2 considered as vectors in R3. If p divides the cross-product, then we consider two cases. - (1) If p divides X I , then p divides X2Zl and X2Y1. Since p cannot divide both Yland Z1, we conclude that p divides X2. NOW,we also know YlZ2 mod p, so we have Y2Zl (2) If p does not divide X I , then For the converse, suppose Pl = P2. We know that p does not divide all three of XI,Yl, Z1, so suppose p { XI. The other two cases will proceed in exactly the same CHAPTER 2. ELLIPTIC CURVES way. Since PI = P2we have p { X2, thus Since the first components of these points are the same, we must have Y1X2 F XlY2 mod p and Z1X2 = X1Z2 mod p. So it only remains to show that Y1Z2 r Y2Z1 mod p. If p divides both Yl and Z1 this is clear, otherwise replacing X1, X2 by Yl, Y2 or Z1, Z2 in the above argument gives the result. 0 Now we are ready to characterize the torsion points of the curve E : y" x3-n2x. Theorem 1. I E(Q)torsl= 4. Proof. We know there are exactly four two-torsion points on E ( Q ) , (0,(0, 0), (fn, 0)). Suppose there is another torsion point on E(Q). Since this point is not a two-torsion point it must have order greater than two. Thus the group E(Q),,,, has a subgroup H of order nL where either m is odd, or m = 8. In fact, h4azur's Theorem (2.6)) lists all possible torsion groups, but we do not need such heavy machinery here. Let {PI,.. . , P,). We now examine for which p the reduction map P H P is injective on H. If we consider P I , . . . , P, as vectors in iR3, since they are distinct in lP2(Q) no H = two are multiples of each other, so the cross product Pi x Pj # 6.If we let nij denote the greatest common divisor of the components of the vector Pi x Pj, by Lemma 3, Pi # P' if and only if p { nij. map P H P is So if we let N = max(nij), we have that the reduction injective on H for all primes p > N. Thus m divides the order of the group E(Fp) for all such p. By Lemma 2, if p F 3 mod 4, then IE(Fp)I = p thus mlp + 1, or p G -1 + 1, mod m. Thus we have shown that for all but finitely many primes p with p = 3 mod 4, we have p = -1 mod m. Now recall Dirichlet's famous theorem that for any a , m with gcd(a, m ) = 1 there are infinitely many primes p with p r a mod m. (See [19] Chapter 16 Theorem 1). If m = 8, we have shown that there are only finitely many primes p =3 mod 8, which contradicts Dirichlet's Theorem. If m is odd, then for all but finitely many primes p =3 mod 4 we also have p = -1 mod m which together give p $ 3 mod 4m which is a contradiction to Dirichlet's Theorem if 3 { m. On the other hand, if 31m, then we have if p r 3 mod 4 then for CHAPTER 2. ELLIPTIC CURVES all but finitely many primes p = -1 18 mod 3 k , so there are only finitely many primes 7 mod 12k which again contradicts Dirichlet's Theorem. of p 0 Calculating the rank of an elliptic curve is significantly more difficult, and currently no general algorithm is known. The bulk of this thesis will be devoted to finding the rank of certain "congruent number curves". Isogenies 2.8 An isogeny, 4 : El 4(OE1) + E2,is = 0E2. In fact a morphisni between elliptic curves El and E2 such that 4 induces a group homomorphism from the group E1(K) to E 2 ( K ) ,this is [29] Chapter 111 Theorem 4.8. We prove a special case of this theorem in 52.9. The map 4 also induces an injection of function fields by the pull-back, 4 we define the degree of 4 to be the degree of the field K ( E 1 ) as an extension of the field 4"K ( E 2 ) . So degree(4) = [I<(El): 4" K (E2)],where K (El) For non-constant is the rational function field of K over E l . If K is a number field, then the Mordell-Weil theorem holds, and we can talk about the rank of the curves El and E2. In this case we have the property that isogenous curves have equal rank. A proof is sketched below. We have already seen the multiplication-by-m map, denoted [m]. It is easy to see that this is in fact an isogeny. This isogeny is particularly important as it exists for all curves E and all positive integers m. For a more thorough discussion of isogenies see [29] Chapter 111, Section 4. The degree of the multiplication-by-nz. map is m2. 4 : El + E2 is an isogeny of degree m , then there is a unique isogeny of degree m , 4 : E2 + El such that 4 o 4 = [m]. The isogeny 4 is called the dual isogeny to 4. If The existence and uniqueness of the dual isogeny is proven in [29] Theorem III.G.l. Since 4 is a homomorphism, 4 takes elements of finite order in E 2 ( K ) to elements of finite order in E1(K). The map [m] takes elements of infinite order to elements of CHAPTER 2. ELLIPTIC CURVES 4 19 q5 = [m], so we must have that q5 takes elements of infinite order elements of infinite order. This implies that infinite order and o Rank(E1( K ) ) 5 Rank(E2 ( K ) ) Then, applying the same argument t,o 4,we have that Rank(E1( K ) ) = Rank(E2(K)). The fact that isogenous curves have equal rank is useful to us, as we can bound the rank of E1(K) by bounding the rank of E 2 ( K ) . We now examine in more detail a type of isogeny that will be of use to us. Specifically, we show how to create a degree two-isogeny from any two-torsion point on a curve. We follow the method outlined in [5] Chapter 14. While it is common, given an elliptic curve, to change variables and write the curve in the form y2 = x3 + ax + b. In this section, to simplify calculations, we write our elliptic curve in the form E : y2 = x(z2 + ax + b). This change of variables has the effect of putting a two-torsion point at (0,O). Consider a map Notice that since (0,O) is a two-torsion point, we have that +(+(x, y)) = (x, y). The map + : (z,y) H (zl,yl) induces an automorphism on the function field K ( E ) . Let us ~alculat~e the fixed field. The line through (0,O) and (z,y) intersects the curve at a third point, (zl, -yl). Solving gives CHAPTER 2. ELLIPTIC CURVES 20 Since (z: y) and ( z l , -yl) are on the same line through the origin, we must have Y = -, -y1 so $ is invariant under $. Using the fact that y2 = x(x2 ax b), we have 5 + + 51 To find another fixed function, it suffices to notice that $(x, y) = ( X I , yl) and $(XI,yl) = (x, y) we have that y + yl is fixed by $. Call this value p. Thus To find a relation between X and ,u notice that So we have the equation We now show that K(X, p ) is the entire fixed field of $. To this end, we solve for x , y in terms of X,p. We begin with CHAPTER 2. ELLIPTIC CURVES We also have X= x2 + a x + b = x + - b+ a . x x Combining these gives x= 2 Thus K ( x , y) C K (A, p , a ) , so [K(x,y) : K(X, p)] 5 2. Clearly the field of invariants contains K(X, p ) , so to show that K(X, p ) is the complete field of invariants, it remains only to show that 11, is not the identity automorphism. This is clear though, since $(x) = # x. Now we are in a position to define the isogeny. Notice that the map 11, has given us another curve El defined be equation (2.2), so we let It remains to show that 4 is, in fact, an isogeny, i.e. 4 preserves the group law. If P and Q are points on E, then we have seen that there is a function f E Q(x, y) with simple poles a t P, Q and simple zeros a t 0 , P can simply multiply f by its conjugate + Q. To create a function in Q(X,p ) , we 11,( f ) . Thus f 11,( f ) is in Q(X, p ) , and f $( f ) + has simple poles a t 4 ( P ) , #(Q) and simple zeros at O,4(P Q) since 4(0) = 0 . Thus f $(f ) corresponds to the equation which gives us that 2.10 4 is a group homomorphism. The Curve y 2 = x 3 - n 2x We are now ready to begin examining t,he curve E defined by equation (1.2). If n is a congruent number then we have shown how to construct a rational point on the elliptic curve E , from the sides of the triangle with area n. If E has a rational point, this is not enough to guarantee that n is a congruent number, and we will examine the necessary and sufficient conditions below. CHAPTER 2. ELLIPTIC CURVES 22 To see why every point on E does not correspond to a right hiangle with area n, notice, for instance, the point,s that we can generate from a right triangle all have a square x-coordinate. Points on E that have a non-square x-coordinate were not generated from a right triangle. In fact, a point P on our curve comes froin a right triangle if and only if P is twice a rational point, i.e. P = 2Q for some point Q E E. This follows from basic calculations using the addition laws on E. We sketch the correspondence here. If Q = (xo,yo), then the addition law for points (equat,ion 2.1) on E gives Using the fact, that y; = xi - n2xo,we have z; = + n,2 2 (yo) Thus we find that n is a congruent number (this is just our second definition of a congruent number). To show the converse, we must find a point (x, y) such that Writing out the addition law gives the two equations and Solving for x and y we find CHAPTER 2. ELLIPTIC CURVES Thus 2 ( x , y ) = ((1)2, (a2 - b 2 ) ; ) . Clearly 2((.z,y) + P ) = 2 ( x ,y) for any two- torsion point P. But we know that the two torsion of the curve E is (0,( 0 , O ) )( n ,0 ) , (-n, 0 ) ) (see Theorem (I))! so, in fact, we can find four points which are "half' of the point coming from a right triangle. To effectively use this correspondence, we begin by analyzing which points on E can be twice another point. For our curve E , the sihation is rather simple, and since by Theorem 1 we know Etors(Q) = (0,(0,O),(f n ,0 ) ) . None of these is twice another point because there is no four-torsion. So a squarefree natural number n is a congruent number if and only if the curve E defined by (1.2) has positive rank. For the remainder of the paper we will concern ourselves with finding upper bounds on the rank of E. For an overview of some of the methods and results in the study of ranks of elliptic curves, see 1271. Chapter 3 Two Descent 3.1 Overview Our goal is to find an upper bound on the rank of the curve E. To that end we will find a bound on the size of t,he group E ( Q ) / 2 E ( Q ) Since . we know E ( Q ) e E(Q)torsx Z T , we have lE(Q)/2E(Q)l= lEtors(Q)/2Etors(Q) I . 12'1 . For the elliptic curve E : y2 = x3 - n2x,we know there are exactly four torsion elements, and E,,,,(Q) = 2/22 x 2/22, so In order t,o bound the size of E ( Q ) / 2 E ( Q )we , construct a homomorpliism p from the group E ( Q ) with ker(p) = 2E(Q). Thus we have an isomorphism In particular ( E ( Q ) / 2 E ( Q )=J Ip(E(Q))I.It will be difficult to calculate the group p(E ( 0 )directly, ) so we will calculate p ( E ( Q ) )for various primes I . While Ip(E ( Q ) ) I# Ip(E(Q))1, we can use information about p ( E ( Q l ) )in an attempt to bound the size of P ( E ( Q ) ) . CHAPTER 3. T W O DESCENT 3.2 Themapp Consider the elliptic curve E : y2 Let K [Q] = K [XI/( f ( x ) ) .Set f = = f ( x ) over a field K ; where f ( x ) = x3 + ax n,"_, be the factorization of f fi + b. into irreducibles in K [ x ] . Since f has no repeated roots, f must square-free. In particular the irreducible components fi are distinct. Then by the Chinese Remainder Theorem, n2, K[Ql = K[xl/(fi). Let AK = K[B],A;( be the group of units in A K , and let A: be the multiplicative subgroup of A> consisting of the squares of elements in A>. Define the map p as follows: We know that (xo - Q ) E A;( when ( x o ,yo) 6 E [ 2 ] ( K because ) then relatively prime t o f (x). (xo- x ) is - When ( x o ,yo) E E [ 2 ] ( K we ) have f ( x o ) = 0, so f ( x ) = ( x - X O ) ~ ( X )and K[B] K x K [ x ] / ( g ( x ) Here ) . we define We now illustrate the important properties of p. Lemma 4. The map p is a homomorphism. Proof. We follow the proof in [19]Chapter 19. For an alternate proof see [5]Chapter 15 Lemma 1. We begin by noting that if P = ( x ,y ) , then because p(x, y ) is independent of y. We wish to show that P(P+Q) = p ( P ) p ( Q ) .Mul- ) , see that this is equivalent t o p(P+Q)p(P)p(Q)= tiplying both sides by p ( p ) ~ ( Qwe CHAPTER 3. T W O DESCENT 26 1. Then using the identity above, we see that it is enough t o prove p(P+Q)p(-P)p(--Q) = 1. Changing notation, we need to show that if A + B + C = 0 , then j ~ ( A ) p ( B ) p ( c = ) 1. Recall that A A = + B + C = 0 is equivalent ( x l ,y l ) , B = distinct, for if p ( P to stating that A, B, C are collinear. Let (x2,y2) and C = (xg,y3). We can also assume that A, B , C are + Q ) = p ( P ) + p(Q) for distinct P,Q, then we have Now, we divide the proof into cases 0 If xl = x2 then since A # B , we must have B = -A, which gives C = 0 , so we have the equation If xl # x2 and none of the points have order two then, since A, B, C are collinear, there is a line y = cx + d passing through A, B, C. Thus we have Since both sides are monic polynomials of degree three with the same roots. Reducing modulo f (x) and recalling that 8 denotes the residue of x, we have Noticing the right hand side is just p ( A ) p ( B ) p ( C ) then gives the result. If exactly one of the points has order two, then without loss of generality, we may assume A = ( x l , 0). This means that xl is a root of f , and writing f (x) = (x - xl)g(x) as above, we have K[8] = K[x]/(x - x l ) x K[x]/(g(x)), and we check each conlponent separately. By the definition of p we have that so the first component of p ( A ) p ( B ) p ( C ) is CHAPTER 3. T W O DESCENT Since y = cx 27 + d goes through the point (zl,O), we have czl differentiating equation (3.1) and evaluating at x = xl +d = 0, thus gives Thus the first con~ponentis just ( f l ( ~ ~ )The ) ~ .fact that the second component, is a square follows as in the previous case by reducing equation (3.1) by g(x) and noting that g (f . 0 If two of the points have order two, then the third point must as well, so that leaves us in the final case that .4, B, C all have order two. Thus yl = y2 = y3 = 0. So K[8] K[z]/(.x K[z]/(n: - z2)x K[z]/(n: - x3). - zl) x We have But differentiating Equation (3.1) and plugging in x = X I , x2 and x3 gives Lemma 5. ker(p) = 2 E ( K ) . Proof. We follow the proof in [19]. For an alternate proof see [5] Chapter 15 Lemma 2. The kernel of p clearly contains 2 E ( K ) because p ( 2 P ) = / I ( P ) ~= 1 E A;(lA4?, so it only remains t o show the opposite inclusion. Let P E ker(p). If P write P = (xo,y o ) Since p(P) = 1, if 2 P # # 0 , we can 0 , it must be that xo - 8 is a square in A;(. On the other hand, if 2 P = 0 , then xo is a root of f , so one of the component of xo - 19 is zero, and the the others must be squares since P E ker(p). Thus we can write xo - 8 = (uo + ulQ+ u2Q2)2 (3.3) CHAPTER 3. T W O DESCENT for some uo, u l , u2 E K Since f ( 0 ) = 0, we have O3 = -a0 - b, so we have So v , w E K . Squaring this equation gives Then substituting equation (3.3) gives We must have u, # 0, for otherwise equation (3.3) would not be satisfied. So dividing by U; we have (10 - + 0)(v2- q2= (v18 ~ 1 + ~ where v2 = u 1 / u 2v1 , = v / u 2 ,w1 = w / u 2 . Thus (vlx ) ~ ) ( x o~- x)(v2 - x ) is~ a 1- ) multiple of f ( x ) . Since they are both nlonic cubic polynon~ials,they must be equal, thus f ( x ) = (ulx + ~ 1 - ()2 0 ~ x)(u2 - x )2 . Now, we can interpret this geometrically. If we consider the line L : y = vlx + wl, we see that L intersects E a t x = xo and x = 212, with the latter intersection being of multiplicity two. Since three points P, Q, R are collinear iff P + Q + R = 0 , this gives for some t . Thus 2(v2,- t ) = ( x o ,y o ) , so P E 2 E ( K ) . With these facts in hand we be describe the method of 2-descent. 3.3 The Two-descent With the above definition of p we have the exact sequence CHAPTER 3. TWO DESCENT We will use the fact that 29 is an injection from E ( K ) / 2 E ( K )to A k / A $ to bound the rank of E over K . To do this we will make heavy use of the commutative diagram There is no known method for computing the image of p in A*/A*2,but it is contained in a finite group which is con~putablein practice. We define the 2-Selmer Group, denot,ed S ( 2 ) ( ~ / as Q) s ( ~ ) ( E / Q=) (6 E A*/A*2: $4(6) E p ( E ( Q I ) )for all 1 ) . The 2-Selmer group contains the image of p, because if ( x ,y) E E ( Q ) ,then ( x ,y ) E E ( Q ) for all 1. Since p ( E ( Q ) )c S ( 2 ) ( ~ / Qwe) ,will attempt to bound # p ( E ( Q ) )by calculating the size of S ( 2 ) ( E / Q ) . In other words, by intersecting the groups p ( E ( Q 1 ) with ) the images Ab/A$ in A & / A g for various 1 we hope to obtain a bound on the size of Ab/Ab2, and hence the rank of E over Q. For a further discussion of the method of 2-Descent see [29]Chapter X, [9]Chapter 3, [3],[19]Chapter 19 or [32]. Chapter 4 Two-descent applied to congruent number curves Let p be a prime number. We will be working with the elliptic curve: Since f (z) = x ( x (Q*/Q*2))". So the map /L - p)(x + p) splits completely over Q , we have A ~ / A $ = becomes Lemma 6. If ( x , y ) E E ( Q ) then p ( x , y ) ( 1 , -1; 2 , - 2 , p , -p, 2p, - 2 p ) , and b1b2b3 E = Q*2 ( b 1 , b 2 ,b3) where each 6, is in the set CHAPTER 4. TWO-DESCENT APPLIED TO CONGRUENT NUMBER CURVES31 Proof. We have that blb2b3 = y2. Consider the l-adic valuation of the hi's. We have ordl(y2) = ordl(b162b3) = ordl (dl) + ordl(b2) + ordl(53) is even for any 1. If one of the hi's has odd valuation, then two must. If bi has odd valuation, then 1 I hi, but the &'s differ by p or 2p, so if I # 2,p, none of the working modulo squares, we see that if 1 # & can have odd valuation. Since we are 2 , p the hi's must have valuation 0 for all 1. So each bi E (1,-1,2,-2,p,-p12p,-2p). 0 Lemma 7. where 1211 is the 1-adic valuation of 2. Proof. This is [6] Lemma 5.1 and [7] equation 7.6.2. For an argument using Haar measure see [12] page 451. First notice that since ker(p(E(Q1)))= 2E(Q1) we have that #p(E(Qi)) = #E(Q)/2E(Q1). We will consider two cases, 1 < oo and 1 = oo. If 1 < oo, then the group E(Ql) has a subgroup H of finite index such that I1 -. Z1. See [29] Theorem VII.6.3. Consider the map Since H is torsion free, the kernel of map [2IH is in one-to-one correspondence with the kernel of the map [2],which is E[2](Q1),thus This gives us that The following equations are true of any abelian groups, CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENT NUMBER CURVES32 combining these three equations we have If I # 2 then #Z1/2Z1 = 1, and if I = 2 then #Z1/2Z1 = 2. This proves the lemma for I < ca. When I = ca, we distinguish two cases depending on how many roots f has in R. If f splits completely over R, we call its roots, a l , a 2 , a3, and we can, without loss of generality, assume a l < a 2 < as. In this case #E[2](R) (1W*/1W*2)3 -. (iZ/2Z)3. If = 4 and p : E ( R ) (x, y) is a point on the curve, t)hen a1 5 x 5 a 2 or a s 5 x, which gives p(x, Y) = ( x - a l , x - a 2 , x - a s ) = (1, -1, -1) or ( I l l , 1). So #p(E(R)) 2. If f has only one root, a l , over R then #E[2](R) = 2 and p(x, y) = --+ = (1,1,1) for all (x, y) E E ( R ) since al 5 x for all x , so #P(E(R)) = 1. 4.1 General Bounds Here we try t o find a bound on the size of p ( E ( Q ) ) , and hence the rank of E. By Lemma 6 we have that Now Ak/Ak2 = {(f1,f 1,f 1)) since 1W*/IP2 and p((0,O)) = (-1, -1,l) SO = Z/2Z. We know #p(E(R)) = 2 we conclude that p ( E ( R ) ) = {(1,1,1),(-1, -1,l)). Then taking I = ca in diagram 3.5, we see that (-1,1, -1) $! p(E(Q)). This shows that p ( E ( Q ) ) is generated by at most 5 elements, but two of these are the images of two-torsion points, so at most three of these generators can be images of points of infinite order, so we conclude that the rank of E ( Q ) is at most 3. CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENT NUMBER CURVES33 4.2 Specific Bounds In the previous section we found a bound on the size of p ( E ( Q ) ) that is independent of p. Now we will impose congruence conditions on p t o further bound the size of P(E(Q)). For odd p, by (4.1): we have #E[2](Qp) = #p(E(Qp)) = 4, SO we have that p(E(QP)) = {(Il 1-11,(-1, -P,P)] (P, 212 ~ 1(-P, , -2~,2)}. Note that this is just the image of the 2-torsion points of E. We only wish to keep elements of ((-1, -1, I ) , (-1,1, -I), (p, p, I ) , (p, 1,p), (2,2, I ) , (2,1,2)) that are equivalent to one of the above elements in 0,. We distinguish 4 cases for p If p 1 mod 8, then -1 If p 5 mod 8, then -1 I f p = 7 mod 8, then 2 = 2 .= -2 = 1 E Qg/Qg2 - 1 and 2 $ -1 E = 1 and -1 SO Qg/q2 we eliminate nothing. SO $ 1 EQg/Qg2 Finally we look at p(E(Q2)). By equation 4.1, #p(E(Q2)) = 8, so we need 3 generators for the group. The image of the 2-torsion provides 2 generators, the point with x = 114 happens to map to an independent generator, (1/4,1/4 P(E(Q~)). - p, 114 + p) E CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENT NUMBER CURVES34 (1/4,1/4 - p, 114 + p) - (1, 1 - 4p, 1 Thisisbecausepisoddandso1-4p= + 4p) = (1,5,5) E Q;/Qa2. 1+4p=5 mod8,so1-4p=1+4p= 5 E Q;/Qj12. This gives us our third generator, so we have that Now we remove the elements that are not equivalent nlodulo squares (in Q2) to one of the above elements. We again distinguish 4 cases for p: This is because (-1, -5,5) . (1,5,5) = (-1, - 1 , l ) E p(E(Q2)). CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENTNUABER CURVES35 4.3 Results Our results so far can be summarized in the following inequality O i f p ~ 3m o d 8 rank(E) 5 1 if p 5 mod 8 From here we can already conclude that if p =3 mod 8, p is not a congruent number. This was already proven in the 19th century by Genocchi in [13] (see [lo]). When p =5 mod 8, or p G 7 mod 8, p is in fact a congruent number. This was stated in [31], and proven in [ll]for p < lo6 (p not necessarily prime), and finally proven for all prime p in [24]. Another derivation of the bounds we have obtained on the size of the 2-Selmer group can be found in [17]. If U I ( E / K ) were known t o be finite, by [29]Chapter X Theorem 4.14, #UI(E/K)[2] would have to be a perfect square, thus it would have an even number of generators. The exact sequence - = )#UI(E/K)[2]. Because S ( 2 ) ( ~ / ~ ) gives us that # s ( ~ ) ( E / K ) / # E ( K ) / ~ E ( K has a t most 3 generators for p =5 mod 8, and p 7 mod 8, this implies that III(E/K)[2] must be the trivial group. If ILLI(E/K)[2]1 = 1 then E ( K ) / 2 E ( K ) 2. S ( 2 ) ( ~ so / ~the ) ,rank of E ( Q ) would have to be exactly 1. For a slightly more in depth discussion of this, see [29] Chapter X, Rernark 6.3. While it is not known that, LU(E/K)[2] is finite, it is conjectured to be so, see [29] Chapter X, conjecture 4.13. We will continue by further analyzing the case when p =1 mod 8. Chapter 5 The Case When p 5.1 - 1 mod 8 A Specific Two-Isogeny We have done a 2-descent on the curve E : y2 = x(x its rank. To further bound the rank in the case p + p)(x - p) in order to bound fi 1 mod 8, we will perform a 2-descent on an isogenous curve. The map is an isogeny of degree two generated by the 2-torsion point (0, 0), from our curve E : y2 = x3 - P Since p 2 to ~ =1 such that p = a2 mod 4, we can write p as the sum of 2 squares in Z.Let a,b E Z + b2. If we assume a , b > 0 and a is odd, then this representation is unique. We will make frequent use of this decomposition of p. Note that, unlike E , El has only t,wo rational 2-torsion points, (0:(0,O)) CHAPTER 5. THE CASE WHEN P 5.2 =1 MOD 8 Method As before, we construct the exact sequence (3.4). Since f ( x ) = x ( x 2 + qp2)does not split completely over Q , we now have two distinct forms for the group Ak/A$ where p is defined as follows. If i = a E K , i.e. f ( x ) splits completely over K , then When i @ K, we have We proceed as before, making use of diagram (3.5),and equation (4.1). 5.3 General Bounds We begin, as before, by identifying a finite set containing the image of p + Lemma 8. p ( E l ( Q ) )c ( ( 2 , l i), ( p , a + bi),( p ,a - bi),(1,i ) ) . CHAPTER 5. THE CASE W H E N P Proof. Suppose =1 MOD 8 38 p ( z , y) = (61, 62). Then we have b1 . N(b2) = y2. Where N(J1) is the norm of b2 as an element of the extension field Q(i), i.e. N(b2) = 6262 where is the ordinary complex conjugate of 62. Consider the I-adic valuation of the 6,s. f2 We have ordl(y2) = ordl(bl . N(62)) = ordl(bl) + ordl(N(b2))is even for any I. If bl or N(b2) has odd valuation, then they both must. If dl has odd valuation, then I 1 dl, but since 62 = dl we have that I I - 2pi, we have N(b2) = b1 + 4p2. So if I 1 dl and I 1 b2 4p2, so the only divisors bl are 1 , 2 , p . In Z[i], 4p2 factors into + i)2(1 - ~ ) ~+( bi)(a a - bi), since every factor of d2 is a factor of 4p2 the only possible factors of 62 are 1 , i , 1 + i , 1 - i , a + bi, a - bi. The fact that p(El(Q)) C ( ( 2 , l + i ) , (p, a + bi), (p, a - bi), (1, i)) follows from the fact that irreducibles as (1 bl . N(d2) = 1 modulo squares. 0 It is also worthwhile t o notice that since we are working modulo squares, 22 (1 + i)2 = 1 and so ( 2 , l + i ) ( l , i) = ( 2 , l - i) 5.4 = mod squares. p-adics Since Ei[21(Qp) = (0, (0, O), (2pi, O), (-2pi, O)), (4.1) gives that #p(El(Qp)) = 4. The map p looks like: The last equivalences follow from the fact that 2 and i are squares in Q,because =1 CHAPTER 5. THE CASE WHEN P p =1 mod 8. Since Ip(E(Qp))l = 4, we must have that Now since ( a + bi)(n - bi) a - MOD 8 bi = p, we have that n + bi - p modulo squares and -- 1 modulo squares in Qp (or vice-versa), and since (1 + i ) ( l- i) = 2, which + i = i - 1 n~odulosquares. Before we can eliminate elements from p ( E ( Q ) ) , we must determine when 1 + i is a square in 0,. is a square in Q,, we see that 1 - 5.4.1 The Quadratic Character of 1 + i We have that p = a2 + b2, SO a bi mod p. Using Quadratic Reciprocity and the laws of the Jacobi symbol we have (;) (E) = Then = (7)+ a2 (y)= (:) (y)= ( ) b2 = (f) = 1. So we just need to calculate (y),which we do using the Jacobi symbol. By Quadratic Reciprocity (5) = ( - quadratic residue mod p exactly when (a by the quadratic charact,er of 1 + i. 1 ) ~ .This gives us that 1 + b)2 E Qp (i.e. 1 p(E(Q)) 2 1 i) + 2 1 - 2) (P, a + bi) (PAP) H p(E(Qp)) C ( l , l , 1) H (1,1 , l ) -+ ++ ++ + i is a 1 mod 16. We distinguish 2 cases, + i is a square mod p) This occurs when (a + b)2 = 1 mod 16. 1. 1 + i =- - (a+b12-1 (Q;/Q;2)3 (P, P, 1) (P,LP). In this case, we cannot further reduce the size of p(E(Q)). CHAPTER 5 . THE CASE WHEN P =1 MOD 8 2. 1 + i f 0 E Qp (i.e. 1 + 2 is not a square mod p) This occurs when (a + b)2 = 9 mod 16. Since El [2](Q2)= (0, (070)),we have, by (4.I ) , that #p(E1 ( 0 2 ) ) =4 and p is defined as follows: Since # p ( E 1(Q2)) = 4 we need to find 2 generators for the group. A little testing yields t,he points x = 5 and x = 2p. To see t,hat these are in fact points on the curve - - 2 / 2 2 x (2/82)*, where an element in Q2 is a square if E1(Q2),recall that Q ; ~ / Q = G~ and only if it has even valuation (the first tern? in the ~ r o d u c t )and , the odd part is a + square mod 8 (i.e. 1 mod 8). Here we see that 5(52 4p2) = 125 + mod 8, and 2 p ( ( 2 ~ 1 ) ~4p2) = gp3 mod 8. + 20p2 5 +4 = 1 + 8p3 = 16p3 which is a square because p 1 C H A P T E R 5. T H E CASE W H E N P =1 p ( x = 5 ) = (5,s- 2 p i ) p(z = 2p) = - MOD 8 ( 1 , 5 - 2p.i) mod squares +i) mod squares. ( 2 p , 2p - 2pi) = ( 2 , l So we distinguish the same 2 cases as before. - 1. 1 + i is a s q u a r e i n Qp i.e. ( a + b)2 1 mod 16 p(E(Q)) (21+) ( p , a + bi) (p,a-bi) (1,~) ++ p(E(Q2)) (2,1+2) ++ (171) -+ + + + + c Q2 x Q2(i) (L1) (1.4 51 p ( E ( Q 2 ) ) . i ) , which gives us that p ( E 1 ( Q ) ) has at most 3 generators, so we find that in this case the rank of El is at most 2. We lose the generator ( 2 , l - - 2. 1 + i is not a square in Qp i.e. ( a + b)2 - 9 mod 16 From this argument we have t'hat if p r 1 mod 8 , with a , b such that a2 and ( a + b)2 when ( a + b2 = p 9 mod 16, then p is not a congruent number. On the other hand, + b)2 = 1 mod 16, we cannot make any conclusions. This result was stated CHAPTER 5. T H E CASE WHEN P =1 MOD 8 42 as early as 1915 by L. Bastien [2]. This result can also be obtained as a consequence of Tunnell's Theorem, as in [34] Proposition 6. Chapter 6 Homogeneous Spaces 6.1 Method In the previous sections, we have examined the image of the map where, as before E : y2 = f ( x ) , A = Q [ x ] / ((fx ) )= Q[0]and 0 is the residue of x. As f is a polynomial of degree three, we have that 1,0,O2 is a basis for A, so any element in A can be written as uo + ule + u2Q2,where uo, ul, u2 E Q. So for any 6 in the image of p we have the equation + + x ( P ) - 0 = b(uo u18 ~ ~ 0 ~ ) ~ . Expanding the square on the right hand side we have for some quadratic forms Q6,iE Q[uo,ul , u2] Equating powers of 0, we have that CHAPTER 6. HOMOGENEOUS SPACES So equat,ions (6.1) gives us a necessary condition for b to be in the image of p. These equations are in fact sufficient as well, and if equations (6.1) are satisfied, then Qs,o is the x-coordinate of a point on the elliptic curve E. We would like to restrict our attention to integral solutions to equations (6.1), and this is easily done by noting that Q6,1is homogeneous of degree two, and clearing denominators, which gives us the two equations We will begin by examining the equation Qa,2= 0. If we set then C is a conic, and if we can find a rational point on C will allow us to parametrize C . Supposing that we were able to parametrize, C as (uo(A),ul (A), u2(A)), we can move on to the second equation, Qa,i = -u$. Using our parametrization for C , we obtain the equation Since Qa,i,QhI2are quadratic forms, the left hand side will be a quartic in A, call it g(X). Then the curve g(A) = -u: image of p. has a rational point exactly when d is in the CHAPTER 6. HOMOGENEOUS SPACES 6.2 Conics We now examine the homogeneous spaces associated to the curve E~: y2 = x3 + 4p2x. We have f (x) = x3 + 4p2x, which gives A = Q[x]/(x3+ 4p2x) = Q x Q(i). In the preceding sections we showed that if p =1 mod 8, then P(E(Q)) c ( ( 2 , l + 4, (P, a + bi), (P, a - b4. Remember, that when we calculated the 6 we used the natural basis (1,O), (0, I ) , (0, i) of Q x Q(i). We have to convert these into the basis 1,9, O2 where 9 is the residue of + x in Q[x]/(x3 4p2x) Y Q x Q(i). TO do this, we make explicit the isomorphism Now we are ready to try to solve the equation Qs,2 = 0 for the allowable values of 6. When 6 = (1,p) = 1 - so2, i.e. 6 is the image of the two-torsion point, (0,O). Then we have 1 Q6,2 = -(I 4 - P)U; - This has a solution (O,2p, 1). 4p5u22 + 2p uou2 + p ul. 3 3 2 (6.3) CHAPTER 6. HOMOGENEOUS SPACES 0 When 6 = ( p ,a + bi) = p + $ 3 + yo2we have When 6 = ( p ,a - bi) = p + - LO $$02 2~ When 6 = ( 2 , l + i ) . ( p , a + bi) = 2 p + we have $o+ we have -Q2 + + Q6.2 = ( a b)puoul+ 2(a - b)p2uou2 ( a - b)p2u? 1 b-a -pu; 4 ( b - a)p4u; - 4 ( a b ) p h l u 2 -uo. 2 4 + 0 + When 6 = ( 2 , l + i ) . ( p , a Qs,2 = When 6 = (a- ~ + + , + 2 0 + T O 2 have + 2(a + b ) p 2 u o u+~ ( a + b)p2u? - bi) = 2p (6.7) 2p-b-a (6.8) ) P U ~ I L ~ ( 2 , 1 + i ) . ( p , a - h i ) . ( p , a + bi) = ( 2 , p ( l + i ) ) = 2 + ;Q+3Q2 have The first equation corresponds to a two-torsion point, so we analyze the only other space with a rational point. From (6.4) for 6 = ( 2 , l + i ) , we have We know that (-2p, 1 , 0 ) is a point on C , so letting L be the line uo = Xu2 - 2p, and considering the intersections of L with Qs,2 as in [25],we arrive a t an equation Setting this t o zero and solving for u2 we get CHAPTER 6. HOMOGENEOUS SPACES Plugging back in to L we have So this gives a parametrization for C as Thus ul(X),u2(X)) = -ug becomes Absorbing squares into u:, the equation Qs,l(uo(X), So if we define f (x) = p(x4 + 16p2x3 - 96p4x2 + 768p6x - 1792p8), we have that f (z)is an irreducible quartic, and we would like t,o determine when the equation y2 = f ( 4 has solutions We can simplify this by making the change of variable x y ++ ++ p2x, and p4y, and dividing both sides by p8, which gives us the equation Making a further change of coordinates x ++ -42 and y ++ 16y, and dividing both sides by 28, we can reduce this to Thus we conclude that (2:1 solution to equation (6.10). + i) is in the image of ,LL exactly when we have a CHAPTER 6. HOMOGENEOUS SPACES 6.3 Follow up The next step would be to find conditions on p such that equation (6.10) has a solution. If we can find a rational solution to equation (6.10), then ( 2 , l + i ) is in the image of p , which means that the #E1/2E1 > 2, so p is a congruent number. On the other hand, if there is no rational solution to equation (6.10), then We know also that so in this case, we must have #LLI[2] > 1. If, as it is conjectured, #LU < m then as we argued before, by [29] Chapter X, Theorem 4.14, the order of IJ..I[2]must be a perfect square, so #LLI[2] 2 4. Since we have shown in the previous sections that # s ( ~ ) ( E ~ / Q 5 )8, we must have #p(EIQ) = 2, which means that Rank(El/Q) = 0. Thus in this case we conclude that p is not a congruent number. This gives us a criterion to determine whether p is a congruent number: p is congruent exactly when equation (6.10) has rational solutions. We have already examined local conditions for the solubility of equation (6.10). In sections 5.4 and 5.5 we examined when ( 2 , 1 + i ) was in the image of p ( E (Q)) and p ( E ( Q 2 ) ) ,and we determined that ( 2 , l + i) was in both images when (a mod 16. Thus equation (6.10) has solutions Q2 and Qp when (a + b)2 E 1 + b)2 = 1 mod 16. After testing for local solubility, the standard approach is to perform a second 2descent on (6.10) as described in [23] and [33] to determine when equation (6.10) has rational solutions. We leave this for a future paper. Bibliography [ I ] Ronald Alter. The congruent number problem. T h e American A4athematical Monthly, 87(1):43-35, 1980. [2] L. Bastien. Nombres congruents. Interme'diaire Des Mathe'maticiens, 22:231-232, 1915. [3] B. J . Birch and H.P.F Swinnerton-Dyer. Notes on elliptic curves. I. Journal fur die Reine un.d Angeuiandte Mathematik, 212:7-25, 1963. [4] B.J. Birch and H.P.F Swinnerton-Dyer. Notes on elliptic curves. 11. Journal fur die Reine und Angewandte Mathematik, 218:79-108, 1965. [5] J.W.S. Cassels. Lectures o n elliptic curves. Number 24 in London Mathematical Society Student Texts. Cambridge University Press, 1991. [6] J.W.S Cassels. Second descents for elliptic curves. Journal fur die Reine und Angewandte Mathematik, 494:101-127, 1998. [7] J.W.S. Cassels and E.V. Flynn. Prolegomena t o a A4iddlebrow Arithmetic of Curves of Genus 2. Number 230 in London Mathematical Society Lecture Note Series. Cambridge University Press, 1996. [8] V . Chandrasekar. The congruent number problem. Resonance, 3(8):33-45, 1998. [9] John Cremona. Algorithms for Modular Elliptic Curves. Cambridge University Press, second edition, 1997. BIBLIOGRAPHY 50 [lo] Leonard Eugene Dickson. History of the Theory of Numbers Volume, volun~e11. American Mathematical Association, 1920. [ll] Noam Elkies. Curves dy2 = x3 - x of odd analytic rank. In Claus Fieker and David R. Kohel, editors, Algorithmic Nwnber Theory, 5th International Sympo- sium, A N T S - V, pages 244-251, 2002. [12] E.V. Flynn, B. Poonen, and E. Schaefer. Cycles of quadratic polynonlials and rational points on a genus-2 curve. Duke Mathematical Journal, 90:435-463, 1997. [13] Angelo Genocchi. Note analitiche sopra tre scritti. Annuli di Scienze Matematich.e e Fisiche, 6, 1855. [14] Richard Guy. Unsolved problems i n number theory. Problem Books in Mathematics. Springer-Verlag, third edition, 2004. [15] Joe Harris. Algebraic Geometry: A First Course. Number 133 in Graduate Texts in Mathematics. Springer-Verlag, 1992. [16] Robin Hartshorne. Algebraic Geometry. Number 52 in Graduate Texts in Mathematics. Springer-Verlag, 1977. [17] D.R. Heath-Brown. The size of selmer groups for the congruent number problem 11. Inventzones Mathematicae, 118:331-370, 1994. [18] Dale Husemoller. Elliptic Curves. Number 111 in Graduate Texts in Mathematics. Springer-Verlag, 1987. [19] Kenneth Ireland and Michael Rosen. A Classical In.troduction To Modern Number Theory. Number 84 in Graduate Texts in Mathematics. 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