On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples


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1 On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, Introduction 1.1 Mazur s Theorem Let C be a nonsingular rational cubic curve, and suppose that C(Q) contains a point of finite order m. Then either 1 m 10 or m = 12. More precisely, the set of all points of finite order in C(Q) forms a subgroup which has one of the following two forms: (i) A cyclic group of order N with 1 N 10 or N = 12. (ii) The product of a cyclic group of order two and a cyclic group of order 2N with 1 N The extreme case According to Mazur s Theorem, C(Q) tor must be isomorphic to one of the listed groups. Of these possible torsion subgroups, has the most elements; = 16. Curves with sixteen torsion points are thus considered to be the most extreme legal elliptic curves. Today we ll look at such a type of elliptic curve for which C(Q) tor = Relation to Pythagorean triples A Pythagorean triple is any element (a, b, c) N 3 such that a 2 + b 2 = c 2. If additionally gcd(a, b, c) = 1, then (a, b, c) is called a primitive Pythagorean triple. We showed on the first day of class that a = m 2 n 2, b = 2mn, and c = m 2 + n 2, for some m, n such that m > n, gcd(m, n)= 1, and m and n are not congruent modulo 2. 1
2 2.1 Claim Let (A, B, C) be a Pythagorean triple. Then let C : y 2 = x(x + A 4 )(x + B 4 ). (1) Then C(Q) tor and C(Q) tor = 16. Before we investigate and prove this claim, let s digress a bit and explore the multitude of curves of this form. 2.2 Algorithm that generates Pythagorean triples Surprisingly, we can use the Euclidean Algorithm to find all Pythagorean triples of the form (x, y, p) where p is prime. Additionally, these Pythagorean triples can be multiplied by scalars or added to generate all other Pythagorean triples. For example, (7,24,25) is a primitive Pythagorean triple that is not of this form Proposition Given a prime p, the following statements are equivalent: a) p 1 (mod 4). b) There exists a Pythagorean triple (x, y, p). c) 1 is a square modulo p Algorithm, first step Suppose p 1 (mod 4). Fermat proved that there infinitely many primes congruent to 1 modulo 4. We can use an algorithm to find the corresponding Pythagorean triple (x, y, p). First, solve the congruence a 2 1 (mod p). (It may help to cheat by using Mathematica for this step.) Additionally, we already know two facts about a: a) Clearly, ord p (a)= 4. b) Congruences of the form x 2 a (mod p), such that a and p is prime, have 0 or 2 incongruent solutions modulo p. By Proposition 2.2.1, a 2 1 (mod p) has 2 incongruent solutions, which can either be written as ±a, or as a and p a. For the algorithm, we need a to be a positive integer less than p Algorithm, second step Next use the Euclidean Algorithm on p and a. Although we usually use the Euclidean Algorithm to find greatest common divisors, we already know that gcd(p, a)=1. In this case, we are interested in the remainders. Let m and n be the two greatest remainders which are less than p, m > n. Then, remarkably, x = m 2 n 2, y = 2mn, and p = m 2 + n 2. 2
3 2.2.4 Examples Here is a demonstrative example. Let p = 89; a is found to be = 34(2) = 21(1) = 13(1) = 8(1) = 5(1) + 3 p Then m = 8 and n = 5. Thus x = ( ) = 39 and y = 2(8)(5) = 80. Hence, the corresponding Pythagorean triple is (39,80,89). This method is especially useful for finding very large Pythagorean triples. Take the one billionth prime, p = We find that a = , p , and the 12th and 13th remainders of the Euclidean Algorithm on p and a are and This generates the Pythagorean triple ( , , ) Operations Clearly, we can multiply a Pythagorean triple by a positive integer to generate another Pythagorean triple. Using an analogue in the complex plane, we can also find an operation that generates a third Pythagorean triple from any two. Using these two operations, we can express all Pythagorean triples in terms of triples generated by our algorithm. Suppose (a, b, c) and (e, f, g) are Pythagorean triples. In the complex plane, let z= a + bi and w= e + fi. (a + bi)(e + fi) = (ae bf) + i(af + be) (ae bf) 2 + (af + be) 2 = (a 2 + b 2 )(e 2 + f 2 ) = z 2 w 2 = (cg) 2. (2) Thus ( ae bf, af + be, cg) is a Pythagorean triple. Define this to be the sum of the Pythagorean triples (a, b, c) and (e, f, g). 3 The Pythagorean elliptic curve 3.1 Claim restated Let (A, B, C) be a Pythagorean triple. Then let C : y 2 = x(x + A 4 )(x + B 4 ). (3) Making use of the corresponding triples, we can consider every curve of this form to have the same additive and multiplicative properties as the corresponding triples. Again we claim that C(Q) tor and C(Q) tor = 16. 3
4 3.2 NagellLutz method Let s try, blindly, to find C(Q) tor using the NagellLutz Theorem. On the curve, f(x)= x(x+a 4 )(x+b 4 ). The theorem tells us that if P =(x, y) C(Q) tor, then y = 0 or y 2 Disc(f). If y = 0, the resulting equation has three solutions: (0, 0), ( A 4, 0) and ( B 4, 0). All three solutions are in fact torsion points of order two. Including the identity, we have found four torsion points. After this, however, things get messy. Using the formula for discriminant, we find that Disc(f) = A 8 B 8 (A 4 B 4 ) 2 (4) This means if y 0, y A 4 B 4 (A 4 B 4 ). For Mathematica, this is good enough, but we would like to have a more efficient method. Take the simplest curve, analogous to (4, 3, 5). Then y , a product with = 270 distinct positive divisors to check. 3.3 Lemma Let C : y 2 = (x a)(x b)(x c), a, b, c, and suppose P = (u, v) is a nonidentity rational point on C. Then there exists a rational point Q Csuch that P = 2Q iff (u a), (u b) and (u c) are perfect squares Proof Suppose u a = A 2, u b = B 2, and u c = C 2. Define the point Q: Q = (u + AB AC BC, (A + B)(A C)(B C)). (5) Then, we have X(2Q) = u. Suppose P = 2Q, where Q = (r, s). Let L : y = mx + n be the tangent line to C at Q. Then Q = (r, s) and P = Q Q = (u, v) are both on L, which means they satisfy the equation (x a)(x b)(x c) = y 2 = (mx n) 2 Moreover, since L is tangent to C at Q, the equation (x a)(x b)(x c) (mx n) 2 = 0 has the three roots r, r, u. Thus it is the case that: (x a)(x b)(x c) (mx n) 2 = (x r) 2 (x u) (6) Without loss of generality, setting x = a in equation 6 gives (ma n) 2 = (u a)(a r) 2 Suppose a = r. Then Q = (a, s) = (a, 0) and P = 2Q is the identity, a contradiction. Thus u a is a perfect square. Using this Lemma, we can find points that have higher order than two from (0,0), which is a point we can easily verify to be a 2torsion point. 4
5 3.4 Finding some torsion points Let s begin with a curve of the form C : y 2 = x(x+r 2 )(x+s 2 ). Applying equation 5 of the Lemma to the 2torsion point (0, 0), we find that P = (rs, rs(r + s)) is a point of order 4 on C. We also know that ( r 2, 0) is a 2torsion point on the curve. Because of Mazur s Theorem, this is enough information to show that the torsion subgroup of C : y 2 = x(x + r 2 )(x + s 2 ). either is 4 or. They are the only two possible torsion subgroups by Mazur s Theorem that have a point of order 4 and two different points of order 2. Moreover, the torsion subgroup must contain Let r = A 2 and s = B 2. Then P = (rs, rs(r + s)) = (A 2 B 2, A 2 B 2 C 2 ). Using equation 5 of the Lemma generates the point of order 8 (AB(A + C)(B + C), ABC(A + C)(B + C)(A + B)). The existence of this torsion point proves 4. our original claim, that C(Q) tor and C(Q) tor = Finding the rest of the torsion points The fact that C(Q) tor is a group implies a very useful fact that should be explicitly stated: C(Q) tor is closed under addition Isomorphism Define a group isomorphism f : C(Q) tor using a table (page 6). Beginning with the chart below, we will use closure under addition to find the others. Note we have several different ways to plug in our points; f is just an example. Other ways will result in other permutations of the 16 torsion points. C(Q) tor (0, 0) O (0, 1) (AB(A + C)(B + C), ABC(A + C)(B + C)(A + C)) (0, 2) (0, 3) (0, 4) (0, 0) (0, 5) (0, 6) (0, 7) (1, 0) ( A 4, 0) (1, 1) (1, 2) (1, 3) (1, 4) ( B 4, 0) (1, 5) (1, 6) (1, 7) 5
6 4 The torsion subgroup stated Addition on C(Q) tor corresponds with addition on, which is much easier. After adding the corresponding points on the curve, we arrive at the following table, the group isomorphism f: C(Q) tor Order (0, 0) O 1 (0, 1) (AB(A + C)(B + C), ABC(A + C)(B + C)(A + C)) 8 (0, 2) (A 2 B 2, A 2 B 2 C 2 ) 4 (0, 3) (AB(C B)(C A), ABC(A + B)(C A)(C B)) 8 (0, 4) (0, 0) 2 (0, 5) (AB(C B)(C A), ABC(A + B)(C A)(C B)) 8 (0, 6) (A 2 B 2, A 2 B 2 C 2 ) 4 (0, 7) (AB(A + C)(B + C), ABC(A + C)(B + C)(A + C)) 8 (1, 0) ( A 4, 0) 2 (1, 1) (AB(C + A)(B C), ABC(C + A)(C B)(B A)) 8 (1, 2) ( A 2 B 2, A 2 B 2 (A 2 B 2 )) 4 (1, 3) (AB(B + C)(A C), ABC(B A)(C A)(B + C)) 8 (1, 4) ( B 4, 0) 2 (1, 5) (AB(B + C)(A C), ABC(B A)(C A)(B + C)) 8 (1, 6) ( A 2 B 2, A 2 B 2 (A 2 B 2 )) 4 (1, 7) (AB(C + A)(B C), ABC(C + A)(C B)(B A)) 8 6
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