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Rational Points on an Elliptic Curve Rational Points on an Elliptic Curve Dr. Carmen Bruni University of Waterloo November 11th, 2015 Lest We Forget Dr. Carmen Bruni Rational Points on an Elliptic Curve Revisit the Congruent Number Problem Congruent Number Problem Determine which positive integers N can be expressed as the area of a right angled triangle with side lengths all rational. For example 6 is a congruent number since it is the area of the 3 − 4 − 5 right triangle. Dr. Carmen Bruni Rational Points on an Elliptic Curve Enter Elliptic Curves The associated equations with the congruent number problem, namely x2 + y 2 = z2 xy = 2N can be converted to an elliptic curve of the form Y 2 = X 3 − N2X : We also saw that we can reduce our problem to considering only squarefree numbers N. Dr. Carmen Bruni Rational Points on an Elliptic Curve Going Backwards The belief now is that solving problems related to elliptic curves might be easier than the originally stated problem. The question now that occurs is can we go from an elliptic curve of the form y 2 = x3 − N2x to a rational right triangle with area N? Dr. Carmen Bruni Rational Points on an Elliptic Curve Key Theorem 1 Theorem 1. Let (x; y) be a point with rational coordinates on the elliptic curve y 2 = x3 − N2x where N is a positive squarefree integer. Suppose that x satisfies three conditions: 1 x is the square of a rational number 2 x has an even denominator 3 x has a numerator that shares no common factor with N Then there exists a right angle triangle with rational sides and area N, that is, N is congruent. Dr. Carmen Bruni Rational Points on an Elliptic Curve Key Theorem 2 Theorem 2. A number N is congruent if and only if the elliptic curve y 2 = x3 − N2x has a rational point P = (x; y) distinct from (0; 0) and (±N; 0). Thus, determining congruent numbers can be reduced to finding rational points on elliptic curves! Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 1 Let (x; y) be a point with rational coordinates on the elliptic curve y 2 = x3 − N2x where N is a positive squarefree integer where x is a rational square, has even denominator (in lowest terms) and has a numerator that shares no common factor with N. Our goal is to trace backwards the proof from last week. p Let u = x which is given to be rational. y Set v = u giving 2 y 2 x3−N2x 2 2 v = u2 = x = x − N : Let d be the smallest integer such that du 2 Z (namely the denominator of u in lowest terms). Note that d is even by assumption and that d4 is the denominator for u2 = x. Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 1 Since v 2 = x2 − N2 and N2 is an integer, then d4 is also the denominator of v 2. Multiplying everything by d4 gives (d2v)2 = (d2x)2 − (d2N)2 . Since (d2v)2 = (d2x)2 − (d2N)2, the triple (d2v; d2x; d2N) forms a Pythagorean triple. Since the numerator of x shares no common factor with N, we have that this is a primitive triple and thus, by problem set 1, there exist integers a and b of opposite parity such that d2N = 2ab d 2v = a2 − b2 d2x = a2 + b2 Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 1 Create the triangle with sides 2a=d, 2b=d and 2u. This satisfies the Pythagorean Theorem since (2a=d)2 + (2b=d)2 = 4a2=d2 + 4b2=d2 = 4=d2(a2 + b2) = 4=d2(d2x) = 4x = (2u)2 and it has area N since 1 2a 2b 2ab A = · · = = N: 2 d d d2 Dr. Carmen Bruni Rational Points on an Elliptic Curve Summary of Key Theorem 1 From the triple, d2N = 2ab d2v = a2 − b2 d2x = a2 + b2; we can add and subtract twice the first to the last equation to get d2(x + N) = a2 + 2ab + b2 = (a + b)2 d2(x − N) = a2 − 2ab + b2 = (a − b)2 Taking square roots yields p p d x + N = a + b d x − N = a − b (where above we've assumed that 0 < b < a). Adding and subtracting and dividing by 2. gives expressions for a and b, namely p p p p a = d=2( x + N + x − N) b = d=2( x + N − x − N) Dr. Carmen Bruni Rational Points on an Elliptic Curve Example of Key Theorem 1 Let's find the triangle for N = 7. On the elliptic curve y 2 = x3 − 72x, we close our eyes and pray we can find a triple that consists of integers. After some trying we see that (x; y) = (25; 120) gives a solution. Adding the point to itself gives 2P = (x2P ; −y2P ) where (using the formulas from last time) 3x2 − 72 913 913 −1685 m = = b = 120 − 25 = 2y 120 120 24 113569 3372 x = m2 − 2x = = 2P 14400 120 −17631503 y = mx + b = 2P 2P 1728000 Hence 2P = (113569=14400; 17631503=1728000). p Now, d is the denominator of x2P = 337=120 and so d = 120. Dr. Carmen Bruni Rational Points on an Elliptic Curve Example of Key Theorem 1 Finding the a and b values gives... p p a = d=2( x + N + x − N) = 120=2(p113569=14400 + 7 + p113569=14400; −7) = 288 and p p b = d=2( x + N − x − N) = 120=2(p113569=14400 + 7 − p113569=14400; −7) = 175 This gives the triangle 2a 2 · 288 24 2b 2 · 175 35 p 337 = = = = 2 x = d 120 5 d 120 12 60 which indeed has area 7 and is a right angle triangle (the side lengths satisfy the Pythagorean Identity) Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 Theorem 3. A number N is congruent if and only if the elliptic curve y 2 = x3 − N2x has a rational point P = (x; y) distinct from (0; 0) and (±N; 0). We have already seen that if N is congruent, then we can find a rational point on the elliptic curve. Now, suppose our elliptic curve has a rational point P = (x; y) where P is not one of (0; 0) and (±N; 0). Our goal will be to show that 2P satisfies the conditions of the previous theorem. Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 Using the results of adding a point to itself from last time, we see that the x-coordinate of P + P on the elliptic curve y 2 = x3 + Ax + B is given by 3x2 + A2 9x4 + 6Ax2 + A2 − 2x = − 2x 2y 4y 2 9x4 + 6Ax2 + A2 = − 2x 4(x3 + Ax + B) 9x4 + 6Ax2 + A2 −2x · 4(x3 + Ax + B) = + 4(x3 + Ax + B) 4(x3 + Ax + B) 9x4 + 6Ax2 + A2 −8x4 − 8Ax2 − 8Bx = + 4(x3 + Ax + B) 4(x3 + Ax + B) x4 − 2Ax2 − 8Bx + A2 = 4(x3 + Ax + B) Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 Using the results of adding a point to itself from last time, we see that the x-coordinate of P + P on the elliptic curve y 2 = x3 + Ax + B is given by 3x2 + A2 x4 − 2Ax2 − 8Bx + A2 − 2x = 2y 4(x3 + Ax + B) Specializing to when A = −N2 and B = 0 (that is, on the elliptic curve y 2 = x3 − N2x) gives us the formula for the x-coordinate of P + P as x4 + 2N2x2 + N4 (x2 + N2)2 = : 4(x3 − N2x) (2y)2 Notice that by our restriction on the rational point P, the denominator is nonzero and the numerator is nonzero. Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 The x-coordinate of P + P, namely x4 + 2N2x2 + N4 (x2 + N2)2 = : 4(x3 − N2x) (2y)2 satisfies that it is the square of a rational number. It is also true that the numerator shares no common factor with N. Suppose p divides x2 + N2 and p divides N for some prime p. Then p j x and hence p3 divides x3 − N2x = y 2. Hence p3 divides y 2. Thus, in the x-coordinate above, we can factor out a p2 in the numerator and cancel it with a p2 in the denominator. By repeating this, the numerator can be reduced so that it shares no common factor with N. So it suffices to show that the number has an even denominator. Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 The x-coordinate of P + P, namely x4 + 2N2x2 + N4 (x2 + N2)2 = : 4(x3 − N2x) (2y)2 immediately appears to have an even denominator but we need to be careful. What happens if the factor of 4 in the denominator cancels with the numerator? In what cases is this possible? Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 The x-coordinate of P + P, namely x4 + 2N2x2 + N4 (x2 + N2)2 = : 4(x3 − N2x) (2y)2 Case 1: x and N are even. Then 2 divides both x and N which means that the numerator and N share a common factor. Applying the previous argument shows that we can reduce the fraction. Dr. Carmen Bruni Rational Points on an Elliptic Curve Proof of Key Theorem 2 The x-coordinate of P + P, namely x4 + 2N2x2 + N4 (x2 + N2)2 = : 4(x3 − N2x) (2y)2 Case 2: x and N are odd.
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