Grade 11/12 Math Circles Rational Points on an Elliptic Curves Dr. Carmen Bruni November 11, Lest We Forget

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1 Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 11/12 Math Circles Rational Points on an Elliptic Curves Dr. Carmen Bruni November 11, Lest We Forget 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 right triangle. Enter Elliptic Curves The associated equations with the congruent number problem, namely x 2 + y 2 = z 2 xy = 2N can be converted to an elliptic curve of the form Y 2 = X 3 N 2 X. We also saw that we can reduce our problem to considering only squarefree numbers N. 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 to a rational right triangle with area N? Key Theorems y 2 = x 3 N 2 x Theorem 1. Let (x, y) be a point with rational coordinates on the elliptic curve y 2 = x 3 N 2 x 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 1

2 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. Theorem 2. A number N is congruent if and only if the elliptic curve y 2 = x 3 N 2 x 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! Proof of Theorem 1 Let (x, y) be a point with rational coordinates on the elliptic curve y 2 = x 3 N 2 x 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. Let u = x which is given to be rational. Set v = y u giving v 2 = y2 u = x3 N 2 x 2 x = x 2 N 2. Let d be the smallest integer such that du Z (namely the denominator of u in lowest terms). Note that d is even by assumption and that d 4 is the denominator for u 2 = x. Since v 2 = x 2 N 2 and N 2 is an integer, then d 4 is also the denominator of v 2. Multiplying everything by d 4 gives (d 2 v) 2 = (d 2 x) 2 (d 2 N) 2. Since (d 2 v) 2 = (d 2 x) 2 (d 2 N) 2, the triple (d 2 v, d 2 x, d 2 N) 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 d 2 N = 2ab d 2 v = a 2 b 2 d 2 x = a 2 + b 2 Create the triangle with sides 2a/d, 2b/d and 2u. This satisfies the Pythagorean Theorem since (2a/d) 2 + (2b/d) 2 = 4a 2 /d 2 + 4b 2 /d 2 = 4/d 2 (a 2 + b 2 ) = 4/d 2 (d 2 x) = 4x = (2u) 2 and it has area N since A = 1 2 2a d 2b d = 2ab d 2 = N. Summary of Theorem 1 From the triple, d 2 N = 2ab d 2 v = a 2 b 2 d 2 x = a 2 + b 2, we can add and subtract twice the first to the last equation to get d 2 (x + N) = a 2 + 2ab + b 2 = (a + b) 2 d 2 (x N) = a 2 2ab + b 2 = (a b) 2 2

3 Taking square roots yields 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 a = d/2( x + N + x N) b = d/2( x + N x N) Example of Theorem 1 Let s find the triangle for N = 7. On the elliptic curve y 2 = x x, 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 = (x 2P, y 2P ) where (using the formulas from last time) m = 3x2 7 2 = 913 ( ) 913 b = = y x 2P = m 2 2x = ( ) = 120 y 2P = mx 2P + b = Hence 2P = (113569/14400, / ). Now, d is the denominator of x 2P = 337/120 and so d = 120. Finding the a and b values gives... and This gives the triangle a = d/2( x + N + x N) = 120/2( / /14400, 7) = 288 b = d/2( x + N x N) = 120/2( / /14400, 7) = 175 2a d = = b d = = x = which indeed has area 7 and is a right angle triangle (the side lengths satisfy the Pythagorean Identity) Proof of Theorem 2 Theorem 2. A number N is congruent if and only if the elliptic curve y 2 = x 3 N 2 x 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. 3

4 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. 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 = x 3 + Ax + B is given by ( 3x 2 ) 2 + A 2x = 9x4 + 6Ax 2 + A 2 2y 4y 2 2x = 9x4 + 6Ax 2 + A 2 2x = 9x4 + 6Ax 2 + A 2 + 2x 4(x3 + Ax + B) = 9x4 + 6Ax 2 + A 2 + 8x4 8Ax 2 8Bx = x4 2Ax 2 8Bx + A 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 = x 3 + Ax + B is given by ( 3x 2 ) 2 + A 2x = x4 2Ax 2 8Bx + A 2 2y Specializing to when A = N 2 and B = 0 (that is, on the elliptic curve y 2 = x 3 N 2 x) gives us the formula for the x-coordinate of P + P as Notice that by our restriction on the rational point P, the denominator is nonzero and the numerator is nonzero. 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 x 2 + N 2 and p divides N for some prime p. Then p x and hence p 3 divides x 3 N 2 x = y 2. Hence p 3 divides y 2. Thus, in the x-coordinate above, we can factor out a p 2 in the numerator and cancel it with a p 2 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. 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? 4

5 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. Case 2: x and N are odd. In this case, write x = 2a+1 and N = 2b+1. Plugging in and simplifying gives (x 2 + N 2 ) 2 = 16(a 2 + a + b 2 + b) (a 2 + a + b 2 + b) + 4 Hence 4 exactly divides the numerator. Since y 2 = x 3 N 2 x, we have that y is even and so at least 16 divides the denominator. Hence the denominator is even. Thus, this point P satisfies the conditions of the previous theorem and so the number N is congruent. Don Zagier To compute a rational point on the elliptic curve y 2 = x x, Zagier noted that if N 5 is divisible by 8 for a prime N, then each of the factors x and x ± N in y 2 = x 3 N 2 x must be of the form ±s 2, ±2s 2, ±ns 2 or ±2ns 2 where s is a rational number. Then, as an example, if x = A 2, x + n = B 2 and x n = C 2, simplifying gives us that we must solve C 2 B 2 = 2A 2 C 2 A 2 = n Similarly to the techniques we used to find Pythagorean triples, it must hold that A = 2RS M, B = R2 2S 2 M and C = R2 +2S 2 M for suitable integers R, S, M. In this way the problem is reduced to the solvability of M 2 N = R 4 + 4S 4. For N = 5 this has a clear solution but for N = 157, we need to apply this idea a few more times to the equation N = U 2 + 4V 2 and find a solution with UV a rational square. Then take U = R 2 /M and V = S 2 /M. The ideas here are difficult to flesh out but can be done which is what Zagier did. (Thanks to Carlos Beenakker!) Given the previous discussion, we have become interested in the following problem: How can we find a rational point on an elliptic curve of the form y 2 = x 3 + Ax + B? This is a complex problem! Points on Elliptic Curves over Z p Given a complex problem, sometimes we try to simplify it! In some sense the problem is that the rational numbers are too big. What we ll consider is dealing with points over a finite field, in this case, Z p. 5

6 We define the field Z p for a prime number p to be the set of integers and we state that two integers are equal in Z p provided they differ by a multiple of p. It will turn out over Z p that the group law will still hold. So for example, in Z 7, 1 and 15 are the same number because they differ by 14 which is divisible by 7. We denote this by 1 15 mod 7 and in general by a b mod p. Exercise 1 Which of the following numbers are equivalent to 1 in Z 5? Solutions to Exercise 1 Which of the following numbers are equivalent to 1 in Z 5? 1. 1 Equivalent since 5 divides 1 1 = Not equivalent since 5 does not divide 2 1 = Equivalent since 5 divides 6 1 = Not equivalent since 5 does not divide 17 1 = Equivalent since 5 divides 4 1 = Equivalent since 5 divides = (3 4 ) 50 1 = = (81 1)( ) (Think: x 1 is a factor of x 50 1 so 81 1 = 80 is a factor of and 5 divides 80 so 5 divides ). Working in Z p When we think about numbers in Z p for a prime p, we can restrict ourselves to just looking at numbers between 0 and p 1 inclusive since every number is equivalent to one of these numbers (this actually follows quickly from long division with remainders!) Hence, we sometimes write Z p = {0, 1, 2,..., p 1}. Points on Elliptic Curves over Finite Fields Now we can start to look at elliptic curves over different finite fields. Let s look at the elliptic curve y 2 = x 3 x. 6

7 In order to reduce this elliptic curve over a finite field, we need to avoid primes dividing the discriminant of the elliptic curve (see the problem sheet). In this case, the discriminant is 64 so we can look for points on finite fields over all Z p for odd primes. Over Z 3, we can look at all the possible x values, namely x = 0, 1, 2. These give the equations y 2 = mod 3 y 2 = = 0 mod 3 y 2 = = 6 0 mod 3 Hence, this elliptic curve when considered over Z 3 has three points (0, 0), (1, 0), (2, 0) and one more point for the point at infinity. What about the elliptic curve y 2 = x 3 x over Z 5? Again we can look at all the possible x values, namely x = 0, 1, 2, 3, 4. These give the equations y 2 = = 0 y 2 = = 0 y 2 = = 6 1 mod 5 y 2 = = 24 4 mod 5 y 2 = = 60 0 mod 5 Thus, we want to know what solutions we have for y 2 0 mod 5, y 2 1 mod 5 and y 2 4 mod 5. We can compute the squares modulo 5 via x x 2 mod Hence, this elliptic curve when considered over Z 5 has seven points (0, 0), (1, 0), (2, 1), (2, 4), (3, 2), (3, 3), (4, 0) and one more point for the point at infinity. What about the elliptic curve y 2 = x 3 4x over Z 5 (If you re quick, change 5 to 7 and 11 and see what happens)? Again we can look at all the possible x values, namely x = 0, 1, 2, 3, 4. These give the equations y 2 = 0 3 4(0) = 0 y 2 = 1 3 4(1) = 3 2 mod 5 y 2 = 2 3 4(2) = 0 y 2 = 3 3 4(3) = 15 0 mod 5 y 2 = 4 3 4(4) = 48 3 mod 5 From the table before, we see that 2 and 3 are not squares in Z 5. Hence the only points here are (0, 0), (2, 0), (3, 0) and the point at infinity. What can we do with these ideas? By taking the information at many primes, we can gain a lot of information about our elliptic curve. Elliptic curves can be used to help factor numbers (Lenstra s Algorithm). Elliptic Curves over finite fields form the basis for a cryptosystem in use today called Elliptic Curve Cryptography. They form a correspondence with certain types of modular forms which are another beautiful mathematical object with many applications. You now have a good foundation to pick up an introductory book on elliptic curves and start to study these objects more deeply. 7

Grade 11/12 Math Circles Elliptic Curves Dr. Carmen Bruni November 4, 2015

Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 11/12 Math Circles Elliptic Curves Dr. Carmen Bruni November 4, 2015 Revisit the Congruent Number