1. Rational Points on Conic Sections We call a point (x,y) in the plane R 2 a rational point if both of its coordinates are rational numbers. There are infinitely many rational points on the curve x 2 + y 2 -1 = 0. It is easy to verify that if t is any rational number then (x,y) given by (x.y) = ((1 – t 2 )/(1 + t 2 ), 2t/(1 + t 2 )) (1) is a point on the curve. In fact, (eqn. 1) gives all the rational points on x 2 + y 2 - 1 = 0. The curve x 2 + y 2 - 2 = 0 has infinitely many rational points as well, and there is a formula closely analogous to (1) that gives all rational points on this curve. However, x 2 + y 2 - 3 =0 has no rational points whatsoever (the proof is in the adjacent column). These examples show that for some integer triples (a,b,c) the curve ax 2 + by 2 - c = 0 has infinitely many rational points and and for others the corresponding curve has no rational points. There cannot be finitely finitely many rational points on the curve since one can generate infinitely many rational points from a single rational point. Thus the question of whether or not there are infinitely many points on the curve boils down to whether there is a single one. A natural quest is the search for a finite algorithm for determining whether there is any rational point at all. Legendre’s theorem provides such an algorithm. Legendre’s Theorem What’s special about (3) is that the existence of nontrivial solutions (mod m) for some m actually implies that there are nontrivial whole solutions to (3). This is the substance of the following theorem of Legendre: Suppose a, b, c are integers that are not all of the same sign, squarefree, and gcd(a,b,c) = 1. Then the equation ax 2 + by 2 + cz 2 = 0 has a nontrivial solution if and only if ax 2 + by 2 + cz 2 = 0 (mod |abc|) has a nontrivial solution) Note that there are only finitely many possible nontrivial solutions to the latter congruence so that this gives us finite algorithm for determining whether (3) has nontrivial solutions. While Legendre’s theorem doesn’t cover all equations of form (3), it’s easily to see that every equation of form (3) can be treated by reducing it to one that satisfies Legendre’s hypotheses.The necessity of solvability of the congruence is straightforward and essentially demonstrated in the box below. Sketch of proof of sufficiency :Show that ax 2 + by 2 + cz 2 splits into linear factors (mod m) with m = |abc|, so that finding a nontrivial solution (mod m) is the same as finding a nontrivial solution to ax + by + cz = 0 (mod m). Use a pigeonhole argument to that there is a nontrivial solution to the latter equation with x, y and z fairly small integers, so that ax 2 + by 2 + cz 2 = 0(mod m) for x, y and z small in magnitude. Use the bounds on magnitude together with the modularity condition to show that ax 2 + by 2 + cz 2 = 0 or -abc for (x,y,z) nonzero. In the former case we’re done, in latter case an ingenious trick shows that another nonzero triple picked in terms of the first one does yield zero. Local Global Principle Legendre’s theorem is the first historical example of how knowledge of the solutions to an equation (mod m) for all m can yield integer or rational solutions to an equation. These theme is one that has great prevalence in modern number theory. The idea is that it is easy to analyze equations (mod m) or “locally” and that this information can be patched together to obtain a “global” picture of integer solutions. In fact, the famous Birch Swinnerton-Dyer conjecture is of this sort - there the idea is that the number of solutions to an elliptic curve (mod p) (for each prime p) determines the rank of the group of rational points on the elliptic curve and can be used to determine the group in entirety. Legendre’s Theorem and Quadratic Reciprocity Jonah Sinick Swarthmore College, Department of Mathematics & Statistics References Grosswald, Emil. 1984. Topics from the Theory of Numbers (2nd Ed.) Birkhauser Boston Weil, Andre. 1984 Number Theory: An approach through history; From Hammurapi to Legendre. Birkhauser Boston Kato, Kazuya et. Al. 2000. Fermat’s Dream. American Mathematical Society McTutor Mathematical Biography Archive Acknowledgements Thanks to Walter Stromquist for his mathematical mentoring over the years. IH’s taught me a lot. Projectivizing the Conics Suppose we have nonzero integers x, y and z such that x 2 + y 2 = z 2 then dividing through by z 2 we obtain the equation (x/z) 2 + (y/z) 2 = 1, so taking u = x/z and v = y/z we obtain a point on u 2 + v 2 = 1. Moreover, u and v are rational because x, y and z are whole. For example, the Pythagorean triple (3, 4, 5) yields the rational point (3/5, 4/5) on the unit circle. In fact, the problem of finding rational points on a conic ax 2 + by 2 = c (2) is equivalent to finding that of finding integer solutions to ax 2 + by 2 = cz 2 (3) To see this, let T be the set of rational solutions to (2) and let S be the set of nonzero integer triples that satisfy (3). Then it can be shown that f: S --- > T given by (x, y, z) -- > (x/z, y/z) is onto and that the inverse image of a point in T under f is just a family of elements of the form (kx, ky, kz) where gcd(x, y, z) = 1 and k ranges over all nonzero integers. So there is a perfect one to one correspondence between rational points on the conic and relatively prime solutions to (3). Thus, determining the existence of a rational solution to (3) is the same as determining the existence of nonzero integer solutions to (3) Connection with Quadratic Reciprocity Solvability of the congruence in Legendre’s theorem is equivalent to the condition that these three congruences are solvable: x 2 = -bc (mod |a|). x 2 = -ac (mod |b|), x 2 = -ab (mod |c|) (Here solvability (mod 1) is understood to hold by default.) Thus Legendre’s theorem has something to do with quadratic congruences. Legendre himself attempted to use his theorem to prove quadratic reciprocity. He succeeded only in obtaining partial results, for example he showed that if p and q are primes that are 1 and 3 (mod 4) respectively then if p is nonsquare (mod q) then q is nonsquare (mod p). He did so by considering the equation x 2 + py 2 -qz 2 = 0, this equation has no nontrivial whole solutions because it has no nontrivial solutions (mod 4). However, the equation meets the hypotheses of Legendre’s theorem, so one of the quadratic congruences below must be unsolvable: x 2 = pq (mod 1) x 2 = q (mod p) x 2 = -p (mod q) The first hold automatically, also the third holds because both p and -1 are nonsquares (mod q), so their product must be a square. So it must be the second congruence that has no solution which is the same as q being nonsquare (mod p). Legendre’s method of proof cannot be used to prove every instance of quadratic reciprocity, however,it contains the germ of the important Hilbert Product formula: (3) fails to be solvable modulo arbitrarily high powers of p for only a finite number of primes p and moreover, the finite number must be even. Hilbert’s Product formula is equivalent to quadratic reciprocity. Nonzero solutions ---> Nonzero solutions (mod m) Suppose we have (x, y, z) in Z 3 such that x 2 + y 2 = 3z 2,, reducing (mod 3) we see that x 2 + y 2 = 0 (mod 3), but squares are 0 or 1 (mod 3) so we must have x = y = 0 (mod 3). But then x 2 = y 2 = 0 (mod 9) so that 0 = x 2 + y 2 = 3z 2 (mod 9) and z = 0 (mod 3), so that gcd(x, y, z) > = 3. So there are no relatively prime solutions to x 2 + y 2 = 3z 2 which means that there are no nonzero solutions whatsoever. More generally, (3) has no nontrivial solution if it has no nontrivial solution (mod m). Genus 0 Curves If we consider allow x and y in (2) to take on values in complex projective space then the solution set to (2) is topologically equivalent to a sphere. More generally, given a polynomial P(x) with integer coefficients the set S = {x, y on the complex projective line such that P(x,y) = 0} is topologically equivalent to a compact connected orientable surface and by the classification theorem from topology it must be a the surface of torus with g holes in it. It turns out that the structure of the rational points in S is determined by g. When g > 1, by Falting’s theorem there are only finitely many rational points, but there is no algorithm for obtaining them at present. When g = 1 the rational points can be analyzed using the theory of elliptic curves. This poster concerns the nontrivial part of the case with g = 0. Diophantus Legendre Hilbert