Congruent Numbers and Cubic Curves

William A. Stein was@math.harvard.edu http://modular.fas.harvard.edu/∼was

3 March 2001 at MIT

1 Congruent Numbers

Definition 1.1 (Congruent number). An integer n is congruent if it is the area of a right triangle with rational side lengths.

Why “congruent”?: One can show that n is congruent if and only if there is an x such that x − n, x and x + n are all squares of rational numbers (n is the “congruence”).

Example 1.2.

• 6 is congruent; it is the area of the triangle with sides 3, 4, and 5. [[Display on laptop.]]

• The numbers 1, 2, 3, and 4 are not congruent, but 5, 6, and 7 are.

Question 1.3. Is there a simple criterion to determine whether or not a number is a congruent number?

This tantalizing problem has remained open for over a millenium, though perhaps you will live to see (or find?) a complete solution!

The following theorem is not very difficult to prove.

Theorem 1.4. An integer n is a congruent number if and only if the equation

y^{2} = x^{3} − n^{2}x

has more than the three obvious solutions.

Theorem 1.5 (Tunnel). If n is a congruent number then the number of triples (x, y, z) such that 2x^{2} + y^{2} + 8z^{2} = n is twice the number of triples (x, y, z) such that 2x^{2} + y^{2} + 32z^{2} = n. If the Birch and Swinnerton-Dyer conjecture (see below) for y^{2} = x^{3} − n^{2}x is true, then the converse is true.

This is almost a complete solution to the ancient congruent number problem. Unfortunately, nobody knows how to prove the Birch and Swinnerton-Dyer conjecture yet.

2 The Birch and Swinnerton-Dyer Conjecture

The Birch and Swinnerton-Dyer conjecture is a conjecture about whether or not y^{2} = x^{3} − n^{2}x has (infinitely many) non-obvious solutions. [[Display pictures of Birch and Swinnerton-Dyer on my laptop.]] Conjecture 2.1. The equation y^{2} = x^{3} − n^{2}x has infinitely many solutions only in the following two cases:

• n is odd and −2 is a square modulo n, or n is even and −2 is a square modulo n/2

• the sum

∑_{n=1}^{∞} \frac{a_{n}}{n} e^{−πn/M}

is 0. Here M = 2n\sqrt{2} if n is odd, or 2n if n is even, and a_{p} is p minus the number of solutions to y^{2} = x^{3} − n^{2}x modulo p. The a_{n} are got from the a_{p} by a simple recipe.

In fact, Birch and Swinnerton-Dyer made a more general conjecture, which applies to any curve of the form y^{2} = x^{3} + ax + b. If you prove it, then you will be famous and the Clay Math Institute will award you one million dollars.

3 Fun examples (using my computer)

Have somebody from the audience choose an n, and determine whether or not the conjecture asserts that it is congruent. If it is congruent, give the rational right triangle.