CONGRUENT NUMBERS

The Congruent Number problem is: Let n be a positive integer. Is there a right triangle, with the lengths of all three of its sides being rational numbers, that has area n?

Fermat showed that n = 1 is not a congruent number. Euler showed that n = 7 is a congruent number. Eventually, it became known that n = 1, 2, 3, 4 are not congruent, but n = 5, 6, 7 are. But an efficient procedure to answer the question for general n seemed hopelessly out of reach.

A terribly inefficient procedure does exist that will eventually answer “Yes” if n is congruent and run forever if not. But if we ask the procedure whether some value of n (say, n = 157) is a congruent number, and the procedure then runs for a century without giving us an answer, then we don’t know if this is because n = 157 congruent but the procedure hasn’t figured this out yet, or because n = 157 is not congruent.

In fact, n = 157 is congruent. Let

a = \frac{411340519227716149383203}{21666555693714761309610}

and

b = \frac{6803298487826435051217540}{411340519227716149383203}

and

c = \frac{224403517704336969924557513090674863160948472041}{8912332268928859588025535178967163570016480830} .

Then

a^{2} + b^{2} = c^{2},

and

\frac{1}{2}ab = 157.

But an example this gigantic must be using ideas beyond brute-force search.

In fact, the following result holds:

Let n be an odd squarefree natural number. Consider the following two conditions:

• The number n is congruent.

• The number of integer triples (x, y, z) satisfying the equation

2x^{2} + y^{2} + 8z^{2} = n

is equal to twice the number of integer triples satisfying the equation

2x^{2} + y^{2} + 32z^{2} = n.

Then the first condition implies the second; and if a weak form of the so-called Birch–Swinnerton-Dyer conjecture is true then also the second condition implies the first.

Note that checking the second condition for a given n terminates in a finite number of steps, and in fact in a number of steps that we can bound explicitly.