A positive square-free integer is said to be *congruent* if appears as the square-free part of the area of an integral right triangle. Thus for instance 5 is congruent, because the (9,40,41) right triangle has area 20. A nice proof-by-picture demonstrates that is congruent if and only if there exists an arithmetic progression of three (integer) squares with common difference for some :

*An arithmetic progression ** of squares. The common difference 720 (= 5 * 144) appears geometrically as the union of the four red triangles.*

If is congruent, Euclid’s classification of (primitive) Pythagorean triples gives the existence of coprime such that

Conversely, any integer expressible in this form is congruent. Taking prime and , we see that . Then divides with multiplicity , so we have found a congruent number with as a factor. In particular, congruent numbers may have arbitrarily large prime factors and there exist infinitely many congruent numbers.

In concrete terms, this establishes the existence of something like congruent numbers up to . This is laughably (and unsurprisingly) weak. Heegner sketched an argument in 1952 that the congruent numbers include each prime and thereby gave congruent numbers up to . (Apparently, the details for some of Heegner’s claims weren’t put into writing until Paul Monsky’s *Mock Heegner Points and Congruent Numbers* in 1990.)

The main purpose of this post is to give a third proof of the infinitude of congruent numbers. This third proof is analytic moreso than algebraic and gives only an ineffective result. Nevertheless I think it has merit, because its techniques may generalize (with a lot of blood, sweat, and tears) to prove the infinitude of congruent numbers in any reasonable congruence class.

The distribution of congruent numbers is an open problem. A heuristic from random matrix theory due to Keating conjectures the following:

**Conjecture** (Keating, 2005):*1. Any square-free integer congruent to 5, 6, or 7 modulo 8 is a congruent number.2. Denote by *

*the set of square-free integers less than or equal to*

*that are congruent to 1, 2, or 3 modulo 8 and that are congruent numbers. Then*

*has density 0; more precisely, there exists a strictly positive constant*

*such that*

For more discussion on this conjecture, see also Henri Cohen’s *Number Theory (Volume I: Tools and Diophantine Equations)*, pg. 453.

**— PART II: A DIRICHLET SERIES —**

The bijection between Pythagorean triples and three-term arithmetic progressions of squares implies that

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀

in which is the square-indicator function and is the set of hypotenuses of primitive integral right triangles with square-free part (sfp) of the area equal to . (In theory, this is counted with multiplicity; in practice, we’ve never found two primitive triangles with equal hypotenuse which determine the same congruent number.) It’s worth remarking that the series on is absolutely convergent in . This follows from dyadic summation and the bound

which itself follows from a bound one of my recent papers, *A Shifted Sum in the Congruent Number Problem*. (Here, is the rank of the elliptic curve .) The Dirichlet series is thus in . I’m not sure if it is possible to make the explicit in .

**— PART III: CUT-OFF INTEGRALS —**

To prove that there exist infinitely many congruent numbers, we apply an integral transform to the Dirichlet series from the previous section. Our function of choice is a smooth cutoff of compact support, with Mellin transform .

**Proposition:** *We have*

*Proof:* We shift contours in a Mellin transform to compute

Since is , the integral over is . On the other hand, the integral transform of splits into the terms

in which is the number of representation of as a sum of two integer squares. Pointwise estimates for bound this error term by . We produce our final answer by balancing errors.

**— PART IV: THE FINAL PROOF —**

We now begin our proof by contradiction in earnest. Suppose that there exist finitely many square-free which are congruent. Summing over these choices of has the effect of removing the at left in the above Proposition. In particular, we obtain

Note that the right-hand side is of size (by finiteness and the existence of at least one congruent number) and that we have removed the -dependence from by finiteness. To further understand the left-hand side, we introduce another Proposition.

**Proposition: ***When* *is square, we have*

It follows from the equation line preceding this Proposition that

We now complete our proof by proving that the associated Dirichlet series has a ** double** pole at . Indeed, we may write

The Dirichlet series has a double pole, which contradicts the upper bound written above. This completes our proof.