# A Proof of the Infinitude of Congruent Numbers via Dirichlet Series

A positive square-free integer $n$ is said to be congruent if $n$ 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 $n$ is congruent if and only if there exists an arithmetic progression of three (integer) squares with common difference $nk^2$ for some $k$:

An arithmetic progression $(31^2, 41^2, 49^2)$ of squares. The common difference 720 (= 5 * 144) appears geometrically as the union of the four red triangles.

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

$nk^2 = ab (a^2-b^2).$

Conversely, any integer $n$ expressible in this form is congruent. Taking $a$ prime and $b=2$, we see that $nk^2 = 2a(a^2-4)$. Then $a$ divides $nk^2$ with multiplicity $1$, so we have found a congruent number with $a$ 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 $\log \log X$ congruent numbers up to $X$. This is laughably (and unsurprisingly) weak. Heegner sketched an argument in 1952 that the congruent numbers include each prime $p \equiv 5 \!\! \mod 8$ and thereby gave $\Omega(X/\log X)$ congruent numbers up to $X$. (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
$C(X)$ the set of square-free integers less than or equal to $X$ that are congruent to 1, 2, or 3 modulo 8 and that are congruent numbers. Then $C(X)$ has density 0; more precisely, there exists a strictly positive constant $c$ such that

$C(X) \sim c\, X^{3/4} \log(X)^{11/8}.$

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

$\displaystyle D(s;t):=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{\tau(m+n)\tau(m)\tau(m-n)\tau(nt)}{m^s}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀$\displaystyle =\sum_{m=1}^\infty \frac{\# \{\Delta : \mathrm{hyp} (\Delta)=m; \, \mathrm{sfp}(A(\Delta))\}}{m^{2s}}=\zeta(2s)\sum_{h \in \mathcal{H}(t)} \frac{1}{h^{2s}},$

in which $\tau$ is the square-indicator function and $\mathcal{H}(t)$ is the set of hypotenuses of primitive integral right triangles with square-free part (sfp) of the area equal to $t$. (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 $h$ is absolutely convergent in $\Re s > 0$. This follows from dyadic summation and the bound

$\displaystyle \sum_{X< h<2X} \frac{1}{h^{2s}} = O_t\! \left( \frac{(\log X)^{r/2}}{X^{2\Re s}}\right),$

which itself follows from a bound one of my recent papers, A Shifted Sum in the Congruent Number Problem. (Here, $r$ is the rank of the elliptic curve $y^2=x^3-n^2 x$.) The Dirichlet series $\sum_{h \in \mathcal{H}(t)} 1/h^{2 s}$ is thus $O_t(1)$ in $\Re s > 0$. I’m not sure if it is possible to make the $O_t$ explicit in $t$.

— 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 $v_y(x)$ of compact support, with Mellin transform $V_y(s)$.

Proposition: We have

$\displaystyle \sum_{m \leq X} \sum_{n \geq 1} \frac{\tau(m+n)\tau(m)\tau(m-n)\tau(nt)}{\sqrt{m}} = 2 X^{\frac{1}{2}} \sum_{h \in \mathcal{H}(t)} \frac{1}{h} + O_t\left(X^{\frac{1}{6}+\epsilon}\right).$

Proof: We shift contours in a Mellin transform to compute

$\displaystyle \frac{1}{2\pi i} \int_{(1)} D(s;t) V_y(s) X^s ds = X^{\frac{1}{2}} V_y(\tfrac{1}{2}) \!\sum_{h \in \mathcal{H}(t)} \frac{1}{h} +\frac{1}{2\pi i} \int_{(\epsilon)} D(s;t) V_y(s) X^s ds.$

Since $D(s;t)$ is $O_t(\vert s \vert^{1/2})$, the integral over $(\epsilon)$ is $O_t(X^\epsilon\sqrt{y})$. On the other hand, the integral transform of $D(s;t)$ splits into the terms

$\displaystyle \sum_{m \leq X} \sum_{n \geq 1} \frac{\tau(m+n)\tau(m)\tau(m-n)\tau(nt)}{\sqrt{m}} + O\Bigg(\sum_{X

in which $r_2(m)$ is the number of representation of $m$ as a sum of two integer squares. Pointwise estimates for $r_2(m)$ bound this error term by $O(\sqrt{X} \log (X)/y)$. We produce our final answer by balancing errors. $\square$

— PART IV: THE FINAL PROOF —

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

$\displaystyle \sum_{m \leq X} \sum_{n \geq 1} \frac{\tau(m+n) \tau(m)\tau(m-n)}{\sqrt{m}} = 2 \sqrt{X} \sum_{t \geq 1} \sum_{h \in \mathcal{H}(t)} \frac{1}{h} + O\left(X^{\frac{1}{6}+\epsilon}\right).$

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

Proposition: When $m$ is square, we have

$\displaystyle \sum_{n=1}^\infty \tau(m+n)\tau(m-n) = \tfrac{1}{8}r_2(m) - \tfrac{1}{2}.$

It follows from the equation line preceding this Proposition that

$\displaystyle \sum_{m \leq X} \frac{r_1(m)r_2(m)}{\sqrt{m}} \ll \sqrt{X}.$

We now complete our proof by proving that the associated Dirichlet series $\sum_{m \geq 1} r_2(m^2)/m^s$ has a double pole at $s=1$. Indeed, we may write

$\displaystyle \sum_{m \geq 1} \frac{r_2(m^2)/4}{m^s} = \prod_p \sum_{k \geq 0} \frac{r_2(p^{2k})/4}{p^{ks}}=\prod_{p \equiv 1(4)}\frac{1+p^{-s}}{(1-p^{-s})^2}\prod_{p \not\equiv 1(4)}\frac{1}{1-p^{-s}}$
$\displaystyle = \zeta(s)^2 \prod_{p \equiv 1 (4)} \frac{1-p^{-2s}}{1-p^{-s}} \prod_{p \not\equiv 1(4)} \frac{1- p^{-2s}}{1+p^{-s}}=\frac{\zeta(s)^2 L(s,\chi)}{(1+2^{-s})\zeta(2s)}.$

The Dirichlet series has a double pole, which contradicts the upper bound $O(\sqrt{X})$ written above. This completes our proof. $\square$