A double series yielding Riemann's $\zeta$ Can you give me some hints to prove equality:
$$\sum_{m,n=1}^{\infty} \frac1{(m^2+n^2)^2} =\zeta (2)\ G-\zeta(4)=\frac{\pi^2}{6}\ G-\frac{\pi^4}{90}$$
where $\zeta (t):= \sum\limits_{n=1}^{+\infty} \frac{1}{n^t}$ is the Riemann zeta function and $G := \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} \approx 0.915 965 594$ is Catalan's constant?
I tried with some reverse engineering, but I wasn't able to solve the problem at all. Even a good reference may be useful.
Thanks a lot in advance, guys.
 A: This problem can be recast in terms of the famous problem of the number of ways to represent a positive integer as a sum of squares.  With this perspective, we can see that the following more general statement is true for any $p > 1$ (so that each of the infinite series actually converges):
$$\sum_{n=1}^{\infty} \frac{1}{(n^2)^p} + \sum_{m,n = 1}^{\infty} \frac{1}{(m^2+n^2)^p}  = \left(\sum_{n=1}^{\infty} \frac{1}{n^p}\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^p}\right).$$
The left-hand side is
$$\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p},$$ where $n_2(s)$ is the number of ways of representing $s$ as the sum of one or of two squares of positive integers, in which order is distinguished (i.e., $1^2 + 2^2$ is counted separately from $2^2+1^2$).
It is known that $n_2(s) = d_1(s) - d_3(s)$ (see eq. 24 in the site linked above), where $d_k(s)$ is the number of divisors of $s$ congruent to $k \bmod 4$.   
The first sum on the right side of the equation has (the $p$th powers of) all positive integers as denominators and the second sum on the right has (the $p$th powers of) all odd numbers as denominators.  After multiplying those sums together, then, $1/s^p$ (ignoring signs) appears on the right as many times as there are odd divisors of $s$.  Each odd divisor of $s$ congruent to $1 \bmod 4$ contributes a $+1/s^p$, and each odd divisor of $s$ congruent to congruent to $3 \bmod 4$ contributes a $-1/s^p$.  Thus the coefficient of $1/s^p$ on the right side is exactly $d_1(s) - d_3(s)$.  Therefore, the right-hand side is also 
$$\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p}.$$
A: For more detail search for Dedekind zeta function (especially of quadratic fields) and for Dirichlet $L$-series.
Let us compute the sum over all $(m,n)\in\mathbb{Z}^2$ such that $(m,n)\neq(0,0)$. Let us denote that sum $S$. We have $(m^2+n^2)^2=|m+in|^4$. Every number $m+in\in\mathbb{Z}[i]$ decomposes uniquely into a product of primes in $\mathbb{Z}[i]$, up to multiplication by $\pm1,\pm i$. We therefore have
$S=4\prod_\pi\frac{1}{1-|\pi|^{-4}}$ where the product runs over all primes of $\mathbb{Z}[i]$ (4 comes from $\pm1,\pm i$). Now $|\pi|^2$ is a either a prime $p$ (in $\mathbb{Z}$) $p\equiv 1 \bmod 4$ (we get $p$ twice, from $\pi$ and from $\bar\pi$) or (if $\pi=q\in\mathbb{Z}$) $|\pi|^2=q^2$, $q$ prime, $q\equiv 3 \bmod 4$, or $|\pi|^2=2$.
We thus get ($p$ runs over all primes 1 mod 4, $q$ over primes 3 mod 4)
$$S=4\frac{1}{1-2^{-2}}\prod_p\frac{1}{(1-p^{-2})^2}\prod_q\frac{1}{1-q^{-4}}=$$
$$=4\frac{1}{1-2^{-2}}\prod_p\frac{1}{1-p^{-2}}\prod_q\frac{1}{1-q^{-2}}\times \prod_p\frac{1}{1-p^{-2}}\prod_q\frac{1}{1+q^{-2}}=4\zeta(2)\times G.$$
Your sum is $S/4-\zeta(4)$.
