Is there any known value of the infinite sum $\sum_{m=1,n=1}^\infty \frac{1}{(m^2+n^2)^2}$?

Question

Let $L=\{m+in\hspace{1mm}|\hspace{1mm}m,n\in \mathbb{Z}\}$. I was trying to find out the value of $\sum_{w\in L\setminus \{0\}}^\infty\frac{1}{w^4}$ where I encountered the infinite sum $\sum_{m=1,n=1}^\infty \frac{1}{(m^2+n^2)^2}$. Any idea to approach this will help a lot.

Answer 1

Since $\mathbb{Z}[i]$ is a UFD we have that for any $n\in\mathbb{N}$ the function $$ r_2(n) = \left|\{(a,b)\in\mathbb{Z}^2: a^2+b^2=n\}\right| $$ is given by $4(\chi_4*1)(n)$, with $\chi_4$ being the non-principal Dirichlet character $\!\!\pmod{4}$ and $*$ being the Dirichlet (multiplicative) convolution. In particular $$ \sum_{\substack{(a,b)\in\mathbb{Z}^2\\(a,b)\neq(0,0)}}\frac{1}{(a^2+b^2)^s} =\sum_{n\geq 1}\frac{r_2(n)}{n^s}=4\,\zeta(s)\,L(\chi_4,s)$$ holds for any $s>1$. This leads to $$ \sum_{\substack{(a,b)\in\mathbb{Z}^2\\(a,b)\neq(0,0)}}\frac{1}{(a^2+b^2)^2}=4\zeta(2)\sum_{n\geq 0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]= \frac{2}{3}\pi^2 G$$ where $G$ is Catalan's constant. In particular $$ \sum_{a,b\geq 1}\frac{1}{(a^2+b^2)^2}=\color{red}{G\,\zeta(2)-\zeta(4)}\approx 0.4243797762. $$ Here $L(\chi_4,s)$ stands for $\sum_{n\geq 1}\frac{\chi_4(n)}{n^s}$. Different proofs of $r_2(n)=4(\chi_4*1)(n)$ (a combinatorial proof through quadratic forms and double-counting and an analytic proof through Jacobi triple product) can be found in D.H. Cox - Primes of the form $x^2+ny^2$.