Wait. Number Theory?

We note that, for any $s \ge 2$ , $\sum_{a,b \in \mathbb{N}} \frac{1}{(a^2+b^2)^s} \; = \; \frac{1}{4}\left\{ \sum_{a,b \in \mathbb{Z} \atop (a,b) \neq (0,0)} \frac{1}{(a^2+b^2)^s} - 4\zeta(2s)\right\} \; = \; \tfrac14\sum_{n=1}^\infty \frac{r_2(n)}{n^s} - \zeta(2s)$ where $r_2(n)$ is the "sum of squares" function, so that $r_2(n)$ is the number of ways of writing $n$ as the sum of the squares of two integers (including zero and negative integers and counting ordering). A standard result tells us that $\sum_{n=1}^\infty \frac{r_2(n)}{n^s} \;= \; 4\zeta(s) \sum_{n=1}^\infty \big(\tfrac{-4}{n}\big) n^{-s}$ for $s \ge 2$ , where $\big(\tfrac{-4}{n}\big)$ is the Kronecker symbol. But $\big(\tfrac{-4}{n}\big)$ is equal to $0$ whenever $n$ is even, and is equal to $(-1)^{\frac12(n-1)}$ when $n$ is odd. Thus $\sum_{n=1}^\infty \big(\tfrac{-4}{n}\big) n^{-s} \; = \; \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$ Putting $s=2$ yields $\sum_{a,b \in \mathbb{N}} \frac{1}{(a^2+b^2)^2} \; = \; \zeta(2)G - \zeta(4) \; = \; \tfrac16\pi^2 G - \tfrac{1}{90}\pi^4$ where $G$ is the Catalan constant, making the answer $2+6+4+90 = \boxed{102}$ .