For sample proof of formula $\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$ with s=2

Question

I'm looking for a simple proof of this result only in case s = 2, $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$

there is proof here de T. Amdeberhan that I don't understand

Answer 1

Since $\mathbb{Z}[i]$ is a Euclidean domain we have that

$$ r_2(n)=\left|\{(a,b)\in\mathbb{Z}^2: a^2+b^2 = n \}\right| $$ is given by four times a multiplicative function, namely

$$ 4(\chi_4*1)(n) = 4 \sum_{d\mid n}\chi_4(d),\qquad \chi_4(d)=\left\{\begin{array}{rcl}1&\text{if}&n\equiv 1\pmod{4}\\ -1&\text{if}&n\equiv -1\pmod{4}\\ 0 &\text{if}& n\equiv 0\pmod{2}.\end{array}\right. $$ Dirichlet's convolution then gives that $$ \sum_{\substack{(m,n)\in\mathbb{Z}\times\mathbb{Z}\\(m,n)\neq (0,0) }}\frac{1}{(m^2+n^2)^s}=\sum_{N\geq 1}\frac{r_2(N)}{N^s}=4\sum_{n\geq 1}\frac{1}{N^s}\sum_{N\geq 1}\frac{\chi_4(N)}{N^s}=4\zeta(s)\beta(s) $$ and the claim easily follows.