How to show the equivalence of these two sums? How can one show the following equation holds?
$$
\sum_{m=1}^\infty\sum_{n=0}^\infty\left(n^2+m^2\right)^{-p}=
\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{\left(2nm+m\right)^{p}}
$$
I have tried breaking $n$ into even and odd parts but I am not sure how to proceed afterwards. Any thoughts?
Why:
I know the equality holds analytically due another result, but I would like to have an idea of one can show the direct equality of these two sums so as to be able to generalise the expression on the left to a general quadratic form.
Any proof that does not involve number theory would be great.
 A: The hidden fact is that $\mathbb{Z}[i]$ (the ring of Gaussian integers) is a UFD, hence
$$ r_2(n)=\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\}$$
is given by a multiple of a multiplicative function, namely
$$ r_2(n) = 4(\chi_4 * 1)(n) = 4\sum_{d\mid n}\chi_4(d) $$
where $\chi_4(d)$ equals $1$ iff $d\equiv 1\pmod{4}$, $-1$ iff $d\equiv 1\pmod{4}$ and zero otherwise.
The factor $4$ is due to the fact that $\mathbb{Z}[i]$ has four invertible elements.
In particular, in terms of Dirichlet's series
$$ \sum_{m\geq 1}\sum_{n\geq 0}\frac{1}{(m^2+n^2)^p} = \sum_{N\geq 1}\frac{(\chi_4*1)(N)}{N^p} = \zeta(p)\sum_{N\geq 1}\frac{\chi_4(N)}{N^p}=\zeta(p)\beta(p) $$
and
$$ \zeta(p)\beta(p)=\sum_{m\geq 1}\frac{1}{m^p}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^p} = \sum_{n\geq 0}\sum_{m\geq 1}\frac{(-1)^n}{(2mn+m)^p} $$
holds for any $p>1$.
A: The answer should be evident from the answer to your previous question:
$$
\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{\left(2nm+m\right)^{p}}=
\sum_{n=0}^\infty\frac{(-1)^n}{\left(2n+1\right)^{p}}\sum_{m=1}^\infty\frac{1} {m^{p}} =\beta(p)\zeta(p)=\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{1}{\left(n^2+m^2\right)^{p}}.
$$
