<s>[INST] Prove $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}$ How may I prove that
$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}?$$
I also discussed the problem in the chat, but no solution so far.  Some hints? Thanks!
 [/INST] For now, here is how we can prove the second equality. Let the second sum be $S.$ Note that by symmetry we also have $$S= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2(m^2+n^2)}.$$ Now adding the two forms gives: $$2S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2n^2}= \left( \sum_{m=1}^{\infty} \frac{1}{m^2} \right)\left( \sum_{n=1}^{\infty} \frac{1}{n^2}\right)= \frac{\pi^4}{36}.$$
As Fabian alludes to in the comments, it appears the first equality does not hold, since the difference between the two sums is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-n}{n^3}\frac{1}{(m^2+n^2)}>0.$$
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