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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.

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  • $\begingroup$ Well the thing is I know the sum on the left equals $\beta(p)\zeta(p)$, where $\beta$ and $\zeta$ are the Dirichlet and Riemann functions. So the right hand side is just the series expression of this product. But the only proof of this I've seen so far involves some number theory which is not really clear to me. So I was hoping that framing the question in terms of the equivalence of the above sums would have maybe led to a more straightforward demonstration... $\endgroup$ – semola Feb 19 '18 at 14:48
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    $\begingroup$ What are the conditions on $p$? $\endgroup$ – Exodd Feb 19 '18 at 14:58
  • $\begingroup$ What do you mean by a "general quadratic form"? $\endgroup$ – user Feb 19 '18 at 15:06
  • $\begingroup$ @user355705 something like $an^2+bnm+cm^2$ $\endgroup$ – semola Feb 19 '18 at 15:10
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    $\begingroup$ From the equivalence of the sums for arbitrary p it can be concluded that there is no other possibility to prove it as by using the number theory. What should be proved is the fact that the difference of the numbers of odd divisors $(2n+1)$ with even and odd parities of $n$ for any number is equal to the number of ways to split it into a sum of two squares. $\endgroup$ – user Feb 19 '18 at 16:08
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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 $p>1$.

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The answer should be evident from the answer to your previous question:

$$ \sum_{m=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{\left(2nm+m\right)^{p}}= \sum_{n=0}^\infty\frac{(-1)^n}{\left(2n+1\right)^{p}}\sum_{m=1}^\infty\frac{1} {m^{p}} =\beta(p)\zeta(p)=\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{1}{\left(n^2+m^2\right)^{p}}. $$

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  • $\begingroup$ Indeed. But since I don't quite understand the number theory approach to demonstrate this, I was trying to see if there is a better way to see the equivalence $\endgroup$ – semola Feb 19 '18 at 14:50
  • $\begingroup$ There must be a more direct way of showing this right? $\endgroup$ – semola Feb 19 '18 at 14:54
  • $\begingroup$ I would suggest you to reformulate the question especially underlying that the equality of the sums is analytic result and not just that of numerical tests. $\endgroup$ – user Feb 19 '18 at 14:56
  • $\begingroup$ Thanks. The thing is I would like to have an idea of how I can generalise the expression to general quadratic forms. Sorry if it looks a bit like a duplicate... $\endgroup$ – semola Feb 19 '18 at 15:01

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