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In Section 2.15. of Titchmarsh's book The Theory of the Riemann Zeta-Function, after reaching the equality $$\int_0^{\infty} \psi^2(\frac{x}{\pi}) x^{w-1} dx = \int_0^{\infty} (\sum_{n=1}^{\infty} e^{-n^2 x})^2 x^{w-1} dx = \Gamma(w) \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w},$$ it follows with two equalities without any proof that I failed to prove them after a long effort:

$1:$ $$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w} = \dfrac{1}{4} \sum_{n=1}^{\infty} \dfrac{r(n)}{n^w} - \zeta(2w),$$ where $r(n)$ is the number of ways of expressing $n$ as the sum of two squares.

$2:$ $$\dfrac{1}{4} \sum_{n=1}^{\infty} \dfrac{r(n)}{n^w} - \zeta(2w) = \zeta(w) \eta(w) - \zeta(2w),$$ where $$\eta(w) = 1^{-w} - 3^{-w} + 5^{-w} - 7^{-w} + \dots .$$

In $1$, I seperated terms $n=m$ but those terms can be included in $r(n)$ as well! Also, in Silverman's book A Friendly Introduction to Number Theory in all formulas related to $r(n)$, in the definition of it negative integers were included as well but I guess from factor $\frac{1}{4}$ only positive integers are included...

In $2$, any rearrangement does not yield the second equality either by beginning with $\dfrac{1}{4} \sum_{n=1}^{\infty} \dfrac{r(n)}{n^w} - \zeta(2w)$ or with $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w}$.

How these two equalities are derived?

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2 Answers 2

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  1. You collect the terms where $m^2+n^2=d$ and there are, by definition, $\frac{1}{4}r(d)$ of them if for example $n$ would start at $0$ (instead of $1$). $\zeta$ arises because we don't count the lattice points on the let's say the real axis (since there $n$ (or $m$ depending how you choose your coordinates) would be $0$). The $\frac{1}{4}$ is there since we are only summing over lattice points in the let's say upper right quarter plane. (The lattice in question is $\mathbb{Z}^2$. So $r(d)$ counts the number of lattice points $a\in\mathbb{Z}^2$ with $\vert a\vert^2=d$.)
  2. There is a video by 3blue1brown on youtube on exactly this. The Gaussian integers are an UFD with $4$ units. An odd rational prime $p$ factors into $2$ different (modulo units) Gaussian primes of norm $p$ if $p$ is $1$ mod $4$ and else it doesn't factor. $2$ factors into $2$ Gaussian primes which are the same (modulo units). From this we can see (see for example the video) that $$\sum_{n=1}^{\infty} \dfrac{r(n)}{n^s}=$$ $$4(1-2^{-s})^{-1}\prod_{(p\equiv_41)}(1-p^{-s})^{-2}\prod_{(p\equiv_43)}(1-p^{-2s})^{-1}=$$$$4\zeta(s)\prod_{(p\equiv_41)}(1-p^{-s})^{-1}\prod_{(p\equiv_43)}(1+p^{-s})^{-1}=4\zeta(s)\eta(s).$$ Also see the zeta function of the Gaussian integers/of number fields if you are interested.
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  • $\begingroup$ A1 is clear. For A2, I watch the video but I have to find a rigorous argument so I am trying to understand your answer: Number of ways that p^n can be written as a sum of two squares for p=4k+3 form is 1. Number of ways that p^n can be written as a sum of two squares for p=4k+1 form for n=1 is 1 and I could not undertsand neither of the answers here: math.stackexchange.com/questions/1177493/… . And how many ways one can write p^n as a sum of two squares for n>1 and p=4k+1? Then I can use Euler product since r(n) is multiplicative. $\endgroup$
    – Ali
    Jan 18 at 12:46
  • $\begingroup$ I could not find an easy-to-read reference for the questions I asked in my comment above, so if you know some I would study from it as well. $\endgroup$
    – Ali
    Jan 18 at 12:48
  • $\begingroup$ Let's make an example. Let's say we have $n=2*5^2*7^2$ to get the number of ways to write it as a sum of two squares we get the coefficient of $n^{-s}$ on the RHS of the sum over $r(n)$ in my answer above. So the term is $4*1/2^s*3/5^{2s}*1/7^{2s}=12/n^s$ so $12$ ways for this $n$. The $4$ stands for the number of units (they permute the $4$ quadrants). Write $5=r s$ where $r,s$ are Gaussian primes. The $3$ comes from the fact that we could pick $r^2$, $rs$ or $s^2$. $\endgroup$
    – user1279061
    Jan 18 at 12:55
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    $\begingroup$ @Ali if $a$ and $b$ are $1$ mod 4 then $ab$ is $1$ mod $4$, if $a$ is $1$ mod $4$ and $b$ is $3$ mod $4$ then $ab$ is $3$ mod $4$ and if $a$ and $b$ are both $3$ mod $4$ then $ab$ is $1$ mod $4$. So you just multiply everything out and get the result. Note that $(1+p^{-s})^{-1}=1-1/p^s+1/p^{2s}-1/p^{3s}+...$. $\endgroup$
    – user1279061
    Jan 18 at 13:47
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    $\begingroup$ Glad to help you. $\endgroup$
    – user1279061
    Jan 18 at 13:50
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$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w}=\frac14\left(\sum^{\prime}_{m,n\in\Bbb Z}\frac{1}{(m^2+n^2)^w}-2\sum_{m=1}^\infty\frac1{(m^2+0^2)^w}-2\sum_{n=1}^\infty\frac1{(0^2+n^2)^s}\right)=\frac14\left(\sum_{n=1}^\infty\frac{r(n)}{n^w}-4\zeta(2w)\right).$

2. $\zeta(w)\eta(w)=\sum_{n=1}^\infty\frac1{n^w} \sum_{n=1}^\infty\frac{\chi_4(n)}{n^w}= \sum_{n=1}^\infty\left(\sum_{d|n}\chi_4(d)\right)\frac1{n^s}$

And the inner sum is $\sum_{d|n}\chi_4(d)=\frac14r(n)$. You may see this.

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    $\begingroup$ +1, Your A1 was useful to me, thanks :) $\endgroup$
    – Ali
    Jan 18 at 13:52
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    $\begingroup$ @Ali I thank you. This evil sum $\sum_{d|n}\chi_4(d)$ bothers me. $\endgroup$
    – Bob Dobbs
    Jan 18 at 13:57
  • $\begingroup$ I left watching the video I wanted a rigorous way or based on what I already proved rigorously. $\endgroup$
    – Ali
    Jan 18 at 13:59

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