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For the real number field $K = \mathbb{Q}(\sqrt{2})$ the ring of integers is $\mathcal{O}_K = \mathbb{Z}[\sqrt{2}] $. We can solve Pell's equation and so there are units $(1 - \sqrt{2})^k$ with $k \in \mathbb{Z}$. One can show that $K$ is a:

  • Euclidean domain
  • principal ideal domain
  • unique factorization domain

Taking their word for it, there a norm $N_K: a + b \sqrt{2} \mapsto |a^2 - 2b^2 | $ and we can write the Dedekind Zeta function: $$ \zeta_K(s) = \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 2b^2)^k} = \frac{1}{1 - 2^{-s}} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-s})^2} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-2s})} $$

If we plug in $s = 2$ I found the numerical result stated without proof. And I'm starting to migrate the existing proofs over $\mathbb{Z}$ to the ring $\mathbb{Z}[\sqrt{2}]$:

$$ \zeta_K(2) = \sum_{x \in \mathbb{Z}[\sqrt{2}]} \frac{1}{N(x)^2} = \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{\big(a^2 - 2b^2\big)^2} = \frac{\pi^4}{48 \sqrt{2}} $$


The definition does not quite make sense over numbers, which explains the shift to ideals. We have $ \mathcal{O}(K)= \mathbb{Z}[\sqrt{2}$ and

$$ \zeta_K(2) = \sum_{I \subseteq \mathcal{O}(K)} \frac{1}{N_{K/\mathbb{Q}}(I)^2} = \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{\big(a^2 - 2b^2\big)^2} = \frac{\pi^4}{48 \sqrt{2}} $$

The Euler product is the product over prime ideals:

$$ \zeta_K(2) = \prod_{P \subseteq \mathcal{O}(K)} \frac{1}{1- N_{K/\mathbb{Q}}(P)^{-2} } = \frac{1}{1 - 2^{-2}} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-2})^2} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-4})} \stackrel{?}{=} \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 2b^2)^2} $$

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  • 2
    $\begingroup$ It's not true that $\zeta_K(2)=\sum_{x\in\mathbb Z[\sqrt{2}]}1/N(x)^2$. You have to take only one generator of each ideal, otherwise infinitely many terms (the units) will constribute $1$ to the sum. $\endgroup$ – Wojowu Nov 2 '17 at 16:57
  • $\begingroup$ Is the second equation correct? And can you prove the equation involving $\pi$ is correct ? $\endgroup$ – cactus314 Nov 2 '17 at 17:12
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We have that $$ L(\chi_8,s) = \sum_{n\geq 0}\left[\frac{1}{(8n+1)^s}-\frac{1}{(8n+3)^s}-\frac{1}{(8n+5)^s}+\frac{1}{(8n+7)^s}\right]$$ equals $$\prod_{p\equiv \pm 1\!\!\pmod{8}}\left(1-\frac{1}{p^s}\right)^{-1}\prod_{p\equiv \pm 3\!\!\pmod{8}}\left(1+\frac{1}{p^s}\right)^{-1} $$ by Euler's product. On the other hand $$ L(\chi_8,1) = \int_{0}^{1}\sum_{n\geq 0} x^{8n}(1-x^2-x^4+x^6)\,dx = \int_{0}^{1}\frac{1-x^2}{1+x^4}\,dx = \frac{1}{\sqrt{2}}\log(1+\sqrt{2}) $$ and similarly $$ L(\chi_8,2)=\int_{0}^{1}\frac{x^2-1}{x^4-1}\log(x)\,dx = \frac{\pi^2}{8\sqrt{2}}.$$ Since $\zeta_K(2) = \zeta(2)\cdot L(\chi_8,2)$ by Euler's product, $\zeta_K(2)=\frac{\pi^4}{48\sqrt{2}}$ is proved.

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  • $\begingroup$ I didn't know that L-function identity $\endgroup$ – cactus314 Nov 2 '17 at 19:01
  • $\begingroup$ Doesn't $r(n) = \infty$ by Pell equation ? $a^2-2b^2=n$ has infinitely many solution. I guess we could count equivalence classes. $p= a^2-2b^2=(a+b\sqrt{2})(a-b\sqrt{2})$ splits and so does half of your Euler product. $\endgroup$ – cactus314 Mar 17 '18 at 9:56
  • $\begingroup$ @cactus314: I removed the intro since you were right, my previous $r(n)$ was nonsense. On the other hand we may start directly from Euler's product and compute $L(\chi_8,1)$ and $L(\chi_8,2)$ through simple integrals. $\endgroup$ – Jack D'Aurizio Mar 17 '18 at 14:08
  • $\begingroup$ you're OK. there must be some multiplicative function there... can you write down the first few terms of $\zeta_K(2)$? are they all of the $\frac{1}{n^2}$ or can we have $\frac{2}{n^2}$ or even $\frac{3}{n^2}$ ? $\endgroup$ – cactus314 Mar 17 '18 at 16:04
  • $\begingroup$ @cactus314: we have terms of the form $\frac{M}{n^2}$ with arbitrarily large $M$s, since $n$ can be the product of an arbitrary number of distinct primes of the form $8k+1$. The analogue for negative discriminants is that we may have an arbitrary large number of lattice points on the circle $x^2+y^2=R$, by carefully choosing $R$. Of course, large $M$s only seldom occur. $\endgroup$ – Jack D'Aurizio Mar 17 '18 at 16:29

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