Show that $\zeta_K(2) = \frac{\pi^4}{48 \sqrt{2}} $, with $K = \mathbb{Q}(\sqrt{2})$ 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:
$$ \zeta_K(2) = \prod_{P \subseteq \mathcal{O}(K)} \frac{1}{1-    N_{K/\mathbb{Q}}(P)^{-2} } = \frac{1}{1 - 2^{-2}} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-2})^2} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-4})}
\stackrel{?}{=}
\sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 2b^2)^2} $$

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.
