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:

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

• 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. – Wojowu Nov 2 '17 at 16:57
• Is the second equation correct? And can you prove the equation involving $\pi$ is correct ? – cactus314 Nov 2 '17 at 17:12

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.

• I didn't know that L-function identity – cactus314 Nov 2 '17 at 19:01
• 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. – cactus314 Mar 17 '18 at 9:56
• @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. – Jack D'Aurizio Mar 17 '18 at 14:08
• 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 $\frac{3}{n^2}$ ? – cactus314 Mar 17 '18 at 16:04
• @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. – Jack D'Aurizio Mar 17 '18 at 16:29