# $L$-function, easiest way to see the following sum?

What is the easiest way to see that$$\sum_{(m, n) \in \mathbb{Z}^2 \setminus \{0, 0\}} (m^2 + n^2)^{-s} = 4\zeta(s)L(s, \chi)?$$Here $\chi$ is the homomorphism $(\mathbb{Z}/4\mathbb{Z})^\times \to \mathbb{C}^\times$ which sends $3$ mod $4$ to $-1$.

• Shouldn't $\chi$ appear somewhere on the LHS as well? – Wojowu Jul 3 '15 at 20:47
• What character are you referring to? If it is the trivial one you should have just written $\zeta(s)$, otherwise there's no obvious choice here. I think you forgot to write something. – Patrick Da Silva Jul 3 '15 at 20:52
• Yeah, I fixed it. – user231529 Jul 3 '15 at 20:56
• For an algebraic method see here, for an analytic solution see here. – Raymond Manzoni Jul 3 '15 at 21:06

Since $\mathbb{Z}[i]$ is an Euclidean domain, it is a UFD domain. So we have that the numbers that are the sum of two squares are a semigroup, due to the Lagrange identity: $$(a^2+b^2)(c^2+d^2) = (ac+bd)^2+(ad-bc)^2 \tag{1}$$ that is equivalent to the norm on $\mathbb{Z}[i]$ being multiplicative.

That also gives that the representation function: $$r(n) = \#\{(a,b)\in\mathbb{Z}^2 : a^2+b^2 = n \} \tag{2}$$ is a constant times a multiplicative function. We have, indeed: $$r(n) = 4\left(\chi_4 * 1\right)(n) = 4\sum_{d\mid n}\chi_4(n) \tag{3}$$ where $*$ is the Dirichlet convolution and $\chi_4$ is the non-principal Dirichlet character $\!\!\pmod{4}$.

Now back to our series. Assuming $\text{Re}(s)>1$, absolute convergence allows us to rearrange the series as: $$S=\sum_{(m,n)\in\mathbb{Z}^2\setminus(0,0)}\frac{1}{(m^2+n^2)^s} = \sum_{n\geq 1}\frac{r(n)}{n^s} = 4\sum_{n\geq 1}\frac{(\chi_4*1)(n)}{n^s}\tag{4}$$ so by Dirichlet convolution we have: $$S = 4 \sum_{n\geq 1}\frac{1}{n^s}\sum_{n\geq 1}\frac{\chi_4(n)}{n^s} = 4\zeta(s) L(\chi_4,s)\tag{5}$$ as wanted.

Footnote: $(3)$ may be proved also by manipulating Lambert series and exploiting the Jacobi triple product, or through modular forms. Anyway, I believe that the algebraic approach is way easier to follow. If we replace $m^2+n^2$ by $m^2+Dn^2$ and $m^2+Dn^2$ is the only reduced binary quadratic form of discriminant $-4D$ (aka $h(-4D)=1$, class number one), then $r(n)$ is still a constant times a multiplicative function and $S$ is still a constant times the product of two $L$-functions.


$$\zeta_K(s) = \sum_{{\frak a} \subseteq {\cal O}_K} \frac{1}{N({\frak a})^{-s}} = \prod_{{\frak p}\in {\rm Spec}({\cal O}_K)} \frac{1}{1 - N({\frak p})^{-s}},$$

where $N=N_{K/\mathbb{Q}}$ and $\fraka$ ranges over the ideals of $\OK$. Now for a rational prime $p$, if $p\equiv 1(\mod 4)$ there are two primes $\frakp$ over $p$, for which $N({\frak p})=p$. If $p\equiv 3(\mod 4)$, $p$ is inert, so $\frakp = p\OK$ and $N(\frakp)=p^2$. Finally if $p=2$ there is one distinct $\frakp=(1+i)$ above it, and $N(\frakp)=2$.

Noting that $(1-p^{-2s})=(1-p^{-s})(1+p^{-s})$ we have

\begin{align*}\zeta_K(s) &= \frac{1}{1-2^{-s}}\cdot\prod_{p \equiv 1(\mod 4)}\frac{1}{(1-p^{-s})^2}\cdot \prod_{p \equiv 3(\mod 4)} \frac{1}{(1-p^{-s})(1+p^{-s})}\\ &=\prod_p \frac{1}{1-p^{-s}} \cdot \left(\prod_{p\equiv 1(\mod 4)} \frac{1}{1-p^{-s}}\cdot \prod_{p\equiv 3(\mod 4)} \frac{1}{1+p^{-s}}\right) \\ &= \zeta(s)\cdot \left(\prod_{p\equiv 1(\mod 4)} \frac{1}{1-\chi(p)p^{-s}}\cdot \prod_{p\equiv 3(\mod 4)} \frac{1}{1-\chi(p)p^{-s}}\right)\\ &= \zeta(s)L(\chi,s) \end{align*}

On the other hand since $K=\mathbb{Q}(i)$ has class number 1 each ideal $\fraka$ is equal to $(a+bi)$ for some non-zero $a+bi\in \OK=\mathbb{Z}[i]$, with $N(\fraka)=a^2+b^2$. As there are four units $\{\pm 1,\pm i\}$ in $\OK$, the map $(a,b) \mapsto (a+bi)$ from $\mathbb{Z}^2\backslash\{(0,0)\}$ to ideals $\fraka\subset \OK$ is four-to-one, therefore

$$\zeta_K(s) = \frac{1}{4} \sum_{(a,b)\in \mathbb{Z}^2\backslash\{(0,0)\}} \frac{1}{(a^2+b^2)^{-s}}.$$