$L$-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 $-1$.
 A: 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.
A: $\renewcommand{\mod}{{\rm mod\ }} \renewcommand{\frakp}{{\frak p}}
\newcommand{\OK}{{\cal O}_K}\newcommand{\fraka}{{\frak a}}$
If $K=\mathbb{Q}(i)$, the Dedekind zeta function of $K$ is
$$ \zeta_K(s) = \sum_{{\frak a} \subseteq {\cal O}_K} \frac{1}{N({\frak a})^{-s}} = \prod_{{\frak p}\in {\rm Spec}({\cal O}_K)} \frac{1}{1 - N({\frak p})^{-s}},$$ 
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
$$ \zeta_K(s) = \frac{1}{4} \sum_{(a,b)\in \mathbb{Z}^2\backslash\{(0,0)\}} \frac{1}{(a^2+b^2)^{-s}}.$$
