Let $n>0$ a real number. I'm looking to bound the series
$$ \sum_{m_1,m_2 \in \mathbb{N}} \dfrac{1}{(m_1+n)^2+(m_2+n)^2} $$ In terms of $n$. The series converges as
$$ \sum_{m_1,m_2 \in \mathbb{N}} \dfrac{1}{(m_1+n)^2+(m_2+n)^2} \le \sum_{m_1,m_2 \in \mathbb{N}} \dfrac{1}{(m_1)^2+(m_2)^2} $$ and $$ \sum_{m_1,m_2 \in \mathbb{N}} \dfrac{1}{(m_1)^2+(m_2)^2} = \sum_{k=1}^n c_{k^2} \frac{1}{k^2} $$ where $c_{k^2}$ is the number of ways $k^2$ can be written as a sum of two squares that is $$ c_{k^2}= D_1(k^2)-D_3(k^2) $$ where $D_1,D_3$ are the number of divisor of $k^2$ congruent to $1,3 \mod(4)$ (see here).
The number of divisors of $k^2$ can be bounded by $k^{2\frac{1.5379 \log(2)}{\log(2\log(k))}}$ (see this aswer) and so $$ \sum_{m_1,m_2 \in \mathbb{N}} \dfrac{1}{(m_1+n)^2+(m_2+n)^2} \le \sum_{k=1}^\infty k^{-2 \left(1-\frac{1.5379 \log(2)}{\log(2\log(k))}\right)} $$ and for $k>67$ the exponent is bigger then $1$.
But I have no idea how to bound the function in terms of $n$. Based on the one dimensional version I think the function czan be bounded by $1/n^2$, but I can't think any way to even start facing the problem
