What is the closed form of this sum? What is the closed form of this sum?
$$
S = \sum_{k\ge1, r>s\ge 1}\frac{1}{k^2(r^2 + s^2)^2}
$$
Note: Though originally posted for Pythagorean triplets, their was an flaw in the question which changed the meaning of the question and was answered accordingly. I will post a separate question with the original question on Pythagorean triangles.
Update: I have posted the related question of Pythagorean triangles in the link belwo
What is the sum of the reciprocal of the square of hypotenuse of Pythagorean triangles?

\begin{align*}
S &= \sum_{k \geq 1} \frac{1}{k^2}  \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2}  \\
    &= \frac{\pi^2}{6} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2}
\end{align*}
Note that $\frac{1}{(r^2 + s^2)^2}$ is unchanged by $\{r \mapsto -r\}$, $\{s \mapsto -s\}$, and $\{r \leftrightarrow s\}$.  Applying these symmetries, we obtain eight copies of the region $r > s \geq 1$ in the $r$-$s$ plane.  It lacks the four diagonals,
each with value $\displaystyle \sum_{s \geq 1} \frac{1}{(s^2 + s^2)^2} = \frac{\pi^4}{360}$, the four coordinate rays starting at $\pm 1$, each with value $\displaystyle \sum_{s \geq 1} \frac{1}{(0^2 + s^2)^2} = \frac{\pi^4}{90}$, and the origin, which we will continue to exclude.  Placing a prime on the summation sign to indicate that we omit the origin, 
$$  \sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 8 \sum_{r > s \geq 1} \frac{1}{(r^2 + s^2)^2} + 4 \cdot \frac{\pi^4}{360} + 4 \cdot \frac{\pi^4}{90}  $$
Borwein and Borwein [1] (p. 291) (see also (38) at Double Series on MathWorld) show
$$  \sideset{}{'}\sum_{r,s = -\infty}^\infty \frac{1}{(r^2 + s^2)^2} = 4 \beta(2) \zeta(2) = \frac{2}{3} \pi^2 K  \text{,}  $$
where $\beta$ is Dirichlet's Beta function, $\zeta$ is the Riemann Zeta function, and $K$ is Catalan's constant.
Utilizing these facts, 
$$  S = \frac{\pi^2}{6} \cdot \frac{12 K \pi^2 - \pi^4}{144} = 0.1264945807\dots \text{.}  $$
[1] Borwein, Jonathan M. and Peter B. Borwein, "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity", New York: Wiley, 1987. 
P.S.  Mathematica 11.3 does not evaluate the triple sum symbolically.  It numerically estimates $0.126494$, which is comfortably close to the above result.
