Consider the double sums :
$$A_3 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{1}{a^3 + b^3}$$
$$A_4 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{1}{a^4 + b^4}$$
Is there a closed form for $A_3$ or $A_4$ ?
Can we prove they are irrational ?
Notice that double sums over rational functions can have a closed form. For instance the similar :
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$
Although some would call them special cases because they relate to a simple norm of a quadratic ring.
Then again my cases could relate to norms too.
Another idea is to use stuff similar to this :
closed form of $\sum \frac{1}{z^3 - n^3}$
and then sum over $z$.
So we use a formula such as
$$ \sum_{n\in\mathbb{Z}}\frac{1}{z^3 + n^3}=\frac{1}{3z^2}\left(\pi\coth(\pi z)+\tfrac{\pi}{\alpha}\coth(\tfrac{\pi}{\alpha}z)+\pi\alpha\coth(\pi\alpha z)\right) $$
and sum over integer $z > 0$.
I noticed similar sums can be done like
Conjectured closed form for $\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}$
and I assume
$ \sum_{n>1} \coth(\pi n)/n^2 $ also has a closed form.
I know There are similar results in spirit of many such formulas like; $$\sum_{k=1}^{\infty}\frac{\coth\pi k}{k^{3}}=\frac{7\pi^{3}}{180},\hspace{.2cm}\sum_{k=1}^{\infty}\frac{\coth\pi k}{k^{7}}=\frac{19\pi^{7}}{56700}$$
Or identities like
\begin{align} \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}} &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \zeta(3) \, x \end{align}
So alot of similarities, and I assume this is the best way to go.
Yet another idea is to use this :
$$\sum_{z=1}^{\infty} \frac{1}{n^3 + z^3} = -1/3 \sum_{ω: n^3 + ω^3 + 3 ω^2 + 3 ω + 1 = 0} \frac{Polygamma(0, -ω)}{ω^2 + 2 ω + 1}$$
and then sum over $n$.
But this is probably a harder way.
Also notice the related integral exists :
$$\int \int \frac{dx dy}{x^3 + y^3} =$$
$$-\frac{(x + y) (2 \log(x + y) - \log(x^2 - x y + y^2)) + 2 \sqrt 3 x \tan^{-1}(\frac{2 x - y}{\sqrt 3 y}) + 2 \sqrt 3 y \tan^{-1}(\frac{2 y - x}{\sqrt 3 x})}{6 x y}$$
or this integral;
$$\int \int \frac{dx dy}{x^4 + y^4} =$$
$$-\frac{ -(-x^2 \log(x^2 - \sqrt 2 x y + y^2) + x^2 \log(x^2 + \sqrt 2 x y + y^2) - (1 + i) y^2 \log(\frac{\sqrt 2 y - (1 + i) x}{x^2}) - (1 - i) y^2 \log(\frac{\sqrt 2 y - (1 - i) x}{x^2}) + (1 - i) y^2 \log(\frac{\sqrt 2 y + (1 - i) x)}{x^2}) + (1 + i) y^2 \log(\frac{\sqrt 2 y + (1 + i) x}{x^2}) + 2 x^2 \tan^{-1}(\frac{\sqrt 2 x}y + 1) + 2 x^2 \tan^{-1}(\frac{\sqrt 2 x - y}y)) }{8 \sqrt 2 x^2 y^2}$$
