I am doing research on perfect cuboids, and I'm looking for values $a,b,c$ such that the following is integer, and I'm not sure how to continue this. Any suggestions are appreciated!
$PED$ is a very large constant=$899231100768000$
$$ \begin{align} &\exp\left(\sigma_1+\sigma_2+\frac{\ln(a^2+b^2+c^2)}{2}-\ln(PED) \right)\in\mathbb Z\\ &\sigma_1=\ln a+\ln b+\ln c\\ &\sigma_2=\frac{\ln(a^2+b^2)}{2}+\frac{\ln(b^2+c^2)}{2}+\frac{\ln(a^2+c^2)}{2} \end{align} $$
This is not a complete answer but too long for a comment.
Since $\exp(a+b)=\exp(a)\cdot\exp(b)$, the expression can be rewritten as
$$ \frac1{PED}\cdot abc\sqrt{a^2+b^2}\sqrt{a^2+c^2}\sqrt{b^2+c^2}\sqrt{a^2+b^2+c^2} $$
So we could start finding integer numerators, since the whole expression is integer when, if $n$ is the numerator, then $n|PED$, since $a,b,c\in\mathbb Z$, the $abc$ factor does not change whether the number is an integer, so it remains to find when the following is integer
$$ \sqrt{a^2+b^2}\sqrt{a^2+c^2}\sqrt{b^2+c^2}\sqrt{a^2+b^2+c^2} $$
If $\alpha,\beta,\gamma$ form a pythagorean triblet, that is $\alpha^2+\beta^2=\gamma^2$ then exactly one of $\alpha,\beta,\gamma$ is odd.
This means that at most 2 of $\sqrt{a^2+b^2},\sqrt{a^2+c^2},\sqrt{b^2+c^2}$ are integer.
However there are several cases where it is true that it's an integer, such as these pairs. This can happen because the roots left from each square root might cancel out each other. I havent been able to find any relations between these pairs either.