Pythagorean triple problem 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=$899231100768000$
$$
\begin{align}
&\exp\left(\sigma_1+\sigma_2+\frac{\ln(a^2+b^2+c^2)}{2}-\ln(PED)
\right)\in\mathbb Z\\
&\sigma_1=\ln a+\ln b+\ln c\\
&\sigma_2=\frac{\ln(a^2+b^2)}{2}+\frac{\ln(b^2+c^2)}{2}+\frac{\ln(a^2+c^2)}{2}
\end{align}
$$
Another way to write this is:
$$
\begin{align}
&\frac {abc*\sqrt{a^2+b^2}*\sqrt{a^2+c^2}*\sqrt{b^2+c^2}*\sqrt{a^2+b^2+c^2}}{899231100768000}\in\mathbb Z\\
\end{align}
$$
 A: If $a=b=1$, $c=4$, then $$(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)=2\times17\times17\times18=102^2,$$ so $$abc\sqrt{a^2+b^2}\sqrt{a^2+c^2}\sqrt{b^2+c^2}\sqrt{a^2+b^2+c^2}=408$$ is an integer. Multiply each of $a,b,c$ by $899231100768000$, and you have an example. 
A: assuming $a^2+b^2$, $a^2+c^2$, $b^2+c^2$, and $a^2+b^2+c^2$ are perfect squares, then it's impossible.

$\sqrt{(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)}$
must be an integer, therefor

$(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)$
must be a square, which allows us to assume that all terms are squares, and that makes the first 3 terms Pythagorean triples.

one method of generating Pythagorean triples is
for $A^2+B^2=C^2$
$A=2mn$
$B=m^2-n^2$
$C=m^2+n^2$
$n$ and $m$ must be integers to insure that $A$, $B$, and $C$ are integers
Note: every Pythagorean triple must include $A$, $B$, and $C$

if $a=A$
and $b=B$
then, because $a^2+c^2$ must equal $A^2+B^2$, $c$ must be $B$
and because $b^2+c^2$ must equal $B^2+A^2$, $c$ must be $A$
therefor $c=A$ and $c=B$. Which means
$A=B$
$2mn=m^2-n^2$
divide both sides by $mn$, simplify,
$2=\frac mn-\frac nm$
and since
$\frac nm = \frac 1{\frac mn}$
you have
$2=\frac mn-\frac 1{\frac mn}$
then if you introduce a new variable $x$
$x=\frac mn$
substitute $\frac mn$
$2=x-\frac 1x$
then multiply both sides by $x$,
$2x=x^2-1$
$0=x^2-2x-1$
and use the quadratic equation, you get.
$x=\frac {2\pm\sqrt{8}}2$
which you can simplify further into
$1+\sqrt{2}$ and $1-\sqrt{2}$
which means
$\frac mn=1+\sqrt{2}$
 or
$\frac mn=1-\sqrt{2}$

if we rearrange $A=2mn$, to get $\frac {A}{2n^2}=\frac mn$
then substitute $\frac mn$ and simplify, we see that
$A=2n^2(1+\sqrt{2})$
or
$A=2n^2(1-\sqrt{2})$
therefor, if $c$ is $A$ and $B$ (which it must be) then $c$ is irrational
you can do the same thing with $a$ and $b$, you will get the same contradiction.

ps. with the method of finding Pythagorean triples that I used, if you find one then you have found an infinite list of them by multiplying $a$, $b$, and $c$ by some integer $k$. If you do this for every integer pare $m$,$n$ you get almost all triples, the only ones missing are when $k=\frac {1}{2}$, but including that gets every triple that exists!
(including the ones that just flip $A$ and $B$ around) here is a video that can explain it better than I could.

ex: $m=2$, $n=1$
$3^2+4^2=5^2$, $k=1$
$6^2+8^2=10^2$, $k=2$
$9^2+12^2=15^2$, $k=3$
.
.
.
