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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:

$$ \begin{align} &\frac {abc*\sqrt{a^2+b^2}*\sqrt{a^2+c^2}*\sqrt{b^2+c^2}*\sqrt{a^2+b^2+c^2}}{899231100768000}\in\mathbb Z\\ \end{align} $$

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    $\begingroup$ It is well known that the diagonals of a parallelepiped cannot have all integer lengths, so I am afraid there are no solution no matter how big is the constant PED. $\endgroup$ – Jack D'Aurizio Jan 11 '15 at 1:29
  • $\begingroup$ @JackD'Aurizio: references? $\endgroup$ – abiessu Jan 11 '15 at 2:04
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    $\begingroup$ ams.org/journals/mcom/2014-83-289/S0025-5718-2013-02791-3/… $\endgroup$ – daniel Jan 11 '15 at 4:32
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    $\begingroup$ @daniel Yes you are correct. There are an infinite amount of perfect parallelepipeds, but perfect cuboids are an even more specific version. They have not been proven. $\endgroup$ – Seth Kitchen Jan 11 '15 at 4:37
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    $\begingroup$ @GerryMyerson sorry finals week delayed my check -- thanks for the help $\endgroup$ – Seth Kitchen May 14 at 23:05
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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.

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    $\begingroup$ Yes this is correct. Thank you -- as this was quite a bit a long time ago I will need to check where this will be used at. I will probably want to create parametric generators for a,b, and c to have all possibilities especially since this case does not have a and b unique $\endgroup$ – Seth Kitchen May 14 at 23:04
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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: $m=2$, $n=1$
$3^2+4^2=5^2$, $k=1$
$6^2+8^2=10^2$, $k=2$
$9^2+12^2=15^2$, $k=3$
.
.
.

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  • $\begingroup$ Your parameterization is not correct. What if some of the Pythagorean triples are not primitive? $\endgroup$ – Carl Schildkraut May 8 at 19:03
  • $\begingroup$ @spydragon not correct (44,117,240) is an example -- generate 44 and 117 from your list $\endgroup$ – Seth Kitchen May 9 at 2:30
  • $\begingroup$ Right (44,117), (44,240), (117,240) all work as "a" and "b" and they didn't fit the (3x)^2 expression you had in a comment above $\endgroup$ – Seth Kitchen May 9 at 3:32
  • $\begingroup$ @CarlSchildkraut I now see the flaw you mentions a^2+b^2 could share a factor with one of the others, I will edit my answer to specify that they cannot share factors. $\endgroup$ – spydragon May 14 at 5:15
  • $\begingroup$ I am not sure which terms you are referring to in your most recent comment. If $a$ is even and $b,c$ are odd then $b^2+c^2$ and $a^2+b^2+c^2$ are both even, while if $a,b,c$ are all odd then $a^2+b^2$ and $a^2+c^2$ are both even. $\endgroup$ – DanielWainfleet May 14 at 10:47

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