# 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}

• 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. – Jack D'Aurizio Jan 11 '15 at 1:29
• @JackD'Aurizio: references? – abiessu Jan 11 '15 at 2:04
• ams.org/journals/mcom/2014-83-289/S0025-5718-2013-02791-3/… – daniel Jan 11 '15 at 4:32
• @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. – Seth Kitchen Jan 11 '15 at 4:37
• @GerryMyerson sorry finals week delayed my check -- thanks for the help – Seth Kitchen May 14 at 23:05

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.

• 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 – Seth Kitchen May 14 at 23:04

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

• Your parameterization is not correct. What if some of the Pythagorean triples are not primitive? – Carl Schildkraut May 8 at 19:03
• @spydragon not correct (44,117,240) is an example -- generate 44 and 117 from your list – Seth Kitchen May 9 at 2:30
• 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 – Seth Kitchen May 9 at 3:32
• @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. – spydragon May 14 at 5:15
• 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. – DanielWainfleet May 14 at 10:47