# Solution to the Diophantine equation $x^4+y^4=2z^2$ [duplicate]

Does there exists a nontrivial positive integer solution with $x\ne y,$ of $$x^4+y^4=2z^2.$$

## marked as duplicate by Dietrich Burde, M Turgeon, Jyrki Lahtonen, Rick Decker, Fly by NightAug 3 '14 at 18:26

• – Dietrich Burde Aug 3 '14 at 16:55
• The answer referred to in the previous comment is simply a reference to a possibly evanescent web site. – André Nicolas Aug 3 '14 at 16:58
• This may be a better duplicate. For the reason pointed out by André. – Jyrki Lahtonen Aug 3 '14 at 18:12
• @AndréNicolas: Sorry, yes. It should have been the other web site. – Dietrich Burde Aug 3 '14 at 20:13
• The fact that $x^4\pm y^4=z^2$ has no non-trivial solutions goes back to Fermat, and is standard first number theory course material. A solution that uses this seems perfectly reasonable. – André Nicolas Aug 4 '14 at 4:20

First, I suppose that you know there no exist $x,y,z\in\Bbb Z^+$ such that $$x^4-y^4=z^2.$$ You might have seen this while proving FLT for $n=4$ and further theorems. I can give you a hint for this if you need.
If $x^4+y^4=2z^2$, we have $$z^4-(xy)^4=\left({x^4-y^4\over2}\right)^2.$$ Due to the above result, $z=0$ or $x=0$ or $y=0$ or $|x|=|y|$, and these are all trivial solutions.