Diophantine equation x^4 - y^4 = 2 z^2

TTob

I need to prove that the equation x^4 - y^4 = 2 z^2 has no positive integer solutions.

I have tried to present this equation in some known from (like x^2+y^2=z^2 with known solutions, or [tex]x^4 \pm y^4=z^2[/tex] that has no integer solutions) without success.

Any hints ?

Petek

I think I have a solution. I'll post a hint and take another look tomorrow. Rewrite

[tex]x^4 - y^4 = 2z^2[/itex]

as

[tex][(x + y)^2 + (x - y)^2](x + y)(x - y) = (2z)^2[/tex].

Then let

[tex] X = x + y, Y = x - y, Z = 2z[/tex]

so we have

[tex]XY(X^2 + Y^2) = Z^2[/tex].

Show that this equation has no non-trivial solutions (several additional steps are still required).

Petek

disregardthat

I dont know if this works, but looking at different residues modulo 8 seems promising.

CRGreathouse

Yes, it works.