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 $$x^4 \pm y^4=z^2$$ 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]<br /> <br /> as<br /> <br /> $$[(x + y)^2 + (x - y)^2](x + y)(x - y) = (2z)^2$$.<br /> <br /> Then let<br /> <br /> $$X = x + y, Y = x - y, Z = 2z$$<br /> <br /> so we have<br /> <br /> $$XY(X^2 + Y^2) = Z^2$$.<br /> <br /> Show that this equation has no non-trivial solutions (several additional steps are still required).<br /> <br /> Petek[/tex]

 

[disregardthat]

I don't know if this works, but looking at different residues modulo 8 seems promising.

 

[CRGreathouse]

I don't know if this works, but looking at different residues modulo 8 seems promising.

Yes, it works.