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

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 ?

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 Recognitions: Gold Member I think I have a solution. I'll post a hint and take another look tomorrow. Rewrite $$x^4 - y^4 = 2z^2[/itex] as [tex][(x + y)^2 + (x - y)^2](x + y)(x - y) = (2z)^2$$. Then let $$X = x + y, Y = x - y, Z = 2z$$ so we have $$XY(X^2 + Y^2) = Z^2$$. Show that this equation has no non-trivial solutions (several additional steps are still required). Petek
 Recognitions: Science Advisor I dont know if this works, but looking at different residues modulo 8 seems promising.

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