I would to solve the Diophantine equation $2x^4 - 2y^4 = z^2$ by descent. This is an exercise from Carmichael Diophantine Analysis but I cannot do it.
Since the LHS is even $2|z$, let z = 2z' and divide both sides by $2$ to get (x-y)(x+y)(x^2+y^2)=2z'^2. Congruence conditions tell us that either x and y are both odd or both even, that tells us both sides are a multiple of $8$ so let z' = 2z'' and we have \frac{x-y}{2}\frac{x+y}{2}\frac{x^2+y^2}{2}=z''^2. The problem now is I don't know if these numbers on the LHS are coprime so I can't claim they are all squares. I don't see any way to continue so maybe I took a wrong turn earlier.
Any hints about how to deal with this equation?