Diophantine equation from Carmichael: $2x^4 - 2y^4 = z^2$ 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?

If $p$ is a prime and it divides both $x$ and $y$, then it divides $z$: if $p$ is odd, then $p^4|z^2$; if $p=2$, then $32|z^2$. Either way, you can factor out the common prime from all of $x$, $y$, and $z$.
Thus, we may assume that $\gcd(x,y)=1$. Since you also know that $x\equiv y\pmod{2}$, you can reduce to the case where $x$ and $y$ are coprime and both odd.
Then $\gcd(x+y,x-y) = 2$, since any common divisor must divide $(x+y)+(x-y)=2x$ and $(x+y)-(x-y)=2y$, and the gcd of $2x$ and $2y$ is $2$. Therefore, $\frac{x-y}{2}$ and $\frac{x+y}{2}$ are coprime.
Any common divisor of $x-y$ and $x^2+y^2$ must divide $2x^2 = (x-y)(x+y) + (x^2+y^2)$, and must also divide $2y^2 = x^2+y^2 - (x-y)(x+y)$. Hence it must divide $\gcd(2x^2,2y^2) = 2$. Thus, $\gcd(x-y,x^2+y^2) = 2$. By the same argument, $\gcd(x+y,x^2+y^2) = 2$. This means that $\frac{x-y}{2}$ and $\frac {x^2+y^2}{2}$ are coprime, as are $\frac{x+y}{2}$ and $\frac{x^2+y^2}{2}$.
