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?

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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 $\frac{x^2+y^2}{2}$.

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That means we can define a,b,c such that x-y = 2a^2, x+y = 2b^2 and x^2+y^2=2c^2. Adding and subtracting the first two gives x = a^2+b^2, y=b^2-a^2 which, substituting into the third equation proves that b^2 = c^2 so x+y=x^2+y^2.. does that imply that the complete solution of the equation is x=0 or 1 and y=0 or 1? –  quanta Apr 1 '11 at 15:45
@quanta: I get $c^2 = a^4 + b^4$, not $b^2=c^2$. –  Arturo Magidin Apr 1 '11 at 15:52
@Arturo, Thanks! I see my mistake. So we have instead a^4+b^4=c^2 -- it's not descent but it's reduction to a Diophantine equation I have already solved so I think that solves the equation! –  quanta Apr 1 '11 at 15:54
@quanta: Yes. You have $a^4+b^4=c^2$ with $a$, $b$, and $c$ coprime, which has no nontrivial solutions. –  Arturo Magidin Apr 1 '11 at 15:55
@quanta: No mismatch: there are only two solution with $x=y$ and $\gcd(x,y)=1$ (which was our first reduction), namely $x=y=\pm1$. Also, there was an implicit assumption that $z'\neq 0$: if $z'=0$, you cannot conclude that each of the factors is a square, just that one of them is equal to $0$ (which gives either $x=y$ or $x=-y$). –  Arturo Magidin Apr 1 '11 at 18:26
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