Fermat solved the Diophantine equation $(x^2)^2 + (y^2)^2 = z^2$ using descent, the key step was using the Pythagorean triples:

- $x^2 = u^2 - v^2$
- $y^2 = 2 u v$
- $z = u^2 + v^2$

but then it is seen that $x,v,u$ is another Pythagorean triple,

- $x = m^2 - n^2$
- $v = 2 m n$
- $u = m^2 + n^2$

then the observation that $y^2 = 2^2 u m n$ (along with some coprimality conditions) implies that $u$,$m$ and $n$ are squares - which gives us the descent step needed.

I wanted to use this technique to solve $(x^2)^2 + (y^2)^2 = 2 z^2$ but it does not seem to work out. Parameterizing the curve through $(1,1,1)$ gives:

- $x^2 = -m^2 - 2 m n + n^2$
- $y^2 = m^2 - 2 m n - n^2$
- $z = m^2 + n^2$

Nothing here seems to match up so well with the original problem (as it did in Fermat's case) and parameterizing through different points doesn't help either.

Is there any way to make descent work here? Was Fermat just lucky or is there a reason why descent can't work out here?

update: I found the parametrization

- $x^2 = 2m^2 - n^2$
- $y^2 = 2m^2 - 4mn + n^2$
- $z = 2m^2 - 2mn + n^2$

which looks like a better candidate for descent but I haven't been able to actually perform it. Since I cannot see how to show that $n$ and $x$ are both squares.