Solve $x^2+y^3=z^2$ in which x,y, and z are relatively prime and y is even.
I have done a similar proof for $x^2+y^2=z^2$; however, the $y^3$ significantly changes the method I used before.
My solution for that was
$x=v^2-u^2$, $y=2uv$, and $z=v^2+u^2$ where $u$ and $v$ are arbitrary relatively prime numbers in which one is even and one is odd.
I figured it out.
Similar to problem I did before,
$x=2u^3-v^3$ and $z=v^3+2u^3$ where $u$ and $v$ are arbitrary relatively prime numbers.
However y is the value that must change since it is a third power. Thus rather than taking a square root I must take a cube root.
Therefore, $y={2uv}$
The proof was very long and tedious. It took me many hours to get, but if you check that parametrization it does work.