Variation of Pythagorean triplets: $x^2+y^2 = z^3$ I need to prove that the equation $x^2 + y^2  = z^3$ has infinitely many solutions for positive $x, y$ and $z$.
I got to as far as $4^3 = 8^2$ but that seems to be of no help.
Can some one help me with it?

There are already infinitely many solutions among $2^k$'s:
$$2^k+2^k=2^{k+1}$$
So, if $k$ is even and $3\,|\,k+1$ (that is, $k\equiv 2\pmod{6}$), then it's a solution.
