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?

Set $z=a^2+b^2=(a+bi)(a-bi)$, $(i=\sqrt{-1})$
$x^2+y^2=(x+yi)(x-yi)=z^3=(a+bi)^3(a-bi)^3,$
$(1)\quad x+yi=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i,$
$x=a^3-3ab^2,y=3a^2b-b^3,z=a^2+b^2.$
$(2)\quad x+yi=(a+bi)^2(a-bi)=(a^2+b^2)(a+bi),$
$x=(a^2+b^2)a,y=(a^2+b^2)b,z=a^2+b^2$
