<s>[INST] Diophantine equation $x^2 + y^2 = z^3$ I have found all solutions to the Diophantine equation $x^2 + y^2 = z^3$ when 
$z$ is odd. I am having some difficulty finding the solutions when $z$ is even. I am asking for a proof that provides the solutions where $z$ is even. I want the proof to be elementary and use only Number theory and perhaps Calculus or basic ideas about groups and rings.
 [/INST] Unfortunately, there isn't (apparently) one complete polynomial parameterization to
$$x^2+y^2 = z^k\tag1$$
when $k>2$. For $k=2$, the complete solution is,
$$x,\,y,\,z = (a^2-b^2)s,\; (2ab)s,\; (a^2+b^2)s$$
where $s$ is a scaling factor. Using complex numbers $a+b i$, one can generalize the method. For $k=3$, it is
$$x,\,y,\,z = (a^3 - 3a b^2)s^3,\; (3a^2 b - b^3)s^3,\; (a^2+b^2)s^2\tag2$$
but you can no longer find rational $a,b,s$ for certain solutions. For example,
$\hskip2.7in$ $9^2+46^2 = 13^3\quad$ Yes 
$\hskip2.7in$ $58^2+145^2=29^3\quad$ No 
(You can click on the Yes/No links for Walpha output.) A related discussion can be found in this post while an alternative method is described here. For the case $k=3$, if $a^2+b^2=c^3$, then an infinite more can be found as,
$$(a u^3 + 3 b u^2 v - 3 a u v^2 - b v^3)^2 + (b u^3 - 3 a u^2 v - 3 b u v^2 + 
        a v^3)^2 = c^3(u^2+v^2)^3\tag3$$
which should provide some solutions not covered by $(2)$.
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