Diophantine equation $x^2 + y^2 = z^3$

Solution 1:

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 $(2)$.

Solution 2:

Fermat's two squares theorem says exactly which integers $n$ can be written as the sum of two squares, and indeed it can be made constructive, with a procedure to find all such representations. I recommend applying that known procedure to $n=z^3$. I don't think there's a significantly easier way; for example, already when $z$ is a high power of $5$ (or twice a high power of $5$), there are many representations.