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?

the equation:
$X^2+Y^2=Z^3$
Has the solutions:
$X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$
$+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$
$+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$
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$Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^3+$
$+2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$
$-2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$
.................................................................................................................................................
$Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k+2(q^2+t^2)^2$
$q,t,k$ - What are some integers any sign.  After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.
