how do you solve $a^2+b^2+c^2=d^3$ let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$

It is a theorem that one can identically solve,
$$x_1^2+x_2^2+\dots+x_n^2 = (y_1^2+y_2^2+\dots+y_n^2)^k$$
for any positive integer n and k. Thus the kth power of n squares is itself the sum of n squares.  For example, for $n =3$, we have,
$k=2:$
$$(a^2-b^2-c^2)^2+(2ab)^2+(2ac)^2 = (a^2 + b^2 + c^2)^2$$
$k=3:$
$$a^2(a^2 - 3b^2 - 3c^2)^2 + b^2(-3a^2 + b^2 + c^2)^2 + c^2(-3a^2 + b^2 + c^2)^2=(a^2 + b^2 + c^2)^3$$
and so on.  See Theorem 1 at https://sites.google.com/site/tpiezas/004.
