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: $$a^2+b^2+c^2=d^3$$
 A: 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.
A: For the equation:
$$x^2+y^2+z^2=r^3$$
Will make a replacement that formula was compact.
$$c=2(q-p-s)t$$
$$d=s^2+t^2-q^2-p^2+2p(q-s)$$
$$k=p^2+t^2-q^2-s^2+2s(q-p)$$
$$n=p^2+t^2+s^2-q^2$$
$$j=p^2+s^2+t^2+q^2-2q(p+s)$$
$p,s,t,q$ - integers asked us. Then decisions can be recorded.
$$x=dn^2+2cnj-dj^2$$
$$y=cj^2+2dnj-cn^2$$
$$z=k(n^2+j^2)$$
$$r=n^2+j^2$$
A: Refer to Gohierre de Longchamps's formula
\begin{gather*}
\left\{
\begin{split}
a&=u\Big(\lambda\,\!u^2+\mu\,\!v^2-3\nu\,\!w^2\Big)\\
b&=v\Big(\lambda\,\!u^2+\mu\,\!v^2-3\nu\,\!w^2\Big)\\
c&=w\Big(3\lambda\,\!u^2+3\mu\,\!v^2-\nu\,\!w^2\Big)\\
d&=\lambda\,\!u^2+\mu\,\!v^2+\nu\,\!w^2
\end{split}
\right.\\
\\
\Downarrow\\
\\
\Large{\lambda\,\!a^2+\mu\,\!b^2+\nu\,\!c^2=d^3}
\end{gather*}
L'Intermédiaire des mathematiciens.
1902-1903. P311-312

A: The simple formula for the equation:
$$x^2+y^2+z^2=q^3$$
You can write this:
$$x=3(p-k-t)(p^2+2k^2-2kt+2t^2)s^3$$
$$y=3(p-k+2t)(p^2+2k^2-2kt+2t^2)s^3$$
$$z=3(p+2k-t)(p^2+2k^2-2kt+2t^2)s^3$$
$$q=3(p^2+2k^2-2kt+2t^2)s^2$$
$p,k,t$ - integers asked us.
If there is a one simple solution, it will be necessary to reduce to the corresponding $s$.
There are other formulas. But they are bulky. Don't know whether it makes sense to write them.
