Suppose $w$, $x$, $y$, and $z$ are all positive integers less than $100$. Find all such solutions to the equation $x^2+y^2+z^2=w^3$.
This problem was in a competition I participated in this past week, and I was unable to determine a clean way to solve it. Is there a more efficient way to approach it (perhaps using a variant of Legendre's three-square theorem?) without checking all values of $w$ less than $100$?
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.
$$x=dn^2+2cnj-dj^2$$
$$y=cj^2+2dnj-cn^2$$
$$z=k(n^2+j^2)$$
$$r=n^2+j^2$$
You can write more compact. You can write this:
$$x=3(p-k-t)(p^2+2k^2-2kt+2t^2)s^3$$
$$y=3(p-k+2t)(p^2+2k^2-2kt+2t^2)s^3$$
$$z=3(p+2k-t)(p^2+2k^2-2kt+2t^2)s^3$$
$$r=3(p^2+2k^2-2kt+2t^2)s^2$$
$p,k,t$ - 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.