Finding solutions to $x^2+y^2+z^2=w^3$ 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.
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.