If $x, y$, and $z$ are distinct positive integers such that $x^{2}+y^{2}=z^{3}$, what is the smallest possible value of $x+y+z$.
Solution 1: Without loss of generality let $x>y$. We must have $z^{3}$ expressible as the sum of two squares, and this first happens when $z=5$. Then $x$ and $y$ can be 10 and 5 or 11 and 2 . If $z>5$ then $z \geq 10$ for $z^{3}$ to be a sum of two distinct squares, so $x^{2}>500, x>22$, so $x+y+z>32$. Thus the smallest possible value of $x+y+z$ is $11+2+5=\mathbf{1 8}$.

Solution 2: If $z>5$, then $z \geq 6$, so $z^{3} \geq 216$. Now $x^{2}+y^{2} \geq 216$, so $x \geq 11$ and $y \geq 1$, thus $x+y+z \geq 18$. Since $x=11, y=1, z=6$ does not work, we must have $x+y+z>18$, and the solution given is the best possible.