10. Find the positive integer solutions of the equation $x^{2}+y^{2}=z!$, where $x \leqslant y, z<14$.
10. $z>1$, when $z=2$, $x=y=1$. When $z \geqslant 3$, $x^{2}+y^{2}=z!\equiv 0(\bmod 3)$, and $x \neq 0(\bmod 3), x^{2}+y^{2} \equiv 0\left(\bmod 3^{2}\right)$, from which it can be deduced that $z \geqslant 6$. When $z=6$, the positive integer solutions can be found as $x=12, y=24$. When $z \geqslant 7$, $x^{2}+y^{2} \equiv 0(\bmod 7)$, then it can be deduced that $x \equiv y \equiv 0(\bmod 7)$, thus $x^{2}+y^{2} \equiv 0(\bmod 2)$, but when $z<14$, $72 \nmid z!$. This indicates that when $z \geqslant 7$, the original equation has no integer solutions. Therefore, the solutions to the original equation are $(1,1,2)$ and $(12,24,6)$.