8.5.3 ** Find all solutions to $a^{2}+b^{2}=n!$ where $a, b, n$ are positive integers, and $a \leqslant b$,
$$
n<14 .
$$
For $n=1$, it is obviously unsolvable.
For $n=2$, there is one solution $a=b=1$.
If $n \geqslant 3$, then $a^{2}+b^{2}=n!\equiv 0(\bmod 3)$. Since a square number $\equiv 0$ or $1(\bmod 3)$, the above equation leads to $a^{2} \equiv b^{2} \equiv 0(\bmod 3)$, thus $3|a, 3| b, 3^{2} \mid a^{2}+b^{2}=n!$. Therefore, $n \geqslant 6$.

If $n=6$, let $a=3 a_{1}, b=3 b_{1}$, then $a_{1}^{2}+b_{1}^{2}=2^{4} \times 5$, which implies that $a_{1}, b_{1}$ are both even (otherwise $a_{1}^{2}+b_{1}^{2} \equiv 1$ or $\left.2(\bmod 4)\right)$. Let $a_{1}=2 a_{2}, b_{1}=2 b_{2}$, then $a_{2}^{2}+b_{2}^{2}=2^{2} \times 5$, similarly, $a_{2}, b_{2}$ are even. Let $a_{2}=2 a_{3}, b_{2}=2 b_{3}$, then $a_{3}^{2}+b_{3}^{2}=5$, yielding $a_{3}=1, b_{3}=2$. Therefore, $a=12, b=24$.

If $n \geqslant 7$, then $a^{2}+b^{2} \equiv 0(\bmod 7)$. Thus $a \equiv b \equiv 0(\bmod 7)$, but when $n<14$, $7^{2} \nmid n!$. Hence, there are no positive integer solutions in this case.
In summary, there are two solutions: $n=2, a=b=1 ; n=6, a=12, b=24$.