Find all pairs of integers $(x, y)$ such that $x^{2}-y!=2001$.
We notice that $3 \mid 2001$ but $9 \nmid 2001$. If $y \geqslant 6$, then $9 \mid y$!, so $3 \mid x^{2}$ and $9 \nmid x^{2}$, thus $3 \mid x$, hence $9 \mid x^{2}$: absurd. Therefore, $y<6$. We handle the remaining cases manually, and we find the unique solution: $(x, y)=(45,4)$