Problem - 2099
Prove that there are infinitely many positive integers $n$ such that $(n^2+1)$ divides $n!$.
Consider the equation $x^2-5y^2 = -1$. Because it has at least one solution $(2,1)$, it has infinitely many positive integer solutions.
Consider those solutions with $y>5$. We have $4y^2 = x^2-(y^2-1) < x^2$, hence $5 < y < 2y < x$.
Therefore, $(x^2+1)=5\cdot y \cdot y $ divides $x!$.