SAUDI ARABIAN MATHEMATICAL COMPETITIONS

We have $n^3-1=(n-1)\left(n^2+n+1\right)$. We shall prove that there are infinitely many positive integers $n$ such that $n^2+n+1$ can be represented as the product of some factors less than $n-1$ and then deduce that $n^3-1$ divides $n!$. In fact, we first choose $n=m^2$ for some positive integer $m$, one has $$n^2+n+1=m^4+m^2+1={\left(m^2+1\right)}^2-m^2=\left(m^2-m+1\right)\left(m^2+m+1\right).$$ Next, we choose $m=3k+1$, then $m^2+m+1=(3k+1{)}^2+(3k+1)+1=9k^2+6k+1+3k+1+1=3\left(3k^2+3k+1\right)$ and $m^2-m+1=(3k+1{)}^2-(3k+1)+1=9k^2+6k+1-3k-1+1=9k^2+3k+1$. It is clear that $$3<3k^2+3k+1<9k^2+3k+1<9k^2+6k=(3k+1{)}^2-1=n-1.$$ Therefore, for $n=(3k+1{)}^2$ with any positive integer $k$, it follows that $n!$ is divisible by $$3\left(3k^2+3k+1\right)\left(9k^2+3k+1\right)\left(9k^2+6k\right)=n^3-1.$$ The problem is proved. $\square$