Find the greatest integer $d$ that divides $n^{5}-n$ for all integers $n \in \mathbb{Z}$.
First, $d$ must divide $2^{5}-2=30$. We will show that $d=30=2 \times 3 \times 5$ works. To do this, it suffices to note that if $p \in\{2,3,5\}$, then $p-1$ divides $5-1=4$, so that $n^{5} \equiv n(\bmod p)$ for all $n \in \mathbb{Z}$. Consequently, $n^{5}-n$ is divisible by $2 \vee 3 \vee 5=30$.