For every integer $n$ the number $n^5-n$ is a multiple of a fixed number. What is the LARGEST number that always divides it?

Factorise: $n^5-n=n(n^4-1)=(n-1)n(n+1)(n^2+1)$. Three consecutive integers guarantee a factor of 2 and a factor of 3. A factor of 5 is always there too: if $n$ is not a multiple of 5 then $n^2$ leaves remainder 1 or 4 on division by 5, so either $n^2-1$ or $n^2+1$ is a multiple of 5. That gives $2\times 3\times 5=30$. Nothing bigger is safe, because $n=2$ gives 32 − 2 = 30 exactly, which is not a multiple of 60 — the very first case rules out the larger options.