What is the largest integer which must evenly divide all integers of the form $n^5-n$ ?
I am stuck on this problem,I don't know how to approach this.
Some scribble I've tried is:
Given that $n^5-n \equiv 0 \mod x $,I have $\left(n- \cfrac{1}{2} \right)^2 \equiv \cfrac{1}{4} \mod x \implies n \equiv 1 \mod x \implies x\cdot q =n-1 $ for some integer $q$.
I know that $n^5-n$ is always a multiple of $5$ by Fermat's little theorem so $x$ is some multiple of $5$ but after that I don't know what to do.