<s>[INST] How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
 [/INST] \begin{align*}
\dfrac{n(n^4-1)}{30}&=\dfrac{n(n^2-1)\color{#c00}{(n^2+1)}}{30}\\&=\dfrac{(n-1)n(n+1)\color{#c00}{(5+(n+2)(n-2))}}{30}\\&=\dfrac{(n-1)n(n+1)}{6}+\dfrac{\color{#c00}{(n-2)}(n-1)n(n+1)\color{#c00}{(n+2)}}{30}\\&=\binom{n+1}{3}+4\binom{n+2}{5}\in\Bbb Z
\end{align*}
</s>