<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] $n^5-n=(n-1)n(n+1)(n^2+1)$. Rewrite $n^2+1$ as $5(n-1)+(n^2-5n+6)$ to obtain $n^5-n=5(n-1)^2n(n+1)+(n-3)(n-2)(n-1)n(n+1)$ and use the fact that $n!$ divides the product of n consecutive numbers.
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