<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] a) $\small n^5-n = n \cdot(n^4-1) $ and $\small (n^4-1) $ is divisible by 5 by little fermat.
b)  $\small n^5-n = n \cdot(n^2-1)(n^2+1) $
$ \small \qquad \qquad =n \cdot (n-1)(n+1)(n^2+1) $
$\small  \qquad \qquad = (n-1)n(n+1) \cdot (n^2+1) $
   $\qquad$ and $\small (n-1)n(n+1) $ is divisble by 6 (because by 2 and by 3)         
Combine a) AND b) to see that $\small n^5-n $ is divisble by 5 AND 6 so also by 30.
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