<s>[INST] How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
 [/INST] Just to give a different approach, we have
$$\begin{align}
n^7-n&=n(n^6-1)\\
&=n(n^3-1)(n^3+1)\\
&=n(n-1)(n^2+n+1)(n+1)(n^2-n+1)\\
&=n(n-1)(n+1)(n^2+n-6+7)(n^2-n-6+7)\\
&=n(n-1)(n+1)\Big((n-2)(n+3)(n+2)(n-3)+7\cdot\text{quadratic in }n\Big)
\end{align}$$
where the quadratic in $n$, to be explicit, is $2(n^2-6)+7$.  What's important, though, is that $n(n-1)$, being the product of two consecutive integers, is always divisible by $2$, $n(n-1)(n+1)$, being the product of three consecutive integers, is always divisible by $3$, and $n(n-1)(n+1)(n-2)(n+3)(n+2)(n-3)$, being the product of seven consecutive numbers, is always divisible by $7$.
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