<s>[INST] Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$. Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$.
$n^7 \equiv n \bmod 2\implies n^6 \equiv 1 \bmod 2$
Using Fermat's little theorem, it is easy to see that:
$n^1 \equiv 1\bmod 2\implies n^a \equiv 1^a\bmod 2, \forall a \in \mathbb{N}$
Algebraic approach:
$(n^3-1)(n^3+1)\equiv 0\bmod 2$.
So, if either of the two factors is divisible by $2$, then fine.
So, now have:

*

*$n^3 \equiv 1 \pmod 2$

*$n^3 \equiv -1 \pmod 2$
So, how to pursue? Cannot get anything by stating $n^3 = 1 + 2k, k\in \mathbb {Z}$, or $n^3 = -1 + 2l, l\in \mathbb{Z}$.
Nor, will gain anything by assuming a form of $n$.
 [/INST] $n^7-n = n(n^6-1) = n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1)$. Observe the product consists of $2$ consecutive integers $n,n-1$ hence divisible by $2$.
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