A problem of divisibility ... If $7\nmid n$, $7\nmid n-1$ and $7\nmid n^{3}+1$ then $7\mid n^{2}+n+1$ $7\nmid n$, $7\nmid n-1\;\;$ and $\;\;7\nmid n^{3}+1$ $\Longrightarrow 7\mid n^{2}+n+1$$$$$Do not quite understand how to do, but do not want to use modular arithmetic, thought of using consecutive numbers, but did not get the idea ..
Just know that$$n^{3}-1=(n-1)(n^{2}+n+1)$$

$$n^7-n=n(n^6-1)=n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1).$$
Ask Little Fermat to help.
If you are really into consecutive numbers, then you may also observe that
$$
(n-2)(n-4)=n^2-6n+8=n^2+n+1-7(n-1),
$$
and
$$
(n+2)(n+4)=n^2+6n+8=n^2-n+1+7(n+1).
$$
