Show that $7 |(n^6 + 6)$ if $7 ∤ n$, $∀ n ∈ ℤ$

Question

Show that $7 |(n^6 + 6)$ if $7 ∤ n$, $∀ n ∈ ℤ$

I need to prove this by the little Fermat's theorem.

My attempt

$n^6 \equiv -6 \pmod 7$

To show $7 ∤ n$ I need to show that $N$ is not congruent to $0$ mod $7$.

as $-6 \equiv 1\pmod 7$

$n^6 \equiv 1\pmod7$

But now, How can I show $N$ is not congruent to $0$ mod $7$ ?

Answer 1

By Fermat's little theorem we know that:
Remark: For every $n$, with $7 \nmid n$ we have : $ \ \ \ \ \ n^6 \overset{7}{\equiv} 1 \ \ .$

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$$ n^6 \overset{7}{\equiv} 1 \Longrightarrow n^6 -1 \overset{7}{\equiv} 0 \Longrightarrow n^6 -1 + 7 \overset{7}{\equiv} 0 \Longrightarrow n^6 + 6 \overset{7}{\equiv} 0 . $$

Answer 2

Because $n^6+6=n^6-1+7$ and $n^6-1$ divisible by $7$.

$$n^6-1=(n-1)(n+1)(n^2-n+1)(n^2+n+1)$$ and now easy to check $n\equiv\pm1,\pm2\pm3(\mod7)$.