Exercise 2.1.19 ($n^6 - 1$ is divisible by $7$ if $(n,7) = 1$.)

Question

Prove that $n^6 - 1$ is divisible by $7$ if $(n,7) = 1$.

Richard Ganaye · Accepted Answer

\begin{proof} By Fermat's theorem (Theorem 2.7) for $p = 7$, if $n\wedge 7 = 1$, that is if $7 
mid n$, then
$$7 \mid n^6 - 1.$$
\end{proof}

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