2.1: Problem 19: Prove that \(n^6 - 1\) is divisible by \(7\) if \((n,7) = 1\).
Proof
Suppose that \((n,7) = 1\). By Fermat’s theorem we know that
$$
\begin{align*}
n^{p-1} \equiv 1 \pmod{p}
\end{align*}
$$
Thus
$$
\begin{align*}
n^6 &\equiv 1 \pmod{7} \\
n^6 - 1 &\equiv 0 \pmod{7} \\
\end{align*}
$$
as desired. \(\ \blacksquare\)