2.1: Problem 19: Prove that \(n^{12} - 1\) is divisible by \(7\) if \((n,7) = 1\).
Proof
Suppose that \((n,7) = 1\). By Fermat’s theorem we know that
$$
\begin{align*}
n^{6} \equiv 1 \pmod{7}
\end{align*}
$$
We know that if \(x \equiv y \pmod{m}\), then \(x^k \equiv y^k \pmod{m}\). Therefore, we can square both sides to see that
$$
\begin{align*}
n^{12} \equiv 1 \pmod{7}
\end{align*}
$$
This implies that \(n^{12} - 1\) is divisible by \(7\) as we wanted to show. \(\ \blacksquare\)