If $n$ is a composite number, then $(7^n-1)/6$ is also composite

Question

Let $n \ge 2$ and $a_n = \dfrac{7^n−1}{6}$. Prove that if $n$ is composite then $a_n$ is composite.

I would normally prove something like this with induction but in this case I don't know how to define a composite number so that I can come to a proper conclusion.

Answer 1

You simply write: $7^n - 1 = 7^{pq} - 1 = (7^p - 1)(7^{p(q-1)} + 7^{p(q-2)} + .. + 1)$, and

$7^p - 1 = (7-1)(7^{p-1} + 7^{p-2} + ...+1) = 6(7^{p-1} + 7^{p-2} + ...+ 1)$, and the conclusion follows from these two equations.