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

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.

• not a very helpful question title.. – user26486 Sep 7 '14 at 23:37
• @mathh haha, I was about to change that but it's pretty cool. Imagine a question titled "Algebra." – Shahar Sep 7 '14 at 23:38
• Do you mean $\frac{ 7^n-1}{6}$? – Calvin Lin Sep 7 '14 at 23:39
• en.wikipedia.org/wiki/Zsigmondy%27s_theorem – Jack D'Aurizio Sep 7 '14 at 23:39

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.