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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.

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

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