Show that if $n>1$ and $a$ is a positive integer with $a^n-1$ prime, then $a=2$ and $n$ is prime

Question

Show that if $n>1$ and $a$ is a positive integer with $a^n-1$ prime, then $a=2$ and $n$ is prime.

Hint: Look at the proposition, if $n$ is composite then $2^n-1$ is composite.

Answer 1

Hint: Firstly, we have: $$ a^n - 1 = (a-1)(a^{n-1} + a^{n-2} + \cdots + a^2 + a + 1) $$ Secondly, if $n$ is not prime, say $n = ij$ with $i, j \ge 2$, then we have (by the same factorization trick): $$ a^n - 1 = a^{ij} - 1 = (a^i - 1)(a^{ij - i} + a^{ij-2i} + \cdots + a^{2i} + a^i + 1). $$

Answer 2

Hint: If $m$ divides $n$ and $1<m<n$, then $a^m-1>1$ also divides $a^n-1$. So $n$ must be prime.