How to use modular arithmetic for $a^b-1 $

Question

The problem is: If $a,b \ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.

My ideas so far:

Since the we want to show that $a = 2$, we can take the the expression modulo $a$.

$a^b-1 \equiv -1 \pmod{a}$, so we know that there's a prime $p>2$ such that $p \equiv -1 \pmod{a}$ However I am not really sure where to go from here.

Answer 1

Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1\mid a^b-1$, $2\leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.

Answer 2

Hmm, If $b \ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a \ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.

The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.

These are the Mersenne primes. Which are well researched.