For $a,n \in \Bbb N$, how to prove if $a^n-1$ is prime then $a=2$ and $n$ is prime? Suppose  $a,n \in \Bbb N $. How do you prove that if $a^n-1$ is prime, then $a=2$ and $n$ is prime .
Is the converse true I.E : (if $p$ is prime then $2^p-1$ is prime)?

If $n=1,a-1$ will be prime $=p$(say) if $a=p+1$ where $p$ is any prime.
For $n>1,$
As here $(a-1)\mid(a^n-1)$
$a-1$ must be $<2\implies a=2$
Now, if $n=rs$ where $r,s$ are integers$>1$,
$(a^r-1)\mid (a^{rs}-1)\implies a^n-1$ can not be prime
So, the necessary condition for $a^n-1$ be prime are $a=2$ and $n$ is prime.
The converse is not true as for example $2^p-1$ is not prime for $p=11$
This using $3$rd theorem of this, if prime $q\mid(2^p-1), 2p\mid(q-1)$ i.e., $q=2\cdot k\cdot p+1$ where integer $k\ge0$
For $p=11, q=22k+1$ and we can verify $2^{11}-1=23\cdot89$
Reference: List of known Mersenne primes
