If $a^{n}-1$ is prime then $a=2$ and $n$ is prime? I was doing some basic Number Theory problems and came across this problem :

Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime

My Solution : (Sloppy)

  
*
  
*$a^{n}-1$ = $(a-1)$ . $(a^{n-1} + a^{n-2} + ... + a + 1)$
  
*This means that $(a-1)$ | $(a^{n}-1)$ 
  
*But $(a^{n}-1)$ is prime
  
*So , $(a-1)$ = 1 $\Rightarrow$ $a = 2$
  
*Now , let $n$ be composite
  
*$n = kl$ , where $1 \lt k \lt n$ and $1 \lt l \lt n$
  
*Now , $a^{kl}-1$ = $(a^{k}-1)$ . $(a^{k.{(l-1)}} + a^{k.{(l-2)}} + ... + a^k + 1)$
  
*This means that $2^{n} -1$ is composite 
  
*Hence , we have achieved a contradiction
  

My Question :
Am I correct ?
 A: The proof is alright there are two or three details though (the same issue twice actually), oe was already pointed out in comments: 


*

*Likely you should exclude the case $a=1$ right away. Just by saying $1^n -1 = 0$ is not prime so assume $a>1$.

*You cannot derive from $a^n -1$ being a prime and  $(a-1) \mid (a^n-1)$ directly that $(a-1)= 1$. What you can do is say $a-1=1$ or $a-1 = a^n-1$. The latter is impossible as $n > 1$ (and $a >1$); note here you use $n>1$. 

*When you assert that $a^{kl}-1$ is composite, you should state explicitly that both factors you exhibit are not $1$. 
These are not major problems, but if one is picky one could insist on them.
A: I'm a little late to the party, but here is a way to justify this using a contrapositive statement.
Let $a,n\in\mathbb{Z}^+$, where $n>1$, and assume $a^n-1$ is prime. Show $n$ is prime by the contrapostive, i.e., show if $n$ is composite, then $a^n-1$ is composite. Since $n$ is composite, $n=ks,$ where $k,s,\in\mathbb{Z}$.
Now, $a^n-1=a^{ks}-1=(a^k)^s-1 =(a^k-1)[(a^k)^{s-1}+(a^k)^{s-2}+\cdots+a+1)$.  Clearly $a^n-1$ is composite.
We can directly show $a=2$. Since $a^n-1$ is prime, $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$ is also prime. Since the R.H.S. expression is prime, $a-1$ must equal $1$, showing $a=2$.
$\therefore$ if $a^n-1$ is prime, $n>1$ is prime and $a=2$.
A: Since $a$ and $n$ are both positive integers so $a>0$ and $n>0$
then taking $a=1$ putting in $a^n-1$ we get $1^n-1=0$ which is not prime hence take $a=2$ where $2^n-1$ is a prime if and only if $n$ is a prime
