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 ?
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 $a&gt;1$.
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 $n &gt; 1$ (and $a &gt;1$); note here you use $n&gt;1$. 
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.
Since $a$ and $n$ are both positive integers so $a&gt;0$ and $n&gt;0$
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