Proving that either $2^n-1 $ or $ 2^n+1$ is not prime Not sure what approach to take with this:
Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$  

As $n>2,2^n-1>3$
If $n$ is even, $=2m$(say),
 $2^n-1=2^{2m}-1=4^m-1$ which is divisible by $4-1=3$ as $(a-b)$ divides $a^n-b^n$ for integer $n\ge 0$ 
If $n$ is odd, $=2m+1$(say), 
$2^{2m+1}+1=2^{2m+1}+1^{2m+1}$ which is divisible by $2+1=3$  as $(a+b)$ divides $a^{2m+1}+b^{2m+1}$ for integer $m\ge 0$

Alternatively,
$(2^n-1)(2^n+1)=4^n-1$ is divisible by $4-1=3,$ 
but $2^n+1>2^n-1>3$
References :  Fermat Prime 1,2,  Mersenne Prime1,2
