Prove that $12 \mid n^2 - 1$ if $\gcd(n,6)=1$ 
Prove that $12 \mid n^2 - 1$ if $\gcd(n,6)=1$. 

I know I have to use Fermat's Little Theorem for this but I am unsure how to do this problem.

As is mentioned in answers above, $gcd(n,6)=1$ clearly implies that $n$ is odd.  But then each of $n+1$ and $n-1$ are even, so $4$ divides $(n+1)(n-1)=n^2-1$.  Thus it is only left to show that $3$ also divides $n^2-1$.  But given any three consecutive integers, we know that $3$ must divide exactly one of them.  Applying this to $n-1,n,n+1$, and using the fact that $3$ cannot divide $n$ (since $gcd(n,6)=1$), we get the desired result. 
