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.

If $\gcd(n,6) = 1$, then $n$ is odd. Hence, $n^2 \equiv 1 \pmod8$. Further, $\gcd(n,3) = 1$ and hence $n^2 \equiv 1 \pmod3$. Now conclude that $n^2 \equiv1 \pmod{24}$.
