TITLE: Prove that $12 \mid n^2 - 1$ if $\gcd(n,6)=1$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: If $\gcd(6,n)=1$ then $n=6k\pm 1$. This yields $$n^2-1=36k^2\pm 12k+1-1=12k(3k\pm 1)$$ So $12\mid n^2-1$.