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.

Can you conclude that $\gcd(n,2)=1$ and $\gcd(n,3)=1$? What about $\gcd(n,4)=1$?
Can you show that $3 \mid n^2-1$ and $4 \mid n^2-1$?   Because if yes, then you're done. Do you know why?
I leave the part $3 \mid n^2-1$ to you, because it's clear I think. Notice that if $\gcd(n,2)=1$ then $n$ must be an odd number. You can write $n^2-1=(n+1)(n-1)$ If $n$ is odd, and it must be if $\gcd(n,2)=1$ then $(n+1)$ and $(n-1)$ will both be even. so $n^2-1$ is divisible by $4$ and $4 \mid n^2-1$.
