TITLE: Suppose $p$ and $p-2$ are both prime numbers where $p>5$. Prove $6$ divides $p-1$.
QUESTION [5 upvotes]: Intuitively, any two prime numbers greater than five that have a non prime in between them is divisible by $6$. $6$ is between $5$ and $7$, $12$ is between $11$ and $13$, etc. But I have no idea how to represent this mathematically. A push in the right direction would be much appreciated.

REPLY [4 votes]: You are essentially being asked to prove that $p-1\equiv0\mod{6}$.  
You also know that the only possible residues$\mod{6}$ are $0,1,2,3,4,5$, or more conveniently $0,\pm1,\pm2,3$. If you think about the divisibility of numbers of the forms $6k, 6k\pm1,6k\pm2,6k+3$, you will find that only certain of them can possibly be primes. Note that since $p>5, k\ge1$. 
Now try letting your number $p-1$ have various residues $\mod{6}$ and see what residues both $p$ and $p-2$ have as a consquence, and decide what that means about their primality. Then conclude what constraints that puts on $p-1\mod{6}$.