# Suppose $p$ and $p-2$ are both prime numbers where $p>5$. Prove $6$ divides $p-1$.

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.

• Clearly $3$ divides one of $p-2,p-1,p$ so... – lulu Jul 10 '18 at 21:37
• Being divisible by $6$ is equivalent to being divisible by both $2$ and $3$ (CRT). – Arnaud Mortier Jul 10 '18 at 21:39

Just a sketch for intuition:

All even numbers (besides 2) are non prime so $p - 1$ is divisible by 2.

One of $p - 2$, $p - 1$, $p$ are divisible by 3, we know $p$ and $p - 2$ are not, so $p - 1$ must be.

If $p - 1$ is divisible by 2 and 3, then $p - 1$ is divisible by 6.

• The question was for a "push in the right" direction $\neq$ providing a full answer. Try more to promote the asker's proficiency, rather than "doing for" or "spoonfeeding" an asker. – Namaste Jul 10 '18 at 23:03
• well asker still has to prove why being divisible by 2 and 3 means p-1 is divisible by 6, which I think is the hard part – Eli Jul 15 '18 at 16:58

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 $p-1\mod{6}$.

Suppose $p$ is of the form $6k + 3$, that is, $p \equiv 3 \bmod 6$.

Then $p$ must be $-3$ or $3$. But you said $p > 5$. So how about $p \equiv 5 \bmod 6$?

But again, you said, $p > 5$, so $p - 2$ is divisible by $3$ and at least one other prime (or one other instance of $3$). That leaves the case $p \equiv 1 \bmod 6$.

I ended each of the previous paragraphs with $6$. Mwahahahahaha!

Hint: Give $k$ consecutive numbers, one of them is divisible by $k$.

.....

So given $p-2, p-1, p$ one of them is divisible by $3$. And one or more of them is divisible by $2$.

and

But $p$ and $p-2$ are both prime and larger than $3$ so... it's not them.

.....

So $p-1$ must be the one that is divisible by $3$ and $p-1$ must be the one that is divisible by $2$. So $p-1$ is divisible by both $3$ and by $2$ so it is divisible by $6$.