## Fall 2017, problem 50

Prove that if $p$ and $p+2$ are prime integers greater than $3$, then $6$ is a factor of $6+1$.

### Comments

Sweet & succinct, Hubert.....love it!

I think that you want to say ..." then $6$ is a factor of $ p+1$ ".

If $p \equiv 1 \pmod{3}$ then $p+2 \equiv 3 \pmod{3}$. But $p+2$ is prime, so $p+2$ can't be a multiple of $3$. Note that $p$ is greater than $3$

So $p \not \equiv 1 \pmod {3}$.

Then $p \equiv 2 \pmod {3}$, and so, $p+1 \equiv 3 \pmod {3}$, which means that $p+1$ is a multiple of $3$.

As $p+1$ is also even , then $3\times 2=6$ is a factor of $p+1$.

p+1 is even. (p+1) mod 3 can not be 1 (p would be multiple of 3) nor 2 (p+2 wold be multiple of 3)

$p$ being prime, its possible congruences modulo $6$ are $r=1$ or $5$

$p+2$ being prime, $r=5\Rightarrow~p+1=6(q+1)$