# Fall 2017, problem 50

### Comments

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)

All primes greater than 3 are either of the form 6k+1 or 6k−1. The proof is straightforward: any number that is not in any of the two aforementioned forms is in one of the four forms 6k, 6k+2, 6k−2 or 6k+3, which have the factors 6, 2, 2 and 3 respectively. Since k>0, all of these numbers are composite. Primes greater than 3 can’t be 3 and can’t be 2. Since they are primes, they can’t be multiples of 3 and can’t be even.

Think mod 6:

p and “p+2” can’t be 0 (mod 6) and can’t be even (mod 6) … so, not 2 (mod 6) and not 4 (mod 6)

p and “p+2” can’t be 3 (mod 6) either… 3+6k would mean it’s a multiple of 3.

So: “p” and “p+2” must be either 1 (mod 6) or -1 (mod 6) But in case p was 1 (mod 6) then “p+2” would be 1+2 (mod 6) = 3 (mod 6)

So p must be -1 (mod 6) and “p+1” must be +1 (mod 6)

And that means “p+1” must be 0 (mod 6)… that is a multiple of 6.

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

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