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⇒ p+1=6(q+1)