Theorem:Originally Posted byTreadstone 71

Let p be a prime >3, then all three of p, p+2, p+4 cannot be prime, because

at least one of them must be divisible by 3.

Proof:

Suppose p>3 and not divisible by 3, then there exist k>0 such that

p=k.3+r, where r is either 1 or 2 (r cannot be 0 since then p would be

divisible by 3, which it is not by supposition).

If r=1 then p+2 is divisible by 3.

If r=2 then p+4 is divisible by 3.

QED

Note: this is stated on page 5 of Hardy and Wright (fifth edition)

but without the restriction that p>3, so failing to exclude 3,5,7!

Your problem is a corollary of the theorem. For if p>3 and p and p+2

are both prime neither of them is divisible by 3, so p+4 must be

divisible by 3.

RonL