**Problem 3 of the Canadian Mathematical Olympiad 1973**

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

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Since p and p + 1 are prime integers, they are not divisible by 2 and we can express

p = 2k + 1 (k is an integer);

p + 2 = 2k +3,

or p + 1 = 2(k + 1),

or 2 is a factor of p + 1

and since they are not divisible by 3

p = 3n + 1, or p = 3n + 2 (n is an integer),

but if p = 3n + 1 then p + 2 = 3(n + 1) which is divisible by 3

so the only option is p = 3n + 2 and p + 2 = 3n + 4 or p + 1 = 3(n + 1) or 3 is also a factor of p + 1

Both 2 and 3 are factors of p + 1 then 2 x 3 = 6 is a factor of p + 1.