We know that there are two prime numbers that have a difference of one: 2 and 3. And we know there is at least one pair of primes with a difference of two: 5 and 7. Same with a difference of three: 2 and 5. This pattern continues until we reach a difference of seven. As far as I can tell (and I have convinced myself) there are no two prime numbers that differ by exactly seven. And I know that the only odd numbered differences that will work will be those differences involving 2. So my question then is:
Given that $P_2$ and $P_1$ are primes and $P_2>P_1$
$$P_2-P_1=2d, \quad \forall d \in Z , \quad d>0. $$
Simply, is there a pair of prime numbers such that their difference is a multiple of two for all multiples of two?
I may not have typed it perfectly, but I think it gets the point across. Whatever the answer, please explain how to go about solving it.