# Difference Between Primes

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.

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There are no two prime numbers that differ by exactly 7. – Gerry Myerson Oct 12 '12 at 3:34
Oh, thank you. I don't know how I missed that one. Thank you for pointing that out, even though it is not answering my question. That is a good catch. I shall change that. – Ben Oct 12 '12 at 3:43
If you want to allow negative primes, then $2$ and $-5$ differ by $7$. If two integers differ by an odd number, then one of them is even and the other odd. So if two prime numbers differ by $7$, then one of them is an even prime number, and there's only one positive prime number that's even, so look at numbers differing from $2$ by $7$. And of course, allowing negative numbers in certain contexts is almost cheating. – Michael Hardy Oct 12 '12 at 16:43

It is conjectured that for every even $d$ there exist an infinity of pairs of prime $p,q$ such that $p-q=d$. However, there is no proof that for all even $d$ there is even one pair of primes $p,q$ with $p-q=d$.