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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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    Yes very good point. I believe that there are no two primesthat differ by seven since it only goes to six differences occurring that many primes will not be accountable for this math.2018-04-16

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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 $p-q=d$.

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    If there were a proof that there is no proof, it would not make much sense to keep the conjecture open, wouldn't it? Apparently, Gerry meant to say "to my knowledge, hoone has found any proof so far".2012-10-12