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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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    There are no two prime numbers that differ by exactly 7.2012-10-12
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    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.2012-10-12
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    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.2012-10-12
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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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