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.

  • $\begingroup$ There are no two prime numbers that differ by exactly 7. $\endgroup$ – Gerry Myerson Oct 12 '12 at 3:34
  • $\begingroup$ 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. $\endgroup$ – Ben Oct 12 '12 at 3:43
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Oct 12 '12 at 16:43
  • $\begingroup$ 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. $\endgroup$ – Jeffery Apr 16 '18 at 0:28

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 $p-q=d$.

  • $\begingroup$ Can we prove there is no proof? Where did you get this information out of curiosity. $\endgroup$ – Ben Oct 12 '12 at 3:40
  • $\begingroup$ 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". $\endgroup$ – Hagen von Eitzen Oct 12 '12 at 4:18

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