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Is it true that for every even natural number $k$ there exists some $n \in \mathbb{N}$ such that $g_n = p_{n+1} - p_n = k$?

I don't know how to approach the problem at all, and in fact I don't even know enough about prime gaps to even form a conjecture as to the answer. I feel like the answer is "yes", but only because that would be "nicer" than having some even integers never appear in sequence of prime gaps.

I hope it's not an unsolved problem!

Edit: My question is distinct from Polignac's Conjecture, since I ask if there is at least one prime gap, instead of infinitely many prime gaps, for every size.

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  • $\begingroup$ Do you have a source? I would like to read more about it if it is an open problem. $\endgroup$
    – feralin
    Jan 31, 2016 at 5:40
  • $\begingroup$ I think that it is an unsolved problem known as Polignac's Conjecture. $\endgroup$
    – user170039
    Jan 31, 2016 at 5:46
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    $\begingroup$ @user170039 my question is distinct from Polignac's Conjecture. I ask if there is at least one prime gap of every size, not infinitely many prime gaps of every size. $\endgroup$
    – feralin
    Jan 31, 2016 at 5:48

2 Answers 2

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It appears to be open if every even number is the difference of two primes, let alone consecutive primes. Here is a m.se question mentioning that and an mo question here

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For $n$ a positive integer, the numbers $$(n+1)!+2,(n+1)!+3,..,(n+1)!+n+1$$ are $n$ consecutive composite integers. Does this help ?

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    $\begingroup$ No, this doesn't help. OP asked about some pair of primes that have a given difference. Finding a long stretch of non-primes does not speak to this at all. $\endgroup$
    – Ross Millikan
    Apr 7, 2018 at 5:25
  • $\begingroup$ I think it's unfair to say that this "does not speak to this at all". This result is clearly related, for instance, anytime (n+1)!-1 and (n+1)! + (n+2) are prime this would give a solution for k = n+3. $\endgroup$
    – Adam Boocher
    Nov 13, 2021 at 15:27

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