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I have the following conjecture:

Let $k\in\mathbb{N}$ be even. Now $k+p$ is prime for infinitely many primes $p$.

I couldn't find anything on this topic, but I'm sure this has been thought of before. I tried to solve this using Dirichlet's theorem on arithmetic progressions and the Green–Tao theorem, but no luck with those. Is this question equivalent to an existing open problem? If not, how can I prove this (I prefer hints, but I appreciate full answers, too)?


- Edit -

As has been pointed out in the comments, this is not a duplicate. I'm asking for infinitely many primes $p$ such that $p+k$ is prime, not only one.

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  • $\begingroup$ Note that this is false for $k$ odd. $\endgroup$ – T. Bongers Feb 3 '16 at 19:22
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    $\begingroup$ For $k=2$ this is equivalent to the twin prime conjecture.... $\endgroup$ – N. S. Feb 3 '16 at 19:25
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    $\begingroup$ @Micah That question doesn't involve the condition that it be expressible in infinitely many ways. $\endgroup$ – David Feb 3 '16 at 19:25
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    $\begingroup$ To be fair that is not a duplicate: this question asks for a much more general problem "can every even integer be expressed as the difference of two primes in infinitely many ways". Anyway, it should be clear by now that this question is an unsolved problem (and a Very Hard$^{\text{TM}}$ one). $\endgroup$ – Winther Feb 3 '16 at 19:41
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    $\begingroup$ @T.Bongers Possibly, but it could have been the case that this conjecture was known to be false. $\endgroup$ – David Feb 4 '16 at 20:06
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This is a consequence of Polignac's conjecture. It also implies the twin prime conjecture. Therefore your conjecture is intermediate in strength between those two.

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    $\begingroup$ Well, I don't require $p$ and $k+p$ to be consecutive primes, so it's not the same. The Plignac's conjecture states that there are infinitely many prime gaps size $k$, I just want the difference of the primes to be $k$, but I don't care whether or not there are primes between $p$ and $k+p$ $\endgroup$ – vrugtehagel Feb 3 '16 at 19:25
  • $\begingroup$ You're correct. I've edited my answer. $\endgroup$ – David Feb 3 '16 at 19:27
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    $\begingroup$ There is another related open problem that has been studied, see here ("Every even number is the difference of two primes."). I don't think it has a proper name. OP's conjecture is even stronger than that since we are asking for infinitely many for each difference, not just a single one. $\endgroup$ – Dan Brumleve Feb 3 '16 at 19:29
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There are some results known. It has been proven that there is at least one $k \leq 246$ that appears infinitely often as a prime gap.

Furthermore, assuming respectively the Elliott–Halberstam conjecture and its generalisation, one can prove that there is at respectively at least one $k \leq 12$ and $k \leq 6$ that appears infinitely often as a prime gap.

Of course, a prime gap is stronger than your condition, since the primes don't have to be consecutive.

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    $\begingroup$ The distinction between "prime gap" and "primes with difference $k$" is immaterial in this "$\le 246$" formulation. In fact I believe the sieve methods used here just establish "there are infinitely many blocks of width 246 containing at least two primes". The equivalence to "some prime gap $\le 246$" is easy. $\endgroup$ – Erick Wong Feb 11 '16 at 20:02
  • $\begingroup$ For a reference, the result that an even $k$ exists is Yitang Zhang's theorem. He proved in 2013 that such a $k$ exists and can be chosen to be at most $70{,}000{,}000$. Before 2013 it was not known if $p_{n+1}-p_n \to \infty$ as $n\to\infty$. $\endgroup$ – Jeppe Stig Nielsen Sep 23 '17 at 11:44

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