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Yesterday I saw this question: A question about divisibility of sum of two consecutive primes (you should read the OP to understand the full problem), it just asks to prove that for all $k\in \mathbb Z^+$, there exist infinitely many consecutive primes such that : $$k\mid p_{n+1}+p_n.$$ the guy who asked this took care of the cases where $k=1,2,3,4,6$. The general case where $k$ is any positive integer is beyond me, but I attempted to prove the case where $k=12$ and I wonder it the proof is correct:

Assume that the twin prime conjecture is true, which says that there exist infinitely many consecutive primes such that $$p_{n+1}-p_n=2$$

Since any prime $\ge 5$ is on the form $6k\pm 1$ and every pair of twin primes is on the form $(6k-1,6k+1)$, Hence $$p_{n+1}+p_n=6k+1+6k-1=12k$$ $$12\mid p_{n+1}+p_n$$

for infinitely many consecutive primes?

Note that the twin prime conjecture also implies the case where $k=1,2,3,4,6$, because $1,2,3,4,6\mid 12$

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  • $\begingroup$ As you mentioned , this proof requires the truth of the twin prime conjecture, but it is a valid proof. $\endgroup$
    – Peter
    May 26, 2021 at 12:02
  • $\begingroup$ if we assume the truth of polignac's conjecture, can we do better than $k=12$? @Peter $\endgroup$
    – PNT
    May 26, 2021 at 13:20
  • $\begingroup$ I think we can extend this result , but I have no good approach yet. Intuitively , I would guess, that we can find infinite many pairs for every $k$. $\endgroup$
    – Peter
    May 26, 2021 at 13:22
  • $\begingroup$ that's exactly what I thought in the beginning but it turns out that you can't do any better than $12$, at least that's what I've found. @Peter $\endgroup$
    – PNT
    May 26, 2021 at 13:27
  • $\begingroup$ Your same argument works for $8$ too. $\endgroup$
    – arbashn
    Jul 13, 2021 at 15:33

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Assuming Schinzel's hypothesis , for every positive integer $k$ , there are infinite many positive integers $\ n\ $ , such that $\ kn-1\ $ and $\ kn+1\ $ are both prime. Those primes are obviously consecutive because they have difference $\ 2\ $ (in fact they are twin primes). The sum is $\ 2kn\ $ which is divisible by $\ k\ $.

Of course, Schinzel's hypothesis is much stronger than the twin prime conjecture, but at least this is some evidence that we can find infinite many pairs for every $\ k\ $.

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  • $\begingroup$ There is a gap in your proof, note that $k$ is a variable, in other words, it's not fixed, so it changes with respect to the primes $kn\pm 1$. The OP asked given a fixed $k\in \mathbb Z^+$ are there infinitely many consecutive primes s.t $k\mid p_{n+1}+p_n$ @Peter $\endgroup$
    – PNT
    May 26, 2021 at 13:42
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    $\begingroup$ You have not read the answer carefully. For every $k$ , we can find infinite many $n$ giving a suitable twin prime pair , if Schinzel's hypotehsis is true. $\endgroup$
    – Peter
    May 26, 2021 at 13:43
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    $\begingroup$ yeah... sorry you're right. @Peter $\endgroup$
    – PNT
    May 26, 2021 at 13:45
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    $\begingroup$ You've said that Schinzel's hypothesis is weaker than the twin prime conjecture, but it looks like a generalization of it? @Peter $\endgroup$
    – PNT
    May 26, 2021 at 13:49
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    $\begingroup$ Schinzel's hypothesis is far too strong. Dickson's conjecture suffices. $\endgroup$
    – arbashn
    Jul 13, 2021 at 15:31

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