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I was curious about the sum of two consecutive primes and after proving that the sum for the odd primes always has at least 3 prime divisors, I came up with this question:

Find the least natural number $k$ so that there will be only a finite number of $2$ consecutive primes whose sum is divisible by $k$.

Although I couldn't go anywhere on finding $k$, I could prove the number isn't $1, 2, 3, 4$ or $6$, just with proving there are infinitely many primes $P_n$ so that $k\mid(P_n+P_{n+1})$ and $k$ is one of $1, 2, 3, 4, 6$:

The cases of $k=1$ and $k=2$ are trivial. The case of $k=3$, I prove as follows:

Suppose there are only a finite number of primes $P_k$ so that $3\mid(P_k+P_{k+1})$. We can conclude there exists the largest prime number $P_m$ so that $3\mid(P_m+P_{m-1})$ and thus, for every prime number $P_n$ where $n>m$, we know that $P_n+P_{n+1}$ is not divisible by $3$. We also know that for every prime number $p$ larger than $3$ we have: $p \equiv 1 \pmod 3$ or $p \equiv 2 \pmod 3$. According to this we can say for every natural number $n>m$, we have either $P_n \equiv P_{n+1} \equiv 1 \pmod 3$ or $P_n \equiv P_{n+1} \equiv 2 \pmod 3$, because otherwise, $3\mid(P_n+P_{n+1})$ which is not true. Now according to Dirichlet's Theorem, we do have infinitely many primes congruent to $2$ or $1$, mod $3$. So our case can't be true because of it.

We can prove the case of $k=4$ and $k=6$ with the exact same method, but I couldn't find any other method for proving the result for other $k$.

VizDracViz
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CODE
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    this is a very nice question and it will seem weird if there exists such a $k$ – Konstantinos Gaitanas Oct 15 '13 at 22:17
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    Thanks, I myself think such a $k$ doesn't exist too, but proving it is another matter :P – CODE Oct 16 '13 at 16:27
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    I think Schinzel's Hypothesis says it's true though. – CODE Oct 17 '13 at 15:22
  • I also doubt that this is going to be answered.why don't you try posting this to http://mathoverflow.net/? – Konstantinos Gaitanas Oct 19 '13 at 09:15
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    @CODE: Indeed, Schinzel's Hypothesis H implies it. In fact you don't even need that much: Dickson's conjecture suffices, with some cleverness. – Charles Oct 24 '13 at 07:06
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    You can make a kind of a prime counting function that counts the pairs of consecutive primes divisible by $k$ below $n$. Maybe call it $\pi_k(n)$. After doing some computations it seems that $\pi_k(n) \sim c Li(n)$ for some constant $c$ which for a lot of $k$'s looks to be close to $1/\phi(k)$ – Alexander Vlasev Oct 24 '13 at 10:41
  • @ghosts not true, $12$ does not divide $103+107$, for instance. – Peter Oct 14 '15 at 11:44
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    A particular case of Dirichlet's Theorem is that there are infinitely many primes congruent to 1 mod 3 and infintely many congruent to 2 mod 3. But Dirichlet's Theoerem does NOT say anything about CONSECUTIVE primes. Your reasoning for k=3 is invalid. ... However it is a good Q, and, I suspect, with no known A. – DanielWainfleet Jan 07 '17 at 18:39
  • 3+ years and no A. Maybe put it on MathOverflow and see what the pros say. – DanielWainfleet Jan 07 '17 at 18:43
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    @user254665 my reasoning for k=3 says that if the hypothesis is false, from some point on we would have all prime numbers are either congruent to 2 mod 3 or 1 mod 3, THAT can't be true according to Dirichlet's Theorem. – CODE Mar 03 '17 at 10:29
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    My apologies. You are right. – DanielWainfleet Mar 04 '17 at 01:24
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    The question amounts to : does there exist a k, such that, only finitely many pairs of consecutive primes are additive inverses mod k. –  Feb 22 '19 at 17:26
  • @CODE has this been crossposted to Math Overflow? – RSpeciel Aug 13 '20 at 18:21
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    @RomainS It has, but no answer. – Daniel Fischer Aug 13 '20 at 18:43
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    The link to Yahoo Answers shut down today. Can you link elsewhere? – David Jonsson May 04 '21 at 12:49
  • Anecdotally, there are a plethora of numbers that never divide the sum of twin primes, which are consecutive primes: $11,13,19,31,43, \dots$. However, these numbers may well divide the sums of other consecutive primes. – Keith Backman Jul 10 '21 at 20:13
  • @CODE: I'm just curious, isn't $k=2$ the answer?, of course this is except for $2 + 3$. – Mathew Mahindaratne Aug 07 '21 at 15:28
  • @KeithBackman late response, but surely that can't be true: Schinzel's hypothesis H implies that for all $n$, there are infinitely many pairs of twin primes of the form $kn \pm 1$. For example, $11$ divides $197+199$, $13$ divides $311+313$, $19$ divides $227+229$, etc. – Ravi Fernando Nov 25 '22 at 05:29
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    OP, your first link goes to some weird Yahoo page. Please fix. Ty – Daniel Donnelly Jun 01 '23 at 07:13
  • Easily to check that each of the described sums for $n\in[2,5000];$ has one or more dividers among $\pi(k),; k\in[2, 999].$

    That's why the required k should be too big.

     – Yuri Negometyanov Jul 28 '23 at 13:31
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    This is equivalent to: Find the least natural number $k$ so that there is no pair of $2$ consecutive primes whose sum is divisible by $k$. Let say $k$ is the least natural number such there is only a finite number of pair of $2$ consecutive primes whose sum is divisible by $k$, and that $p+q=kn$ is the largest of those sums. Then no sum of $2$ consecutive primes is divisible by $k\left(n+1\right)$ ... – François Huppé Mar 03 '25 at 07:01
  • Can it be proved that $k=12$ is impossible to be the answer? – Tong Lingling May 02 '25 at 08:07

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