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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 $k$.

Noor
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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 – EffingLoveMath 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
  • According to the prime number theorem, the number of primes decreases as we move further along the number line. This implies that the number of consecutive primes also decreases as we go ahead. This could be a logic supporting the main answer that came to my mind. This is not the whole answer, but a logic that could help you proceed. Good luck solving it. – Atharv Rege Nov 25 '25 at 14:38

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