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 $k|P_n+P_{n+1}$ 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 $3|P_k+P_{k+1}$ .We can conclude there exists the largest prime number $P_m$ so that $3|P_m+P_{m-1}$ and thus, for every prime number $P_n$ where $n>m$, we know that $P_n+P_{n+1}$ does not divide $3$. 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, $3|P_n+P_{n+1}$ 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 $k$.

  • 2
    $\begingroup$ Thanks, I myself think such a $k$ doesn't exist too, but proving it is another matter :P $\endgroup$
    – CODE
    Oct 16, 2013 at 16:27
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    $\begingroup$ @CODE: Indeed, Schinzel's Hypothesis H implies it. In fact you don't even need that much: Dickson's conjecture suffices, with some cleverness. $\endgroup$
    – Charles
    Oct 24, 2013 at 7:06
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    $\begingroup$ 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 $1/\phi(k)$ $\endgroup$ Oct 24, 2013 at 10:41
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    $\begingroup$ @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. $\endgroup$
    – CODE
    Mar 3, 2017 at 10:29
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    $\begingroup$ The question amounts to : does there exist a k, such that, only finitely many pairs of consecutive primes are additive inverses mod k. $\endgroup$
    – user645636
    Feb 22, 2019 at 17:26


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