# A question about divisibility of sum of two consecutive primes

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 $$k$$.

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