# 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 $$kk$$ so that there will be only a finite
number of $$22$$ consecutive primes whose sum is divisible by $$kk$$.

Although I couldn’t go anywhere on finding $$kk$$, I could prove the number isn’t $$1,2,3,41, 2, 3, 4$$ or $$66$$, just with proving there are infinitely many primes $$PnP_n$$ so that $$k|Pn+Pn+1k|P_n+P_{n+1}$$ and $$kk$$ is one of $$1,2,3,4,61, 2, 3, 4, 6$$:

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

Suppose there are only a finite number of primes $$PkP_k$$ so that $$3|Pk+Pk+13|P_k+P_{k+1}$$ .We can conclude there exists the largest prime number $$PmP_m$$ so that $$3|Pm+Pm−13|P_m+P_{m-1}$$ and thus, for every prime number $$PnP_n$$ where $$n>mn>m$$, we know that $$Pn+Pn+1P_n+P_{n+1}$$ does not divide $$33$$. We also know that for every prime number $$pp$$ larger than $$33$$ we have: $$p \equiv 1 \pmod 3p \equiv 1 \pmod 3$$ or $$p \equiv 2 \pmod 3p \equiv 2 \pmod 3$$. According to this we can say for every natural number $$n>mn>m$$, we have either $$P_n \equiv P_{n+1} \equiv 1 \pmod 3P_n \equiv P_{n+1} \equiv 1 \pmod 3$$ or $$P_n \equiv P_{n+1} \equiv 2 \pmod 3P_n \equiv P_{n+1} \equiv 2 \pmod 3$$, because otherwise, $$3|P_n+P_{n+1}3|P_n+P_{n+1}$$ which is not true. Now according to Dirichlet’s Theorem, we do have infinitely many primes congruent to $$22$$ or $$11$$, mod $$33$$. So our case can’t be true because of it.

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