# A question about the divisibility of sum of 2 consecutive primes.

Well as I was curious about the sum of $2$ consecutive primes, 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 which their sum is divisible by $k$.

Well, 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 $1, 2, 3, 4, 6$:
For $1$ and $2$ it is trivial. For $3$ I do the following:
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: $p \equiv 1 \bmod 3$ or $p \equiv 2 \bmod 3$. According to this we can say for every natural number $n>m$, we have either $P_n \equiv P_{n+1} \equiv 1 \bmod 3$ or $P_n \equiv P_{n+1} \equiv 2 \bmod 3$, 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 for $k=4$ and $6$ with the exact same method, but i couldn't find any other method for proving the case for other amounts of $k$. I would appreciate any help.

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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
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
I'm having trouble proving that such a $k$ doesn't exist, since little is known about consecutive primes. – user45878 Oct 17 '13 at 14:56
I think Schinzel's Hypothesis says it's true though. – CODE Oct 17 '13 at 15:22
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 $1/\phi(k)$ – Aleks Vlasev Oct 24 '13 at 10:41