# 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 $k$ such that there will be only a finite number of 2 consecutive primes whose sum is divisible by $k$.

Well, Although I couldn't get anywhere with 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.

• For any natural number $\ell \ge 3$ one expects that there are infinitely many pairs of consecutive primes with $p_k \equiv -1\pmod \ell$ and $p_{k+1}\equiv 1\pmod \ell$. Indeed one would expect infinitely many twin primes of this form. So the number that you want most likely does not exist. I don't think one can prove this yet. A lovely related result of Daniel Shiu (Strings of congruent primes) shows that there are infinitely many pairs of congruent primes lying in the same residue class $a\pmod q$ for any $(a,q)=1$. – Lucia Oct 19 '13 at 13:47
• In the previous comment on Shiu's result I meant to say "consecutive primes lying in the same residue class $a \pmod q$ ... " – Lucia Oct 19 '13 at 13:54
• In fact, for any integer $k$, one expects that $1/\phi(k)$ of the time, the sum of two consecutive primes will be divisible by $k$ - far from only finitely many. – Greg Martin Oct 19 '13 at 16:44
• So, if we say it happens $1/\phi(k)$ of the time its not sufficient to say there are infinitely many of them, is it? – CODE Oct 20 '13 at 15:07