Yesterday I saw this question: A question about divisibility of sum of two consecutive primes (you should read the OP to understand the full problem), it just asks to prove that for all $k\in \mathbb Z^+$, there exist infinitely many consecutive primes such that :
$$k\mid p_{n+1}+p_n.$$
the guy who asked this took care of the cases where $k=1,2,3,4,6$. The general case where $k$ is any positive integer is beyond me, but I attempted to prove the case where $k=12$ and I wonder it the proof is correct:
Assume that the twin prime conjecture is true, which says that there exist infinitely many consecutive primes such that $$p_{n+1}-p_n=2$$
Since any prime $\ge 5$ is on the form $6k\pm 1$ and every pair of twin primes is on the form $(6k-1,6k+1)$, Hence $$p_{n+1}+p_n=6k+1+6k-1=12k$$
$$12\mid p_{n+1}+p_n$$
for infinitely many consecutive primes?
Note that the twin prime conjecture also implies the case where $k=1,2,3,4,6$, because $1,2,3,4,6\mid 12$
Best Answer
Assuming Schinzel's hypothesis , for every positive integer $k$ , there are infinite many positive integers $\ n\ $ , such that $\ kn-1\ $ and $\ kn+1\ $ are both prime. Those primes are obviously consecutive because they have difference $\ 2\ $ (in fact they are twin primes). The sum is $\ 2kn\ $ which is divisible by $\ k\ $.
Of course, Schinzel's hypothesis is much stronger than the twin prime conjecture, but at least this is some evidence that we can find infinite many pairs for every $\ k\ $.