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 Pn so that k|Pn+Pn+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 Pk so that 3|Pk+Pk+1 .We can conclude there exists the largest prime number Pm so that 3|Pm+Pm−1 and thus, for every prime number Pn where n>m, we know that Pn+Pn+1 does

notdivide 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.

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