Is the set of primes which can be represented as sum of different primes infinite?

Question

Although everything is clear from a title of a question I will just add some notation and two simple examples.

If we denote the set of all primes which are sum of different primes as $P$ then, for example $29 \in P$ since $29=2+3+11+13$ but $11 \notin P$ since $11=2+2+7$ and $11=3+3+5$.

Is $P$ an infinite set?

Edit: As mentioned in the comments, if the twin-primes conjecture is true then also this question is answered in the affirmative but I firmly believe that this can be solved without first solving twin-primes conjecture because this is much weaker than twin-primes conjecture since here there are no restrictions on the number of summands, which equals two in the twin-prime conjecture.

Answer 1

Probably.

As several commenters have noted, the result would follow from the twin prime conjecture.

Improvements to the bound in Zhang's prime gap theorem say that there are infinitely many pairs of primes separated by at most $246$.

Unfortunately, you can't express all the positive integers less than $246$ as a sum of distinct primes. But there may be a way to use this bound to reduce the problem to a doable finite computation.

http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes

Answer 2

Yes, in fact something stronger holds: Every integer $n\ge12$ can be written as a nontrivial sum of distinct primes (nontrivial means here that at least 2 summands are involved).

This holds unconditionally (i.e., it is not dependent on the twin prime conjecture or any of its variants), and follows easily from Bertrand's postulate. See here for details.

It would be nice to strengthen the above so that there is an absolute bound on the number of required summands. But such a thing seems very close now to the twin prime conjecture itself.