For every integer $n$, does $\{2n+p \mid| p \text{ prime }\}$ contain infinitely many primes?

Question

True or false: For every integer $n$, $\{2n + p \mid p \text{ prime} \}$ contains infinitely many primes?

A conjecture, a theorem or just false?

Answer 1

A sensible interpretation asks if:

• Every positive even number $2n$ is the difference of two primes infinitely often.

This is a weaker claim than Polignac's conjecture:

• Every positive even number $2n$ is the difference of two consecutive primes infinitely often.

because the latter asks about an infinite number of prime gaps of every size $2n$ (credit to @CODE for pointing out the article). That is, here we ask merely that $2n$ appears as the difference of two (positive) primes infinitely often, not necessarily as the difference of consecutive prime numbers. For the special case $n=1$ (twin prime conjecture) there is no distinction to be drawn, since for odd prime $p$ there cannot be a prime between $p$ and $p+2$.

Certainly the conjecture, even in this weakened form or even specialized to $n=1$, is unsettled. However there has been tremendous progress recently, such that for some $n \gt 0$ it is known that there are infinitely many prime pairs $p$ and $p+2n$. First Zhang Yitang announced a proof that for some $2n \lt 70,000,000$, there are infinitely many prime gaps of size $2n$. The upper bound on gap size has been reduced by James Maynard and others, so that currently we know that there are infinitely many prime gaps of size $2n$ for some $2n \le 270$. Terence Tao's blog is a particularly good place to find the latest breaking news in this area.

Abandoning the restriction to consecutive primes should make proving a result for all even numbers easier, but to date little progress has been made beyond:

Chen's Theorem II (1973) Every positive even number $2n$ is $m - p$ for infinitely many primes $p$ where $m = p+2n$ has at most two prime factors.