I'd like to know if there is some proof on this.

Suppose we have an even integer $n$, then the conjecture tells us that you can express it in the form of $n=2p+g$, where $p$ and $g$ are some prime and some prime gap $\geq0$(respectively). When saying prime gap, I'm talking about the distance between $p$ and another equal or greater prime, not just to the following prime.

Ex. $4=2\cdot2+0$(two times the prime $2$ plus the gap between $2$ and $2$), $6=2\cdot3+0$, $8=6+2$ ($6$ from $2\cdot 3$, and $2$ from the gap between $3$ and $5$).

Is this conjecture true?

The reversed version, proving that $2p+g\in 2\mathbb{N}$ is an easy step, just knowing that $g$ is a difference between $p$ and a greater prime, and hence even for all $p>2$, and that $2$ times an integer is always even, we can conclude that $2p+g$ is even, but the conjecture states the reverse of this, namely, to prove that every even number has this form.

  • 1
    $\begingroup$ I think a more interesting question (and stronger statement) would be "does every even number appear as a difference between two (not necessarily consecutive) primes?" $\endgroup$
    – Arthur
    Jan 10, 2018 at 11:47
  • $\begingroup$ @Arthur It well may be a more interesting question, but i have a proof that proving this conjecture directly proves the goldbach conjecture. $\endgroup$
    – Garmekain
    Jan 10, 2018 at 12:10
  • $\begingroup$ @Peter Just set a prime $q$ as $q=p+g$. $\endgroup$
    – Garmekain
    Jan 10, 2018 at 12:44
  • 1
    $\begingroup$ @Garmekain OK, now I see it. If there are primes $p$ and $q$ with $q-p=g$, then we have $2p+g=2p+q-p=p+q$ , so $n=2p+g$ implies $n=p+q$ $\endgroup$
    – Peter
    Jan 10, 2018 at 13:28
  • $\begingroup$ @Peter By definition, $g$ is the difference between $p$ and a greater or equal prime.(I edited the question) $\endgroup$
    – Garmekain
    Jan 10, 2018 at 13:36


You must log in to answer this question.

Browse other questions tagged .