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Prove that the following conjecture is equivalent to the strong Goldbach conjecture:

Every integer $n>3$ is halfway between $2$ primes.

I'm able to prove it, but i don't have much experience in writing proofs, witch is why i need help to find a proper way to explain it. I'd like to have a proof that is as "short and sweet" as the conjecture itself. The shorter the better!

What i have so far:

If $p$ and $q$ are a Goldbach's partition of an even integer $2n$, then: $$ 2n=p+q $$ The midpoint between $p$ and $q$ is: $$\frac{p+q}{2}=\frac{2n}{2}=n$$
Therefore, if an even integer $2n$ can be written as the sum of $2$ primes, $n$ is halfway between those $2$ primes.

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  • $\begingroup$ If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer? $\endgroup$ – user334732 Jan 2 at 15:28
  • $\begingroup$ Hard to do this without seeing the proof you already have. $\endgroup$ – Randall Jan 2 at 15:28
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    $\begingroup$ If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you. $\endgroup$ – Arthur Jan 2 at 15:30
  • $\begingroup$ yes i'm adding what i have so far! should not be long $\endgroup$ – François Huppé Jan 2 at 15:31
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So, let's do the equivalence.

Say Goldbach's conjecture is true, and take an integer $n>3$. Then there are primes $p, q$ such that $p+q = 2n$, and therefore $n = \frac{p+q}2$ is the midpoint between $p$ and $q$.

On the other hand, let's say your conjecture is true, and let $2n>6$ be an even number. Then there are primes $p, q$ such that $n$ is the midpoint between $p$ and $q$. In other words, $\frac{p+q}2 = n$, which transforms into $p+q = 2n$, and we have shown that the arbitrary even number $2n$ is the sum of two primes.

Thus either conjecture may be used to prove the other, and they are equivalent.

(I'm assuming that the specifics of whether Goldbach's conjecture starts at $4$ or $6$ or $8$ isn't the important part of the conjecture. If you include those cases, then no, the two aren't entirely equivalent.)

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  • $\begingroup$ The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :) $\endgroup$ – François Huppé Jan 2 at 15:55
  • $\begingroup$ What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ? $\endgroup$ – François Huppé Jan 5 at 1:22

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