# Goldbach conjecture: Every integer $n>3$ is halfway between $2$ primes.

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.

• If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer? – user334732 Jan 2 at 15:28
• Hard to do this without seeing the proof you already have. – Randall Jan 2 at 15:28
• 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. – Arthur Jan 2 at 15:30
• yes i'm adding what i have so far! should not be long – François Huppé Jan 2 at 15:31

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.
(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.)