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The Goldbach conjecure is not yet proved. But, when an even number is represented as a sum of two primes, is there any knwon result about the difference of the two primes?

That is, if $2n$ is a sum of two primes, then can we choose primes $p$ and $q$ such that $2n = p+q$ and the differene $p-q$ is bounded by some formula about $n$?

It is clear that the difference is not bounded by a constant, because consecutive composite numbers can be made to be arbitrarily long.

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    $\begingroup$ As a first heuristic, if we start at $p = q = n$ and walk outwards, the expected time to hit primes would be $O(\log^2 n)$. So an upper bound is probably that plus another $\log \log n$ factor? $\endgroup$
    – Geoffrey Irving
    Commented Jun 12, 2023 at 9:38
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    $\begingroup$ Almost certainly there are no actually proven results in this direction. But one can ask for heuristics, or about what is known for most numbers (since we know Goldbach holds for most even numbers, I imagine results of this sort might be available). $\endgroup$
    – Wojowu
    Commented Jun 12, 2023 at 10:03
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    $\begingroup$ If you look at the ternary Goldbach problem, it is known that every sufficiently large odd number can be written as $p_1+p_2+p_3$ with $p_i = n/3 + O(n^{11/20+\epsilon})$. arxiv.org/abs/1610.02017 $\endgroup$
    – David E Speyer
    Commented Jul 16, 2023 at 11:08
  • $\begingroup$ I think there has been more (computational) work on how to make $p-q$ as big as possible, than as small as possible. $\endgroup$
    – Gerry Myerson
    Commented Jul 16, 2023 at 12:12
  • $\begingroup$ Relevant OEIS sequence: oeis.org/A047160, "a(n) = smallest number m >= 0 such that n-m and n+m are both primes" - I was hoping for a pointer into the literature, but nothing there. $\endgroup$
    – Michael Lugo
    Commented Nov 13, 2023 at 14:29

2 Answers 2

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The same method used to prove Chen's theorem produces the following:

Let $N$ be any even integer, then we have for sufficiently large $x$ that

\begin{aligned} \#\{p\le x:p+N&\text{ is a product of at most two primes}\} \\ &>0.67{x\over\log^2x}\prod_{\substack{p|N\\p>2}}{p-1\over p-2}\prod_{p>2}\left(1-{1\over(p-1)^2}\right). \end{aligned}

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    $\begingroup$ A result more along the OPs interest should be along the lines of, "for any sufficiently large $n$, we can write $n=p+q$ with $p$ prime and $q$ prime or semiprime such that $p-q=O(f(n))$" for some $f(n)$. Do you know what kind of result Chen's method can give in that direction? Can we at least get $f(n)=o(n)$? $\endgroup$
    – Wojowu
    Commented Jun 16, 2023 at 10:05
  • $\begingroup$ Let $p_n$ be $n$'th prime. Then it is known that $p_{n+1}-p_n=O(p_n^\theta)$ for some $\theta<1$. The value of $\theta$ depends on subconvexity estimates of $\zeta\left(\frac12+it\right)$. More can be found in the wiki page $\endgroup$
    – TravorLZH
    Commented Jun 16, 2023 at 13:57
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    $\begingroup$ If this was meant as a response to me - I don't see how that's relevant. Sure we have bounds on prime gaps but we want specifically pairs of primes which sum to the given $n$. $\endgroup$
    – Wojowu
    Commented Jun 16, 2023 at 15:20
  • $\begingroup$ For that case, it seems unlikely to obtain any global bound better than $O(n)$. As $p+3$ is always even whenever $p$ is odd. $\endgroup$
    – TravorLZH
    Commented Jun 18, 2023 at 2:41
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    $\begingroup$ Sure, there will be some pairs for which $p-q$ is large. The question is whether there will always be pairs for which it is small. $\endgroup$
    – Wojowu
    Commented Jun 18, 2023 at 17:09
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This is only a partial answer. Write $r_{0}(n):=\inf\{r\geq 0,(n-r,n+r)\in\mathbb{P}^{2}\}$, which is well defined whenever $n>1$ if you assume Goldbach's conjecture. Then if I'm not mistaken $r_{0}(n)$ is reached when you apply this algorithm recursively: $u_{0}(n):=0$ and $u_{k+1}(n):=u_{k}(n)+2^{1_{2\not\mid n}}.3^{1_{3\not\mid n}}\frac{(\Omega(n-u_{k}(n))+\Omega(n+u_{k}(n)-2))+\vert\Omega(n-u_{k}(n))-\Omega(n+u_{k}(n))\vert}{2}$ and the latter fraction equals $0$. If you can prove rigorously that this algorithm stops in $O((\log n)^2)$ steps, you're done.

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