Assume Goldbach's conjecture. Then for every integer $n$ greater than one there exists a non negative integer $r$ such that both $n-r$ and $n+r$ are prime numbers. For a given $n$, let's denote $r_{0}(n)$ the smallest of these numbers $r$. Following the ideas developped in About Goldbach's conjecture, my question is: which non trivial upper bound can be otained for $r_{0}(n)$ through the Chinese remainder theorem?

Actually, I conjecture that $r_{0}(n)=O(\log^{2}n)$ (it's worth noticing that this would imply Cramer's conjecture) but several experts told me it was very unlikely that we would be able to establish such an upper bound in the near future. Nevertheless, maybe a weaker but still interesting bound can be obtained through a clever analysis. Thanks in advance for any insight.

  • 2
    $\begingroup$ I don't see any way to get such an upper bound other than the trivial $r_0(n)\le n-3$. More precisely, the Chinese remainder theorem alone certainly doesn't provide such an upper bound - because if it did, we'd've solved Goldbach's conjecture. Adding in the assumption that Goldbach's conjecture is true doesn't seem to help anything here - without knowing anything more about the hypothesized proof or a quantitative version of it, we don't have any more tools to attack the $r_0(n)$ problem. $\endgroup$ – Greg Martin Nov 29 '13 at 21:48
  • $\begingroup$ I only assumed Goldbach's conjecture to give a rather simple definition of $r_{0}(n)$. Otherwise one can define it unconditionally as the smallest "potential typical primality radius" of $n$, but this requires to know quite a bit of the approach mentionned in mathoverflow.net/questions/61842/about-goldbachs-conjecture. I expected the Chinese remainder theorem to be rather useful as one can use it in a constructive way to solve a given system of congruences through the algorithm mentionned in Olivier Bordellès, Arithmetic Tales, Springer, p.39-40. $\endgroup$ – Sylvain JULIEN Nov 29 '13 at 22:00
  • 1
    $\begingroup$ I believe I understand. I think the fact that the Chinese remainder theorem cannot be used to give any nontrivial information about $r_0(n)$ is a manifestation of the phenomenon "information about divisibility by small primes gives some information about numbers being prime, but not enough information". In other words, the prime numbers are not fully characterized by (finite) systems of congruents. This is the same phenomenon that would lead one to define "potential typical primality radius" in the first place. $\endgroup$ – Greg Martin Nov 29 '13 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.