Goldbach's conjecture states that every even integer greater than $3$ is the sum of two primes. I'm interested in a weaker assertion: has it been proven that every positive integer $n$ such that $6\vert n$ is the sum of two primes? Thanks in advance.

  • $\begingroup$ How did this question arise? $\endgroup$ – Steven Landsburg Jan 24 '13 at 21:56
  • 1
    $\begingroup$ I strongly doubt it. If someone had a proof of the weaker assertion, I can't see any plausible reason that the method wouldn't extend to Goldbach's conjecture itself. $\endgroup$ – zeb Jan 24 '13 at 21:56
  • 4
    $\begingroup$ I think it would be a major breakthrough to prove that every number divisible by $100!$ is a sum of two primes. $\endgroup$ – Gerry Myerson Jan 24 '13 at 22:02
  • $\begingroup$ @Steven Landsburg: the question arose tonight, when I came to consider what I call "n-symmetric sequences", which are, for any positive integer $n$ non multiple of $3$, finite sequences $(u_k)_{0<k\leq N}$ the first term of which is an integer coprime with $n$ and less than $n$, such that $u_{k+1):=u_{k}+6 \ \ mod \ \ n$ and such that $u_{N+1-k}=n-u_{k}$. See les-mathematiques.net/phorum/read.php?5,810982 if you read French. $\endgroup$ – Sylvain JULIEN Jan 24 '13 at 22:10
  • $\begingroup$ @Mahdi, you can find any number of claimed proofs of Goldbach, Riemann, P = NP, etc., etc., on the web and even on the arxiv. They aren't hoaxes, but that doesn't mean they are correct. People make mistakes. $\endgroup$ – Gerry Myerson Jan 25 '13 at 4:13

No. This would imply that every odd number at least $7$ is the sum of $3$ primes, since you can subtract $3$, $5$, or $7$ according to its residue mod 3. But that is not known. The strongest results known are that every sufficiently large odd number is the sum of $3$ primes, and that every odd number at least $11$ is the sum of $5$ primes.

  • $\begingroup$ Doesn't GRH imply that every odd number at least $7$ is the sum of $3$ primes? $\endgroup$ – Sylvain JULIEN Jan 24 '13 at 22:02
  • $\begingroup$ An aside, from someone who doesn't know any number theory: are there any estimates of what "sufficiently large" means in the quoted theorem? I mean, do we know any explicit upper bound on the number of cases left to check? Presumably, this number of cases, if known, is larger than the storage capacity of the universe ... $\endgroup$ – Theo Johnson-Freyd Jan 24 '13 at 22:21
  • 1
    $\begingroup$ @Sylvain JULIEN: I think so, ams.org/mathscinet-getitem?mr=1469323 @Theo: ams.org/mathscinet-getitem?mr=1932763 this was the best known a year ago, according to Terry Tao. Looks like a pretty large exponential. $\endgroup$ – Will Sawin Jan 24 '13 at 22:53
  • 1
    $\begingroup$ Specifically, without GRH the second paper lowers the upper bound of $\exp(\exp(11503)$ to the dramatically lower $e^{3100}.$ With GRH the bound can be further lowered to $10^{20} \approx e^{46}$ which is low enough that further theory and much computer time can finish the job. $\endgroup$ – Aaron Meyerowitz Jan 25 '13 at 3:00
  • 2
    $\begingroup$ Patience, please... $\endgroup$ – H A Helfgott Feb 6 '13 at 18:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.