# Is every positive multiple of 6 the sum of two primes?

Hello,

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.

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How did this question arise? – Steven Landsburg Jan 24 '13 at 21:56
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. – zeb Jan 24 '13 at 21:56
I think it would be a major breakthrough to prove that every number divisible by $100!$ is a sum of two primes. – Gerry Myerson Jan 24 '13 at 22:02
@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. – Sylvain JULIEN Jan 24 '13 at 22:10
@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. – Gerry Myerson Jan 25 '13 at 4:13

## 1 Answer

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.

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Doesn't GRH imply that every odd number at least $7$ is the sum of $3$ primes? – Sylvain JULIEN Jan 24 '13 at 22:02
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 ... – Theo Johnson-Freyd Jan 24 '13 at 22:21
@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. – Will Sawin Jan 24 '13 at 22:53
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. – Aaron Meyerowitz Jan 25 '13 at 3:00
Patience, please... – H A Helfgott Feb 6 '13 at 18:37