Talk:Goldbach's conjecture

Goldbach equivalent to Lawson

The following argument should show that Lawson's conjecture is equivalent to Goldbach. Assume Goldbach, and let n be the given integer in Lawson. Then 2n is even, and there exists two primes p and q such that 2n=p+q. Assume p is less than or equal to q, and take l=(q-p)/2. Noting that n=(p+q)/2, observe that n-l=p, and n+l=q. Assume Lawson, and let 2n be the given even number in Goldbach. Then there is an l such that n-l=p and n+l=q are primes. Clearly, 2n=p+q. Also note that l need not be non zero, hence garyW's objection. If 2n=p+q and p is even, note that that requires q to be even, and 2n to be 4. Goldbach might be restated as, every even number greater than 4 is the sum of two odd primes, and Lawson might be given as every n larger than 3 has an l such that n-l and n+l are odd primes. — Preceding unsigned comment added by 67.5.135.253 (talk • contribs) 17:15, 24 April 2004‎

Proof: p,q odd primes ; n,l out of N with n>l

p*q = n²-l² = (n+l)(n-l) => p=(n-l) and q=(n+l)

=> (n+l)+(n-l) = 2n = p+q

For every (n+l) prime is (n-l) also prime. — Preceding unsigned comment added by 84.135.3.157 (talk) 13:09, 7 April 2014 (UTC)

You refer to Talk:Goldbach's conjecture/Archive 1#Lawson's Conjecture. It was added to the article by a "Bill Lawson" in 2003 [1] and quickly removed. I guess he named it after himself and I haven't found mention of it elsewhere. If the pair of primes is allowed to be two identical primes then it's trivially equivalent to Goldbach's conjecture as you show. It's a non-notable reformulation and shouldn't be mentioned in the article. PrimeHunter (talk) 13:50, 7 April 2014 (UTC)

Yeah and, you can have infinitely many equivalents, necessary conditions, etc. eg. Bertrand's postulate is a necessary condition for Goldbach, p+q=2n=n+n -> p-n=n-q, n>q>=3 implies 2n-3>=p>n . If n is prime, it implies that if 2n=n+n isn't the only partition of 2n into primes that an arithmetic of 3 primes exists, so disproving n+n being the only one for all but a finite number of cases, will be a base case for the Green-Tao Theorem. — Preceding unsigned comment added by 96.30.157.85 (talk) 23:11, 13 February 2019 (UTC)

Heuristic for primes adding to 2n₃

Only if no solution exist to:

$q_{2}-n_{2}=n_{2}-p_{2}=q_{1}-n_{1}=n_{1}-p_{1}=d$

where

$n_{1}+n_{2}=2n_{3}$ will no solution for $2n_{3}$ exist. Roderick MacPhee (talk) 16:06, 1 June 2023 (UTC)

Inconsistency

The article is not totally consistent about Helfgott's supposed proof. 2A00:23CC:B59B:6B01:E914:3ED4:E410:C8A5 (talk) 15:38, 9 September 2023 (UTC)

I have rewritten the paragraph for clarifying the history of this paper. This history suggests that the first proof was incomplete, that the referee was unable to check the details of the proof, and thus asked for a revision without rejecting the paper. The number of revisions suggests that the demand of the referee revealed to Helfgott a gap in the proof that he is unable to fill. D.Lazard (talk) 16:37, 9 September 2023 (UTC)

Moved here until peer review available

An asymptotic proof has now been claimed by Agama and Gensel using their method of circle of partitions

--Kmhkmh (talk) 04:48, 4 January 2024 (UTC)

References

Hmmm

$\forall {k}\in \mathbb {N} \cap (1,\infty ),\exists {n}\in \mathbb {N} /k-n{\text{ and }}k+n{\text{ are prime}}$ This looks like a good alternate formulation of this conjecture Serouj2000 (talk) 14:51, 28 March 2024 (UTC)