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. It was added to the article by a "Bill Lawson" in 2003 [//en.wikipedia.org/w/index.php?title=Goldbach%27s_conjecture&diff=880110&oldid=880087] 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)

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)

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)


 * This formula can rather easily be corrected for becomimg a correct formulation of Goldbach's conjectire. This would certainly not be a "good alternative", since, for most readers, reading prose is much easier than deciphering a cryptic formula. D.Lazard (talk) 16:28, 28 March 2024 (UTC)


 * Why is the qualifier $$\cap(1,\infty)$$ in the statement?—Anita5192 (talk) 16:54, 28 March 2024 (UTC)
 * Because it is only for integers strictly bigger than 1, which isn't all of $$\mathbb{N}$$ Serouj2000 (talk) 21:11, 28 March 2024 (UTC)
 * This is probably the real open interval that is denoted $$(1,\infty).$$ Since the real numbers are out of scope here, "$$\cap(1,\infty)$$" should be replaced with "$$k>1 \implies$$". Also, there is a confusion between "$$/$$" (division operator) and "$$|$$" (separator). These are the two points that need to be fixed for getting a correct formula. This is the work needed to understand notation that I called deciphering. D.Lazard (talk) 21:14, 28 March 2024 (UTC)
 * $$k>1$$ can be understood as all numbers bigger than 1, when I just mean the integers Serouj2000 (talk) 21:17, 28 March 2024 (UTC)
 * Is this better then:
 * $$\forall{k}\in\mathbb{N}_{>1},\exists{n}\in\N|k-n\text{ and }k+n\text{ are prime}$$ Serouj2000 (talk) 21:29, 28 March 2024 (UTC)


 * What is the point of putting the conjecture into symbols?—Anita5192 (talk) 21:35, 28 March 2024 (UTC)
 * How about I ask you the same question but for theorems Serouj2000 (talk) 21:37, 28 March 2024 (UTC)
 * No version of this is better than a clear statement in mathematical English. People do not communicate mathematical ideas by this kind of cryptic formula, for good reason.  --JBL (talk) 23:24, 28 March 2024 (UTC)
 * "Every integer bigger than 1 is either prime or the midpoint between two primes" same statement but in "mathematical English". Serouj2000 (talk) 09:22, 29 March 2024 (UTC)
 * This is a correct formulation, except for an ambuiguity: for most readers, "midpoint" may suggest that the points are different. But the main problem is that this is not the standard way of stating the conjecture. So, people who have already heard of the conjecture may be confused by not recognizing the statement they know. Also, if you are unable to provide a WP:reliable source for this statement, this statement results from your own WP:original research, and including it would go against basic Wikipedia policies. D.Lazard (talk) 10:57, 29 March 2024 (UTC)
 * Whatever you'd call $$\dfrac{a+b}{2}$$ for integers a and b Serouj2000 (talk) 14:44, 29 March 2024 (UTC)
 * It's their [arithmetic] mean. But so what?  Is there anything more to this than noodling about (the very many) alternate ways that any mathematical statement can be expressed?  If so, could you please be more explicit about it?  And if not, could you please observe WP:NOTFORUM? --JBL (talk) 17:06, 29 March 2024 (UTC)
 * well, instead of taking a random even integer and hope you could find 2 primes that add up to it, you could just take any integer and prove it is the arithmetic mean of a pair of prime numbers Serouj2000 (talk) 20:16, 29 March 2024 (UTC)

Please review the following proof
Prime number is can expressed to " ab + 1 "

Even number is can expressed to " 2k + 1 + 1"

The sum of two prime number is expressed to

" (ab +1 ) + (ab +1) " = 2ab + 1 + 1

threfore,

Prime number ( 2k + 1 + 1) is equal to the sum of two prime number ( 2ab + 1 + 1). Dreamtexter (talk) 03:05, 11 May 2024 (UTC)


 * Your grammar is so poor, your proof is unreadable.—Anita5192 (talk) 03:14, 11 May 2024 (UTC)
 * Not to mention farcically wrong. (2k + 1 + 1) is an even number, not a prime. All this does is show the trivially obvious result that doubling a prime gives an even number. Doubling any integer gives an even number. Meters (talk) 04:13, 11 May 2024 (UTC)