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Re: Vindication of Goldbach's conjecture
Posted:
Jul 24, 2012 9:20 AM


On 22 juil, 04:14, Ben Bacarisse <ben.use...@bsb.me.uk> wrote: > mluttgens <lutt...@gmail.com> writes: > > On 21 juil, 15:32, Ben Bacarisse <ben.use...@bsb.me.uk> wrote: > >> lutt...@gmail.com writes: > > >> <snip> > > >> > Both terms of 6 = 3 + 3 are primes. > >> > I considered the case where at least one of the terms is not prime. > >> > Your example is irrelevant! > > >> Your claim is essentially the same as GC. I thought you'd missworded > >> it which is why I thought there was a counter example. Correctly worded > >> (as I think it is) proving it is equivalent to proving GC. > > >> <snip> > > >> >> ... counter examples may > >> >> be very hard to find, but that does not constitute a sound argument: you > >> >> can't prove X by noting that X follows from Y and challenging people to > >> >> disprove Y (but you know that, yes?). > > >> > No, you did not. > > >> What does that mean? Is it a comment on my remark about your "proof by > >> you can't contradict me" method? > > > Not at all.I was referring to some quibbling of you... > > Ah, then better to put it next to the quibble. In my opinion, the > comment about you proof structure ("look, you always get two primes if > you add and subtract some even number  show me a counter example") was > much more than a quibble. > > > > >> (By the way, can you get you newsreader to stop turning plain 7bit > >> characters into HTML entities?) > <snip> > > Sorry, the new Goggle interface was responsible. For that reason, > > I have just went back to the older interface. > > Thanks. Much better. > > <snip> > > > Proof of the validity of Goldbach's conjecture > > _______________________________________ > > > According to the conjecture, every even integer greater than 4 can be > > expressed as the sum of two primes. > > > Let?s consider the infinite series of uneven integers. > > Such series contains an infinite number of products p = ab, where a > > and b are primes. > > To each product p corresponds a single sum s = a + b, s being of > > course an even integer. > > This approach leads to all possible sums of two primes. > > There's no point to this preamble. It adds nothing to the discussion > and just looks like padding. > > > By the way, some even integers can be the sum of two uneven integers, > > at least one of them not being a prime. > > All even integers other than zero can be written as the sum of two odd > integers, at least one of them not being prime: 2k = 1 + (2k1). It > comes over as a bit odd to say "some" when you are stating an obvious > property of all numbers != 0. > > > This leads to the bold assumption, that one or more even numbers > > greater than 4 could not necessarily be expressed as the sum of 2 > > primes. > > I'd start the argument here... You don't need (or use) any of the > above. > > > A sum s of two primes a and b greater than 3 can always be written as > > s = (a + n) + (b  n) or s = (a ? n) + (b + n), where n is an even > > integer. > > The obtained terms (a +/ n) and/or (b /+ n) can be prime numbers, > > but being ordinary uneven numbers does not imply that an even integer > > cannot be a sum of two primes. > > Let?s notice that such method, which consists of adding or > > subtracting the successive elements of the series of even numbers n, > > can be applied for arbitrarily large sums s. > > It leads to all possible pairs of numbers: two primes, a prime and a > > uneven number, that is not a prime, or two uneven numbers, which are > > not prime. > > ...and you are assured of getting two primes for all s, only if GC is > true. > > > On the other hand, a sum s? of two uneven integers, where at least one > > of its terms is not prime, can be transformed into a sum s of primes > > by adding some even integer n to one of its terms and subtracting the > > same n from its other term. > > This statement needs a proof. If GC is true is it's obviously true; if > GC is false, it's false. > > > To determine n, it suffices to apply the above method to the sum s = > > s?. Then, one straightforwardly gets the value of n leading to the > > uneven terms of sum s?. > > The above is not a method of getting two primes  it's a method of > getting all pairs of odd numbers that sum to s. One of these will > always be a pair of primes only if GC is true. > > > > > > > Example: > > > s? = 13 + 15 = 28 (s? is not the sum of two primes). > > > From s = s? = 28, one gets > > s = 5+23 and 11+17, and also > > s = (5+8) + (238) = 13+15 = s? > > s = (11+2) + (172) = 13+15 = s? > > QED! > > > The assumption that one or more even numbers greater than 4 could not > > be expressed as the sum of 2 primes is thus refuted. > > > This leads to the conclusion that any even integer can indeed be > > expressed as the sum of two primes. > > > Marcel Luttgens > > > July 22, 2012 > > No need to date your posts. Usenet records the date of posting in the > header. > >  > Ben. > >
Thank you! You are of course right.
But my aim was to show that a sum s? = a + b of two uneven numbers, at least one of them not being a prime, could easily be transformed into a sum of two primes, simply by adding and subtracting some even number from its terms:
The chosen example was:
s? = 13 + 15 = (138) + (15+8) = 5+23 = (132) + (15+2) = 11+17
It has been claimed that such transformation could sometimes not be possible. I am wondering about which terms a and b should be chosen to justify that claim. Till now, I did not find a clue in the litterature, but you have perhaps a reference?
Marcel Luttgens



