Math Forum Discussions

mluttgens

Posts: 70
Registered: 3/3/11

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

Plain Text

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 miss-worded
> >> 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 7-bit
> >> 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 pre-amble. 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 + (2k-1). 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) + (23-8) = 13+15 = s?
> > s = (11+2) + (17-2) = 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 = (13-8) + (15+8) = 5+23
= (13-2) + (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