Goldbach's Conjecture - Proof (better link)

rofler

Then you must prove that he cannot create a counterexample.

So first and foremost, if you are using the fact that his counterexample does not work because it is larger than n^2, then you must then prove the following statement for your proof to be valid:

We cannot have n consecutive numbers between 1 and n^2 so that each is divisible by at least one of 3,5,...,n.

After that, the other points of the proof can be discussed.

Cheers,

Rofler

CRGreathouse

Do you seriously think that my lack of a counterexample to Goldbach's Conjecture means that you have proved it?

Your proof is wrong for the reasons I have outlined. I have given a counterexample to your method, which is the best I can do since I don't have a counterexample to Goldbach's conjecture itself.

And you still haven't even suggested a reason that M <= sqrt(2N).

Andy Lee

I never suggested that your lack of a counter example means I proved Goldbach!

You attempted an example to counter a particular element of my proof, in an effort to demonstrate the proof as invalid. You were unable to do so. So it is that the proof was not destroyed by your example. Your example has not verified the proof true or false.

CRGreathouse

Actually, this wouldn't be enough to make his proof work. But a proof of this statement (as quoted, or replacing "numbers" with "odd numbers" as Andy is wont to do) would be of interest to me.

Sloane's A049300 seems relevant, but sadly doesn't list any papers.

CRGreathouse

Wrong. Reread what I wrote.

rofler

In the proof, you explicitly use the theorem that there is no string of n consecutive numbers below n^2 so that they are all divisible by at least one of 3,5,...,n, and you must prove this, or it will not be valid.

Furthermore, even if CRGreathouse cannot construct a counterexample (although I think he did construct a counterexample to your original proof), you have given no reason for one not to exist, so the proof is not "complete".

Cheers,

Rofler

CRGreathouse

A counterexample to that would be a counterexample to the full GC. I can't provide that.

Andy Lee

I know. So it will be necessary to show error in the proof by some other fashion (or accept it).

Andy Lee

In fact, I did not use this conjecture as part of the proof. Rather I suggest it is a result that has been proved (though unintentionally) by my proof of GC.

CRGreathouse

I've already done that in posts #6 and #20.

In post #29 you agreed that I found a hole. In post #31 you attempt to fix it, but your fix doesn't work. I suggest carefully studying the transition from {3} to {3, 5} to {3, 5, 7} in order to see why this is.

For reference, the first mistake in that post is
though this may perhaps be rescued. The first serious mistake is

which, unsurprisingly, is not justified in any way.

Andy Lee

I defined M as the maximum divisor in the process.

Take my example in post 26 where N=37. There is no reason to extend the
iteration beyond sqrt(74). Sundaram's sieve is illustrative...

9, 15, 21, ....
25, 35, 45, ....
49, 63, 77, ...
p^2+2px gives all odd composites.

CRGreathouse

I didn't ask what it was -- though I'll admit that even after that explanation I don't know what you mean. I was asking how you knew that it was less than sqrt(2N).

Your claim is that there are at least (1/sqrt(2N))*((N-M)/2)) Goldbach pairs for 2N. This is really the only context in which I care about M. If M is defined in a way that makes it obvious that it is less than sqrt(2N), then I want to know how you came to this statement; if it's defined in a way that makes this statement obvious, I want to know how you can bound it below sqrt(2N).

Andy Lee

Sorry, I realize now from your post 27 in which you said:
"In fact, out of N consecutive odd numbers, as few as (N - 2)/3 might remain."

Your N was my (N-1)/2, so your (1/3)(N-2) becomes (1/3)((N-5)/2) for me.
It seems I was in fact too generous allowing for M to be the maximum divisor in the process, when 5 will do just fine.

I believe then my proof shows there are at least (1/sqrt(2N))*((N-5)/2)) Goldbach pairs for 2N. This is greater than 1 for N>=17.

CRGreathouse

Ah. Then you entirely misunderstood. My bound (which is tight) was only for divisibility by 3 for n or 2N - n. It's worse for divisibility for 3 and 5, and worse yet for divisibility by 3, 5, and 7.

Andy Lee

Yes. Thanks for your time on this.