goldbach conjecture proof

Raphie

First of all, Happy Thanksgiving everybody.

Second, a little bit of research to share with any who find the recent discussion to be of interest, and also to demonstrate the relevancy of the question: "Are all 2n constructible as either the sum or difference of a twin prime and another integer with less than or equal to 2 divisors?"...

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via Wolfram Mathworld
Twin Primes
It is conjectured that every even number is a sum of a pair of twin primes except a finite number of exceptions whose first few terms are 2, 4, 94, 96, 98, 400, 402, 404, 514, 516, 518, ... (Sloane's A007534; Wells 1986, p. 132).
http://mathworld.wolfram.com/TwinPrimes.html

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via OEIS
A007534 Even numbers which are not the sum of a pair of twin primes.

2, 4, 94, 96, 98, 400, 402, 404, 514, 516, 518, 784, 786, 788, 904, 906, 908, 1114, 1116, 1118, 1144, 1146, 1148, 1264, 1266, 1268, 1354, 1356, 1358, 3244, 3246, 3248, 4204, 4206, 4208

No other n < 10^9. - T. D. Noe, Apr 10 2007

REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 132.

EXTENSIONS
Conjectured to be complete (although if this were proved it would prove the "twin primes conjecture"!)
http://oeis.org/A007534
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Since CRGreathouse has already checked the question to 100,000,000, then the answer is "yes, at least to 10^9, and unknown thereafter.".

Thirdly, the question: "Are all 2n constructible as either the sum or difference of two twin primes?" might reasonably be referred to as "The Extended Wells Conjecture" (unless Wells was merely passing on someone else's conjecture, which I have not been able to determine...) The question I initially asked, which, in the manner it is presented, allows one to segment the even numbers into partitions of "length" equal to the gap between the arithmetic average of consecutive twin primes...

e.g.
======================================
For...

z_n = p +/- 1, where p is a twin prime
q' = 1 or a prime number

Then...

# of 2n to 1000 not constructible as:

z_(n) + q' or z_(n+1) - q' for 2n > z_(n) and 2n < z_(n+1)
(Twin Prime Pair "Gap" = 1)
= 8

488, 496 [=(31^2 + 31)/2], 686 [=26^2], 694, 724, 746, 776, 784 [=28^2]

# of 2n to 1000 not constructible as either:

z_(n) + q' or z_(n+1) - q' for 2n > z_(n) and 2n < z_(n+1)
z_(n) + q' or z_(n+2) - q' for 2n > z_(n) and 2n < z_(n+2)
(Twin Prime Pair "Gap" = 2)
= 0
======================================

... could be construed as a corollary to that, one I have playfully dubbed the "Qis-Qin + p Corollary" to "The Extended Wells Conjecture." "Qis-Qin" is pronounced "Kiss Kin," suggestive of twin primes, and the letters stand for "q*i^Squared + p" and "q*i^Null + p." Thus, -q + p and +q + p.

Important to note is that a) The "The Wells Conjecture" could be wrong, but "The Extended Wells Conjecture" could be right, and b) Both could be wrong, in which case we would, by default, be back at the original question: Are all 2n constructible as either the sum or difference of a twin prime and another integer with less than or equal to 2 divisors? In such a case, then "The Qis-Qin + p Corollary" (Personal Nickname: "The Kissing Kin Corollary") would become the "The Qis-Qin + p Conjecture," unless, of course, the question has already been asked and given a name by another or a better name were to be suggested.

Question: Does anybody have a copy of "The Penguin Dictionary of Curious and Interesting Numbers"? (1986) If so, would they be kind enough to take a look at p.132 and see if it was Wells who came up with the conjecture discussed in this post or if he was just reporting the findings of someone else?

Best,
Raphie

P.S. Fourthly, I want to make the point that the question I initially asked was based on a simple construction of 2n up to 488 (The lower bound on the answer to this question, yay or nay, with the assistance of CRGreathouse, T.D. Noe, digital computing and digital access to information has been upped by a factor of over 2 million in just a couple of days...). The point being that (I believe) one can ask meaningful mathematical questions, especially in the Information Age, based upon very small sample sizes, not incongruent, in spirit, with research conducted by Usability Expert Jacob Nielsen:

Jakob Nielsen's Alertbox, March 19, 2000
Why You Only Need to Test with 5 Users
http://www.useit.com/alertbox/20000319.html

P.P.S. @ CRGreathouse: Thanks for the nudge to research this further. I'd like to note that what I am calling "The Wells Conjecture," in essence, "predicts" both the truth of "The Goldbach Conjecture" and the truth of "The Twin Prime Conjecture," and it does so with a subset of the primes, the relative scarcity of which is determinable via Brun's Constant. If one were able to exploit some logical loophole (which I suggest only in the abstract, not as a likelihood) in order to prove it in isolation (as an existence theorem), then one would also, by extension, be proving two of the most famous mathematical conjectures. Such proof would also, I might add, answer the question I asked in the affirmative. I personally view such manner of conjecture as may or may not have originated with Wells, even if not the original question that led me to it, as worthy of having a name.

metheor

I can prove the goldbach conjecture with a simple method
but first I prove square of any prime -1 can be divided by 12
primexprime -1 = 12k k is an integer .

Boorglar

This is not really hard (and works only for primes greater than 3):

Let p be a prime number.
p[itex]^{2}[/itex]-1 = (p+1)(p-1)

Since p is prime, it is not divisible by 2 (unless p is 2 but 2[itex]^{2}[/itex]-1=3 is not 12k).
So p = 2n+1 for some positive integer n.

Any integer number can be written in the form 3k, 3k+1 or 3k+2.

Try n = 3k. Then p = 6k+1

Try n = 3k+1. Then p = 6k+3 = 3(2k+1) = 3m is impossible since p is prime. (unless k = 0 then p = 3 is prime, but 3[itex]^{2}[/itex]-1 = 8 is not 12k)

Try n = 3k+2. Then p = 6k+5.

We see that p must be of the form 6k+1 or 6k+5.
Then (p+1)(p-1) = (6k+2)(6k) = 12(3k+1)k = 12m
or (p+1)(p-1) = (6k+6)(6k+4) = 12(k+1)(3k+2) = 12m

Therefore, for any prime number p greater than 3, p[itex]^{2}[/itex]-1 is divisible by 12.

lostcauses10x

The complication of all of this: is proving there is no space between primes through infinity: such that for the binary part, the last prime before the spacing is such that it added to itself ( or times 2): is greater than or equal to the spacing been it and the next prime.

Of course this comes down to prime numbers distribution, and the maximum spacing between primes.

ode_to_joy

FYI, someone submitted a proof of Goldbach Conjecture and Riemann's Hypothesis in Arxiv. It's under General Mathematics section.

lostcauses10x

ode_to_joy
got a link? .
.

fedaykin

http://arxiv.org/ftp/math/papers/0005/0005185.pdf

Is this the paper you were talking about ode_to_joy?

ode_to_joy

http://arxiv.org/abs/1110.3465
goldbach conjecture

http://arxiv.org/abs/1110.2952
riemann hypothesis

micromass

On the first look, there seems to be some kind of mistake. He proves the Goldbach conjencture for big n. Thus if [itex]n\rightarrow \infty[/itex], then it is true. But he never states what "big" exactly is. He reasons that it has been numerically checked for n going until [itex]\sim 10^{18}[/itex] and that this large so it holds for larger integers. This is not enough.

PAllen

I'm confused. Suppose the ternary is true. Then, for an even integer, you have that it is the sum of 3 primes. For n>6, one of these must be 2, the other two being odd primes. So then, for every even n>6, the ternary implies n-2 satisfies the binary. Isn't that enough to say that ternary implies binary?

PAllen

According to:

http://mathworld.wolfram.com/GoldbachConjecture.html

what Vinogradov proved for sufficiently large odd n was the ternary form of odd numbers. Restricted to odd numbers only, the ternary does not imply the binary. However, it is correct, as I argued above, that the ternary for all numbers implies the binary; and the binary trivially implies the ternary.

raymo39

the proof of the riemann hypothesis is pretty interesting, although at first glance I'm not sure about one of his derivatives, and there seems to be ALOT of assumptions...

micromass

From the first glance, the proof of the Riemann hypothesis suffers from the same problem. He checks it for the first 23 trillion values and he proves it for sufficiently large numbers. That is not enough...

fibonacci235

I have a question concerning Goldbach's Conjecture. First, it is easy to demonstrate that the sum of any two odd integers will always be even.
For example,
let an odd integer q=2k+1 and an odd integer p=2m+1
It then follows that q+p= 2(k+m+1) = 2n which is even.

Now, it is true that any prime number >2 is odd. So, couldn't you simply use this fact to prove a substantial amount of Goldbach's conjecture? I don't understand people who sit down and add two arbitrary primes >2 to find an even number as the sum thinking that they may find an exception to the conjecture, but they won't because of what I just proved.

I know that prime numbers don't have a neat little general form like 2k+1 or 2n, but they are a subgroup of the odd numbers. All prime numbers >2 are odd but not all odd numbers are prime. Primes >2 are odd simply by definition of a prime number.

Then after you prove that the only time you will get an even number as a sum by using the prime number 2 is when you add 2 to itself. I'm probably missing some subtle logical step, and if so, please enlighten me.

In conclusion, wouldn't it be simpler to focus on parity instead of the primality of numbers to prove the conjecture?

dodo

Hi, fibonacci235,
there are many ways of adding two primes and obtaining an even number. The point of the conjecture is, can *all* even numbers (> 2) be written as the sum of two primes? In other words, given *any* arbitrary even number greater than 2, say, 123456, why is it mandatory that there exists a prime "p" such that 123456-p is also prime? That's the problem.

fibonacci235

Ok, that does make more sense. Now, I understand why this is so hard to prove.

Sievert

Goldbach Conjecture is 2n = Prime (a+n)+ Prime (a-n), 1 is here prime


2 = (0+1)+(1-0)

4 = (1+2)+(2-1)

6 = (2+3)+(3-2)

8 = (3+4)+(4-3)

10= (2+5)+(5-2)

12= (1+6)+(6-1)

14 =(4+7)+(7-4)

16 =(3+8)+(8-3)

18 =(4+9)+(9-4)

20=(3+10)+(10-3)

22=(6+11)+(11-6)
.
.
.
2n=(a+n)+(n-a)

Proof:

(a+n) = 2n+(a-n)=2n-(n-a)

q.e.d.