Is the Goldbach Conjecture Finally Proven?

[praharmitra]

is there a proof for the goldbach's conjecture?

that the every number can be written as the sum of three primes...
or, every even integer can be written as the sum of two primes??

 

[Gokul43201]

Not yet.

 

[CRGreathouse]

I think Vinogradov proved that there is some (huge) n beyond which ternary Goldbach holds. I don't think we know even that in the binary case.

 

[dodo]

Correct me if I'm wrong, but there is no need to refer to the sum-of-three-primes case, since a proof of the sum-of-two would immediately provide a sum of three primes for any integer >= 6. Just subtract 2 or 3, depending on whether your number is even or odd respectively, and then use the sum-of-two on the resulting even number. Right?

 

[LukeD]

Sure, but the ternary case doesn't immediately imply the binary one (as far as I know), so they're not equivalent, and the ternary case could be easier to prove.

 

[CRGreathouse]

Correct me if I'm wrong, but there is no need to refer to the sum-of-three-primes case, since a proof of the sum-of-two would immediately provide a sum of three primes for any integer >= 6. Just subtract 2 or 3, depending on whether your number is even or odd respectively, and then use the sum-of-two on the resulting even number. Right?

The binary implies the ternary, that's why they're both considered versions of the same problem. But the binary version is much harder.

 

[metheor]

Conjecture: ANY EVEN NUMBER CAN BE REPRESENTED AS THE DIFFERENCE OF TWO PRIME NUMBERS.

The proof:
Set of Prime Numbers
P= { 3,5,7,11,13,17,19,23,.} = P(x)

Set of Even Numbers
E={2,4,6,8,10,12,..} =
E(x) = 2x ; x€ N+

Prime number function could be represented as P(x)
P(0) =2 ;
P(1) =3,P(2) =5 , P(3)=7,...
In more general way;
P(x+y) = P(x) + 2*k ; x,y,k € N+ and k >= y
For any k € N+ you can find an appropriate y and x .

Then, you can produce all the set of even numbers by giving the appropraite values to x,y and k .

k=1 then y must be equal to 1;y=1 and you can find x=1 .
for k = 2 then y must be equal to 1 or 2 and you can find x = 1 or 3 (or any appropriate number) .
for k = 3 , you can find x and y .
..
..
So, you have all the numbers in even number's set.
Then 'any even number can be represented as the difference of two prime numbers' is correct

 

[al-mahed]

For any k € N+ you can find an appropriate y and x .

prove it

 

[metheor]

if you want a more formal paper.It will take at least 10 pages.but you have to admit that my approximation is correct.
Think of that:
another function F(k) = 2*k = P(x+y) - P(x) x,y € N+ and x,y>=1 and k >=y
just give the numbers to x and y from 1 to infinity and you will have all the even number.

As I sad , if you want more formal proof , I can do it , too.
Don't worry.

 

[CRGreathouse]

As I sad , if you want more formal proof , I can do it , too.

Please do. The approximation is worthless, it's been done before (and better). A proof would be extremely impressive.

 

[Raphie]

Below is a "neat" little symmetrical construction that combines the ideas of prime sums to 2n (i.e. The Goldbach Conjecture) and prime differences equal to 2n (name for this, anyone?).

For first terms 29 & 31, a twin prime pair (it could be any twin prime pair...) and second term a prime not greater than 23, then...

29 + 01 = 30; 31 - 01 = 30
--------------------------------
29 + 03 = 32; 31 - 03 = 28
31 + 03 = 34; 29 - 03 = 26
29 + 07 = 36; 31 - 07 = 24
31 + 07 = 38; 29 - 07 = 22
29 + 11 = 40; 31 - 11 = 20
31 + 11 = 42; 29 - 11 = 18
31 + 13 = 44; 29 - 13 = 16
29 + 17 = 46; 31 - 17 = 14
29 + 19 = 48; 31 - 19 = 12
31 + 19 = 50; 29 - 19 = 10
29 + 23 = 52; 31 - 23 = 08
31 + 23 = 54; 29 - 23 = 06

Using this construction method, which involves nought but two sign changes, 30 + 1 becomes 30 - 1, and + becomes - (and which could be easily extended, but without "full coverage" of 2n), it is easy to see that all even numbers up to 372 (349 + 23) are constructible either as the sum or difference of a twin prime (upper or lower, depending) and 1 or another prime (the set of integers with less than two divisors). 374 (= 348 + 26) is not constructible in this (overlapping symmetrical) manner (note that 347 + 27 and 349 + 25 both include composite terms, thus the "not greater than 23" proviso...) because the next twin prime pair is (419, 421) more than 50 distant from 348. (although 421 - 47 = 374, 347 + 29 = 376, 347 + 31 = 378, 349 + 31 = 380, 419 - 37 = 382, 347 + 37 = 384, 349 + 37 = 386, 419 - 31 = 388, 347 + 43 = 390, 421 - 29 = 392, 347 + 47 = 394 and all other 2n at least to 486 are constructible as per above. Related sequence: A014574 Average of twin prime pairs http://oeis.org/A014574)

Hardly a "proof" of anything other than that, for z = the arithmetic average of twin primes, then all z + 2x (for x = 1--> 12) are constructible as the sum of two primes and all z - 2x are also constructible as the difference of two primes (for x = 1--> 12).

Admittedly, this example is rather trivial, and only addresses 2n +/- 24 from the arithmetic average of twin primes, but I do wonder if the general line of thinking could be, or already has been, extended by mathematicians, because, personally, I view sums and differences of primes to 2n as related, with the Goldbach Conjecture being but a special case of a more general and expansive theorem yet to be, to the best of my knowledge, properly formulated, much less proven.

Insights from the number theorists hereabouts more than welcome, especially insofar as anyone might lead me to other work that has been done in regards to prime differences equal to 2n...Raphie

P.S. I can't help but wonder how the Green-Tao theorem might fit into the overall jigsaw puzzle...

 

[CRGreathouse]

Below is a "neat" little symmetrical construction that combines the ideas of prime sums to 2n (i.e. The Goldbach Conjecture) and prime differences equal to 2n (name for this, anyone?).

Lacking a better name, I'd call them Sloane's A020483.

Insights from the number theorists hereabouts more than welcome, especially insofar as anyone might lead me to other work that has been done in regards to prime differences equal to 2n...

What kind of insight are you looking for?

 

[Raphie]

What kind of insight are you looking for?

First, CRGreathouse, in terms of interest and to provide a bit more context, let me note that I conceptually view the arithmetic average of twin primes (= z = 0') as "nodes" and/or "relative zeroes" along the integer number line. Insofar as this is the case, every step "back," relative to z, can be viewed (relatively speaking) as a negative number and every step "forward" can be viewed as a positive number. (Thus, in my thinking p/q --> z- --> -1', z --> 0', q/p --> z+ --> +1').

Framed in this manner, then p - q can be thought of (at least by me) as being on similar footing with p + q. So, for instance, one question I have that relates to this and merges p+ q (which has a name) with p - q (which, apparently, has a number :-)) is the following:

What is the first 2n not constructible as the sum or difference of a twin prime and another integer with fewer than or equal to 2 divisors?

In reference to what I posted previously, up to 486, 2n can be constructed from adjacent twin prime pairs, but for 488 this routine doesn't work because:

521 - 33 = 488
523 - 35 = 488
461 + 27 = 488
463 + 25 = 488

... and 33, 35, 25, 27 are all composite. But, for instance, 31 + 457 = 488 and 31 is a member of twin prime pair.

Where does this pattern of 2n constructibility using twin primes as one's reference point end? Or does it never end? Surely someone has asked this question before, no? So, in terms of "insights," then for starters:

Has this question already been asked and resolved by mathematicians?Raphie

P.S. The answer to the above, of course, is at least partly dependent on whether or not the twin prime conjecture is true.

 

[Raphie]

Related to recent discussion...
--------------------------------------------------------------------------------------------------
via Wikipedia...
POLIGNAC'S CONJECTURE
In number theory, Polignac's conjecture was made by Alphonse de Polignac in 1849 and states:

For any positive even number n, there are infinitely many prime gaps of size n. In other words: There are infinitely many cases of two consecutive prime numbers with difference n.

... For n = 2, it is the twin prime conjecture.
http://en.wikipedia.org/wiki/Polignac's_conjecture
--------------------------------------------------------------------------------------------------

 

[CRGreathouse]

What is the first 2n not constructible as the sum or difference of a twin prime and another integer with fewer than or equal to 2 divisors?

I assume here that "twin prime" means either the higher or the lower member, i.e., A001097.

It's clear that this conjecture is beyond the ability of modern mathematics to prove; it's harder than the twin prime conjecture. (Clearly it requires the twin prime conjecture: if there are only finitely many twin primes, then by the prime number theorem there is some number that is not the sum or difference of a twin prime and a prime or 1.)

But it is surely true. Just for kicks I verified it up to one hundred million.

 

[Raphie]

It's clear that this conjecture is beyond the ability of modern mathematics to prove; it's harder than the twin prime conjecture. (Clearly it requires the twin prime conjecture: if there are only finitely many twin primes, then by the prime number theorem there is some number that is not the sum or difference of a twin prime and a prime or 1.)

But it is surely true. Just for kicks I verified it up to one hundred million.

Seems this conjecture should have a name, if it doesn't already have one, as it is distinct from both the Polignac and Goldbach Conjectures...

 

[CRGreathouse]

Seems this conjecture should have a name, if it doesn't already have one, as it is distinct from both the Polignac and Goldbach Conjectures...

Well, then, start researching it to see who first asked it. Personally, I don't think it needs a name; I could make up hundreds of similar conjectures: "There are cofinitely many odd integers equal to the difference between a Sophie Germain prime and twice an odd prime", "Almost all numbers are the sum of (1) a square, (2) a square, (3) a prime number, and (4) one less than a power of two", etc.

 

[Raphie]

Well, then, start researching it to see who first asked it. Personally, I don't think it needs a name

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?"...

---------------------------
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

---------------------------
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
---------------------------

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?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]

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 .

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

Let p be a prime number.
p$^{2}$-1 = (p+1)(p-1)

Since p is prime, it is not divisible by 2 (unless p is 2 but 2$^{2}$-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$^{2}$-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$^{2}$-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]

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

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

 

[PAllen]

The binary implies the ternary, that's why they're both considered versions of the same problem. But the binary version is much harder.

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]

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.

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.

 

[PAllen]

Forget proof, you don't even posit any argument at all that an 'a' with right properties exists. Listing that it does for N cases is irrelevant, unless you list all infinity cases. Certainly, there is no induction in your (total non) argument.