Allen Knutson

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133 pages!

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Busy day in analytic number theory; Harald Helfgott has complemented his previous paper http://arxiv.org/abs/1205.5252 (obtaining minor arc estimates for the odd Goldbach problem) with major arc estimates, thus finally obtaining an unconditional proof of the odd Goldbach conjecture that every odd number greater than five is the sum of three primes. (This improves upon a result of mine from last year http://terrytao.wordpress.com/2012/02/01/every-odd-integer-larger-than-1-is-the-sum-of-at-most-five-primes/ showing that such numbers are the sum of five or fewer primes, though at the cost of a significantly lengthier argument.) As with virtually all successful partial results on the Goldbach problem, the argument proceeds by the Hardy-Littlewood-Vinogradov circle method; the challenge is to make all the estimates completely effective and to optimise all parameters (which, among other things, requires a certain amount of computer-assisted computation). [EDIT: the proof also relies on extensive numerical verifications of GRH that were performed by David Platt.]

Abstract: The ternary Goldbach conjecture, or three-primes problem, asserts that every odd integer N greater than 5 is the sum of three primes. The present paper proves this conjecture. Both the ternary Goldbach conjecture and the binary, or strong, Goldbach conjecture had their origin in an ...

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Allen Knutson

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133 pages!

roux cody

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Just to be clear: there is little to no hope that this technique could be extended to prove the Goldbach conjecture right? Because of the "parity problem"?

Terence Tao

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Yes, the circle method is very unlikely to be able to settle the even Goldbach conjecture by itself; I have a post about this at http://terrytao.wordpress.com/2012/05/20/heuristic-limitations-of-the-circle-method/ . (And yes, the parity problem also blocks sieve-theoretic attacks on the even Goldbach problem.)

Why does the abstract in the sample that is displayed on G+ say that the ternary Goldbach conjecture asserts that every odd integer N is greater than **5**? That's not right is it?

Allen Knutson

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5 isn't the sum of exactly three primes.

Oh gotcha. That little odd word there changes things.

It's still strange though because the conjecture is stated differently in the actual abstract.

It's still strange though because the conjecture is stated differently in the actual abstract.

Eric Pouhier

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There must exist a much simpler proof

Noel Niles

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What three prime factors divide 7?

Noel Niles

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+Tatyana Beketova, you are right. I'm a moron. I completely misread that. Thanks.

+Allen Knutson , it says every odd integer GREATER than 5... So not 5.

+rishi sadhir That's what Allen said; "five _isn't _ the sum of exactly 3 primes."

Thomas Egense

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What a busy week for prime-number results! Just a few days ago it was the Twin-Prime conjecture that was improved radically.

Awesome!

this great mind.

Awesome!

cool

Vadim Lebedev

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+Noel Niles Maybe i'm really dumb and missing something but:

5 = 1 + 1 + 3

5 = 1 + 1 + 3

Tobias Mühlbauer

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+Vadim Lebedev ... but 1 is not prime

then 3+2=5 because both are prime

3,2 are just two numbers not three

oh, see i didnt read it all, only like, a few comments

+Vadim Lebedev : the conjecture holds true for odd numbers greater than 5 !

+Kaley Wood : 3+2=5 is true but we're talking about an odd number greater than 5 and can be represented as a sum of three prime numbers! in this case 3 and 2 is ok, but where is the third number !

+Kaley Wood : 3+2=5 is true but we're talking about an odd number greater than 5 and can be represented as a sum of three prime numbers! in this case 3 and 2 is ok, but where is the third number !

If you read the conjecture it states for numbers > 5 (I know I'm repeating others)

The ternary Goldbach conjecture, or three-primes problem, asserts that every odd integer N greater than 5 is the sum of three primes.

The ternary Goldbach conjecture, or three-primes problem, asserts that every odd integer N greater than 5 is the sum of three primes.

you need three digits... 3+2 is only two.

Leona Machado, I'm not sure whether you've read what I wrote, the conjecture states for integers greater than 5, the first prime being 7.

colorful

Martin Källström

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Fascinating how quickly a discussion about a mathematical discovery dilutes into a pillow fight over who misread who.

Congratulations, I thought this had already been proven as using in a pet project for few years now. Though not using numbers upto infinity, so I guess that may explain why.

You might also wish to do another paper and state:

"All even numbers over 5 are the sum of no more than four primes - see my previous paper and add +1 to each value" One is a funny prime too many. That would save a few tree's as well }->.

You might also wish to do another paper and state:

"All even numbers over 5 are the sum of no more than four primes - see my previous paper and add +1 to each value" One is a funny prime too many. That would save a few tree's as well }->.

Paul, 1 Isn't prime.

Congrats to +Harald Helfgott

+Allen Knutson 1,2,2

Khoa Nguyen

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Now I can easily prove every odd number greater than 7 is the sum of four primes ^_^

+Allen Knutson

Dude , never went full retard. the teory says "GREATER THAN 5 ".

Dude , never went full retard. the teory says "GREATER THAN 5 ".

+Allen Knutson That's why the conjecture says greater than 5

+Eric Pouhier there is. proof follows: all odd integers greater than 5 can be expressed as a sum of 3 primes. See Helfgott Theorem. QED. :)

All even numbers greater than 2 can be expressed as (2n + 1) + 3 where (2n+1) is an odd number ;).

+Thomas Egense

Can you please post the link to the new result on the twin prime conjecture that you referred to in your comment?

Can you please post the link to the new result on the twin prime conjecture that you referred to in your comment?

What I don’t understand is how one can check even using a supercomputer all odd numbers up to 10^30? Suppose one can check one trilion number per second. Then it would still take 10^21 seconds, or 3.17e+13 years…

+Bill Bertucci Parallel processing dude.

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That's awful! Three breakthrough of mathematics in this week from 2013-05-12 to 2013-05-18:

1. First proof that infinitely many prime numbers come in pairs.

2. Proof of the weak Goldbach Conjecture:Major arcs for Goldbach's theorem.

3. Prime abc Conjecture b== (a-1)/(2^c) put forward. http://oeis.org/A225759

1. First proof that infinitely many prime numbers come in pairs.

2. Proof of the weak Goldbach Conjecture:Major arcs for Goldbach's theorem.

3. Prime abc Conjecture b== (a-1)/(2^c) put forward. http://oeis.org/A225759

As always, thank you so much, Dr Tao, for being the go-to person regarding a disproportionate amount of mathematical stuff!

Yúber Ramiro

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congratulations Harald Andrés Helfgott ........ nos has dado algo mas que estudiar en colegios y universidades ......... AQP-Perú

Translate

Categorizing each positive integer by its divisibility type is very useful. Since 1 cannot be regarded unambiguously as either prime or composite, and allowing it to be both would violate the fundamental theorem of arithmetic, it is the lone member of a third (unnamed) divisibility type.

i have arived at the following proof and need your suggestions(sudarsh0706@gmail.com)whether it is right or not i cant believe it as the steps are too simple but still i thought posting it here ...

Let l =2q where q is any integer between 2 to infinity therefore l would be any even integer greater than 2. Let us assume that l is not equal to m+h where m and h are odd prime numbers

Let m = x+1 therefore here since m is odd x would be even

Similarly let h=y+1 where y would be even

m+h=(x+1)+(y+1) = x+y+2 . since here x y and 2 are even integers therefore m+h would be even

thus our assumption would be wrong and 2q=m+h i.e 2 (2……infinity)=m+h

Let l =2q where q is any integer between 2 to infinity therefore l would be any even integer greater than 2. Let us assume that l is not equal to m+h where m and h are odd prime numbers

Let m = x+1 therefore here since m is odd x would be even

Similarly let h=y+1 where y would be even

m+h=(x+1)+(y+1) = x+y+2 . since here x y and 2 are even integers therefore m+h would be even

thus our assumption would be wrong and 2q=m+h i.e 2 (2……infinity)=m+h

i have sum resultat of twin Goldbach conjecture if you have some time.

i'm in morocco

+Noel Niles 2+2+3

The limit is very likely to be 16. Here is why:

After doing quick study of the statistical distribution of prime gap numbers, it seems to me that the gaps are a geometric distribution, which makes a lot of sense. The probability of hitting a prime number seems to equal to 1/exp(exp(1)), or the inverse of e to the power of e. The inverse of the probability is 15.15426. Hence the smallest gap must be 16.

Regards,

Charlie

After doing quick study of the statistical distribution of prime gap numbers, it seems to me that the gaps are a geometric distribution, which makes a lot of sense. The probability of hitting a prime number seems to equal to 1/exp(exp(1)), or the inverse of e to the power of e. The inverse of the probability is 15.15426. Hence the smallest gap must be 16.

Regards,

Charlie

Read this +Chery Dang

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