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The Solution for Goldbach Conjecture

The following is the solution for Goldbach's Conjecture, the oldest unsolved math problem around. Can anyone find any stupid mistake in it?

Title: Solution for Goldbach’s Conjecture

Author: Wilber Valgusbitkevyt

Abstract: For any given even number 2X, there exists prime numbers which can be noted as X – A. If X – A (mod Pq) =/= 2X for all q’s, then X + A is a prime number, and the sum of X – A and X + A is 2X for all even numbers bigger than 4. This is shown with Arbitrary Modular Arithmetic and Fermat’s Infinite Descent Method. Then, it is shown that the number of possible X – A such that guarantees X + A to be a prime number is at least 1 for any X >= 4. In other words, all even numbers can be represented as the sum of two odd prime numbers.

Proof: I am going to start by creating another theorem Victoria Hayanisel Theorem 4: For any positive integer X >= 4, there exists at least one prime number smaller than X that does not divide X. Although this is not used in my proof for Goldbach’s Conjecture, it is still a new theorem with a proof that helps describe the problem.

Suppose that X is the multiple of all prime numbers smaller than X. Then, X – 1 is a prime number. But if X – 1 is included in the set of the prime factors of X, then it cannot be X – 1 (obviously, some positive integer multiplied by X – 1 cannot be X). Hence, there is at least one prime number smaller than X that doesn’t divide X for all X >= 4.

Let’s get back to the proof for Goldbach Conjecture:

Consider all even numbers 2X such that X > = 4. For 2X when X = 2 and X = 3, 4 = 2 + 2 and 6 = 3 + 3 hence the conjecture is true. For 2X when X >= 4, consider Victoria Hayanisel Theorem 4. Consider the following arbitrary modular arithmetic. Consider e + 1 prime numbers less than X + A. Let:

X – A (mod P1) = J1 X (mod P1) = I1 A (mod P1) = I1 – J1

………………………… ………………………… …………………………

X – A (mod X - A) = 0 X (mod X - A) = 0 A (mod X – A) = A

……………………….. ……………………….. …………………………

X – A (mod Pe) = Je X (mod Pe) = Ie A (mod Pe) = Ie - Je

X + A (mod P1) = 2I1 – J1

…………………………

X + A (mod X - A) = 2A

………………………..

X + A (mod Pe) = 2Ie – Je

Now, the question is whether there exists X + A which is a prime number given some X – A which is a prime number, for some A at given X. In other words: Given j1, …, je are not 0, 2Iq – Jq =/= 0. In other words, 2Iq (mod Pq) =/= Jq.

Suppose 2Iq (mod Pq) = Jq. If this were true, 2Iq (mod Pq) = X – A since X – A (mod Pq) = Jq.

Let’s rearrange this equation: X – A (mod Pq) = 2Iq.

But, X (mod Pq) = Iq, which means 2X (mod Pq) = 2Iq.

Hence, X – A (mod Pq) = 2X.

Hence, if there exists X – A such that X – A (mod Pq) =/= 2X for all q, then Goldbach’s Conjecture is true.

The last piece of the puzzle: Is there such X – A for a given X such that X – A (mod Pq) =/= 2X for all q’s?

I am going to create a three-dimensional sequence called Victoria Hayanisel Sequence:

T 1 2 3 4 5 6 7 8 9 10 …
2 1 0 1 0 1 0 1 0 1 0 …
3 1 2 0 1 2 0 1 2 0 1 …
5 1 2 3 4 0 1 2 3 4 0 …
…………………………………………………………………………………………………………………………………… ……………
…………………………………………………………………………………………………………………………………… ……………
…………………………………………………………………………………………………………………………………… ……………

The first row is just the sequence of positive integers. The first column is the sequence of prime numbers. The other rows are the sequence of the remainders when the first column divides the first row.

Given i prime numbers, there are P1 x P2 x … x Pi number of combinations of the remainders. Given a specific set of remainders, there are (P1 – 1) x (P2 – 1) x … x (Pi – 1) number of different possible combination of remainders.

The last piece of the puzzle was “is there such X – A for a given X such that X – A (mod Pq) =/= 2X for all q’s?” The answer is yes, and there are (P1 – 1) x (P2 – 1) x … x (Pi – 1) number of them. Since 2X is an even number, we are interested in odd prime numbers, and odd prime numbers are always not divisible by 2, we can forget about 2. Hence, let’s consider only from 3. Now, the question is “is one of them a prime number?” The answer is yes. There are at least (P2 – 2) x … x (Pi – 2) of them (to be accurate we would exclude itself, but it is irrelevant for this as it makes no difference because it is still bigger than 1), which is always bigger than 1, because whatever the given set of remainders are, exclude them and 0’s from each rows. so that it would be prime because its remainders are set of non-zeros and each element is different from the given set of remainders.
Hence, given X – A (mod Pq) =/= 2X for all q’s, X + A has to be a prime number, and their sum is 2X for all X >= 4. For 2X when X = 2 and X = 3, 2X can be assigned 2 + 2 and 3 + 3 to it. For the rest, the proof is applicable. There exists at least one such X – A for all X. Hence, Goldbach’s Conjecture is true for all even numbers.

(edited 12 years ago)

Scroll to see replies

Reply 1

capital S

I think this question is more suitable placed at
http://math.stackexchange.com/
or
http://mathoverflow.net/

Reply 2

Jake22

Original post by capital S

I think this question is more suitable placed at
http://math.stackexchange.com/
or
http://mathoverflow.net/

Definately the latter - that is for research level questions of which this is not.

Without reading it all, it is fairly apparent that it uses no more than high school math (or first year undergrad - I forget when modular arithmetic is normally introduced)

It looks like a fairly typical crank proof.

Reply 3

Jake22

Lol. By the same author: http://vixra.org/pdf/1204.0012v2.pdf

And I quote: "The following solution is dedicated to Princess Eugenie of York of United Kingdom while waiting for her response to my marriage proposal letter."

Also: http://www.filmcrewpro.com/uk/view.php?uid=419541

(edited 12 years ago)

Reply 4

eugeniehayanisel

Mathoverflow is useless. They are not willing to accept any questions beyond their standard questions. They don't want to deal with research papers. I will try the later.

Reply 5

nuodai

Original post by eugeniehayanisel

Mathoverflow is useless. They are not willing to accept any questions beyond their standard questions. They don't want to deal with research papers. I will try the later.

This isn't a research paper. I'm sure if you thought you had a valid proof of Goldbach's conjecture then they'd be happy to look over it, but to them this is their equivalent of the "oh look, here's a proof of 1=0" threads we get on TSR.

Reply 6

around

Original post by eugeniehayanisel

odd prime numbers are always not divisible by 2, we can forget about 2. Hence, let’s consider only from 3.

this bit at least is true

Reply 7

Jake22

Original post by eugeniehayanisel

Mathoverflow is useless. They are not willing to accept any questions beyond their standard questions. They don't want to deal with research papers. I will try the later.

Are you a comedian?

If you are trying to be funny - I would push the more unusual aspects of mathematical/scientific crankery.

For instance, your comments on your proposal to Princess Eugenie were quite amusing so insert more of that.

Also, if you really want to make it look cool - I would make your crank papers more genuine looking i.e. make it so it takes a minute to spot that it is total bull**** rather than 2 seconds. For a start, you could produce your files with latex and for seconds you could quote some more genuine and non-high school theorems and so forth however unrelated to the problems you claim to solve are. I am assuming from a web search that you did at least part of a degree with significant maths content so it shouldn't be hard to do a bit better than this.

(edited 12 years ago)

Reply 8

nuodai

Original post by Jake22

That was brutal!

Reply 9

Jake22

Original post by nuodai

That was brutal!

Brutal?

I feel that it is much kinder to assume that this guy is someone having some kind of joke rather assume that they are a delusional person who believes that they have solved several major unsolved problems with high school maths and feels that unsolicited correspondence with minor royals proposing marriage is normal enough to report about anecdotally in an academic paper...

Reply 10

L'art pour l'art

Impressive the number of unsolved problems this genius solved! :biggrin:

(edited 12 years ago)

Reply 11

gff

Riemann's hypothesis is not there? I'm shocked! :biggrin:

Reply 12

nuodai

Original post by Jake22

Oh it wasn't necessarily a criticism. He's either joking or is sticking it to the mathematical community after dropping out of his degree. Or he actually thinks he's solved all of the above conjectures.

I suppose to come across as less of a douche I should find an error in the proof. It's very hard to read without being scripted though, and it's not worth any effort in the likely event that it's a joke.

(edited 12 years ago)

Reply 13

f1mad

Wow, you guys are taking it far...

Reply 14

gff

Original post by f1mad

Wow, you guys are taking it far...

All right.

Original post by eugeniehayanisel

Proof: I am going to start by creating another theorem Victoria Hayanisel Theorem 4: For any positive integer X >= 4, there exists at least one prime number smaller than X that does not divide X. Although this is not used in my proof for Goldbach’s Conjecture, it is still a new theorem with a proof that helps describe the problem.

Suppose that X is the multiple of all prime numbers smaller than X. Then, X – 1 is a prime number. But if X – 1 is included in the set of the prime factors of X, then it cannot be X – 1 (obviously, some positive integer multiplied by X – 1 cannot be X). Hence, there is at least one prime number smaller than X that doesn’t divide X for all X >= 4.

Although the author said he's not going to use it, two lines below says consider the following theorem... I would start with that.

Spoiler

However, assuming this is a proposition of yours (referring to author), you may want to look up Bertrand's Postulate (theorem).

Spoiler

Reply 15

Jake22

He followed my advice yesterday and latexed his document:

http://vixra.org/pdf/1204.0021v2.pdf

However, he also extracted some of the humourous references, i.e. theorems named after Fergie's kids.

Now it is just an average crank rather than an amusing one.

Reply 16

gff

Original post by Jake22

Now it is just an average crank rather than an amusing one.

It's still funny though, and he still puts effort to make his work readable. If he didn't care whether you get it or not, he wouldn't bother LaTeX-ing it. :tongue:

Original post by eugeniehayanisel

I am going to create a three-dimensional sequence called Victoria Hayanisel Sequence:

T 1 2 3 4 5 6 7 8 9 10 …
2 1 0 1 0 1 0 1 0 1 0 …
3 1 2 0 1 2 0 1 2 0 1 …
5 1 2 3 4 0 1 2 3 4 0 …

I looked at the PDF, but didn't get what is three-dimensional about this sequence.
It certainly makes sense, but don't see why it is so special. :biggrin:

I think there is also a need for an elaboration on those combinations of remainders stuff.

Few lines below, there is the following claim.
"Given a specific set of remainders, there are (P1 - 1) x ... x (Pi - 1) number of different possible combination of remainders."

Surely, given any set there is a certain number of possible permutations on it, and this number not need do anything with prime numbers.

Haha!!

I see no gold.

Was this not solved already?
Break 2k down into n + (2k-n) for all n<2k
have a look at all of them and look for two such that you can't change either from a line into a rectangle (pebbles work well for small k)
there they are! like every time!
QED