Is my proof correct?

[michusid]

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[swansont]

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[michusid]

15 minutes ago, swansont said:

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Is my proof correct?

finalengl.pdf

[taeto]

How can you take \(p_2\) and \(p_4\) from [8] to use in [9]?

[michusid]

 

since it is proved in paragraph 1 that 2N can have any value. I can choose 2p2 + 2p4 +2. This is the sum of two adjacent ones. adjacent even numbers.

[taeto]

1 hour ago, michusid said:

since it is proved in paragraph 1 that 2N can have any value. I can choose 2p2 + 2p4 +2. This is the sum of two adjacent ones. adjacent even numbers.

Very well then. But your paragraphs are not numbered. Which one of your paragraphs do you number by 1, what does it say?

In [8] what is N? You did not specify what N means. 

If \(2N_1\) is the "previous even number" (what does that mean?) and [8] says that \(p_1+p_2+p_3+p_4 = 2N\) holds, then why is \(p_2+p_4=2N_1\) true?

[michusid]

[3] - [8] point 1, previous for example 10. Next 12

[taeto]

Again, what is your paragraph number 1, please explain what it says.

If \(2N = 12,\) and \(p_1=p_2=p_3=p_4=3,\) then if \(2N_1=10\) is the previous even number 10 before 12, then \(p_2+p_4 = 3 + 3 = 6 \neq 2N_1  = 10.\)  So why do you say that \(p_2+p_4=2N_1?\) 

 

[studiot]

 

You are doing a grand job checking the detail.

Edited February 7, 2019 by studiot

[michusid]

1. For the first even number 12 = 3+3+3+3.
 
We allow justice for the previous N> 5:                        p1 + p2 + p3 + p4 =2N                               [3]      We will add to both parts on 1                  p1 + p2 + p3 + p4 +1 =2N +1                          [4]       where on the right the odd number also agrees [1]        p1 + p2 + p3 + p4 +1= p5 + p6 + p7                         [5]
      Having added to both parts still on 1 

          p1 + p2 + p3 + p4 + 2= p5 + p6 + p7 +1               [6]          We will unite    p6 + p7 +1    again we have some odd number,
 
which according to [1] we replace with the sum of three simple and as a result we receive:            p1 + p2 + p3 + p4 + 2= p5 + p6 + p7 + p8              [7]   at the left the following even number is relative [3], and on the right the sum
 
four prime numbers.           p1 + p2 + p3 + p4 = 2N                                          [8]
 
Thus obvious performance of an inductive mathematical method. As was to be shown.

This is item 1. Previous 12 next14, previous 14 next 16 i mat. induction.

[taeto]

Thank you: I see now which part of the explanation you refer to as "paragraph 1". The confusion comes from the fact that it does not appear to explain the things that come later.

My question remains the same. In [8] you say \(p_1+p_2+p_3+p_4 = 2N. \) And (skipping the intermediate [9]) in [10] you say in effect \( N-1 = p_2+p_4.\) Are the primes not supposed to be odd? What if \(N\) is an even number?

[michusid]

Автор нашего вопроса и снимает свой вопрос с форума. 

The author found his mistake and removes his question from the forum.

[michusid]

The author proposes another version of the proof, as I found an error in the previously laid out one.

http://new-idea.kulichki.net/pubfiles/190210115350.pdf

[taeto]

The author is still being very careless.

[1] is taken to mean the version of the weak Goldbach problem in which every odd number \(\geq 7\) is a sum of three primes. This holds for the number 7, by writing 7 as the sum \(2+2+3.\) Sometimes you have to use the even prime 2. Now in equation [5] some of the primes may be even. Therefore it is wrong to say that \(p_6+p_7+1\) is an odd number. And it is also wrong to say that it is at least as large as 7.

To get to [7] from [6] is possible if and only if \(p_8=1.\) This is false however. Similarly it is possible to deduce [11] from [10] if and only if \(p_7+p_8=2.\) This is absurd.

The logic is very flawed. It is very easy to find primes \(p_1,\ldots, p_6\) for which [9] is true. So it is meaningless to try to deduce a contradiction from [9] alone.

[michusid]

The prime even number 2.4 is already composite. 1 is not a prime number. It is not defined. The sum of three simple odd numbers starting from 7. The sum of two simple two does not include 2 otherwise
get an even number except
  2 + 2 = 4
It does not interest us and does not affect
on the proof.

[michusid]

Goldbach's weak problem is formulated as follows:

Each odd number greater than 7 can be represented as a sum of three odd prime numbers.

[taeto]

3 hours ago, michusid said:

Goldbach's weak problem is formulated as follows:

Each odd number greater than 7 can be represented as a sum of three odd prime numbers.

Why would you say that? The equation [1] clearly talks about the weaker conjecture, which does not require odd primes. Or am I looking at the wrong version of the paper now? Please replace it by the correct version.

[michusid]

http://ru.math.wikia.com/wiki/Проблема_Гольдбаха

[taeto]

5 hours ago, michusid said:

http://ru.math.wikia.com/wiki/Проблема_Гольдбаха

But the author of http://new-idea.kulichki.net/pubfiles/190210115350.pdf chose the equivalent wording: Each odd number greater than 5 can be represented as a sum of three primes.

Are you discussing this paper, or a different paper?

 

[michusid]

It is known that the weak problem of Goldbach is finally solved.
 [one]
where on the left is the sum of three prime numbers
more than 7(you are right in the text
  mistake)

[michusid]

http://new-idea.kulichki.net/pubfiles/190212200340.pdf

[taeto]

When we compare [10] and [11] it only makes sense if we assume \(p_7+p_8=2.\)

 

[michusid]

 

Thank . You really are right [11] should be p5 + p6 + 1 +1 = p7 + p8 + p9 +1 = p9 + p10 + p11 + p12.
 Here it is a matter of renaming. From the side may not understand. It is necessary to correct.

 

[taeto]

Not quite. In [9], [10] and [11] we have inequalities \(\neq.\) 

[michusid]

[11] equality as [7] is the case in the designation of -speed.
Thanks again.