michusid

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

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.

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

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)

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

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.

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.

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

Автор нашего вопроса и снимает свой вопрос с форума. 

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

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.

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

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.

15 minutes ago, swansont said:

!

Moderator Note

If you want to discuss something, post it here. (please see rule 2.7 in the guidelines) 

 

Is my proof correct?

finalengl.pdf

url deleted