michusid

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Everything posted by michusid

  1. [11] equality as [7] is the case in the designation of -speed. Thanks again.
  2. 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.
  3. http://new-idea.kulichki.net/pubfiles/190212200340.pdf
  4. 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)
  5. http://ru.math.wikia.com/wiki/Проблема_Гольдбаха
  6. 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.
  7. 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.
  8. 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
  9. Автор нашего вопроса и снимает свой вопрос с форума. The author found his mistake and removes his question from the forum.
  10. 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.
  11. [3] - [8] point 1, previous for example 10. Next 12
  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.