Wetzlar, Germany, pensioner, e-mail: michusid@mail.ru Mykhaylo Khusid Representation of even number in the form of the sum of four simple.

Abstract: Harald Andres Helfgott finally solved 2013 a weak problem of Goldbach.

` [1] `

where at the left the sum of three prime numbers, on the right odd numbers, since 7 The author provides the proof in this work, being guided by the decision

weak problem of Goldbach that: p1 + p2 + p3 + p4 = 2N [2] where on the right sum of four prime numbers, at the left any even number,

since 12, by method of mathematical induction. Keywords: On this basis, solves two actual problems of the theory numbers Decision.

- 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.

2.Any even number starting with six is representable in the form of the sum of two

prime numbers. Goldbach-Euler’s hypothesis.

We will break [8] for two sums: p1 + p2 =2K [9] p3+ p4 =2K1 [10] [10] – possible even number. .

Let there is an even number which isn’t representable in the form of the sum of two the simple 2K2 . Then exists [8]: 2N2 = 2K2 +2K1 [11] 2K2 = 2K =p1 + p2 [12] that completely contradicts the assumption of existence of the even

numbers of two simple which aren’t provided in the form of the sum. They do not exist! 3.Thus we proved: Any even number since 6 is representable in the form of a bag of two odd

the simple. p1 + p2 =2N [13]

Any even number is representable in the form of the sum of two simple. In total

even numbers, without exception, since 6 are the sum of two prime numbers.

Goldbakha-Euler’s problem is true and proved! 4. On the basis above the proved we solve one more fundamental

task. 5. Any even number, since 14, is representable in the form of the sum of four

odd prime numbers, from which two twins. p1 + p2 + p3 + p4 = 2 N [14] Let p3, p4. . – prime numbers twins, then a difference of any even,

since 14, and the sums of twins too even number which agrees,

the proved Goldbach-Euler’s hypothesis it is equal to the sum of two simple. Further we will arrange prime numbers from left to right in decreasing order.

6.And in case even number, 2 N =2 p2 +2 p4 + 4 then p1 , p2 .

inevitably also twins. We will subtract [14] sum from both parts 2 p2 +2 p4 : p1− p2 + p3 − p4 = 4 [15] From [15], obviously, p1 , p2 . – inevitably twins. 7. Prime numbers of twins infinite set. Let their final number and last prime numbers twins p3, p4. . We will designate two prime large numbers than p3, p4. . as p1 , p2 . .

We will summarize all four prime numbers and then according to item punkt6 there is even number 2 N, at which inevitably big p1 , p2 . twins. And further substituting in the sum instead p3, p4. . of numerical values p1 , p2 . process becomes infinite.

Literature: 1. Wikipedia.