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Mathematics, theory of numbers, Wetzlar, Germany, pensioner, e-mail: michusid@meta.ua Mykhaylo Khusid Two problems of number theory and additions. Abstract: The Goldbach-Euler binary problem is formulated as follows: Any even number, starting from 4, can be represented as the sum of two primes. The ternary Goldbach problem is formulated as follows: Every odd number greater than 7 can be represented as the sum of three odd primes, which was finally solved in 2013. The second problem is about the infinity of twin primes. The author carries out the proof by the methods of elementary number theory. Keywords: solving , problems, in ,number, theory 1. Introduction, literature review and scope of work. In 1742, the Prussian mathematician Christian Goldbach sent a letter to Leonard Euler, in which he made the following conjecture: Every odd number greater than 5 can be represented as the sum of three prime numbers. Euler became interested in the problem and put forward a stronger conjecture: Every even number greater than two can be represented as the sum of two prime numbers. The first statement is called the ternary Goldbach problem, the second the binary Goldbach problem (or Euler problem). As of July 2008, Goldbach's binary conjecture has been tested for all even numbers not exceeding 1.2×1018[2]. The binary Goldbach conjecture can be reformulated as statement about the unsolvability of a Diophantine equation of the 4th degree some special kind.[3][4] The question of whether there are infinitely many twin primes has been one of the most open questions in theory numbers for many years. This is the content of the twin prime conjecture, which states that there are infinitely many primes p such that that p+2 is also prime. In 1849 de Polignac advanced more the general conjecture that for every natural number k there exists infinitely many primes p such that p + 2k is also simple. [5] The case k = 1 of the de Polignac conjecture is the conjecture about twin prime numbers. 2. Content (main part) 2.1.Lemma. Any even number starting from 12 is representable as a sum four odd prime numbers. 1. For the first even number 12 = 3+3+3+3. We allow justice for the previous N> 5: p1 + p2 + p3 + p4 =2N (1) We will add to both parts on 1 p1 + p2 + p3 + p4 + 1 = 2N + 1 (2) where on the right the odd number also agrees p1 + p2 + p3 + p4 + 1 = p5 + p6 + p7 (3) Having added to both parts still on 1 p1 + p2 + p3 + p4 + 2 = p5 + p6 + p7 + 1 (4) 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 (5) at the left the following even number is relative (1), and on the right the sum four prime numbers. p1 + p2 + p3 + p4 = 2N (6) Thus obvious performance of an inductive mathematical method. As was to be shown. 2.2.Corollary1: Possible value of one of the four primes odd equals in the sum of 2N from 3 to 2N-9 inclusive. This follows from the finally solved Goldbach's ternary problem. 2.3.Corollary2: The difference between the sums of two simple odd numbers is any even, starting from 2. Proof. Let's prove that: p1+p2−( p3+p4 )=2N (7) by mathematical induction: ( p1+p2 )−( p3+ p4 )+2=( p5+p6 )−( p7+ p8 ) (8) ( p1+p2 )+( p7+ p8 )+2=( p5+ p6 )+( p3+p4 ) (9) By Lemma : ( p1+p2 )+( p7+ p8 )+2=2N1+2 (10) ( p5+p6 )+( p3+ p4 )=2N2 (11) and with N2=N1+1 (7) it is true starting from , from 14 , N=7,8... ∞ Up to 14 we confirm arithmetically. 2.4.Theorem. There is no even number starting from 12 representable as the sum of two odd prime numbers. Let's say: p1+p2≠2K1 , 2( p1+ p2 )≠4K1 (12) but: p1+p3=2K2 (13) p2+ p4=2K3 (14) p1+p5=2K4 (15) p2+ p6=2K5 (16) add the left and right sides of (13),(14),(15),(16) and subtract (12). p3+p4+ p5+ p6≠2K2+2K3+2K4+2K5−4K1 (17) 2K2+2K3+2K4+2K5−4K1=2N (18) Thus the sum of any four odd prime numbers is not equal to 2N, which contradicts Lemma and means (18) and (!2) the equality is inevitable! The sum of two primes equal to even up to 12 is shown arithmetically, which, as a result, the solution of the binary Goldbach-Euler problem means: p1+p2=2N and p1+p2+2= p3+ p4 (19) 2.5. Corollary3. The difference of two primes is any even number. Let's show that: p1−p2+2= p3−p4 (20) p1+p4+2= p3+ p2 (21) which is proved by (19). 2.6. Corollary 4. If the sum of four simple odd , then the sum two-even number from 6 to 2N-6 inclusive. What follows from the solved Goldbach-Euler conjecture. 3. Twin primes are infinite. Any even number starting from 14 can be represented as a sum four odd primes, two of which are twins. p1 + p2 + p3 + p4 = 2 N (22) Let two prime numbers be p3, p4. . twins, then the other two numbers are also prime numbers, which follows from Corollary 2. We then arrange prime numbers from left to right in descending order. And if 2 N = 2 p2 + 2 p4 + 4 , then inevitably p1 , p2 . -twins. Subtract from both parts 2 p2 + 2 p4 : p1− p2 + p3 − p4 = 4 (23) From (22) it is obvious p1 , p2 . inevitable twins . Let them p3, p4. . last prime numbers are twins. Let's denote two large prime numbers as p1 , p2 . . And according to Lemma (22), there exists an even number 2 N such that inevitably p1 , p2 . large - twins. Then substituting instead p3, p4. . numerical values p1 , p2 . etc. In (22),(23) we have an infinite number of twins. 4. Additions to the ternary Holbach problem. The difference between the sum of two odd prime numbers and an odd prime number as well as the difference between the odd prime number and the sum of two -any odd numbers. Proof. It is necessary to prove: p1+p2−p3=2K+1 , p1−p2−p3 =2K+1 (24) where K=1,2... ∞ According to the method of mathematical induction: p1+p2−p3+2= p4+p5−p6 (25) p1+p2+ p6+2= p4+p5+p3 (26) and p1−p2−p3+2= p4−p5−p6 (27) p1+p5+p6+2=p4+ p2+p3 (28) And according to the finally solved Goldbach's ternary problem: where: 2K1+3=2K2+1 (29) K2=K1+1 (30) Note: Since in the ternary problem odd numbers start with 9, then for 3, 5, and 7 where K =1,2,3 we confirm arithmetically. Q.E.D. 5. Conclusion. the solution of these problems opens up opportunities for solving a number of problems in number theory. Literature [1]Int[Log[10,3^(3^15)]] — Wolfram Alpha [2]Weisstein, Eric W.Goldbach Conjecture(англ.)на сайте Wolfram MathWorld. [3]Yuri Matiyasevich. Hilbert's Tenth Problem: What was done and what is to bedone. [4] Yu. V. Matiyasevich, Hilbert's tenth problem. - Science, 1993. [5] de Polignac, A. (1849). Recherches nouvelles sur les nombres premiers

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