A simple solution to the Goldbach's conjecture

A simple solution to the Goldbach’s conjecture Ing. Mg. Carlos A. CORREA (carcorrea@hotmail.com.ar) In this work a very simple solution to the Goldbach’s conjecture is proposed, starting from a modified version of the Sieve of Eratosthenes. Introduction In 1912, at the V Congress of Mathematics held in Cambridge, the German mathematician Edmund Landau proposed a conjecture, among others, which can be stated using basic mathematical knowledge but which has remained unsolved for centuries: the Goldbach conjecture. Goldbach conjecture In June 1742, the German mathematician Christian Goldbach wrote a letter to Leonhard Euler, in which he proposed the following conjecture: “Every even integer greater than 2 can be expressed as the sum of two primes” Comment1. Before starting the demonstration, it is worth mentioning that we will limit ourselves to using those types of even numbers (N) that most restrict the possibility of being represented as the sum of two primes. The clearest example of this is to avoid the choice of N = 2.pn, since it is obvious that 2.pn can always be expressed as the sum of two primes. We anticipate that the even numbers that most restrict the appearance of primes pairs whose sum coincides with them, are those pairs with value N = 4.pn, (where will it be pn>√N, Ɐ pn>3). Let's explain what has been said using an example: To analyze a case (e.g., N = 4.3.5) we will present in a double column to the rows of odd numbers “candidates” whose sum is N N=4.3.5=60 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 N/2= 30 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 As we see, every multiple of 3 in the left column shares the row in which it is with another multiple of 3 located in the right column. The same goes for multiples from 5 to left and right. However, it is different for the multiples of 7, which do not align in pairs to the left and right, they are located in different rows on each side. This is because both 3 and 5 are factors of N, but not the 7. N/2 = 2.3.5=30 ∴ as 3|N/2, it will also divide to N/2±n.3, corresponding N/2-n.3 to values in the column on the left, and, N/2+n.3 to values, aligned with the previous ones, in the right column. The same goes for 5. In the modified sieve of Eratosthenes which we will present, a complete row will be eliminated when any of its numbers, whether on the left or right, be a multiple of the prime number with which it is to sieve. As we see, the sieve capacity of a prime will be reduced to half when its multiples appear aligned on both sides of the row that integrate. The elimination of rows that produces the multiples of 3 is shown in blue with bidirectional arrows, symbolizing the redundancy of these sieve. The same is shown in green for the multiples of 5. On the other hand, the sieve of 7 can be more effective since it is not “redundant”, which is symbolized by the unidirectional red arrows, therefore, it would mark twice the rows that would mark if said sieve coincided in the same rows to the left and right. N=4.3.5=60 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 This justifies the choice of N=4.pn as the case which most restricts the appearance of pairs of primes whose sum is equal to N We also point out that, unlike the Eratosthenes sieve, in the version presented here when sieving with p, the row of p is also eliminated. Without considering that this row could be part of the solution if the number that accompanies p turns out to be a prime. An example of this is row 7, 53 which we will eliminate due to the sieve of multiples of 7. Why do we do this? For two reasons: the first is that we do not have the certainty that the row partner of said prime is also prime, and the second reason is to strengthen our demonstration, dismissing some possible solutions. N/2= 30 Accepting that the even numbers N that most restrict the appearance of pairs of primes, whose sum match them, are those of value N= 4.pn; we will formalize a little more our modified Eratosthenes sieve using an example of the aforementioned type. N=4.31=124 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 To show the 3 sieves, we separate the double column of odd candidates in 3 double columns, whose rows will distance themselves in d = 2.3 = 6. Constituting each of these single columns a finite mathematical succession. . 1 7 13 19 25 31 37 43 49 55 61 123 117 111 105 99 93 87 81 75 69 63 3 9 15 21 27 33 39 45 51 57 121 115 109 103 97 91 85 79 73 67 5 11 17 23 29 35 41 47 53 59 119 113 107 101 95 89 83 77 71 65 As a result of the sieve with 3, all the rows of the first 2 double columns are eliminated, the non -eliminated double column must now be sieved by the prime 5, for this we separate this double column into 5 double columns, whose rows will be distanced at d = 2.3.5 = 30. 3−2 ) 3 Remaining rows ≅ Candidate rows. ( 5 119 11 113 17 107 23 101 29 95 35 89 41 83 47 77 53 71 59 65 5 119 35 89 11 113 41 83 23 101 53 71 17 107 47 77 29 59 95 65 3−2 5−2 ). ( 5 ) 3 Remaining rows ≅Candidate rows. ( As a result of the sieve with the 5, all the rows of two of the double columns are removed, each of the three non-eliminated double columns must be sieved by the following prime (7), using the procedure shown; until sieving with the greatest prime < √N ′ , being N ′ the greatest odd number surviving the previous sieve. For the example we would have to sieve up to 7. It is important to point out that, in this our "worst case", when the prime pi participates in the sieve process, under the conditions raised, the number of Remaining rows (𝑅𝑟) survivors is reduced to: 3−2 5−2 7−2 𝑝𝑖−2 ). ( 5 )( 7 ) … . . ( 𝑝𝑖 ); being Candidate rows = N/4 3 𝑅𝑟 ≅Candidate rows. ( 𝑅𝑟 ≅ 𝑁/4 . 1 3 . 3 5 . 5 7 . 9 … 𝑝𝑖−2 ; 11 𝑝𝑖 If we sieve with primes greater than 7, we can change in the previous expression the symbol ≅ for >, doing the following: we replace the prime P5-2 = 11-2 = 9 for p4 = 7 to continue simplifying; then we do the same for all the sieve primes which is not twin prime with the previous one, since in that case the numerator can be simplified, without having to reduce its value, with the previous prime in the denominator. Therefore, we can write: 𝑅𝑟 > 𝑅𝑟 > 𝑁 4.𝑝𝑖 𝑁 , being 𝑝𝑖 the biggest prime <√N ′<=√𝑁 . √𝑁 , therefore, the number (𝑅𝑟) of primes whose sum is equal to the number N is: 4.√𝑁 √𝑁 𝑅𝑟 > √𝑁 4 Note: This result was tested by comparing the value Q.E.D. √𝑁 with the reality, navigating a table with more than 4 15 million primes without finding any contradiction to what is raised.