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Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’S Conjecture for Cousin Primes) Marko Jankovic

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Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’S Conjecture for Cousin Primes) Marko Jankovic Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’s Conjecture for Cousin Primes) Marko Jankovic To cite this version: Marko Jankovic. Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’s Conjec- ture for Cousin Primes). 2020. hal-02549967v2 HAL Id: hal-02549967 https://hal.archives-ouvertes.fr/hal-02549967v2 Preprint submitted on 27 May 2020 (v2), last revised 18 Aug 2021 (v12) HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Marko V. Jankovic Department of Emergency Medicine, Bern University Hospital “Inselspital”, and ARTORG Centre for Biomedical Engineering Research, University of Bern, Switzerland Abstract. In this paper proof of the twin prime conjecture is going to be presented. In order to do that, the basic formula for prime numbers was analyzed with the intention of finding out when this formula would produce a twin prime and when not. It will be shown that the number of twin primes is infinite. Originally very difficult problem (in observational space) has been transformed into a simpler one (in generative space) that can be solved. The same approach is used to prove the Polignac’s conjecture for cousin crimes. 1 Introduction In number theory, Polignac’s conjecture states: For any positive even number n, there are infinitely many prime gaps of size n. In other words: There are infinitely many cases of two consecutive prime numbers with the difference n [1]. For n = 2 it is known as twin prime conjecture. Conditioned on the truth of the generalized Elliott-Halberstam Conjecture [2], in [3] it has been shown that there are infinitely many primes’ gaps that have value of at least 6. In this paper gap 2 is analyzed. Gap 4 can be analyzed in a very similar manner. The problem is 1 addressed in generative space, which means that prime numbers are not going to be analyzed directly, but rather their representatives that can be used to produce them. Remark 1: Prime numbers 2 and 3 are in a sense special primes, since they do not share some of the common features of all other prime numbers. For instance, every prime number, apart from 2 and 3, can be expressed in the form 6l + 1 or 6l - 1, where l ϵ N. So, in this paper most of the time, prime numbers bigger than 3 are analyzed. Remark 2: In this paper any infinite number series in the form c1 * l ± c2 is going to be called a thread, defined by number c1. Here c1 and c2 are constants that belong to the set of natural numbers (c2 can also be zero and usually is smaller than c1) and l represents an infinite series of consecutive natural numbers in the form (1, 2, 3, …). 2 Proof of the twin prime conjecture It is well known that every two consecutive odd numbers (psn, pln), between two consecutive odd numbers divisible by 3 (e.g. 9 11 13 15, or 39 41 43 45), can be expressed as 푝푠 =6푘 − 1 푝푙 =6푘 + 1, 푘 휖푁. Twin prime numbers are obtained in the case when both psn, and pln are prime numbers. If any of the psn or pln (or both) are a composite number, then we cannot have twin primes. So, the strategy is to check in which cases (for which k) it would not be possible to have twin primes. 2 We denote any composite number (that is represented as a product of prime numbers bigger than 3) with CPN5. A number in the form 6l + 1 is marked with mpl, while a number in the form 6s - 1 is marked with mps (l, s ϵ N). That means that any composite number CPN5 can be expressed in the form mpl x mpl, mps x mps or mpl x mps. If psn represents a composite number, the following equation must hold 퐶푃푁5+1 푘 = 6 . (1) Since CPN5 should be in the mps form, CPN5 can be generally expressed as a product mpl x mps, or mpl = 6x + 1 and mps = 6y - 1(x, y ϵ N), which leads to CPN5 = mpl x mps = 6(6xy - x + y) – 1, (2) or, due to symmetry mpl = 6y + 1 and mps = 6x - 1, which leads to CPN5 = mpl x mps = 6(6xy + x - y) – 1. (3) If (2) or (3) is replaced in (1) forms of k that cannot produce the twin primes will be obtained. Those forms are expressed by the following equations 푘 = (6푥 −1)푦 + 푥 (4a) 푘 = (6푥 +1)푦 − 푥, (4b) 3 where x, y ϵ N. These equations are equivalent (they will produce the same numbers) and can be used interchangeably. Using similar procedure, we can see that pln, that represents CPN5 number in the mpl form, will correspond to the composite number in the case 퐶푃푁5−1 푘 = . 6 (5) In this case, CPN5 can be expressed in the form mpl1 x mpl2, or as mps1 x mps2. Two possibilities exist (x, y ϵ N): mpl 1= 6x + 1 and mpl2 = 6y + 1, which leads to CPN5 = mpl1 x mpl2 = 6(6xy +x + y) + 1, (6) or, mps1 = 6x - 1 and mps2 = 6y - 1, which leads to CPN5 = mps1 x mps2 = 6(6xy - x - y) + 1. (7) When (6) and (7) are replaced in (5), together with (4a) the forms for all k that cannot produce a twin prime pair, are obtained. Those forms are expressed by the following equation (6푥 −1)푦 + 푥 푘 = (6푥 −1)푦 − 푥, (8) (6푥 +1)푦 + 푥 4 where x, y ϵ N. This is a sufficient and necessary condition for k, so that it cannot be used for the generation of twin primes. In other words, at least one of the twin odds generated by a k in any of the forms (8) will be a composite number, and if any of the odds generated by a k is a composite number, then k must be in one of the forms (8). Alternatively, it is possible to use the equation (4b) instead of (4a). In that case a different set of equations that produce the same numbers as the equation (8), is obtained. Here, a list of the k s(first 7) that cannot be presented in the form (8) and that generate all twin primes bigger than 3 and smaller than 100, is presented. k 1 2 3 5 7 10 12 Twin prime 1 5 11 17 29 41 59 71 Twin prime 2 7 13 19 31 43 61 73 In order to prove that there are infinitely many twin prime pairs we need to prove that infinitely many natural numbers that cannot be expressed in the form (8) exists. First, the form of (8) for some values of x will be checked. Case x=1: k = 5y - 1, k = 5y + 1, k = 7y + 1, Case x=2: k = 11y - 2, k = 11y + 2, k = 13y + 2, Case x=3: k = 17y - 3, k = 17y + 3, k = 19y + 3, Case x=4: k = 23y - 4, k = 23y + 4, k = 25y + 4 = 5(5y +1) – 1, Case x=5: k = 29y - 5, k = 29y + 5, k = 31y + 5, Case x=6: k = 35y – 6 = 7(5y – 1) + 1, k = 35y + 6 = 5(7y + 1) +1, k = 37y + 6, 5 Case x=7: k = 41y - 7, k = 41y + 7, k = 43y + 7, Case x=8: k = 47y – 8, k = 47y + 8, k = 49y + 8 = 7(7y + 1) +1. It can be seen that k is represented by the threads that are defined by prime numbers bigger than 3. From examples (cases x = 4, x = 6 and x = 8), it can be seen that if (6x - 1) or (6x + 1) represent a composite number, k that is represented by that number also has representation by one of the prime factors of that composite number. This can be proved easily in the general case, by direct calculation, using representations similar to (2) and (3). Here, only one case is going to be analyzed. All other cases can be analyzed analogously. In this case, assume that (6푥 −1) = (6푙 +1)(6푠 −1) where (l, s ϵ N), and that leads to 푥 =6푙푠 − 푙 + 푠. Considering that and using the following representation of k that includes the form (6x - 1) 푘 = (6푥 −1)푦 + 푥, the simple calculations leads to 푘 = (6푙 +1)(6푠 −1)푦 +6푙푠 − 푙 + 푠 = (6푙 +1)(6푠 −1)푦 + 푠(6푙 +1) − 푙, or 푘 = (6푙 +1)(6푠 −1)푦 + 푠 − 푙 which means 푘 = (6푙 +1)푓 − 푙 6 and that represents the already existing form of the representation of k for the factor (6l+1), where 푓 = (6푠 −1)푦 + 푠. It can be seen that all patterns for k can be represented by the thread defined by the prime number bigger than 3. Now, it is going to be proved that the number of natural numbers that cannot be represented by the models (8) is infinite. In order to do it, a method similar to the sieve of Eratosthenes [4] is going to be used. When all numbers that can be represented in the form 5y − 1 and 5y + 1, are removed from the set of natural numbers N, it can be seen that a ratio of r1 = 2/5 of all natural numbers is removed.
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