PAijpam.eu SOME RESULTS ON PRIME NUMBERS B. Martin Cerna Maguiña 1, Héctor F. Cerna Maguiña 2 and Harold Blas 3

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1 Internationa Journa of Pure and Appied Mathematics Voume 118 No , ISSN: (printed version); ISSN: (on-ine version) ur: doi:.173/ijpam.v118i3.9 PAijpam.eu SOME RESULTS ON PRIME NUMBERS B. Martin Cerna Maguiña 1, Héctor F. Cerna Maguiña and Harod Bas 3 1 Departamento de Matemática Universidad Naciona Santiago Antúnez de Mayoo Campus Shancayán, Av. Centenario 00, Huaraz, PERÚ Departamento Academico de Contabiidad Universidad Naciona Mayor de San Marcos Av. Universitária 6, Lima, PERÚ 3 Instituto de Física Universidade Federa de Mato Grosso Av. Fernando Correa, N Bairro Boa Esperança, Cep , Cuiabá - MT - BRAZIL Abstract: In this artice using the functions f 1(k) = k + 1, k 3+; f (k) = k +3, k 3 {0}; f 3(k) = k + 7, k 3+, k 7 {0}; and f 4(k) = k + 9, k 3, where k N 0, we obtain two important resuts on prime numbers. The first resut indicates that if p is a prime number that ends in 7, then p + wi be a prime number under certain conditions. The second resut states that if k is a number ending in 7, then k+ wi be aso a prime number under certain conditions. AMS Subject Cassification: 11A41, 11A51, 11D7 Key Words: prime numbers, diophantine equations 1. Introduction There are no genera methods in the iterature, to our knowedge, on how to generate a prime number starting from a given prime number. We know of the existence of arge prime numbers and that to determine the Received: Apri, 017 Revised: March 31, 018 Pubished: Apri, 018 c 018 Academic Pubications, Ltd. ur: Correspondence author

2 846 B.M.C. Maguiña, H.F.C. Maguiña, H. Bas primaity of these numbers there are specia agorithms, and that there are aso prizes for those who show the argest prime number; encouraged by these facts, in this work, we obtain two important resuts on prime numbers. The first resut states that if p is a arge number that ends in 7, then for N, +p is a prime number provided that p 7 90, p 7 p+ 7,, 90 p+ 7, p+ 7, p , are not integers and the (x, y) natura numbers beonging to the intervas p 7 p+ 7 p 7 p are not integer soutions of the diophantine equation: p+ = (x+9)(y +3) and, in addition, the (x, y) natura numbers beonging to the intervas [ ] p 7 p+ 7 p 7 p are not soutions of the diophantine equation: p+ = (x+1)(y +7). The second resut states that if k is a arge number ending in 7, k 3, k 7, k 7 / N and there exists an unique (a,b) N N such that 90 k = (a+9)(b+3), then k+ is a prime number provided that the equation k+ = (x+1)(y +7),x 3+,y 3+,y 7, does not have an integer soution. In this work N represents the set of natura numbers, n represents the set of mutipes of n, and N 0 = N {0}. If (X,d) is a metric space, the distance between subsets of X is defined as: d(a,b) = inf d(x,y). x A y B

3 SOME RESULTS ON PRIME NUMBERS 847. Some resuts on prime numbers In this section we state the Lemma (.1) and the Lemma (.), and then we obtain the resuts mentioned in the introduction. Lemma.1. Let f 1,f,f 3,f 4 : N 0 N be functions defined by: f 1 (k) = k +1, k 3+ f (k) = k +3, k 3 {0} f 3 (k) = k +7, k 3+, k 7 {0} f 4 (k) = k +9, k 3. So, et p beanatura numberendingin seven, and thefoowing diophantine equations p = (x+1)(y +7) ; x 3+,x 1,y 7 {0},y 3+ p = (z +3)(w +9) ; z 3 {0},w 3, do not possess integer soutions, then p is a prime number. Proof. Is inmediate, see [1]. Lemma.. If p and are fixed numbers, C 1 = { (x,y) R : (x+9)(y +3) = p,x 0,y 0 }, C = { (x,y) R : (x+9)(y +3) = p+,x 0,y 0 }, C 3 = { (x,y) R : (x+1)(y +7) = p,x 0,y 0 }, C 4 = { (x,y) R : (x+1)(y +7) = p+,x 0,y 0 }, subsets of R and d the eucidean metric, then d(c 1,C ) = inf x C 1 y C d(x,y) = d(c 3,C 4 ) = inf x C 3 y C 4 d(x,y) = Proof. Just use the Lagrange mutipiers, see []. ( p+ p ). ( p+ p ).

4 848 B.M.C. Maguiña, H.F.C. Maguiña, H. Bas Theorem 1. Let p be a arge prime number that ends in 7, then p+ is a prime number, where N, is a fixed natura number, smaer than p, provided that p 7 90, p 7 p+ 7,, 90 p+ 7, p+ 7, p , are not integers, and the (x, y) natura numbers beonging to intervas p 7 p+ 7 p 7 p are not soutions of the equation p+ = (x+9)(y +3), in addition, the (x, y) natura numbers beonging to the intervas [ ] p 7 p+ 7 p 7 p are not integer soutions of the equation p+ = (x+1)(y +7). Proof. Suppose that p +, 1, N, is not a prime number. Then there exists (A,B) N N such that the next two eqs. (1) or () wi happen p+ = (A+9)(B +3) (1) p+ = (A+1)(B +7). () If it happens to occur (1) we have that the (x,y) natura numbers beonging to the intervas p 7 p+ 7 p 7 p are not integer soutions of the equation p+ = (x+9)(y +3), thus A 0, p 7,B 0, p The straight ine y = B intersects the equation p = (x+9)(y +3), so, we have p = (x+9)(b +3). (3)

5 SOME RESULTS ON PRIME NUMBERS 849 From reations (1), (3) and emma (.) we have [ ] p+ p A x = Simiar anaysis shows us that [ ] p+ p B y = B +3. (4) A+9. (5) From the equation (1) we have two possibiities p+ A+9 (6) p+ B +3. (7) If the reation (6) is true, then from this reation and (5) we have [ ] p+ p B y. (8) p+ If the reation (7) is true, then from this reation and (4) we have [ ] p+ p A x (9) p+ Therefore we have that either (8) or (9) is true. Since p is a arge prime number, we have that (A,B) N N woud be a soution of the equation (3), which is fase, since p is a prime number. Simiary, if it happens to be the case () we wi have A 0, p 7 70,B 0, p 7 ( ) p+ p A x ( ) p+ p B y, p+ or p+. For arge p we have that (A,B) woud be a soution of the equation p = (x+1)(y +7) which is fase, since p is a prime number.

6 850 B.M.C. Maguiña, H.F.C. Maguiña, H. Bas Theorem. Let k be a arge natura number ending in 7, k 3. If k 7, k 7 90, k 17, k 17 / N and there exists an unique (a,b) N N 90 such that k = (a+9)(b+3), then k+ is a prime number, provided that the equation k+ = (x+1)(y +7),x 3+,y 3+,y 7, does not possess an integer soution. Proof. Supposek+ is not a prime number, so there exists (A,B) N N such that k+ = (A+9)(B +3). () So one has k 7, k 7, k , k 17 / N 0 < A < k 7 The straight ine y = B intersects the equation which impies that From () and (11) we have Simiary we have k = (x+9)(y +3),,0 < B < k k = (x+9)(b +3). (11) A x = B y = 1 B +3. (1) 1 A+9. (13) Given that k A+9 or k B+3 then from (1) and emma (.) or (13) and emma (.) we have that [ k+ ] k A x 1 [ k+ ] or k B y 1 ; k k so, ask isargewemightconcudethat(a,b)isanothersoutionoftheequation k = (x+9)(y +3), which is a contradiction, since there exists an unique soution by hypothesis.

7 SOME RESULTS ON PRIME NUMBERS 851 Theorem 3. Let k be a arge natura number, ending in 7, k 3. If k 7 70, k+3 / N and there exists an unique (a,b) N N such that 70 k = (a+1)(b +7),a 1, then k+ is a prime number, provided that the equation k + = (x+9)(y +3), x 3, y 3 {0}, does not possess an integer soution, and y 0 = k 7 equation k+ = (x+1)(y +7). is not a soution of the Proof. Anaogous to the above. Acknowedgements B.M. CERNA thanks for partia financia support to VICERRECTORÍA DE INVESTIGACIÓN DE LA UNASAM, CONCYTEC and his famiy for support and encouragement. References [1] I. N. Hertein Topics in Agebra, Wiey Editoria, India, (006). [] L. D. Kudriávtsev, Curso de Anáisis Matemático, Editoria MIR, Moscú (1984).

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