Semi Twin Prime Numbers

Transcription

1 Semi Twin Prime Numbers Mohammed Ali Faya Ibrahim Alwah Saleh Ahmad Algarni Najran University College of Arts and Sciences Department of mathematics Abstract In this paper we introduce the concept of n semitwin primes which is generalizes of the definition of twin primes A pair of primes (x, y) is called n-semitwin primes where n is even natural number if and only if x y = n and no primes between x and y It is easy to show that a pair of primes (x, y) is 4 semitwin if and only if x 1 mod 3 In this paper we show that a pair of primes (x, y) is 6 semi twin whenever ((x 1 mod 6) and (x 1 mod 5)) ((x 5 mod 6) and (x 3 mod 5)) In der to study other complicated cases we introduce the concept of isolated primes and apply it f the cases of 8, 10 and 12-semi twin primes We use the computation program GAP(Groups, Algithms, Programming- a System f Computational Discrete Algebra) as tool to exhibi some of the numerical and algebric properties of the mentioned cases keywds : n semitwin primes;, isolated primes; congruence; units digit 1 Introduction Twin primes are pairs of primes of the fm (p, p + 2) All twin prime 5 are of the fm (6K 1, 6K + 1) one of the oldest problems in the they of numbers, and indeed in the whole of mathematics is the twin primes conjecture, which states that there are an infinite number of pairs of twin primes Proving this conjecture remains one of the great unsolved problems in mathematics [9] The term twin prime was coined by Paul Stackel in the late nineteenth century and the set of twin primes pairs has been studied by Brun (1919), Hardy and Littlewood (1922), Selmer (1942), Froberg (1961), Weintraub (1973), Bohman (1973), Shanks and Wrench (1974), and Brent (1975, 1976) Currently M Kutrib and J Richstein (1995) In 1919 Brun showed that the sum of the reciprocals of the twin primes converges to a sum now called Brun s Constant: B = ( ) + ( ) + ( ) + 13 Had this series diverged, then we would have a proof of the twin primes conjecture But since it convereges, we do not yet know if there are infinitely many twin primes 1

2 By calculating the twin primes up to 10 14, Thomas R Nicely heuristically estimated Brun s constant to be Me recently he has improved this estimate to by using the twins to [11] The number of terms has since been calculated using twin prime up to (Sebah 2002), giving the result B In 1923, Hardy and Littlewood conjectured that the number of twin primes less than x is Where C 2 = p prime>2 Li(x) 2C 2 x 2 dt (logt) 2 ( ) 1 1 (P 1) 2 is the twin prime constant, This has been shown to give excellent results f x up to about 10 6 [4] In this paper we introduce the concept of n semitwin primes as a generalization of twin primes A pair of primes (x, y) is called n- semitwin prime where n is even natural number if and only if x y = n and no primes between x and y Then, we introduce the concept of isolated prime which helps to exhibit the properties of n semi twin primes f n 8 2 n-semtiwin primes As a generalization of the well-known twin primes we introduce the n-semitwin primes which is defined as follows : Definition 21 A pair of two primes (x, y) is called n semitwin where n is an even natural number if and only if the following conditions hold: 1 x y = n 2 There is no prime p with min{x, y} + 1 p max{x, y} 1 Example : The first 4-semitwin primes pair is (7, 11) While (23, 29) is the first 6 semitwin primes 21 4-Semitwin Primes In this section, we investigate the properties of 4-semitwin primes Lemma 21 Let p is a prime such that p 1 mod 3 Then the next possible prime is p + 4 Assume p is prime such that p 1 mod 3 Then p = 3k + 1 : k Z Clearly, p + 1, p + 3 are even, and p + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) = Hence p + 4 is the first possible prime Theem 21 Let p be a prime number Then the pair (p, p + 4) is 4-semitwin primes if and only if ( =) Follows Immediately by Lemma 21 (= ) Assume (p, p, p + 4) is 4-semitwin primes, and p 1( mod 3) Then p 7 and thus p 2( mod 3) which means that p = 3k : k 1 Z Thus p + 4 = (3k 1 + 2) + 4 = 3k = 3(k 1 + 2) and 3 p + 4 This contradicts our assumption that (p, p + 4) is 4-semitwin primes Lemma 22 Let p and q be 4-semi twin primes Then (p, q) = (3k 2, 3k + 2) : k Z Theem 21 implies that p 1( mod 3) p = 3k = 3k 1 + (3 2) = 3k 2 where k = k 1 + 1, k 1 Z Therefe (p, q) = (p, p + 4) = (3k 2, 3k + 2) The proof of the next lemma is immediate : 2

3 Lemma 23 Let (p, q) be 4-semitwin primes Then p 1( mod 5) 22 6-Semitwin Primes As mentioned in the introduction, the first pair of 6-semitwin primes is (23, 29) We begin this section with the next lemma Lemma 24 Let p be a prime such that p 5 Then p ±1 mod 6 Assume tha p is prime and p ±1( mod 6) Then either p 2( mod 6) = p = 6k + 2 p = 2(3k + 1) 2 p, p 3( mod 6) = p = 6k + 3 p = 3(2k + 1) 3 p, p 4( mod 6) = p = 6k + 4 p = 2(3k + 2) 2 p p ±1( mod 6) By Lemma 24, it is enough to investigate the next two cases n 1 5 mod 6 The next lemma is of great help : Lemma 25 Let (p, q) be 6-semitwin primes Then p 4 mod 5 Assume that (p, q) 6-semitwin primes and p 4 mod 5 Then p 23 and p = 5k + 4 : k Z Thus q = p + 6 = 5k + 10 = 5(k + 2) 5 p + 6 which contradicts the fact that q is prime Theem 22 Let p be a prime such that either p 1( mod 6) p 1( mod 5) p 5( mod 6) p 3( mod 5) then the pair (p, p + 6) is 6-semitwin primes 1 Assume p 1 mod 6 and p 1 mod 5 Then, p + 1, p + 3, p + 5 are even, p + 2 is divisible by 3 and p + 4 is divisible by 5 2 Assume p 5 mod 6 and p 3 mod 5 Then, again p+1, p+3, p+5 are even, p+2 is divisible by 5 and p + 4 is divisible by 3 Thus, (p, p + 6) is 6-semitwin primes 23 8-Semitwin Primes One can easily verify that (89, 97) is the first pair of 8-semitwin primes The next lemma shows that it is enough to consider primes of congruence 2 mod 3 Lemma 26 If p, p + 8 are primes, then p 2( mod 3) Assume that p 2( mod 3) Then p 1( mod 3) = p = 3k + 1, so p + 8 = 3k p + 8 which contradicts our assumption that p + 8 is prime Lemma 27 If p, p+8 are primes, then the unit digits of p is either 1, 3 9 Suppose that the units digit of p 1, 3, 9 Then the units digit of p is either If the units digit of p is 5 then the resulting number will be divisible by 5 this is a contradiction with the assumption (p is prime) 2 If the units digit of p is 7 5 p + 8 this is a contradiction with the assumption (p + 8 is prime) So, the units digit of p should be 1, 3 9 3

4 Lemma 28 let P be the set of all prime numbers and p, p + 8 are primes where the units digit of p 1 Then q P : p < q < p If the unit digit of p is 3, then p 3 mod 5 which implies that 5 divedes p + 2 Also, Theem 26 implies that 3 divides p + 4 This concludes that the only possible prime q where p < q < p+8 is p+6 and the result follows 2 If the unit digit of p = 9, then p 9 mod 5 which implies that 5 divedes p + 6 Also, Theem 26 implies that 3 divides p + 4 This concludes that the only possible prime q where p < q < p + 8 is p + 2 and the result follows We summarize results of this section in the next theem : Theem 23 Let p be a prime such that p 1 mod 10 Then (p, p + 8) is 8-semitwin primes whenever (p 3 mod 10 and p + 6 is not prime ) (p 1 mod 10 and p + 2 is not prime) Semitwin Primes Lemma 29 If p and p + 10 are primes, then Theem 24 If p is a prime and p 1( mod 210), then the first possible prime is p + 10 We know that p + 1, p + 3, p + 5, p + 7, p + 9 are even, then p + 2 = 210k p + 2 p + 4 = 210k p + 4 p + 6 = 210k p + 6 p + 8 = 210k p + 8, hence p + 10 is the first possible prime Theem 25 If p is a prime and p 199( mod 210), then the first possible prime is p + 10 We know that p + 1, p + 3, p + 5, p + 7, p + 9 are even, then p + 2 = 210k p + 2 p + 4 = 210k p + 4 p + 6 = 210k p + 6 p + 8 = 210k p + 8, hence p + 10 is the first possible prime 3 Isolated primes The first pair of 10-semitwin primes is (139, 149) In der to study the n-semitwin primes f n 10, we need to establish concept of isolated prime of specific degree as defined below Definition 31 A prime number p is said to be isolated of degree (q, r) where q and r satisfy the following conditions: (1) The first next prime to p is p + q p 1( mod 6) (2) r is the minimal prime satisfying that f each α where p < α < p + q, there exists a prime fact f such that Suppose that p 1( mod 6) Then p 5( mod 6) = p = 6k + 5 p + 10 = 6k p + 10 this is a contradiction with the assumption ( p + 10 is a prime ) Hence p 1( mod 6) 2 f r Example 7, 13, 19 and 37 are the first isolated primes of degree (4, 3)While 23 and 47 are isolated primes of degree (6, 5) and (6, 7) respectively The proof of the next lemma is immediate 4

5 Lemma 31 let x be an integer such that x 0( mod 7) and x 7( mod 10), then x 7( mod 70) Lemma 32 If p is an isolated prime of degree (10, 7), then p 1 9( mod 10) Clearly, p + 4 is not divisable by 2 3 and since p is isolated prime of degree (10, 7), we must have 5 p p + 4 Now assume that neither p 1 mod 10, n p 9 mod 10 Then either p 3 mod 10 p 7 mod 10 (1) If p 3( mod 10) and then 5 p + 4 and thus 7 p + 4 Lemma 31 then implies that p mod 70 which means that p + 4 = 70k + 7 : k Z Clearly 2 p + 6, 3 p + 6 and 5 p + 6, thus 7 p + 6 This implies that p + 6 = 7t = 70k = 7(t 10k) 7 9 which is a contradiction (2) Similarly, if p 7( mod 10), we end with the contradiction that p + 6 has no prime fact less than equal to 7 This conclude that p 1 9( mod 10) The proof of the next lemma is not hard but tedious Lemma 33 Let p be an isolated prime of degree (10, 7) Then If p 1( mod 10), then p 1( mod 7) If p 9( mod 10), then p 3( mod 7) Theem 31 Let p be an isolated prime of degree (10, 7), then p 1( mod 210) p 199( mod 210) Remember that if (p, p + 10) is 10- semitwin primes, then, otherwise p + 10 will not be prime Also, Lemma 32 and Lemm 33 imply that either (p 1 mod 10 and p 1 mod 7) (p 9 mod 10 and p 3 mod 7) Therefe, we will have one of the following systems of congruences: p 1 mod 7 p 1 mod 10 p 3 mod 7 p 9 mod 10 Applying the chinese remainder theem [13] to each of the above systems leads to the desired result Now we state the main theem of this paper which is an immediate consequence of the above theem Theem 32 Let (p, p + 10) be a 10-semitwin primes Then, (1) If p + 6 = 7 a p a1 1 pan n where a N, a i NU{0} and p 1 > p 2 > > p n, then p 1 mod 210 (2) If p + 4 = 7 a p a1 1 pan n where a N, a i NU{0} and p 1 > p 2 > > p n, then p 199 mod 210 5

6 (1) Since p 1mod7, we must have p 1 mod 10 Otherwise p 9 mod 10 and thus p 3 mod7 This gives the system of congruences: p 1 mod 7 p 1 mod 10 which has the solution p 1 mod 210 (2) Similarly, p 3 mod 7 and thus p 9 mod 10 which lead to the following system of congruences : p 3 mod 7 p 9 mod 10 Therefe p 199 mod 210 References [1] C Aebi and G Cairns, A Property of Twin Primes La Trobe University, Melbourne, Australia, 2012, pp [2] A Berke, An introduction to the twin prime conjecture, 2006, pp [3] T Buchert, On the twin prime conjecture,adam Mickiewicz University in Poznan, 2011, p 74 [4] H Dubner, Twin prime statistics,journal of Integer Sequences, Vol 8, 2005 [5] D P Gabriele, About twin primes and distribution of primes Bollettino UMI, 2010, pp 1-16 [6] D A Goldston, Large Differences Between Consecutive Prime Numbers, Doctal thesis, University of Califnia, 1(1981), pp [7] D A Goldston, Are There Infinitely Many Primes, San Jose State University, 2007, p 21 [8] Y I Manin, Introduction to Modern Number They, Encyclopaedia of Mathematical Sciences, Vol 49, 1990 [9] M R Murty, The Twin Prime Problem and Generalizations Queens University, Kingston, Ontario, 2013, pp [10] S Nazardonyavi, Some histy about twin prime conjecture, Cnell University, 2012, pp [11] A M Nguyen, Odd perfect numbers, San Jose State University, 2000, p 2059 [12] H Riesel, Prime Numbers and Computer Methods f Factization,Series Progress in Mathematics, Vol 126, 1994 [13] C O Rourke, The prime number theem: Analytic and elementary proofs, Masters thesis, National University of Ireland Maynooth, 2013, pp [14] B B Samir and Y A Y Rezk, Generation of Large Prime Numbers From a Sequence of Previous Prime Numbers Universiti Teknologi Petronas,Vol 1482, 1( 2012) [15] J Senson, An Introduction to Prime Number Sieves, University of Wisconsin, 1990, pp