Subsets of Prime Numbers

Iteratioal Joural of Mathematics ad Computer Sciece, 7(2012), o. 2, 101 112 M CS Subsets of Prime Numbers Badih Ghusayi Departmet of Mathematics Faculty of Sciece-1 Lebaese Uiversity Hadath, Lebao email: badih@future-i-tech.et (Received February 10, 2012, Accepted October 14, 2012) Abstract The Fudametal Theorem of Arithmetic shows the importace of prime umbers. A well-kow result is that the set of prime umbers is ifiite (the subset of eve prime umbers is obviously fiite while that of odd prime umbers is therefore ifiite). The subset of Ramauja primes is ifiite. The set of triplet prime umbers is fiite while it is ot kow whether or ot the subset of twi prime umbers is ifiite eve though it is so cojectured. We give may results ivolvig the differet types of prime umbers. 1 Itroductio A prime triplet is a triplet of the form (p, p +2,p+ 4) cosistig of three primes. The set of prime triplets is fiite. Ideed, it is a sigleto set cotaiig oly (3, 5, 7). To see this, otice that (3, 5, 7) is a prime triplet. To prove it is the oly oe, we use the cotradictio method. Suppose there was a prime triplet (P, P +2,P +4)withP 3. The P ca be writte as 3k, 3k +1or3k +2. for some iteger k =2, 3, 4,... However, If P =3k, the P is ot prime. If P =3k +1, the P + 2 is ot prime beig 3(k +1). Key words ad phrases: Prime umbers, Ramauja Primes, Bertrad Postulate, Twi Prime Cojecture, Goldbach Cojecture, Differet kids of primes, Riema Hypothesis. AMS (MOS) Subject Classificatios: 11M06, 11M38, 11M50.

102 B. Ghusayi If P =3k +2, the P + 4 is ot prime beig 3(k +2). I 1845 Bertrad [3] cojectured that if a iteger 1, the there is at least oe prime umber p such that <p 2. Due to its elegace, this was prove at differet times by may promiet mathematicias icludig Tschebyschef [8] i 1850, Ladau [7] i 1909, Ramauja [11] i 1919, ad Erdös [9] i 1932. I cotrast to Ladau s ad Tschebyschef s aalytic proofs, Erdös gave a elemetary proof (it is worth metioig here that this was ot oly Erdös first result but he proved it whe he was oly 18 years old). As we shall see i the ext sectio Ramauja proved a extesio of Bertrad Postulate. 2 Ramauja extesio of Bertrad Postulate. Ramauja Primes Theorem 2.1. (Ramauja) Let π(x) deote the umber of primes x. The if x 2, 11, 17, 29, 41, 47, 59, 67, 71, 97, 101, 107, 127, 149, 151, 167,... we have the respective results π(x) π( 1 x) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,... 2 Remark 2.2. I particular, the first case is iterestig beig Bertrad Postulate: If x 2, the π(x) π( 1 x) 1. 2 To see this, simply let x =2. The coverse of Ramauja Theorem motivates the followig defiitio of Ramauja Primes the first few of which are 2, 11, 17, 29, 41, 47, 59, 67, 71, 97, 101, 107, 127, 149, 151, 167: Defiitio 2.3. For a atural umber, the th Ramauja Prime is the smallest iteger R such that if x R, the π(x) π( 1 x). That is, 2 wheever x R, there exist prime umbers betwee x ad x. The smallest 2 such R must be a prime umber as the fuctio π(x) π( 1 x) ca icrease 2 oly at a prime. This easily implies that there are ifiitely may Ramauja Primes.

Subsets of Prime Numbers 103 Note that Bertrad Postulate is R 1 =2. Recall that if p ad p+2 are prime umbers, the this pair is said to be a Twi Prime Pair. The smallest such pair is (3, 5). The Twi Prime Cojecture states that there are ifiitely may Twi Prime Pairs. Similarly, we defie a Twi Ramauja Prime Pair. The smallest such pair is (149, 151) as we ca see from the begiig of this sectio. As we might expect there are fewer Twi Ramauja Prime Pairs tha Twi Prime pairs. As a matter of fact, amog the first 1100 prime umbers, there are 70 Twi Ramauja Prime 70 Pairs ad 186 Twi Prime Pairs. Note that > 1. This observatio is 186 4 importat for the ext sectio. 3 Results The followig cojecture was stated i [12]: If x 571, the the umber of pairs of Twi Ramauja Primes ot exceedig x is greater tha oe quarter of the umber of pairs of Twi Primes ot exceedig x. I particular, if there are ifiitely may Twi Primes, the there are also ifiitely may Twi Ramauja Primes. O the other had, if there are fiitely may Twi Ramauja Primes, the there are fiitely may twi primes. Therefore, if Sodow Cojecture is prove ad if it is prove also that there are oly fiitely may Twi Ramauja Primes, the that would settle the famous Twi Prime Cojecture. I a letter to Leohard Euler i 1742, Christia Goldbach stated that he believes that: G: Every iteger greater tha 5 is the sum of three primes. Euler replied: E: That every eve umber 4 is a sum of two primes, I cosider a etirely certai theorem i spite of that I am ot able to demostrate it. As a matter of fact, if (G) is satisfied ad 2 4, the 2 +2=p 1 + p 2 + p 3 with primes p 1,p 2, ad p 3 that caot all be odd, for otherwise their sum would be odd. So at least oe of them, say p 3, is 2. The (E) obtais. Coversely, if (E) is satisfied ad k > 5, the 2k 6 ad we cosider

104 B. Ghusayi two cases: Case 1: k is a eve iteger. k =2, say. The k 2=2 2 4, ad so k 2=p 1 + p 2, with primes p 1 ad p 2. Cosequetly, k = p 1 + p 2 +2 ad so (G) obtais i this case. Case 2: k is a odd iteger. k = 2 +1, say. The by case 1 applied to 2 we get k = p 1 + p 2 + 3 ad so (G) also obtais i this case. Notice that (E) is true for ifiitely may eve itegers of the form 2p where p is a prime umber (sice 2p = p + p ad the set of primes is ifiite). But the questio remais whether (E) is true for all eve itegers 4. Util today, this has ot bee proved ad is kow as the Goldbach cojecture. The best result i cojuctio with this cojecture is by Jig-Ru Che [1] who proved i 1966 (aoucig results, but the detailed proofs appeared i two parts oe i 1973 ad the other i 1978) that from some large umber o, every eve umber is a sum of a prime ad a umber that is either a prime or a product of at most two primes. The Goldbach Cojecture crept ito the ews as a millio-dollar questio. Ideed, as a publicity stut, British publisher Toy Faber aouced the prize for the first perso to prove the Goldbach Cojecture before a appearace of Ucle Petros ad Goldbach s Cojecture, a fictio ovel by the Greek writer Apostolos Doxiados ceterig o a ma devotig his life to fid a proof of the Goldbach Cojecture. No oe succeeded but it is worth metioig that the ovel was a huge success ad got traslated ito 15 laguages. The followig theorem provides a equivalet statemet to Goldbach cojecture: Theorem 3.1. For each N, there is at least oe k N such that + k ad ( +2) k are prime umbers if ad oly if every iteger 4 is the sum of two primes. Proof. ( ): Let j 4 be a eve iteger. Write j =2( +1) where N. The there is at least oe k N such that + k ad ( +2) k are prime umbers with j =2( +1)=[ + k]+[(2+) k]. ( ): Let N. Clearly 2( +1)isaeveiteger 4. The there are primes p ad q such that 2 +2=p + q.

Subsets of Prime Numbers 105 Therefore for all k N we have [ + k]+[( +2) k] =p + q. Sice 2 +2=p + q, we ca ow choose k N such that + k = p ad ( +2) k = q. I additio, the followig theorem provides a equivalet statemet to the Twi Prime cojecture: Theorem 3.2. For each N, there is at least oe k N such that + k ad ( +2)+k are prime umbers if ad oly if there are ifiitely may pairs (p, p +2) such that p ad p +2 are both prime umbers. Proof. ( ): This follows sice N is ifiite ad the represetatio [(2 + )+k] [ + k] =2 holds with + k ad (2 + )+k prime umbers (Notice that distict values 1 ad 2 i N yield distict values of k N, for if there is k N such that ( 1 + k, 1 + k +2)=( 2 + k, 2 + k +2), the 1 = 2.) ( ): Let N. Usig the hypothesis, we ca choose k N such that + k ad ( +2)+k are prime umbers. Remark 3.3. Lookig at the statemets o the left-had sides i the previous two theorems, we observe the uusual feature that provig oe of the Goldbach or the Twi Prime cojecture proves the other. 4 Coectio betwee various kids of primes With π(x) deotig the umber of primes x ad p deotig the th prime umber, let us state ad prove the followig results Lemma 4.1. lim x π(x)logx x =1 lim p log =1. π(p Proof. ( ) : Let x = p. The lim ) p / log p = 1. Sice π(p ) =, lim p / log p =1. The there is 0 N such that The p 0 1. 2 log p 0 p 2 log p.

106 B. Ghusayi Therefore 0 log p log 2 + log +loglogp. Hece 0 log p (1 log log p ) log 2 + log log p or 0 log p log (1 log log p ) log 2 log p log +1. log log p Now, sice lim log 2 log p = 0 ad lim, give ɛ>0, there exists log 1 N such that Sice ɛ was arbitrary it follows that Usig this we ca ow write That is, 1 1 ɛ log p log 1+ɛ. log p lim log =1. lim lim p / log =1. p log =1. ( ) :Forayx, there exist prime umbers p ad p +1 such that p x< p +1. Hece π(x) =. Sice the fuctio f(x) = is icreasig, p log p x log x x log x p +1 log p +1. Multiplyig by 1 we get p log p x π(x)logx p +1 log p +1. Takig limits as (ad hece x ) ad usig the hypothesis, the result follows ad the proof of the lemma is complete.

Subsets of Prime Numbers 107 Remark 4.2. The Prime Number Theorem states that π(x)logx lim =1. x x By the above lemma it follows that the Prime Number Theorem is equivalet to p log,. Defiitio 4.3. (a) A set S i a vector space is said to be covex if wheever x, y S ad 0 <λ<1, (1 λ)x + λy S. (b) The covex hull or covex evelope of a set of poits S i dimesios is the itersectio of all covex sets cotaiig S. Theorem 4.4. [10] Let 0 <a 1 <a 2 <... be a sequece of umbers with lim a =0. The there are ifiitely may for which 2a <a i + a +i for all positive i<. Proof. Clearly the boudary of the covex hull of the set {(, a ): = 1, 2,...} is polygoal. The additioal assumptio lim a = 0 implies that the overtical portio of this polygoal boudary is covex ad has ifiitely may vertices. Sice each of these vertices is of the form (, a ) for some, the result follows. Corollary 4.5. There are ifiitely may for which 2p <p i + p +i for all positive i<. Proof. Clearly 0 <p 1 <p 2 <... I additio, usig the above lemma, it follows that lim p =0. Theorem 4.6. Let 0 < a 1 < a 2 <... be a sequece of umbers with a lim =0. The there are ifiitely may for which 2a >a i + a +i for all positive i<. Proof. Clearly the boudary of the covex hull of the set {(, a ): = 1, 2,...} is polygoal. The additioal assumptio lim a = 0 implies that the ohorizotal portio of this polygoal boudary is cocave ad has ifiitely may vertices. Sice each of these vertices is of the form (, a ) for some, the result follows. p Remark 4.7. Sice lim is ot ecessarily 0, we caot hope to get that there are ifiitely may for which 2p >p i + p +i for all positive i<from the above theorem. However, we ca prove the followig

108 B. Ghusayi Corollary 4.8. There are ifiitely may for which p 2 >p i p +i for all positive i<. Proof. Clearly 0 < log p 1 < log p 2 <... I additio, usig the above lemma, we have p log as ad so log p log +loglog as log p from which it follows that lim = 0. Thus there are ifiitely may such that 2 log p > log p i +logp +i for all positive i<.upo expoetiatio we get our result. Defiitio 4.9. A balaced prime umber is a prime which is the average of the earest primes surroudig it. That is, a balaced prime p satisfies the equality p = p 1+p +1. 2 The first few balaced primes are 5, 53, 157, 173, 211, 257, 263, 373, 563, 593, 607, 653, 733, 947, 977, 1103. It is cojectured that there are ifiitely may balaced primes. Defiitio 4.10. A related defiitio is that of a strog prime umber (weak prime umber) which is greater (less) tha the average. The first strog primes are: 11, 17, 29, 37, 41, 59, 67, 71, 79, 97, 101, 107, 127, 137, 149, 163, 179, 191, 197, 223, 227, 239, 251, 269, 277, 281, 307, 311, 331, 347, 367, 379, 397, 419, 431, 439, 457, 461, 479, 487, 499. The first weak primes are: 3, 7, 13, 19, 23, 31, 43, 47, 61, 73, 83, 89, 103, 109, 113, 131, 139, 151, 167, 181, 193, 199, 229, 233, 241, 271, 283. Note that the umber of strog primes ad the umber of weak primes are equal (26) for the first time at the prime P 60 = 281. It follows from a particular case of corollary 4.5 that the umber of weak primes is ifiite. But, ufortuately, we caot say the same thig for strog prime umbers at least usig the above results ad the questio remais ope. However, i a twi prime pair (p, p +2)withp>5,p is always a strog prime umber because 3 must divide p 2 which caot be prime. Cosequetly, if the twi prime cojecture is true, the the umber of strog primes is ifiite. Defiitio 4.11. A good prime umber is a prime whose square exceeds the product of ay two primes at the same umber of positios before ad after it i the sequece of primes; that is, a good prime p satisfies the iequality p 2 >p i p +i for 1 i 1.

Subsets of Prime Numbers 109 The first good primes are 5, 11, 17, 29, 37, 41, 53, 59, 67, 71, 97, 101, 127, 149. It follows from a particular case of corollary 4.8 that the umber of good primes is ifiite. Defiitio 4.12. We ow defie sexy prime umbers. Those are pairs of primes differig by 6 (sex is a Lati word for six). Obviously oe-digit sexy pairs of primes do ot exist. There are seve 2-digit sexy prime pairs: (23, 29), (31, 37), (47, 53), (53, 59), (61, 67), (73, 79), (83, 89). Remark 4.13. Notice the similarity to twi primes. As i the case of twi primes, it is ot kow whether or ot there are ifiitely may sexy primes. Defiitio 4.14. Naturally, a Che prime umber is a prime p such that p +2 is either a prime or a product of two primes. Remark 4.15. Jig-Ru Che himself [1] proved that there are ifiitely may Che primes. Defiitio 4.16. A permutable prime umber is a prime umber which remais prime i all permutatios of its digits. All the permutable primes with less tha 49081 digits are: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 199, 311, 337, 373, 733, 919, 991, 1111111111111111111, 11111111111111111111111,R 317,R 1031, where R = 10 1 9 is the umber with oes. I cojecture that the umber of permutable primes is fiite. Defiitio 4.17. A pseudo-prime permutable umber is a umber where at least a permutable portio of it is prime. Clearly, the umber of pseudo-prime permutable umbers is ifiite because, for istace, 2, 22, 222, are pseudo-prime permutable umbers. Defiitio 4.18. A Fiboacci prime umber is a Fiboacci umber that is prime. The first Fiboacci primes are 2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073 ad, as expected, it is ot kow if there are ifiitely may Fiboaci primes.

110 B. Ghusayi 5 Ladau Ope Problems with Remarks At the 1912 Iteratioal Cogress of Mathematicias, Edmud Ladau [7] raised the followig four major ope questios (cojectures) about prime umbers which were characterized i his speech at that time as uattackable at the preset state of sciece : 1. Goldbach Cojecture. 2. Twi Prime Cojecture. 3. Legedre Cojecture: For each atural umber, there is always a prime umber p such that 2 <p<( +1) 2. 4. There are ifiitely may prime umbers of the form m 2 +1, where m is a iteger. Almost a cetury later, these problems have remaied defiat. Ideed, as we metioed earlier the first oe had bee ope sice Goldbach raised it i 1742 i a letter to Euler ad the secod oe is a 2300-year-old mystery ow. The third oe resisted all attempts eve though it is i the same spirit as Bertrad Postulate (Theorem) ad the followig theorem due to Igham [6]: For every large eough atural umber, thereisaprimeumberp such that 3 <p<( +1) 3. Moreover, i [4] M. El Bachraoui raised a questio which we rephrase as: Is it true that for all itegers >1adafixeditegerk there exists a prime umber i the iterval (k, (k + 1)]? The case k = 1 is Bertrad Postulate. El Bachraoui [4] gave a positive aswer for the case k =2. Ady Loo [5] gave a positive aswer for the case k =3. A positive aswer to the problem would imply a positive aswer to Legedre Cojecture (by takig k = ). Fially, cocerig the fourth problem, it ca be obviously be phrased as: There are ifiitely may prime umbers of the form 2 +1, where is a atural umber. it is kow that there are ifiitely may prime umbers of the forms x 2 +1, where x is a real umber. O the other had, i 1978, Hedrik Iwaiec showed that there are ifiitely may values of m 2 + 1 that are either primes or a product of two primes. Fially, I show a method which could lead to the proof of problem 4. Suppose, to get a cotradictio, that there were oly fiitely may primes of that form: p 1 = m 2 1 +1,m 1 Z p 2 = m 2 2 +1,m 2 Z... p k = m 2 k +1,m k Z.

Subsets of Prime Numbers 111 Clearly p 2,p 3,...p k 5. Now m 2 1m 2 2...m 2 k +1=(p 1 1)(p 2 1)...(p k 1) + 1. Thus it suffices to show that p =(p 2 1)...(p k 1) + 1 is a prime umber differet from each of p 2,..., p k. Suppose, to get a cotradictio, that p = p j for some j =2, 3,..., k. The So p j 1=(p 2 1)(p 3 1)...(p k 1). 1=(p j2 1)(p j3 1)...(p jk 1) which is impossible as the right-had side is eve. Cosequetly, a positive aswer to problem 4 higes o just provig that p is a prime umber. Refereces [1] J.R.Che,O the represetatio of a large eve iteger as the sum of a prime ad the product of at most two primes I ad II, Sci. Siica, 16, (1973), 157-176; 21 (1978), 421-430. [2] Waclaw Sierpiski, Adrej Schizel, Elemetary Number Theory, 1988. [3] Joseph Bertrad, Mémoire le ombre de valeurs que peut predre ue foctio quad o y permute les lettres qu elle referme, J. l École Roy. Polytech., 17 (1845), 123-140. [4] M. El Bachraoui, Primes i the iterval [2, 3], It. J. Cotemp. Math. Sci., 1, o. 13, (2006), 617-621. [5] Ady Loo, O the primes i the iterval [3, 4], It. J. Cotemp. Math. Sci., 6, o. 37-40, (2011), 1871-1882. [6] Albert Edward Igham, O the differece betwee cosecutive primes, Quarterly J. Math. Oxford, 8 (1937), 255-266. [7] Edmud Ladau, Hadbuch der Lehre vo der Verteilug der Primzahle, Teuber, Leipzig, 1909. [8] Pafuty Tschebyschef, Mémoire les ombres premiers, Mém. Acad. Sci. St. Pétersbourg, 7 (1850), 17-33.

112 B. Ghusayi [9] Paul Erdös, Beweis eies Satzes vo Tschebyschef, Acta Litt. Sci. Szeged, 5 (1932), 194-198. [10] Carl Pomerace, The Prime Number Graph, Math. Computatio, 33 (1979), 399-408. [11] Sriivasa Ramauja, A proof of Bertrad Postulate, J. Idia Math. Soc., 11 (1919), 181-182. [12] Joatha Sodow, Ramauja Primes ad Bertrad s Postulate, Amer. Math. Mothly, 116 (2009), 630-635.