The 4-Nicol Numbers Having Five Different Prime Divisors

1 2 3 47 6 23 11 Joural of Iteger Sequeces, Vol. 14 (2011), Article 11.7.2 The 4-Nicol Numbers Havig Five Differet Prime Divisors Qiao-Xiao Ji ad Mi Tag 1 Departmet of Mathematics Ahui Normal Uiversity Wuhu 241000 P. R. Chia tagmi@mail.ahu.edu.c Abstract A positive iteger is called a Nicol umber if ϕ()+σ(), ad a t-nicol umber if ϕ() + σ() = t. I this paper, we show that if is a 4-Nicol umber that has five differet prime divisors, the = 2 α1 3 5 α3 p α4 q α 5, or = 2 α1 3 7 α3 p α4 q α 5 with p 29. 1 Itroductio For ay positive iteger, let φ(), ω() ad σ() be the Euler fuctio of, the umber of prime divisors of ad the sum of divisors of, respectively. We call is a Nicol umber if ϕ() + σ(), ad a t-nicol umber if ϕ() + σ() = t. It is well-kow that t 2, ad is prime if ad oly if ϕ()+σ() = 2. I 1966, Nicol [4] cojectured that Nicol umbers are all eve, ad proved that if α is such that p = 2 α 2 7 1 is prime, the = 2 α 3 p is 3-Nicol umber. I 1995, Mig-Zhi Zhag [6] showed that if = p α q the caot be a Nicol umber, where p ad q are distict primes ad α is a positive iteger. I 1997, Li ad Zhag [2] showed that if ω() = 2, the caot be a Nicol umber. I 2008, Luca ad Sador [3] showed that if is a Nicol umber ad ω() = 3, the either {560, 588, 1400} or = 2 α 3 p with p = 2 α 2 7 1 prime. I 2008, Wag [5] studied the Nicol umbers that have four differet prime divisors. I 2009, Harris [1] showed that the Nicol umbers that have four differet prime divisors must be oe of the followig forms: 1 Correspodig author. This work was supported by the Natioal Natural Sciece Foudatio of Chia, Grat No. 10901002 ad the SF of the Educatio Departmet of Ahui Provice, Grat No. KJ2010A126. 1

1. = 2 3 3 3 5 2 11, 2 4 3 3 5 11, 2 7 5 11 79, 2 3 3 3 5 3 13 2, 2 2 3 2 17 241, 2 2 3 2 17 2 2243; 2. { 2 a 3 p 3 p 4 p 4 = (7 2a 2 1)p 3 + 9 2 a 2 1, where p p 3 (7 2 a 2 3, p 4 are distict 1) primes. } Moreover, Harris [1] proved that all but fiitely may Nicol umbers that have 5 differet prime divisors are divisible by 6 ad ot 9. I this paper, we study the 4-Nicol umbers that have five differet prime divisors ad obtai the followig result: Theorem 1. If is a 4-Nicol umber with ω() = 5, the either = 2 α 1 3 α 2 5 α 3 p α 4 q α 5, or = 2 α 1 3 α 2 7 α 3 p α 4 q α 5 with p 29, where p, q are distict primes, ad α i (i = 1, 2,, 5) are positive itegers. By the Harris result ad Theorem 1, we have the followig result: Corollary 2. All but fiitely may 4-Nicol umbers with 5 differet prime divisors have the form = 2 α1 3 5 α3 p α4 q α 5, or = 2 α1 3 7 α3 p α4 q α 5 with p 29. Throughout this paper, let a ad m be relatively prime positive itegers, the least positive iteger x such that a x 1 (mod m) is called the order of a modulo m. We deote the order of a modulo m by ord m (a). Let V p (m) be the expoet of the highest power of p that divides m. 2 Lemmas The followig three lemmas are motivated by the work of Luca ad Sádor [3]. Here we make some mior revisios. Lemma 3. Let a, b be two atural umbers ad p be a odd prime. If V p (a 1) 1, the V p (a b 1) = V p (b) + V p (a 1). Proof. Let V p (b) = m ad V p (a 1) =. We may assume that b = p m t with p t ad a = 1 + p a 0 with p a 0. Sice 1, we have Thus a t = (1 + p a 0 ) t = 1 + C 1 t p a 0 + + C t t(p a 0 ) t = 1 + p c, p c. a tp = (1 + p c) p = 1 + C 1 pp c + + C p p(p c) p = 1 + p +1 a 1, p a 1. By iductio o m, for all m 0 we have a b = a tpm = 1 + p m+ a m with p a m. Hece This completes the proof of Lemma 3. V p (a b 1) = m + = V p (b) + V p (a 1). 2

Lemma 4. Let t be a atural umber ad p, q be two primes. We have V p (q t 1) V p (q f 1) + V p (t), where f = ord p (q), if p 2; ad f = 2, if p = 2. Proof. (i) p = 2. By [3, Lemma 1], we have V 2 (q t 1) V 2 (q 2 1) + V 2 (t). (ii) p > 2. Now cosider the followig two cases: Case 1. q t 1 (mod p). The above iequality is obvious. Case 2. q t 1 (mod p). The ord p (q) t ad ord p (q) p 1. Let V p (t) = m. We may assume that t = ord p (q) p m k with p k. Thus V p (q t 1) = V p ( (q ord p(q) ) pm k 1 ) This completes the proof of Lemma 4. = V p (q ordp(q) 1) + V p (p m k) = V p (q ordp(q) 1) + m = V p (q ordp(q) 1) + V p (t). Lemma 5. Let = p α 1 1 p α 2 2 p α k k be the stadard factorizatio of ad X = max{α j j = 1, 2,,k}. We fix i {1,,k} such that X = α i. If is a Nicol umber, the we have X 1 where f j = ord pi (p j ), if p i 2; ad f j = 2, if p i = 2. Proof. Sice φ() + σ() ad p X 1 i p X 1 i σ() = V pi (p f j j 1) + k 1 log p i log(x + 1), φ(), we have k ( α p j +1 j=1 j 1 p j 1 ). Hece The above relatio implies that p X 1 i k j=1 (p α j+1 j 1). X 1 j=1 V pi (p α j+1 j 1) = V pi (p α j+1 j 1). 3

By Lemma 4 X 1 This completes the proof of Lemma 5. V pi (p f j j 1) + V pi (p f j j 1) + V pi (α j + 1) log(α j + 1) log p i V pi (p f j j 1) + k 1 log p i log(x + 1). Lemma 6. If is a 4-Nicol umber ad ω() = 5, the must be oe of the followig three forms: 1. = 2 α1 3 α2 5 α3 p α4 q α 5, p,q are distict primes ad 7 p < q. 2. = 2 α1 3 α2 7 α3 p α4 q α 5, p < q are distict primes ad p 29. 3. = 2 α1 3 α2 11 α3 13 α4 p α 5, p 23 is prime. Proof. Let = p α 1 1 p α 2 2 p α 3 3 p α 4 4 p α 5 5 be the stadard factorizatio of. Put l = σ(). Notig that is a 4-Nicol umber ad ϕ()σ() < 2, we have 4 = ϕ() + σ() < l+l 1, hece l > 2 + 3. By ϕ() > σ() = l, we have If p 2 5 the ϕ() = p 1 ϕ() = p 1 a cotradictio. Thus p 2 = 3 ad p 1 = 2. If p 3 13 the ϕ() = p 1 p 2 p 3 p 4 p 2 1 p 3 1 p 4 1 p 5 1 > l > 2 + 3. p 2 p 3 p 4 p 5 p 2 1 p 3 1 p 4 1 p 5 1 2 5 4 7 6 11 10 13 12 < 2 + 3, p 2 p 3 p 4 p 5 p 2 1 p 3 1 p 4 1 p 5 1 2 3 2 13 12 17 16 19 18 < 2 + 3, a cotradictio, thus p 3 11. Case 1. p 3 = 7. The p 4 29. I fact, if p 4 31 the ϕ() = p 1 p 2 p 3 p 4 p 5 p 2 1 p 3 1 p 4 1 p 5 1 2 3 2 7 6 31 30 37 36 < 2 + 3, a cotradictio. Case 2. p 3 = 11. The p 4 = 13 ad p 5 23. I fact, if p 4 17 the ϕ() = p 1 p 2 p 3 p 4 p 5 p 2 1 p 3 1 p 4 1 p 5 1 2 3 2 11 10 17 16 19 18 < 2 + 3, 4 p 5

a cotradictio. If p 5 29 the ϕ() = p 1 a cotradictio. This completes the proof of Lemma 6. p 2 p 3 3 Proof of Theorem 1 p 4 p 5 p 2 1 p 3 1 p 4 1 p 5 1 2 3 2 11 10 13 12 29 28 < 2 + 3, By Lemma 6, it is eough to show that there is o 4-Nicol umbers = 2 α1 3 α2 11 α3 13 α4 p α 5 with p 23. Assume that = 2 α1 3 α2 11 α3 13 α4 p α 5 with p 23 be a 4-Nicol umber, the by ϕ() + σ() = 4 we have: 2 α 1+6 3 α 2+1 5 11 α 3 1 13 α 4 1 p α 5 1 (133p + 10) (p 1) = (2 α 1+1 1) (3 α 2+1 1) (11 α 3+1 1) (13 α 4+1 1) (p α 5+1 1). Case 1. p = 17, = 2 α1 3 α2 11 α3 13 α4 17 α 5. The thus 2 α 1+10 3 α 2+2 5 11 α 3 1 13 α 4 1 17 α 5 1 757 = (2 α 1+1 1)(3 α 2+1 1)(11 α 3+1 1)(13 α 4+1 1)(17 α 5+1 1). (1) By Lemma 5 we have X 35. Notig that ord 757 (2) = 756, ord 757 (3) = 9, ord 757 (11) = ord 757 (13) = ord 757 (17) = 189, 757 2 α 1+1 1, 757 11 α 3+1 1, 757 13 α 4+1 1, 757 17 α 5+1 1. By (1) we have 757 3 α 2+1 1, thus α 2 + 1 = 9k,k Z. By X 35, we have k = 1, 2, 3. Subcase 1: k = 1, α 2 + 1 = 9. The 2 α 1+9 3 10 5 11 α 3 1 13 α 4 2 17 α 5 1 = (2 α 1+1 1)(11 α 3+1 1)(13 α 4+1 1)(17 α 5+1 1). (i) α 4 = 2. By ord 61 (13) = 3 we have 61 13 3 1, this is impossible. (ii) α 4 > 2. The 13 (2 α 1+1 1)(11 α 3+1 1)(17 α 5+1 1). O the other had, we have the followig facts: If 13 2 α 1+1 1, by ord 13 (2) = 12, thus 12 α 1 + 1, ad otig that ord 7 (2) = 3 we have 7 2 α 1+1 1, which is impossible. If 13 11 α 3+1 1, by ord 13 (11) = 12, thus 12 α 3 + 1, ad otig that ord 7 (11) = 3 we have 7 11 α 3+1 1, which is impossible. If 13 17 α 5+1 1, by ord 13 (17) = 6, thus 6 α 5 + 1, ad otig that ord 7 (17) = 6 we have 7 17 α 5+1 1, which is impossible. Thus 13 (2 α 1+1 1)(11 α 3+1 1)(17 α 5+1 1), a cotradictio. 5

Subcase 2: k = 2, α 2 + 1 = 18. By ord 7 (3) = 6, we have 7 3 18 1, thus 7 3 α 2+1 1, which cotradicts (1). Subcase 3: k = 3, α 2 + 1 = 27. By ord 757 (3) = 9, we have 757 3 27 1, thus 757 3 α 2+1 1, which cotradicts (1). Case 2. p = 19, = 2 α1 3 α2 11 α3 13 α4 19 α 5. The 2 α 1+7 3 α 2+3 5 11 α 3 1 13 α 4 1 19 α 5 1 43 59 = (2 α 1+1 1)(3 α 2+1 1)(11 α 3+1 1)(13 α 4+1 1)(19 α 5+1 1). (2) By Lemma 5 we have X 33. Notig that we have ord 59 (2) = ord 59 (11) = ord 59 (13) = 58, ord 59 (3) = ord 59 (19) = 29, 59 2 α 1+1 1, 59 11 α 3+1 1, 59 13 α 4+1 1. By (2) we have 59 3 α 2+1 1 or 59 19 α 5+1 1. If 59 3 α 2+1 1, the 29 α 2 + 1. Sice ord 28537 (3) = 29, we have 28537 3 α 2+1 1, which cotradicts with (2), thus 59 3 α 2+1 1. If 59 19 α 5+1 1, the 29 α 5 + 1. Sice ord 233 (19) = 29, we have 233 19 α 5+1 1, which cotradicts with (2), thus 59 19 α 5+1 1. Case 3. p = 23, = 2 α1 3 α2 11 α3 13 α4 23 α 5. The 2 α 1+7 3 α 2+3 5 11 α 3+1 13 α 4 1 23 α 5 1 31 = (2 α 1+1 1)(3 α 2+1 1)(11 α 3+1 1)(13 α 4+1 1)(23 α 5+1 1). (3) By Lemma 5 we have X 34. Notig that ord 31 (2) = 5, ord 31 (3) = ord 31 (11) = ord 31 (13) = 30, ord 31 (23) = 10, by ord 31 (3) = ord 31 (11) = ord 31 (13) = 30, we kow that 30 α i + 1,i = 2, 3, 4. Notig that ord 61 (3) = 10, ord 19 (11) = ord 61 (13) = 3, we have 61 3 α 2+1 1, 19 11 α 3+1 1, 61 13 α 4+1 1. Which cotradicts with (3), the we have 31 3 α 2+1 1, 31 11 α 3+1 1, 31 13 α 4+1 1. By (3) we kow that 31 23 α5+1 1 or 31 2 α1+1 1. If 31 23 α5+1 1, the by ord 31 (23) = 10, we kow that 10 α 5 + 1. Notig that ord 41 (23) = 5 we have 41 23 α5+1 1, which cotradicts (3). If 31 2 α1+1 1, the α 1 + 1 = 5k,k Z. By X 34, we have k = 1, 2, 3, 4, 5, 6. Subcase 1: k = 1, α 1 +1 = 5, = 2 4 3 α2 11 α3 13 α4 23 α 5. Put m = 3 α2 11 α3 13 α4 23 α 5. The σ(m) m < m ϕ(m) = 3 2 11 10 13 12 23 22 = 299 160 = 1.86875. O the other had, otig that ϕ() + σ() = 4, the 8ϕ(m) + 31σ(m) = 64m, thus σ(m) m > 1.9264 > 1.86875, 6

a cotradictio. Subcase 2: k = 2, α 1 + 1 = 10. Thus 2 16 3 α 2+2 5 11 α3 13 α 4 1 23 α 5 1 = (3 α 2+1 1)(11 α 3+1 1)(13 α 4+1 1)(23 α 5+1 1). Notig that the followig facts: (i) If 2 α 3 α 2+1 1, the α 4. I fact, if α 5, the by ord 32 (3) = 8, we have α 2 + 1 = 8s,s Z. Notig that ord 41 (3) = 8, thus 41 3 α 2+1 1, this is impossible. (ii) If 2 α 11 α 3+1 1, the α 3. I fact, if α 4, the by ord 16 (11) = 4, we have α 3 + 1 = 4s,s Z. Notig that ord 61 (11) = 4, thus 61 11 α 3+1 1, this is impossible. (iii) If 2 α 13 α 4+1 1, the α 3. I fact, if α 4, the by ord 16 (13) = 4, thus α 4 + 1 = 4s,s Z. Notig that ord 7 (13) = 2, thus 7 13 α 4+1 1, this is impossible. (iv) If 2 α 23 α 5+1 1, the α 4. I fact, if α 5, the by ord 32 (23) = 4, we have α 5 + 1 = 4s,s Z. Notig that ord 53 (23) = 4, thus 53 23 α 5+1 1, this is impossible. Let A = (3 α 2+1 1)(11 α 3+1 1)(13 α 4+1 1)(23 α 5+1 1), B = 2 16 3 α 2+2 5 11 α3 13 α 4 1 23 α 5 1. We have V 2 (A) 14 ad V 2 (B) = 16, this is impossible. Subcase 3: k = 3, α 1 +1 = 15. By ord 7 (2) = 3, we have 7 2 α 1+1 1, which cotradicts (3). Subcase 4: k = 4, α 1 + 1 = 20. By ord 41 (2) = 20, we have 41 2 α 1+1 1, which cotradicts (3). Subcase 5: k = 5,α 1 + 1 = 25. By ord 601 (2) = 25, we have 601 2 α 1+1 1, which cotradicts (3). Subcase 6: k = 6,α 1 + 1 = 30. By ord 151 (2) = 15, we have 151 2 α 1+1 1, which cotradicts (3). This completes the proof of Theorem 1. Refereces [1] K. Harris, O the classificatio of iteger that divide ϕ() + σ(), J. Number Theory 129 (2009), 2093 2110. [2] Da-Zheg Li ad Mig-Zhi Zhag, O the divisibility relatio ϕ() + σ(), J. Sichua Daxue Xuebao 34 (1997), 121 123. [3] F. Luca ad J. Sádor, O a problem of Nicol ad Zhag, J. Number Theory 128 (2008), 1044 1059. [4] C. A. Nicol, Some Diophatie equatios ivolvig arithmetic fuctios, J. Math. Aal. Appl. 15 (1966), 154 161. [5] Wei Wag, The research o Nicol problems, Master Degree Theses of Najig Normal Uiversity, 2008. 7

[6] Mig-Zhi Zhag, A divisibility problem, J. Sichua Daxue Xuebao 32 (1995), 240 242. 2010 Mathematics Subject Classificatio: Primary 11A25. Keywords: Nicol umber; Euler s totiet fuctio. Received April 12 2011; revised versio received July 7 2011. Published i Joural of Iteger Sequeces, September 4 2011. Revised, April 11 2012. Retur to Joural of Iteger Sequeces home page. 8