Commentatones Mathematcae Unverstats Carolnae Uzma Ahmad; Syed Husnne Characterzaton of power dgraphs modulo n Commentatones Mathematcae Unverstats Carolnae, Vol. 52 (2011), No. 3, 359--367 Persstent URL: http://dml.cz/dmlcz/141608 Terms of use: Charles Unversty n Prague, Faculty of Mathematcs and Physcs, 2011 Insttute of Mathematcs of the Academy of Scences of the Czech Republc provdes access to dgtzed documents strctly for personal use. Each copy of any part of ths document must contan these Terms of use. Ths paper has been dgtzed, optmzed for electronc delvery and stamped wth dgtal sgnature wthn the project DML-CZ: The Czech Dgtal Mathematcs Lbrary http://project.dml.cz
Comment.Math.Unv.Caroln. 52,3(2011) 359 367 359 Characterzaton of power dgraphs modulo n Uzma Ahmad, Syed Husnne Abstract. A power dgraph modulo n, denoted by G(n, k), s a drected graph wth Z n = {0, 1,..., n 1} as the set of vertces and E = {(a, b) : a k b (mod n)} as the edge set, where n and k are any postve ntegers. In ths paper we fnd necessary and suffcent condtons on n and k such that the dgraph G(n, k) has at least one solated fxed pont. We also establsh necessary and suffcent condtons on n and k such that the dgraph G(n, k) contans exactly two components. The prmalty of Fermat number s also dscussed. Keywords: teraton dgraph, solated fxed ponts, Charmchael lambda functon, Fermat numbers, Regular dgraphs Classfcaton: 11A07, 11A15, 20K01, 05C20, 11A51 1. Introducton Power dgraphs provde a lnk between graph theory and number theory. By usng graph theoretc propertes of Power dgraphs, we can nfer many number theoretc propertes of the congruence a k b (mod n). Some characterstcs of power dgraph G(n, k), where n and k are arbtrary postve ntegers, have been nvestgated by C. Lucheta et al. [2], Wlson [1], Somer and Křížek [7], [8], [9], [10], Kramer-Mller [5], S.M. Husnne, Uzma and Somer [15]. We contnue ther work by generalzng prevous results. The exstence of solated fxed pont for k = 2 s studed n [7] and for k = 3 n [16]. In ths paper we study the exstence of solated fxed ponts n G(n, k) for any postve ntegers n and k. We obtan necessary and suffcent condtons on n and k such that the dgraph G(n, k) has at least one solated fxed pont. We also establsh necessary and suffcent condtons on n and k such that the dgraph G(n, k) contans exactly two components. Let g : Z n Z n be any functon, where Z n = {0, 1,...n 1} and n 1. An teraton dgraph defned by g s a drected graph whose vertces are the elements from Z n, such that there exsts exactly one edge from x to y f and only f g(x) y (mod n). In ths paper, we consder g(x) x k (mod n). For the fxed values of n and k the teraton dgraph s represented by G(n, k), where k 2 and s called power dgraph modulo n. Each x G(n, k) corresponds unquely to a resdue modulo n. The research of the frst author s partally supported by the Hgher Educaton Commsson, Pakstan.
360 U. Ahmad, S. Husnne A component of G(n, k) s a subdgraph whch s the largest connected subgraph of the assocated nondrected graph. The ndegree of x, denoted by ndeg n (x) s the number of drected edges comng nto a vertex x, and the number of edges comng out of x s referred to as the outdegree of x denoted by outdeg n (x). A dgraph G(n, k) s sad to be regular f every vertex of G(n, k) has same ndegree. We note that a regular dgraph does not contan any vertex of ndegree 0. We can see that a dgraph G(n, k) s regular f and only f each component of G(n, k) s a cycle and for each vertex x, ndeg n (x) = outdeg n (x) = 1. A dgraph G(n, k) s sad to be sem-regular of degree j f every vertex of G(n, k) has ndegree j or 0. A cycle s a drected path from a vertex a to a, and a cycle s a z-cycle f t contans precsely z vertces. A cycle of length one s called a fxed pont. It s clear that 0 and 1 are fxed ponts of G(n, k). Snce each vertex has outdegree one, t follows that each component contans a unque cycle. A vertex a s sad to be an solated fxed pont f t s a fxed pont and there does not exst a non cycle vertex b such that b k a (mod n). In other words a has ndegree 1. The Carmchael lambda-functon λ(n) s defned as the smallest postve nteger such that x λ(n) 1 (mod n) for all x relatvely prme to n. The values of the Carmchael lambda-functon λ(n) are λ(1) = 1, λ(2) = 1, λ(4) = 2, for any odd prme p and k 1 and λ(p e1 1 pe2 2...per λ(2 k ) = 2 k 2 for k 3, λ(p k ) = (p 1)p k 1, r ) = lcm(λ(pe1 1 ), λ(pe2 2 ),..., λ(per r )), where p 1, p 2,..., p r are dstnct prmes and e 1 for all. The subdgraph of G(n, k), contanng all vertces relatvely prme to n, s denoted by G 1 (n, k) and the subdgraph contanng all vertces not relatvely prme to n s denoted by G 2 (n, k). It s obvous that G 1 (n, k) and G 2 (n, k) are dsjont and there s no edge between G 1 (n, k) and G 2 (n, k) and G(n, k) = G 1 (n, k) G 2 (n, k). Let n = ml, where gcd(m, l) = 1. We can easly see wth the help of the Chnese Remander Theorem that correspondng to each vertex x G(n, k), there s an ordered par (x 1, x 2 ), where 0 x 1 < m and 0 x 2 < l and x k corresponds to (x k 1, xk 2 ). The product of dgraphs, G(m, k) and G(l, k) s defned as follows: a vertex x G(m, k) G(l, k) s an ordered par (x 1, x 2 ) such that x 1 G(m, k) and x 2 G(l, k). Also there s an edge from (x 1, x 2 ) to (y 1, y 2 ) f and only f there s an edge from x 1 to y 1 n G(m, k) and there s an edge from x 2 to y 2 n G(l, k). Ths mples that (x 1, x 2 ) has an edge leadng to (x k 1, x k 2). We then see
Characterzaton of power dgraphs modulo n 361 that G(n, k) = G(m, k) G(l, k). We can further assert that f ω(n) denotes the number of dstnct prme dvsors of n and (1.1) n = p e1 1 pe2 2... per r, where p 1 < p 2 < < p r and e > 0,.e. r = ω(n), then (1.2) G(n, k) = G(p e1 1, k) G(pe2 2, k) G(per r, k). Let N(n, k, b) denote the number of ncongruent solutons of the congruence x k b (mod n). Then N(n, k, b) = ndeg n (b) and by the Chnese Remander Theorem, we have (1.3) 2. Some prevous results N(n, k, b) = ndeg n (b) = r =1 N(p e, k, b). Theorem 2.1 (Carmchael [14]). Let a, n N. Then a λ(n) 1 (mod n) f and only f gcd(a, n) = 1. Moreover, there exsts an nteger g such that ord n a = λ(n), where ord n g denotes the multplcatve order of g modulo n. Lemma 2.2 ([1]). Let n = n 1 n 2, where gcd(n 1, n 2 ) = 1 and a = (a 1, a 2 ) be a vertex n G(n, k) = G(n 1, k) G(n 2, k). Then N(n, k, a) = N(n 1, k, a 1 ) N(n 2, k, a 2 ). Theorem 2.3 ([1]). Let n be an nteger havng factorzaton as gven n (1.1) and a be a vertex of G 1 (n, k). Then ndeg n (a) = N(n, k, a) = r r N(p e, k, a) = ε gcd(λ(p e ), k), =1 =1 where ε = 2 f 2 k and 8 p e, and ε = 1 otherwse. or N(n, k, a) = 0, Theorem 2.4 ([1]). There exsts a t-cycle n G 1 (n, k) f and only f t = ord d k for some factor d of u, where λ(n) = uv and u s the hghest factor of λ(n) relatvely prme to k. Theorem 2.5 ([9]). Let n 1 and k 2 be ntegers. Then (1) G 1 (n, k) s regular f and only f gcd(λ(n), k) = 1; (2) G 2 (n, k) s regular f and only f ether n s square free and gcd(λ(n), k) = 1 or n = p, where p s prme; (3) G(n, k) s regular f and only f n s square free and gcd(λ(n), k) = 1.
362 U. Ahmad, S. Husnne Lemma 2.6 ([10]). Let p be a prme and α 1, k 2 be ntegers. Then N(p α, k, 0) = p α α k. Theorem 2.7 ([10]). Let n be an nteger havng factorzaton as gven n (1.1). Then [ r A t (G(n, k)) = 1 (δ gcd(λ(p e t ), kt 1) + 1) ] da d (G(n, k)), =1 d t,d t where δ = 2 f 2 k t 1 and 8 p e, and δ = 1 otherwse. Theorem 2.8 ([10]). Let n = n 1 n 2, where gcd(n 1, n 2 ) = 1 and a = (a 1, a 2 ) be a vertex n G(n, k) = G(n 1, k) G(n 2, k). Then a s a cycle vertex f and only f a 1 s a cycle vertex n G(n 1, k) and a 2 s a cycle vertex n G(n 2, k). Lemma 2.9 ([5]). Let n = n 1 n 2, where gcd(n 1, n 2 ) = 1 and J(n 1, k) be a component of G(n 1, k) and L(n 2, k) be a component of G(n 2, k). Suppose s s the length of L(n 2, k) s cycle and let t be the length of J(n 1, k) s cycle. Then C(n, k) = J(n 1, k) L(n 2, k) s a subdgraph of G(n, k) consstng of gcd(s, t) components, each havng cycles of length lcm(s, t). 3. Exstence of solated fxed ponts We know that f n s square free then 0 s an solated fxed pont of G(n, k). Now f G 1 (n, k) s regular then 1 s an solated fxed pont of G(n, k). We also know that for k = 1, the dgraph G(n, k) conssts of solated fxed ponts only. However, the crtera for the exstence of solated pont for other cases are yet not studed by any other author. In the followng secton we attempt to sort out ths problem for the case when G 1 (n, k) s not regular and n s not square free. Lemma 3.1. Let n = ml, where gcd(m, l) = 1 and x = (x 1, x 2 ) be a vertex n G(n, k) = G(m, k) G(l, k). Then x s an solated fxed pont of G(n, k) f and only f x 1 and x 2 are solated fxed ponts of G(m, k) and G(l, k), respectvely. Proof: Let x be an solated fxed pont. Then x s cycle of length one and N(n, k, x) = 1. From Theorems 2.8 and 2.9, x 1 and x 2 are fxed ponts of G(m, k) and G(l, k), respectvely. Also by Theorem 2.2, N(m, k, x 1 ) = 1 = N(l, k, x 2 ). Hence, x 1 and x 2 are solated fxed ponts n G(m, k) and G(l, k), respectvely. Converse s smlar. Theorem 3.2. The power dgraph G(n, k), where n s defned as n (1.1) and k 2, has at least one solated fxed pont f and only f ether e = 1 or gcd(λ(p e ), k) = 1 for all 1 r n prme factorzaton of n. Proof: Suppose G(n, k) has an solated fxed pont a. For all p e n, where 1 r, ether e = 1 or e > 1. Suppose to the contrary that there exsts 1 j r such that gcd(λ(p ej j ), k) 1 and e j > 1. Snce a s a fxed pont, by Theorems 2.8,
Characterzaton of power dgraphs modulo n 363 Theorem 2.9 and equaton (1.2) there exst fxed ponts a G(p e, k) for all 1 r such that a = (a 1,...,a j,..., a r ). Now from Theorem 2.2, we can wrte (3.1) r N(n, k, a) = N(n, k, a ). =1 If a j G 1 (p ej j, k) then N(pej j, k, a j) = gcd(λ(p ej j ), k) 1. Thus n ths case from equaton (3.1), N(n, k, a) 1, whch contradcts the fact that a s an solated fxed pont. Hence, we may suppose a j G 2 (p ej j, k). Now we know that G 2(p ej j, k) conssts of one component contanng fxed pont 0. Thus a j 0 (mod p ej j ). From ej Lemma 2.6, N(p ej j, k, a j) = N(p ej k j, k, 0) = pej j. Snce e j > 1 and k 2, N(p ej j, k, a j) 1. Now from equaton (3.1) t follows that N(n, k, a) 1 whch agan s a contradcton. Conversely, suppose for all p e n, where 1 r, ether e = 1 or gcd(λ(p e ), k) = 1. If e = 1, 0 s an solated fxed pont n G(p, k). If e > 1 and gcd(λ(p e ), k) = 1, 1 s an solated pont n G(pe, k). Now consder a = (a 1, a 2,...,a r ), where a = 0 f e = 1, = 1 f e > 1. From Lemma 3.1, a s an solated fxed pont of G(n, k). Corollary 3.3. Suppose k s even and n > 2 s defned as n (1.1). The power dgraph G(n, k) has at least one solated fxed pont f and only f n s square free. Proof: We know that 2 λ(p e ) for all 1 r. Snce k s even, gcd(λ(pe ), k) 1 for any 1 r. Hence, from Theorem 3.2, e = 1 for all 1 r whch mples n s square free. Conversely, f n s square free, 0 s an solated fxed pont of G(n, k). Corollary 3.4. Suppose G 1 (n, k) s not regular and n s not square free. The power dgraph G(n, k), where n s defned as n (1.1) and k 2, has an solated fxed pont f and only f the followng statements are satsfed. (1) k must be odd. (2) The sets l = {p e e > 1 and gcd(λ(p e are non empty. Also G(n, k) = G(l, k) G(m, k). (3) The dgraph G 1 (m, k) s not regular., k) = 1)} and m = {pej j e j = 1} Proof: Suppose G(n, k) has an solated fxed pont a. If k s even then from Corollary 3.3, n s square free whch s a contradcton. Now from Theorem 3.2, ether e = 1 or gcd(λ(p e ), k) = 1 for all 1 r n the prme factorzaton of n. Snce G 1 (n, k) s not regular and n s not square free, there must exst 1 s < r such that e = 1 for all 1 s and gcd(λ(p e ), k) = 1 for all > s. Hence, the sets l and m are non empty. Snce l and m are dsjont, from equaton (1.2), we get G(n, k) = G(l, k) G(m, k).
364 U. Ahmad, S. Husnne Now f G 1 (m, k) s regular then from equaton (1.2) and Theorem 2.5, G 1 (n, k) = G 1 (l, k) G 1 (m, k) s also regular whch s a contradcton. Conversely, suppose all three condtons are true. Snce l s non empty and G 1 (l, k) s regular, 1 s an solated fxed pont n G(l, k). Agan snce m s nonempty, 0 s an solated fxed pont of G 2 (m, k). Thus from Lemma 3.1, a = (1, 0) s an solated fxed pont of G(n, k) = G(l, k) G(m, k). Example 3.5. Let n = 28 = 2 2 7 and k = 15. Here we can see that the sets l = {2 2 } and m = {7} are non empty. Snce gcd(λ(4), 15) = 1 and gcd(λ(7), 15) = 3 1, from Theorem 2.5, G 1 (l, k) s regular and G 1 (m, k) s not regular. Thus G(28, 15) satsfes condtons 1, 2 and 3 of Theorem 3.2. Hence, G(28, 15) contans an solated fxed pont. It s shown n Fgure 1. Fgure 1. The solated fxed ponts of G(28,15) are 7 and 21 4. Power dgraphs of Fermat numbers Theorem 4.1. The power dgraph G(n, k), where n > 2 and k 2 are postve ntegers exhbts the followng propertes: (1) G(n, k) conssts of exactly two components contanng fxed ponts 0 and 1, (2) G 1 (n, k) s sem-regular of degree 2 d for some d 1 f and only f k s even and n = 2 l or n = F m, where l 2, m 1 are ntegers and F m = 2 2m + 1 s Fermat prme. Proof: Suppose that a power dgraph G(n, k) exhbts the above propertes (1) and (2). Snce 0 and 1 are fxed ponts of G(n, k), G 2 (n, k) and G 1 (n, k) both consst of one component contanng fxed ponts 0 and 1, respectvely. Frst suppose k s odd; then 2 k 1. Snce n > 2, 2 dvdes λ(p e ) for all 1 r. Thus from Theorem 2.6, A 1 (G(n, k)) 3. Ths along wth the fact that each component of G(n, k) contans a unque cycle mples that the number of components of G(n, k) s greater than or equal to 3 whch contradcts (1). We know that the Euler functon φ(n) s a power of 2 f and only n = 2 l F m1 F m2... F ms. Also t s easy to show that φ(n) = 2 f and only f λ(n) = 2 j, where
Characterzaton of power dgraphs modulo n 365 j. Now we clam that n must be of the form 2 l F m1 F m2... F ms, where l 0 and F m are Fermat prmes for all. For f n 2 l F m1 F m2... F ms then λ(n) s not a power of 2. Therefore, there exsts an odd prme dvsor p of λ(n). Then by defnton of λ(n) there exsts, where 1 r such that p s a prme dvsor of λ(p e ). If p k, by Theorem 2.3, ether N(n, k, a) = 0 or p N(n, k, a) for all a G 1 (n, k) whch contradcts (2). Thus we may suppose p k. Now p s a factor of λ(n) whch s relatvely prme to k. Thus from Theorem 2.4 there exsts a cycle of length t n G 1 (n, k) such that k t 1 (mod p). If t = 1 then p k 1. Now from Theorem 2.6, A 1 (G(n, k)) p + 1 whch contradcts (1). Hence, we may suppose t > 1. But then there exsts a component contanng a cycle of length t > 1 whch agan contradct (1). Thus n any case, we get a contradcton. Hence, n = 2 l F m1 F m2... F ms, where l 0 and F m are Fermat prmes for all. Now snce G 2 (n, k) conssts of only one component contanng the fxed pont 0, n must be of the form p α, where p s any prme and α 1. Thus n = 2 l or n = F m, where l 2, m 1 are ntegers and F m = 2 2m + 1 s Fermat prme. Conversely, suppose k s even and n = 2 l or n = F m, where l 2, m 1 are ntegers and F m = 2 2m + 1 s Fermat prme. It s easy to see that λ(n) s a power of 2. Property (2) can be proved from Theorem 2.3. To prove property (1), we frst show that G 1 (n, k) does not contan any cycle of length greater than 1. From Theorem 2.4 and the fact that the greatest dvsor of λ(n) whch s relatvely prme to k s 1, t follows that all cycles of G 1 (n, k) are fxed ponts. Now from Theorem 2.6, A 1 (G(n, k)) = 1. Snce the number of components n G 1 (n, k) s equal to the number of cycles n G 1 (n, k), G 1 (n, k) conssts of only one component contanng 1. Ths along wth the fact that G 2 (n, k) always conssts of one component whenever n s a power of a prme, completes the proof. Remark 4.2. In Theorem 4.1, we have taken n > 2 as for n = 2, the power dgraph G(2, k) always conssts of two components whch are solated fxed ponts. It does not depend on value of k. We also note that property (2) s not satsfed n ths case. Corollary 4.3. Let n be a postve nteger and k = 2 s, where s 1. The power dgraph G(n, k) conssts of exactly two components contanng fxed ponts 0 and 1 f and only f n = 2 l or n = F m, where F m = 2 2m + 1 s Fermat prme for all 1 s and l 1. Proof: Snce k = 2 s, from Theorem 2.3 N(n, k, a) = r =1 gcd(λ(pe ), k) = 2d for some d 1 or N(n, k, a) = 0. Hence, G 1 (n, k) s sem-regular of degree 2 d for some d 1. Corollary follows from Theorem 4.1. Corollary 4.4. Let k be an even nteger (k 2). A Fermat number F m = 2 2m +1 s prme f and only f followng are satsfed: (1) G(F m, k) conssts of two components contanng fxed ponts 0 and 1,
366 U. Ahmad, S. Husnne (2) G 1 (F m, k) s sem-regular of degree 2 d for some 1 d 2 m. Proof: It s straght forward from Theorem 4.1. Corollary 4.5. Let n be a postve nteger and k = 2 s, where s 1. A Fermat number F m = 2 2m +1 s prme f and only f G(F m, k) conssts of two components contanng fxed ponts 0 and 1. Proof: It can be proved from Theorem 2.3 and Corollary 4.4. Corollares 4.3 and 4.5 for s = 1 has been proved n [7]. Theorem 4.6. Let n > 2 be a postve nteger and k = q β1 1...qβs s be the prme decomposton of k. The power dgraph G(n, k) conssts of two components f and only f k s even and n has one of the followng forms: (1) n = p, where p = 1 + 1 s qγ s prme and γ 0 for all ; (2) n = qj α for some 1 j s and q j = 1 + 1 s, j qγ, where γ 0 for all. Proof: Suppose the power dgraph G(n, k) conssts of two components. Now f k s odd then 2 k 1. Also snce n > 2, 2 λ(p e ) for all 1 r. Hence, from Theorem 2.6, A 1 (G(n, k)) 3. Ths along wth the fact that the number of components s equal to the number of cycles n power dgraphs mples that the number of components of G(n, k) s greater than or equal to 3 whch s a contradcton. Hence, k must be even. As the vertces 0 and 1 belong to G(n, k), both of ts components contan fxed ponts and there does not exst any other component contanng a cycle of length greater than 1. Snce G 2 (n, k) tself s a component contanng 0, n must be of the form n = p α, where p s any prme. Suppose on the contrary that n does not satsfy the condtons gven n (1) and (2). The followng cases arse: Case 1. If n = p α, where p q for any 1 s and α > 1, then p λ(n) = λ(p α ) = p α 1 (p 1). We can see that p k whch shows that p s a factor of λ(n) relatvely prme to k. Thus from Theorem 2.4, there exsts a cycle of length t such that (4.1) k t 1 (mod p). The fact that there does not exst any other component contanng the cycle of length greater than 1 forces t = 1. But then p k 1 from (4.1). Consequently from Theorem 2.7, A 1 (G(n, k)) p + 1. Ths further mples that the number of components of G(n, k) s greater than or equal to p + 1 whch s a contradcton. Case 2. Now suppose n = p, where p s any prme or n = qj α for some 1 j s, but there exst prme dvsors p 1 q and p 2 q for any such that p 1 p 1 and p 2 q j 1. Then p 1 and p 2 are prme dvsor of λ(n) relatvely prme to k. Now agan by the same argument as n Case 1, we fnd the contradcton. Conversely, suppose k s even and n has one of the forms gven n (1) and (2). We note that n ether case λ(n) does not contan any prme factor relatvely prme
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