Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS

Degree project NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS Author: MD.HASIRUL ISLAM NAZIR BASHIR Supervisor: MARCUS NILSSON Date: 2012-06-15 Subject: Mathematics and Modeling Level: Master Course code:5ma11e

Abstract In this thesis we study the number of periodic points of congruential monomial dynamical system. By concept of index calculus we are able to calculate the number of solutions for congruential equations. We give formula for the number of r-periodic points over prime power. Then we discuss about calculating the total number of periodic points and cycles of length r for prime power. Keywords: periodic points, monomial dynamical system, index calculus, cycle of length r. 2

Acknowledgments We would like to express special gratitude to our supervisor Marcus Nilsson for his invaluable suggestions and thought - provoking discussions during the entire thesis work. He has been enormously supportive in shaping our ideas to carry out this work within the time frame. We owe special thanks to Professor Andrei Khrennikov from which we have learnt a lot. We are deeply and forever indebted to our parents and family for their love, support and encouragement throughout our entire student life. We owe a great many thanks to a great many people who helped and supported us during the writing of this thesis. We also very thankful to Swedish Government for giving golden opportunity to stay and study in Sweden. At last we would like to dedicate our whole thesis work to great mother late Ulfat Bi Bi. 3

Contents 1 Introduction 4 2 Index Calculus 5 3 Solution of Equations Modulo a Prime Number 6 3.1 Solution of Equation x 2 1 (mod p)................ 6 3.2 Solution of Equation x m 1 (mod 2)............... 8 3.3 Solution of Equation x m 1 (mod p)............... 8 4 Solution of Equations Modulo Composite Number 9 4.1 Solution of Equation x m 1 (mod p k )............... 9 4.2 Solution of Equation x m 1 (mod 2 k )............... 10 4.3 Solution of Equation x m 1 (mod p q).............. 13 4.4 Solution of Equation x m 1 (mod p k q k )............. 16 5 Number of r Periodic Points to Monomial Dynamical System 19 6 Number of Total Periodic Points to Monomial Dynamical System 23 4

1 Introduction Carl Friedrich Gauss was a mathematician who published Disquisitiones Arithmeticae in 1801, which was a text book of number theory. Carl Gustav Jacobi published a book of all indices modulo any prime less than equal to 1000 in 1839. After creation of the Diffe-Hellman cryptosystem in 1976 index calculus became important[7]. Residue arithmetic has close relation to theory of dynamical systems. Such systems describe time evolution of states of physical, biological and financial system, e.g. the stock exchange, the weather conditions directly depend on time. So we can describe these phenomena with the aid of dynamical system. There are two types of dynamical systems: Continuous and discrete. A discrete system is generated by iteration of a function. By discrete dynamical system we can describe phenomena which are related to certain moments[4], e.g. measurements at certain moments of time. They have many applications in biology and physics[8]. Continuous dynamical systems describe phenomena, which evolve continuously with time [4]. We have studied monomial dynamical system f(x) = x m for the finite set of prime powers. This paper is organized as follows: Section 2 defines some basic definitions and properties of index calculus. Section 3 gives solution of congruential equation for prime numbers and some theorems. We will solve the congruence equation modulo 2. Section 4 we will find the solution for prime power 2 k and also for composite numbers. Section 5 introduces number of cycles to the monomial dynamical system. By using Möbius inversion formula, we have proved a formula for the number of cycles of length r. Section 6 we will derive a formula to calculate total periodic points for prime power. We will state some theorems and examples to understand it. 5

2 Index Calculus In this section, we will discuss index calculus. We will also define some basic definitions and state some theorems. For this section, we have taken the material from [1],[2] and [3]. Definition 2.1. Suppose a and b are non zero integers. Then the largest integer d = (a, b) that divides both a and b is called greatest common divisor of a and b. Definition 2.2. The least positive integer x such that a x 1 (mod n) is called the order of modulo n, where (a, n) = 1. The order of a modulo n is denoted by ord n a. Definition 2.3. Suppose n is a positive integer. The Euler phi-function φ(n) is said to be the number of positive integers not greater than n and also relatively prime to n. Definition 2.4. Suppose r and n are relatively prime integers with n > 0. Then r is called a primitive root modulo n if ord n r = φ(n). Definition 2.5. If unique integer x with 1 x φ(n) and a is positive integer with (a, n) = 1, then r x a (mod n) is called index of a to base r modulo n. Where n be a positive integer with primitive root r. Theorem 2.1. Chinese Remainder Theorem. Suppose n 1, n 2... n k are pairwise relatively prime positive integers. Then Congruence system x a 1 (mod n 1 ), x a 2 (mod n 2 ),. x a k (mod n k ), has a unique solution modulo n = n 1 n 2... n k. Theorem 2.2. Let a, b, d and n are integers with n > 0. If d b,then ax b (mod n) has no solution and if d b, then ax b (mod n) has exactly d incongruent solutions modulo n. Where d = (a, n). Theorem 2.3. Let n be a positive integer with primitive root r and a, b be integers relatively prime to n. Then, (i) ind r 1 0 (mod φ(n)), (ii) ind r (ab) ind r a + ind r b (mod φ(n)), (iii) ind r a m m ind r a (mod φ(n)), where m is a positive integer. 6

Theorem 2.4. Suppose n is a positive integer with primitive root. Let m is positive integer and a is an integer relatively prime to n. Then congruence x m a (mod n) has exactly d incongruent solution modulo n if and only if 1 (mod n), where d = (m, φ(n)) is a positive integer. a Φ(n) d Proof. Suppose m and n are positive integers. So the congruence equation is x m a (mod n). Let r be a primitive root modulo n. Now taking index on both sides with base r, ind r x m ind r a (mod φ(n)), m ind r x ind r a (mod φ(n)). Let d = (m, φ(n)) and y = ind r x. Then, x r y (mod n). According to Theorem 2.2 If d b, then ax b (mod n) has no solution and if d b, then ax b (mod n) has exactly d incongruent solutions modulo n. Where a, b and n be integers n > 0 and (a, n) = d. So by theorem 2.2.if d ind r a then the linear congruence my ind r a (mod φ(n)) has no solution. If d ind r a then my ind r a (mod φ(n)) has exactly d incongruent solution modulo φ(n). Now d ind r a if and only if (φ(n)/d) ind r a 0 ind r a (φ(n) d) ind r 1 (mod φ(n)), a φ(n) d 1 (mod n). (mod φ(n)), So theorem is true. 3 Solution of Equations Modulo a Prime Number 3.1 Solution of Equation x 2 1 (mod p) Let r be the least positive primitive root modulo p, where p > 2 is a prime number. x 2 1 (mod p). (3.1) Taking index on both sides of above congruence equation to base r modulo p. We obtain a congruence modulo φ (p) = p 1, ind r x 2 ind r 1 (mod p 1). 7

By Theorem 2.3 ind r 1 = 0. So, 2 ind r x 0 (mod p 1). According to Theorem 2.4, this equation has d = (2, p 1) = 2 solutions. Since p is odd. ind r x l 1, l 2 (mod p 1), where ind r c = l 1 and ind r g = l 2. So, x c, g (mod p). Hence equation x 2 1 (mod p) has two incongruent solutions x c, g (mod p). Theorem 3.1. Suppose p be a prime. The equation x 2 1 (mod p) has (2, φ(p)) = 2 solutions. Example 3.1. Let p = 3, then equation (3.1) become, x 2 1 (mod 3). We find that 2 is a least positive primitive root of 3 ( 2 2 1 (mod 3) ). Now taking index on both sides of the congruence to base 2 modulo 3. We a 1 2 ind 2 a 0 1 Table 1: Indices to the Base 2 Modulo 3 obtain a congruence modulo φ(3) = 2, ind 2 x 2 ind 2 1 (mod 2). By Theorem 2.3 ind 2 1 = 0. So, 2 ind 2 x 0 (mod 2). According to Theorem 2.4 d = (2, 2) = 2. Then, ind 2 x 0, 1 (mod 2), so there are two incongruent solutions x 1, 2 (mod 3). Example 3.2. Let p = 11 then equation (3.1) become, x 2 1 (mod 11). We find that 2 is the least positive primitive root of 11 ( 2 10 1 (mod 11) ). We 8

a 1 2 3 4 5 6 7 8 9 10 ind 2 a 0 1 8 2 4 9 7 3 6 5 Table 2: Indices to the Base 2 Modulo 11 obtain following table. Taking index on both sides of the congruence to base 2 modulo 11. We obtain a congruence modulo φ(11) = 10, ind 2 x 2 ind 2 1 (mod 10), 2 ind 2 x 0 (mod 10). According to Theorem 2.4 d = (2, φ(11)) = 2, so ind 2 x 0, 5 (mod 10), then there are two incongruent solutions x 1, 10 (mod 11). 3.2 Solution of Equation x m 1 (mod 2) Let m be a positive integer and p = 2. Then by Theorem 2.4 (m, φ(2)) = 1 for every m Z +, so there exixt a unique solution which is x 1 (mod 2). Example 3.3. If p is a prime integer (p = 2) and m Z +. solution of congruence, x m 1 (mod 2). Let m = 3, then equation become, Then find the x 3 1 (mod 2). There are only two possibilities which are x 0 (mod 2) and x 1 (mod 2). But x 0 (mod 2) is not a solution. Again by Theorem 2.4, (3, φ(2)) = 1, so we have a unique solution x 1 (mod 2). 3.3 Solution of Equation x m 1 (mod p) Suppose m be a positive integer and p > 2 be a prime number, x m 1 (mod p). (3.2) Let r is the least positive primitive root modulo p. Taking index on both sides of the congruence to base r modulo p. We obtain a congruence modulo φ (p) = p 1, ind r x m ind r 1 (mod p 1). By Theorem 2.3 ind r 1 = 0. So, m ind r x 0 (mod p 1). 9

According to Theorem 2.4 This equation has k = (m, p 1) solutions, so ind r x l 1, l 2... l k (mod p 1), if ind r a 1 = l 1... ind r a k = l k. Then solutions of equation are, x a 1, a 2... a k (mod p). Theorem 3.2. The equation x m 1 (mod p) has (m, φ(p)) incongruent slutions, where m be a positive integer and p is a prime. Example 3.4. Let m = 3 and p = 11, then equation (3.2) become, x 3 1 (mod 11). We find that 2 is the least positive primitive root of 11 ( 2 10 1 (mod 11) ). So we obtain following table. a 1 2 3 4 5 6 7 8 9 10 ind 2 a 0 1 8 2 4 9 7 3 6 5 Table 3: Indices to the Base 2 Modulo 11 Taking index on both sides of the congruence to base 2 modulo 11. We get a congruence modulo φ(11) = 10, By Theorem 2.3 ind 2 1 = 0. Then, ind 2 x 3 ind 2 1 (mod 10). 3 ind 2 x 0 (mod 10). According to Theorem 2.4 (3, 10) = 1. We have unique solution, since ind 2 x 0 (mod 10). So there is one incongruent solution x 1 (mod 11) of the equation. 4 Solution of Equations Modulo Composite Number 4.1 Solution of Equation x m 1 (mod p k ) Let m, k be positive integers and p a prime number, x m 1 (mod p k ). (4.1) Theorem 4.1. When p is an odd prime then either r or r + p is the primitive root modulo p 2. Where r is a primitive root. 10

Theorem 4.2. Let p be an odd prime. Then p k has a primitive root for all positive integers k. Moreover if r is a primitive root modulo p 2. Then r is a primitive root modulo p k for all integers k. Example 4.1. Let m = 5, p = 7 and k = 2. Then equation (4.1) become. x 5 1 (mod 7 2 ), x 5 1 (mod 49). By Theorem 4.2 r p 1 = 3 6 1 (mod 49). It follows that ord 49 3 = 42, hence 3 is also primitive root of p 2 = 49. Taking index on both sides of congruence to base 3 modulo 49. We obtain congruence modulo φ(49) = 42, By Theorem 2.3 ind 3 1 = 0. Then, ind 3 x 5 ind 3 1 (mod 42). 5 ind 3 x 0 (mod 42). According to the Theorem 2.4, (5, 42) = 1. We get unique solution, ind 3 x 0 (mod 42), so there is one solution of the equation which is x 1 (mod 7 2 ). Theorem 4.3. Let m and k are positive integers and p be a prime. Then the number of incongruent solutions of the congruence x m 1 (mod p k ) is (m, φ(p k )). 4.2 Solution of Equation x m 1 (mod 2 k ) Suppose m and k are positive integers, x m 1 (mod 2 k ). (4.2) Theorem 4.4. (i) Every odd integer x satisfies this congruence x 2β 2 1 (mod 2 3 ), (4.3) where β 3. (ii) 5 has order 2 β 2 of equation (4.3), when β 3. (iii) The numbers ±5, ±5 2, ±5 3,..., ±5 2β 2 form a reduced residue system modulo 2 β, where β 3. 11

Theorem 4.5. Consider the congruence, x q a (mod 2 α ). (4.4) Where a, m be an odd prime and q = 2m. The exponent α 3. Then the equation (4.4) has four incongruent solutions if a 1 (mod 8) otherwise it has no solution. Proof. If α 3, we use theorem (4.4) and get following equations, a ( 1) h 5 k (mod 2 α ). (4.5) x ( 1) µ 5 y (mod 2 α ). (4.6) Where h, k, µ and y are integers 0. We have given q = 2m, m is odd and α 3. Then by using (4.4),(4.5) and (4.6) we get, 5 2my ( 1) h 5 k (mod 2 α ). Hence k is even, thus a 1 (mod 4). So 2my k (mod 2 α 2 ). That means k 0 (mod 2) and a 1 (mod 8). When this condition is fulfilled, then we get two incongruent solutions y modulo 2 α 2. Finally four incongruent solutions x modulo 2 α. Definition 4.1. Universal exponent U of the positive integer n such that a U 1 (mod n), where (a, n) = 1. Example 4.2. Let m = 2 and k = 3, then equation (4.2) become, x 2 1 (mod 8). (4.7) Every element of U 8 has unique expression of the form ±5 i. If x = ±5 i then, If x = 5 i, then from equation (4.7), 5 2 25 1 (mod 8), so 1 5 2 (mod 8) and 1 5 0 (mod 8). x ±5 i (mod 8). (4.8) (5 i ) 2 1 (mod 8), 5 2i 1 (mod 8), 5 2i 5 2 (mod 8). Where φ(8) 2 = 2, because it is the order of 2 β 2. Then, 2i 2 (mod 2), 12

i 0, 1 (mod 2). If x = 5 i, then from equation (4.7), ( 5 i ) 2 1 (mod 8), 5 2i 1 (mod 8), 5 2i 5 2 (mod 8). Then, 2i 2 (mod 2), i 0, 1 (mod 2). From equation (4.8), x 5 i (mod 8). Put the values of i = 0, 1 then we get, x 1, 5 (mod 8). Again from equation (4.8), x 5 i (mod 8). Put the values of i = 0, 1 then we get, x 3, 7 (mod 8). So final solutions are x 1, 3, 5, 7 (mod 8). Example 4.3. Let m = 2 and k = 4, then equation (4.2) become, x 2 1 (mod 16). (4.9) Every element of U 16 has unique expression of the form ±5i, 5 4 1 (mod 16), so, 1 5 4 (mod 16) and 1 5 0 (mod 16). If x = ±5 i then, x ±5 i (mod 16). (4.10) If x = 5 i, then from equation (4.9), (5 i ) 2 1 (mod 16), 5 2i 1 (mod 16), 5 2i 5 0 (mod 16). 13

Where φ(16) 2 = 4, because it is the order of 2 β 2. Then, If x = 5 i, then from equation (4.9), Then, From equation (4.10), Put the values of i = 0, 2 then we get, Again from equation (4.10), Put the values of i = 0, 2 then we get, 2i 0 (mod 4), i 0, 2 (mod 4). ( 5 i ) 2 1 (mod 16), 5 2i 1 (mod 16), 5 2i 5 0 (mod 16). 2i 0 (mod 4), i 0, 2 (mod 4). x 5 i (mod 16). x 1, 9 (mod 16). x 5 i (mod 16). x 7, 15 (mod 16). So we get final solutions x 1, 7, 9, 15 (mod 2 4 ). 4.3 Solution of Equation x m 1 (mod p q) Let m is a positive integer and q, p be the distinct prime numbers, x m 1 (mod p.q). (4.11) Example 4.4. Let m = 2,p = 3 and q = 5. Where p and q are distinct prime numbers. Then equation (4.11) become, x 2 1 (mod 15), x 2 1 (mod 3 5). We can write above equation into two equations. Then, x 2 1 (mod 3). (4.12) 14

a 1 2 ind 2 a 0 1 Table 4: Indices to the Base 2 Modulo 3 x 2 1 (mod 5). (4.13) First we solve equation(4.12). We find that 2 is the least positive primitive root of 3 (2 2 1 (mod 3)). Now, x 2 1 (mod 3). Taking index on both sides of congruence to base 2 modulo 3. We obtain a congruence modulo φ(3) = 2, By Theorem 2.3 ind 2 1 = 0. So, ind 2 x 2 ind 2 1 (mod 2). 2 ind 2 x 0 (mod 2). According to Theorem 2.4 (2, φ(3)) = 2, ind 2 x 0, 1 (mod 2), x 1, 2 (mod 3). Now we solve equation (4.13). We find that 2 is the least positive primitive root of 5(2 4 1 (mod 5)). Then, a 1 2 3 4 ind 2 a 0 1 3 2 Table 5: Indices to the Base 2 Modulo 5 x 2 1 (mod 5). Taking index on both sides of congruence to base 2 modulo 5. We obtain a congruence modulo φ(5) = 4, By Theorem 2.3 ind 2 1 = 0. So, ind 2 x 2 ind 2 1 (mod 4). 2 ind 2 x 0 (mod 4). According to Theorem 2.4 (2, φ(5)) = 2, ind 2 x 0, 2 (mod 4), x 1, 4 (mod 5). 15

By solving equations (4.12) and (4.13). We get following equations, x 1 (mod 3) (4.14) x 2 (mod 3) (4.15) x 1 (mod 5) (4.16) x 4 (mod 5) (4.17) Now we apply Theorem 2.1 on equation (4.14),(4.15),(4.16) and (4.17) to find the required result. x 1 (mod 3), x = 1 + 3k. (4.18) Put the value of x = 1 + 3k in equation (4.16). Then we get, 1 + 3k 1 (mod 5), 1 + 3k = 1 + 5l, 3k = 5l, k = 5. Put the value of k = 5 in equation (4.18), x = 1 + 3 5l, x 1 (mod 15). Put the value of x = 1 + 3k from equation (4.18) in equation (4.17), 1 + 3k 4 (mod 5), 3k 3 (mod 5), k 1 (mod 5), k = 1 + 5l 1. Now put the value of k = 1 + 5l 1 in equation (4.18), Now use equation (4.15), x = 1 + 3(1 + 5l 1 ), x = 1 + 3 + 15l 1, x = 4 (mod 15). x 2 (mod 3), x = 2 + 3k 1. (4.19) Put the value of x = 2 + 3k 1 from (4.19) in equation (4.16), 2 + 3k 1 1 (mod 5), 3k 1 1 (mod 5), k 1 3 (mod 5), k 1 = 3 + 5l 2. 16

Put the value of k 1 = 3 + 5l 2 in equation (4.19), x = 2 + 3(3 + 5l 2 ), x = 11 + 15l 2, x 11 (mod 15). Put the value of x = 2 + 3k 1 from equation (4.19) in equation (4.17), 2 + 3k 1 4 (mod 5), 3k 1 2 (mod 5), k 1 4 (mod 5), k 1 = 4 + 5l 3. Put the value of k 1 = 4 + 5l 3 in equation (4.19). Then we get, Now we get the required result, x = 2 + 3(4 + 5l 3 ), x = 14 + 15l 3, x 14 (mod 15). x 1, 4, 11, 14 (mod 15). Theorem 4.6. Suppose m be a positive integer and n = p 1 p 2 p k. Then equation x m 1 (mod pq) has (m 1, φ(p 1 ))(m 2, φ(p 2 )) (m k, φ(p k )) number of solutions. 4.4 Solution of Equation x m 1 (mod p k q k ) Let m, k are positive integers and q, p be the distinct prime numbers, x m 1 (mod p k q k ). (4.20) Example 4.5. Let m = 2, p = 3, q = 5 and k = 2. Then equation (4.20) become, x 2 1 (mod 3 2 5 2 ). We can write above equation into two equations. Then, x 2 1 (mod 9). (4.21) x 2 1 (mod 25). (4.22) First we solve equation(4.21). We find that 2 is the least positive primitive root of 3 (2 2 1 (mod 3)). So by Theorem 4.2 it is also primitive root of 9. Taking index on both sides of congruence to base 2 modulo 9 of equation(4.21). We obtain congruence modulo φ(9) = 6, ind 2 x 2 ind 2 1 (mod 6). 17

a 1 2 4 5 7 8 ind 2 a 0 1 2 5 4 3 Table 6: Indices to the Base 2 Modulo 9 By Theorem 2.3 ind 2 1 = 0. So, 2 ind 2 x 0 (mod 6). According to Theorem 2.4 (2, φ(7)) = 2. Then, ind 2 x 0, 3 (mod 6), x 1, 8 (mod 9). Now we solve the equation(4.22). We find that 2 is the least positive primitive root of 5. So by Theorem 4.2 it is also primitive root of 25. a 1 2 3 4 6 7 8 9 11 12 ind 2 a 0 1 7 2 8 5 3 14 16 9 a 13 14 16 17 18 19 21 22 23 24 ind 2 a 19 6 4 13 15 18 12 17 11 10 Table 7: Indices to the Base 2 Modulo 25 Taking index on both sides of congruence to base 2 modulo 25 of equation (4.22). We obtain congruence modulo φ(25) = 20, By Theorem 2.3 ind 2 1 = 0. So, ind 2 x 2 ind 2 1 (mod 20). 2 ind 2 x 0 (mod 20). According to Theorem 2.4 (2, φ(25)) = 2, ind 2 x 0, 10 (mod 20), x 1, 24 (mod 25). By solving equations (4.21) and (4.22), we get following equations, x 1 (mod 9). (4.23) x 8 (mod 9). (4.24) x 1 (mod 25). (4.25) 18

x 24 (mod 25). (4.26) Now we apply Theorem 2.1 on equation (4.23),(4.24),(4.25) and (4.26) to find the required result. First we use equation (4.23), x 1 (mod 9), x = 1 + 9k. (4.27) Put the value of x = 1 + 9k in equation (4.25). Then we get, 1 + 9k 1 (mod 25), 1 + 9k = 1 + 25l, 9k = 25l, k = 25l. Put the value of k = 25l in equation (4.27), x = 1 + 9 25l, x 1 (mod 3 2 5 2 ). Put the value of x = 1 + 9k from equation (4.27) in equation (4.26), 1 + 9k 24 (mod 25), 9k 23 (mod 25), k 22 (mod 25), k = 22 + 25l 1. Now put the value of k = 22 + 25l 1 in equation (4.27), Now use equation (4.24), x = 1 + 9 (22 + 25l 1 ), x = 1 + 198 + 9 25l 1, x = 199 (mod 3 2 5 2 ). x 8 (mod 9), x = 8 + 9k 1. (4.28) Put the value of x = 8 + 9k 1 from (4.28) in equation (4.25), 8 + 9k 1 1 (mod 25), 9k 1 7 (mod 25), k 1 2 (mod 25), k 1 = 2 + 25l 2. 19

Put the value of k 1 = 2 + 25l 2 in equation (4.28), x = 8 + 9(2 + 25l 2 ), x = 26 + 9 25l 2, x 26 (mod 3 2 5 2 ). Put the value of x = 8 + 9k 1 from equation (4.28) in equation (4.26), 8 + 9k 1 24 (mod 25), 9k 1 16 (mod 25), k 1 24 (mod 25), k 1 = 24 + 25l 3. Put the value of k 1 = 24 + 25l 3 in equation (4.28). Then we get, Now we get the required result, x = 8 + 9(24 + 25l 3 ), x = 224 + 9 25l 3, x 224 (mod 3 2 5 2 ). x 1, 26, 199, 224 (mod 3 2 5 2 ). Theorem 4.7. Let m be a positive integer and n = p 1 l1 p 2 l2 p k l k. Then equation x m 1 (mod n) has (m 1, φ(p 1 l 1 ))(m 2, φ(p 2 l 2 )) (m k, φ(p k l k )) number of solutions. 5 Number of r Periodic Points to Monomial Dynamical System In this section, we will discuss about number of cycles to monomial dynamical system x x m (mod q), where q = p k and p is prime. We have derived a formula for r periodic points. The basic material for this section has been taken from [4] and [5]. Definition 5.1. Let r be a positive integer. The set γ = {x o, x 1... x r 1 } of periodic points of period r is said to be a cycle of f is x o = f(x r 1 ) and x j = f(x j 1 ) for 1 j r 1. The length of the cycle is the number of elements in γ. Definition 5.2. Let x r = f r (x o ). If x r = x o for some positive integer r, then x o is said to be a periodic point of f. If r is the least natural number with this property, then we call r the period of x o and x o an r-periodic point. A periodic point of period 1 is called a fixed point of f. 20

Definition 5.3. Let q = p k, where k Z +. Denote q (m) by the number which is the largest divisor of φ(q) and relatively prime to m. Möbius Inversion Formula: Suppose F and f be an arithmetic functions defined for each n Z +. Then, F (n) = d n f(d), if and only if, f(n) = d n µ(d)f (n d). See for more details [4]. Theorem 5.1. If q = p k, where p > 2 and k Z + then the number r-periodic points of dynamical system f(x) = x m is P r (x m, q) = ( ) µ(d) m r/d 1, φ(q), d r and the number of cycles of the length r of f(x) = x m is C r (x m, q) = P r(x m, q) r = 1 µ(d) r d r ( m r/d 1, φ(q) Proof. Suppose f(x) = x m (mod q) monomial dynamical system has r-periodic points. Then, x mr x (mod q), x mr x 0 x(x mr 1 1) 0 (mod q), (mod q). This above equation has (m r 1, φ(q)) solutions, they are d-periodic points for some d r. So (m r 1, φ(q)) = d r P d(x m, q). Now we follow Möbius inversion formula, P r (x m, q) = ( ) µ(d) m r/d 1, φ(q). d r ). Example 5.1. How to calculate the number of cycle for f(x) = x 2 (mod 5 2 ). We know, C r (x m, q) = 1 ( ) µ(d) m r/d 1, φ(q). r Assume r = 4, C 4 (x 2, q) = 1 { } µ(1)(2 4/1 1, 20) + µ(2)(2 4/2 1, 20), 4 C 4 (x 2, q) = 1. d r 21

f(x) = x 2 (mod 5 2 ) has one cycle of length 4 and two fixed points, which we can see by Figure 1. x 0 1 2 3 4 5 6 7 8 9 10 11 12 f(x) = x 2 0 1 4 9 16 0 11 24 14 6 0 21 19 x 13 14 15 16 17 18 19 20 21 22 23 24 f(x) = x 2 19 21 0 6 14 24 11 0 16 9 4 1 Table 8: f(x) = x 2 (mod 5 2 ) Figure 1: f(x) = x 2 (mod 5 2 ) 22

Example 5.2. How to calculate the number of cycle (of length r). Let us consider a monomial dynamical system f(x) = x 2 (mod 11). We know the formula, C r (x m, q) = 1 ( ) µ(d) m r/d 1, φ(q). r Suppose r = 4, d r C 4 (x 2, q) = 1 { } µ(1)(2 4/1 1, 10) + µ(2)(2 4/2 1, 10), 4 C 4 (x 2, q) = 1. Infact the monomial dynamical system f(x) = x 2 has one cycle of length 4 and two fixed points, we can also see from Figure 2. x 0 1 2 3 4 5 6 7 8 9 10 f(x) = x 2 0 1 4 9 5 3 3 5 9 4 1 Table 9: f(x) = x 2 (mod 11) Figure 2: f(x) = x 2 (mod 11) 23

Example 5.3. Second way to find the periodic points of f(x) = x 2 (mod 11) by using x o = x mr o. If x o is an r periodic point of f(x) = x m then x o = f r (x o ). x mr x mr o = x o, o x o = 0, { } x o x mr 1 o 1 = 0, { } x o x 24 1 o 1 = 0, x o { x 15 o 1 } = 0. x 0 (mod 11). (5.1) x 15 1 (mod 11). (5.2) By using the index calculus method we find the following solution. Hence the final solutions are, x 1, 3, 4, 5, 9 (mod 11). x 0, 1, 3, 4, 5, 9 (mod 11). 6 Number of Total Periodic Points to Monomial Dynamical System In this section, we will discuss formula for total number of periodic points for monomial dynamical system. We state some theorems and example to understand it. The material for this section has been taken from [1],[4],[5] and [6]. Definition 6.1. The r(m) is the order of least integer such that m r(m) 1 (mod q (m)) and r(m) φ(q (m)) Theorem 6.1. Let p > 2 be a fixed prime number and m 2 be a natural number. If R r(m). Then, For more detail see [4] R P r (x m, p) = p (m) + 1. r=1 Lemma 6.2. Iff (m r 1) (m g 1) then r g. Proof. If r g then we have to prove (m r 1) (m g 1). Here r g, there q Z such that g = rq. Now m g 1 = (m r ) q 1 = (m r 1)(m r(q 1) +m r(q 2) +m r +1), so (m r 1) (m g 1). Conversely, (m r 1) (m g 1) then we have to prove r g. Here (m r 1) (m g 1), then m g 1 = s(m r 1), where s is an integer. It follows m g 1 0 24

(mod m r 1). By division algorithm g = qr + c, where 0 c < r. Now m qr+c 1 0 (mod m r 1). Then (m r ) q m c 1 0 (mod m r 1) where (m r ) q 1 (mod m r 1). So m c 1 0 (mod m r 1) therefore (m r 1) (m c 1). Since c < r so c = 0, when m c 1 < m r 1. Hence g = q r and r g. Lemma 6.3. Let q = p k for some prime p > 2. Then for each r N, (m r 1, φ(q)) = (m r 1, q (m)). Proof. We know m r 1 1 (mod v 1 ), where v 1 m. Then we can remove the prime factors from φ(q), which divides m. Because they would not change the value of (m r 1, φ(q)). Lemma 6.4. We have (m r 1, q (m)) = q (m)). Proof. Since (m, q (m)) = 1, then from Euler theorem we have m φ(q (m)) 1 (mod q (m)). There exist a smallest r(m) integer such that m r(m) 1 (mod q (m)) and r(m) φ(q (m)). Infact r(m) φ(q (m)). Theorem 6.5. Let q = p k be a prime and m 2 be a natural number. If r r(m). Then, R P r (x m, q) = P r (x m, q) = q (m) + 1. r=1 r r(m) Proof. We are finding the r periodic points of the monomial dynamical system by using the equation, First we solve, x(x mr 1 1) 0 x mr x (mod q), (mod q). x mr 1 1 (mod q). (6.1) The equation (6.1) has (m r 1, q (m)) solutions from Lemma 6.3 Let u be a primitive root of q = p k. For all x that are relatively prime to q, x = u i where i {0, 1, φ(q)}. Now use x = u i in equation (6.1). Then it becomes, u i(mr 1) 1 0 (mod q), this is equivalent to i(m r 1) 0 (mod φ(q)). There are (m r 1, φ(q)) solutions to the above equation. From Lemma 6.3 it is (m r 1, q (m)) solutions. There are q (m) possible values of i. When r = r from Lemma 6.4 we have exactly q (m) solutions. The solutions are among the elements u l1, u l2 u l q (m). All the solutions we get 25

from x mr 1 1 (mod q) are either r periodic points or r periodic points for some r r. Assume r > r and a be an r periodic point. Then a mr 1 1 (mod q). But a must belong to set of solutions of x m r 1 1 (mod q (m)), which is impossible because r > r. If r r then r r. So, if r r then there are no periodic points. Here x 0, if p x, then p x mr 1 1. So we have q (m)+1 periodic points. Theorem 6.6. If w = p k and v = q l are relatively prime numbers. Then, P (x m, wv) = P (x m, w)p (x m, v), = (1 + w (m))(1 + v (m)). Where w (m),v (m) are the largest divisors of φ(w) and φ(v) respectively and both relatively prime to m. Proof. Let w and v are relatively prime. Now we suppose a dynamical system f(x) = x m (mod n), where n is composite number with n = wv. We have to prove P (x m, wv) = P (x m, w)p (x m, v). Here P (x m, wv) are the periodic points of f(x) = x m (mod n). But P (x m, w), P (x m, v) are the periodic points of f(x) = x m (mod w), and f(x) = x m (mod v) respectively. We consider f(x) = x m (mod w). If x is an r periodic points of f(x) = x m (mod w) then x(x m r 1 1) 0 (mod w). By solving x a 1, a 2, a 3 a k1 (mod w) = R 1 (total number of periodic points for w). Again, for f(x) = x m (mod v), we get x b 1, b 2, b 3 b k2 (mod v) = R 2 (total number of periodic points for v). Now using Theorem 2.1 on R 1 and R 2 and we get x c 1, c 2, c 3 c k3 (mod wv) = R 3 (total number of periodic points for wv) then R 3 = R 1 R 2. But, for f(x) = x m (mod n), we obtain x c 1, c 2, c 3 c k3 (mod n) = R 3 (total number of periodic points for n). So we can write R 3 = R 1 R 2. Then, By using Theorem 6.5 we get, P (x m, wv) = P (x m, w)p (x m, v). P (x m, wv) = (1 + w (m))(1 + v (m)). Example 6.1. Find the r-periodic point of a monomial dynamical system f(x) = x 3 (mod 15). f(x) = x 3 (mod 3 5). Now we split it into two parts, f(x) = x 3 (mod 3). (6.2) f(x) = x 3 (mod 5). (6.3) First solve equation (6.2). According to definition 5.3 f(x) = x 3 (mod 3) has w (3) = 2 periodic points and by this 3 1 1 (mod 2), we get r = 1. 26

Now we solve (6.3), by the definition 5.3 f(x) = x 3 (mod 5) has v (3) = 4 periodic points and by 3 2 1 (mod 4), r = 2. We use formula to find the r-periodic points for dynamical system f(x) = x 3 (mod 15). P (x 3, 3 5) = P (x 3, 3) P (x 3, 5), P (x 3, 3 5) = (1 + w (3))(1 + v (5)). Put the values of w (3) = 2 and v (3) = 4, P (x 3, 3 5) = (1 + 2))(1 + 4), P (x 3, 3 5) = 15. Hence monomial dynamical system f(x) = x 3 (mod 15) has 15 periodic points. Example 6.2. Second way to find the r-periodic points of f(x) = x 3 (mod 15). x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 f(x) = x 3 0 1 8 12 4 5 6 13 2 9 10 11 3 7 14 Table 10: f(x) = x 3 (mod 5) Where φ(15) = 8, p (3) = 8 and by 3 2 1 (mod 8), we get r = 2. If x is an r periodic point of f(x) = x 3 (mod 15). Then, Now, x(x 32 1 1) = 0 (mod 15), x(x 8 1) = 0 (mod 3 5). x 0 (mod 3). (6.4) x 8 1 (mod 5). (6.5) x 0 (mod 5). (6.6) x 8 1 (mod 3). (6.7) After solving the equation (6.5) by index calculus method, we get following solutions x 1, 2, 3, 4 (mod 5). Now solve the equation (6.7) by index calculus method and we get following solutions x 1, 2 (mod 3). So we find the solutions, x 0, 1, 2 (mod 3). (6.8) x 0, 1, 2, 3, 4 (mod 5). (6.9) Applying Theorem 2.1 on equation (6.8), (6.9) and we get the final solutions, We can also see by Figure 3. x 0, 1, 2..., 14 (mod 15). 27

Figure 3: f(x) = x 3 (mod 15) Example 6.3. Find the periodic point of dynamical system f(x) = x 2 (mod 2 2 3 2 ). Now we split this equation into two parts, f(x) = x 2 (mod 2 2 ). f(x) = x 2 (mod 3 2 ). First we solve equation f(x) = x 2 (mod 2 2 ). By definition 5.3 f(x) = x 2 (mod 2 2 ) has w (2) = 1 periodic points and r = 1 by 2 1 1 (mod 1). Now we solve the equation f(x) = x 2 (mod 3 2 ). According to definition (5.3) f(x) = x 2 (mod 3 2 ) has v (2) = 3 periodic points and r = 2 by 2 2 1 (mod 3). We use formula to find the r-periodic points for dynamical system f(x) = x 2 (mod 2 2 3 2 ). P (x 2, 2 2 3 2 ) = P (x 2, 2 2 ) P (x 2, 3 2 ), P (x 2, 2 2 3 2 ) = (1 + w (2))(1 + v (2)). Put the values of w (2) = 1 and v (2) = 3 in above formula, P (x 2, 2 2 3 2 ) = (1 + 1))(1 + 3), P (x 2, 2 2 3 2 ) = 8. So monomial dynamical system f(x) = x 2 (mod 2 2 3 2 ) has 8 periodic points, which we can see from Figure 4. 28

x 0 1 2 3 4 5 6 7 8 9 10 11 f(x) = x 2 0 1 4 9 16 25 0 13 28 9 28 13 x 12 13 14 15 16 17 18 19 20 21 22 23 f(x) = x 2 0 25 16 9 4 1 0 1 4 9 16 25 x 24 25 26 27 28 29 30 31 32 33 34 35 f(x) = x 2 0 13 28 9 28 13 0 25 16 9 4 1 Table 11: f(x) = x 2 (mod 2 2 3 2 ) References Figure 4: f(x) = x 2 (mod 2 2 3 2 ) [1] Kenneth H. Rosen,Elementary Numbers Theory and Its Application,5th edition,greg Tobin,Boston Sun Francisco New York in year 2005. [2] T.Nagell,Introduction To Numbers Theory Almqvist and Wiksells Boktryckeri AB,Uppsala in year 1951. [3] Gareth A. Jones and J. Mary Jones, Elementary Number Theory, Printed in Great Britain, Spriner-Verlag London Limited 1998. [4] Marcus Nilsson, Monomial Dynamical Systems in the Field of p adic Numbers and Their Finite Extensions, Printed by: Intellecta Docusys, Gőteborg 2005. [5] Andrei Yu. Khrennikov and Marcus Nilsson, P-adic Deterministic and Random Dynamics Published by Kluwer Academic Publisher, P.O.Box 17, 3300 AA Dordrecht, The Netherlands 2004. 29

[6] Min Sha and Su Hu,15 August 2011, http://arxiv.org/pdf/0910.5550.pdf [7] W.Diffe and M.E.Heilman,New Direction in Cryptography, IEEE Trans.Info, Theory,IT-22, 644-654. [8] A. Yu. Khrennikov, Non-archimedean analysis: Quantum paradoxes, dynamical systems and bilogical models, Kluwer, Dordrecht, 1997.

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