NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS

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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

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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.

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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 sup-ported 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.

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1 Introduction

Carl Friedrich Gauss was a mathematician who published Disquisitiones Arith-meticae 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) = xm _{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 2k _{and also for composite}

numbers.

Section 5 introduces number of cycles to the monomial dynamical system. By using M¨obius 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.

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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 ax_{≡ 1 (mod n) is called}

the order of modulo n, where (a, n) = 1. The order of a modulo n is denoted by ordna.

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 ordnr = φ(n).

Definition 2.5. If unique integer x with 1 ≤ x ≤ φ(n) and a is positive integer with (a, n) = 1, then rx_{≡ 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 n1, n2. . . nk are

pairwise relatively prime positive integers. Then Congruence system x ≡ a1 (mod n1),

x ≡ a2 (mod n2),

.. .

x ≡ ak (mod nk),

has a unique solution modulo n = n1n2. . . nk.

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 incon-gruent 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) indr1 ≡ 0 (mod φ(n)),

(ii) indr(ab) ≡ indra + indrb (mod φ(n)),

(iii) indram≡ m indra (mod φ(n)),

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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 xm ≡ a (mod n) has exactly d incongruent solution modulo n if and only if aΦ(n)d ≡ 1 (mod n), where d = (m, φ(n)) is a positive integer.

Proof. Suppose m and n are positive integers. So the congruence equation is xm≡ a (mod n).

Let r be a primitive root modulo n. Now taking index on both sides with base r,

indrxm≡ indra (mod φ(n)),

m indrx ≡ indra (mod φ(n)).

Let d = (m, φ(n)) and y = indrx. Then,

x ≡ ry (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 - indra then

the linear congruence my ≡ indra (mod φ(n)) has no solution. If d| indra then

my ≡ indra (mod φ(n)) has exactly d incongruent solution modulo φ(n).

Now d| indra if and only if

(φ(n)/d) indra ≡ 0 (mod φ(n)),

indra(φ(n)|d)≡ indr1 (mod φ(n)),

aφ(n)d ≡ 1 (mod n).

So theorem is true.

3 Solution of Equations Modulo a Prime

Num-ber

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.

x2≡ 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,

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By Theorem 2.3 indr1 = 0. So,

2 indrx ≡ 0 (mod p − 1).

According to Theorem 2.4, this equation has d = (2, p − 1) = 2 solutions. Since p is odd.

indrx ≡ l1, l2 (mod p − 1),

where indrc = l1and indrg = l2. So,

x ≡ c, g (mod p).

Hence equation x2_{≡ 1 (mod p) has two incongruent solutions x ≡ c, g (mod p).}

Theorem 3.1. Suppose p be a prime. The equation x2 _{≡ 1 (mod p) has}

(2, φ(p)) = 2 solutions.

Example 3.1. Let p = 3, then equation (3.1) become, x2≡ 1 (mod 3).

We find that 2 is a least positive primitive root of 3 22≡ 1 (mod 3).

Now taking index on both sides of the congruence to base 2 modulo 3. We

a 1 2

ind2a 0 1

Table 1: Indices to the Base 2 Modulo 3

obtain a congruence modulo φ(3) = 2,

ind2x2≡ ind21 (mod 2).

By Theorem 2.3 ind21 = 0. So,

2 ind2x ≡ 0 (mod 2).

According to Theorem 2.4 d = (2, 2) = 2. Then, ind2x ≡ 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,

x2≡ 1 (mod 11).

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a 1 2 3 4 5 6 7 8 9 10 ind2a 0 1 8 2 4 9 7 3 6 5

obtain following table. Taking index on both sides of the congruence to base 2 modulo 11. We obtain a congruence modulo φ(11) = 10,

ind2x2≡ ind21 (mod 10),

2 ind2x ≡ 0 (mod 10).

According to Theorem 2.4 d = (2, φ(11)) = 2, so ind2x ≡ 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+_{. Then find the}

solution of congruence,

xm≡ 1 (mod 2). Let m = 3, then equation become,

x3≡ 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,

xm≡ 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,

indrxm≡ indr1 (mod p − 1).

By Theorem 2.3 indr1 = 0. So,

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According to Theorem 2.4 This equation has k = (m, p − 1) solutions, so indrx ≡ l1, l2. . . lk (mod p − 1),

if indra1= l1. . . indrak = lk. Then solutions of equation are,

x ≡ a1, a2. . . ak (mod p).

Theorem 3.2. The equation xm _{≡ 1 (mod p) has (m, φ(p)) incongruent}

slu-tions, 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, x3≡ 1 (mod 11).

We find that 2 is the least positive primitive root of 11 210_{≡ 1 (mod 11). So}

we obtain following table.

a 1 2 3 4 5 6 7 8 9 10

ind2a 0 1 8 2 4 9 7 3 6 5

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 ind21 = 0. Then,

3 ind2x ≡ 0 (mod 10).

According to Theorem 2.4 (3, 10) = 1. We have unique solution, since ind2x ≡ 0 (mod 10).

So there is one incongruent solution x ≡ 1 (mod 11) of the equation.

4 Solution of Equations Modulo Composite

Num-ber

4.1 Solution of Equation x

m

_{≡ 1 (mod p}

k

₎

Let m, k be positive integers and p a prime number,

xm≡ 1 (mod pk). (4.1)

Theorem 4.1. When p is an odd prime then either r or r + p is the primitive root modulo p2. Where r is a primitive root.

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Theorem 4.2. Let p be an odd prime. Then pk has a primitive root for all positive integers k. Moreover if r is a primitive root modulo p2. Then r is a primitive root modulo pk for all integers k.

Example 4.1. Let m = 5, p = 7 and k = 2. Then equation (4.1) become. x5≡ 1 (mod 72),

x5≡ 1 (mod 49). By Theorem 4.2 rp−1 _{= 3}6_{6≡ 1 (mod 49).}

It follows that ord493 = 42, hence 3 is also primitive root of p2= 49.

Taking index on both sides of congruence to base 3 modulo 49. We obtain congruence modulo φ(49) = 42,

By Theorem 2.3 ind31 = 0. Then,

5 ind3x ≡ 0 (mod 42).

According to the Theorem 2.4, (5, 42) = 1. We get unique solution, ind3x ≡ 0 (mod 42),

so there is one solution of the equation which is x ≡ 1 (mod 72_).

Theorem 4.3. Let m and k are positive integers and p be a prime. Then the number of incongruent solutions of the congruence xm _{≡ 1 (mod p}k_{) is}

(m, φ(pk_)).

4.2 Solution of Equation x

m

≡ 1 (mod 2

k

₎

Suppose m and k are positive integers,

xm≡ 1 (mod 2k). (4.2)

Theorem 4.4. (i) Every odd integer x satisfies this congruence

x2β−2 ≡ 1 (mod 23), (4.3) where β ≥ 3.

(ii) 5 has order 2β−2 _{of equation (4.3), when β ≥ 3.}

(iii) The numbers ±5, ±52_{, ±5}3_{, . . . , ±5}2β−2

form a reduced residue system mod-ulo 2β_{, where β ≥ 3.}

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Theorem 4.5. Consider the congruence,

xq ≡ a (mod 2α). (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· 5k _{(mod 2}α_). _(4.5)

x ≡ (−1)µ· 5y _{(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,

52my≡ (−1)h· 5k _{(mod 2}α_).

Hence k is even, thus a ≡ 1 (mod 4). So 2my ≡ k (mod 2α−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 2α−2. Finally four incongruent solutions x modulo 2α.

Definition 4.1. Universal exponent U of the positive integer n such that aU _{≡ 1}

(mod n), where (a, n) = 1.

Example 4.2. Let m = 2 and k = 3, then equation (4.2) become,

x2≡ 1 (mod 8). (4.7)

Every element of U8 has unique expression of the form ±5i.

52≡ 25 ≡ 1 (mod 8),

so 1 ≡ 52 (mod 8) and 1 ≡ 50 (mod 8). If x = ±5i then,

x ≡ ±5i (mod 8). (4.8)

If x = 5i, then from equation (4.7),

(5i)2≡ 1 (mod 8), 52i≡ 1 (mod 8), 52i≡ 52 _{(mod 8).}

Where φ(8)₂ = 2, because it is the order of 2β−2. Then, 2i ≡ 2 (mod 2),

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i ≡ 0, 1 (mod 2). If x = −5i_{, then from equation (4.7),}

(−5i)2≡ 1 (mod 8), 52i≡ 1 (mod 8), 52i≡ 52 _{(mod 8).} Then, 2i ≡ 2 (mod 2), i ≡ 0, 1 (mod 2). From equation (4.8), x ≡ 5i (mod 8). Put the values of i = 0, 1 then we get,

x ≡ 1, 5 (mod 8). Again from equation (4.8),

x ≡ −5i (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,

x2≡ 1 (mod 16). (4.9)

Every element of U16 has unique expression of the form ±5i,

54≡ 1 (mod 16), so,

1 ≡ 54 (mod 16) and 1 ≡ 50 (mod 16). If x = ±5i _then,

x ≡ ±5i (mod 16). (4.10)

If x = 5i, then from equation (4.9),

(5i)2≡ 1 (mod 16), 52i≡ 1 (mod 16), 52i≡ 50 (mod 16).

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Where φ(16)₂ = 4, because it is the order of 2β−2_{. Then,}

2i ≡ 0 (mod 4), i ≡ 0, 2 (mod 4). If x = −5i_{, then from equation (4.9),}

(−5i)2≡ 1 (mod 16), 52i≡ 1 (mod 16), 52i≡ 50 _{(mod 16).} Then, 2i ≡ 0 (mod 4), i ≡ 0, 2 (mod 4). From equation (4.10), x ≡ 5i (mod 16). Put the values of i = 0, 2 then we get,

x ≡ 1, 9 (mod 16). Again from equation (4.10),

x ≡ −5i (mod 16). Put the values of i = 0, 2 then we get,

x ≡ 7, 15 (mod 16). So we get final solutions x ≡ 1, 7, 9, 15 (mod 24).

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,

xm≡ 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,

x2≡ 1 (mod 15), x2≡ 1 (mod 3 · 5). We can write above equation into two equations. Then,

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a 1 2 ind2a 0 1

x2≡ 1 (mod 5). (4.13)

First we solve equation(4.12).

We find that 2 is the least positive primitive root of 3 (22_{≡ 1 (mod 3)). Now,}

x2≡ 1 (mod 3).

Taking index on both sides of congruence to base 2 modulo 3. We obtain a congruence modulo φ(3) = 2,

2 ind2x ≡ 0 (mod 2).

According to Theorem 2.4 (2, φ(3)) = 2,

ind2x ≡ 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(24≡ 1 (mod 5)). Then,

a 1 2 3 4

ind2a 0 1 3 2

x2≡ 1 (mod 5).

Taking index on both sides of congruence to base 2 modulo 5. We obtain a congruence modulo φ(5) = 4,

2 ind2x ≡ 0 (mod 4).

According to Theorem 2.4 (2, φ(5)) = 2 ,

ind2x ≡ 0, 2 (mod 4),

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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 + 5l1.

Now put the value of k = 1 + 5l1 in equation (4.18),

x = 1 + 3(1 + 5l1),

x = 1 + 3 + 15l1,

x = 4 (mod 15). Now use equation (4.15),

x ≡ 2 (mod 3),

x = 2 + 3k1. (4.19)

Put the value of x = 2 + 3k1from (4.19) in equation (4.16),

2 + 3k1≡ 1 (mod 5),

3k1≡ −1 (mod 5),

k1≡ 3 (mod 5),

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Put the value of k1= 3 + 5l2 in equation (4.19),

x = 2 + 3(3 + 5l2),

x = 11 + 15l2,

x ≡ 11 (mod 15).

Put the value of x = 2 + 3k1from equation (4.19) in equation (4.17),

2 + 3k1≡ 4 (mod 5),

3k1≡ 2 (mod 5),

k1≡ 4 (mod 5),

k1= 4 + 5l3.

Put the value of k1= 4 + 5l3 in equation (4.19). Then we get,

x = 2 + 3(4 + 5l3),

x = 14 + 15l3,

x ≡ 14 (mod 15). Now we get the required result,

x ≡ 1, 4, 11, 14 (mod 15).

Theorem 4.6. Suppose m be a positive integer and n = p1p2· · · pk. Then

equation xm_{≡ 1 (mod pq) has (m}

1, φ(p1))(m2, φ(p2)) · · · (mk, φ(pk)) 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,

xm≡ 1 (mod pkqk). (4.20) Example 4.5. Let m = 2, p = 3, q = 5 and k = 2. Then equation (4.20) become,

x2≡ 1 (mod 3252). We can write above equation into two equations. Then,

x2≡ 1 (mod 9). (4.21)

x2≡ 1 (mod 25). (4.22)

First we solve equation(4.21).

We find that 2 is the least positive primitive root of 3 (22 _{≡ 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,

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a 1 2 4 5 7 8 ind2a 0 1 2 5 4 3

2 ind2x ≡ 0 (mod 6).

According to Theorem 2.4 (2, φ(7)) = 2. Then, ind2x ≡ 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

ind2a 0 1 7 2 8 5 3 14 16 9

a 13 14 16 17 18 19 21 22 23 24

ind2a 19 6 4 13 15 18 12 17 11 10

Taking index on both sides of congruence to base 2 modulo 25 of equation (4.22). We obtain congruence modulo φ(25) = 20,

2 ind2x ≡ 0 (mod 20).

According to Theorem 2.4 (2, φ(25)) = 2,

ind2x ≡ 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)

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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 3252).

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 + 25l1.

Now put the value of k = 22 + 25l1 in equation (4.27),

x = 1 + 9 · (22 + 25l1),

x = 1 + 198 + 9 · 25l1,

x = 199 (mod 3252). Now use equation (4.24),

x ≡ 8 (mod 9),

x = 8 + 9k1. (4.28)

Put the value of x = 8 + 9k1from (4.28) in equation (4.25),

8 + 9k1≡ 1 (mod 25),

9k1≡ −7 (mod 25),

k1≡ 2 (mod 25),

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Put the value of k1= 2 + 25l2 in equation (4.28),

x = 8 + 9(2 + 25l2),

x = 26 + 9 · 25l2,

x ≡ 26 (mod 3252).

Put the value of x = 8 + 9k1from equation (4.28) in equation (4.26),

8 + 9k1≡ 24 (mod 25),

9k1≡ 16 (mod 25),

k1≡ 24 (mod 25),

k1= 24 + 25l3.

Put the value of k1= 24 + 25l3 in equation (4.28). Then we get,

x = 8 + 9(24 + 25l3),

x = 224 + 9 · 25l3,

x ≡ 224 (mod 3252). Now we get the required result,

x ≡ 1, 26, 199, 224 (mod 3252).

Theorem 4.7. Let m be a positive integer and n = p1l1 · p2l2· · · pklk. Then

equation xm_{≡ 1 (mod n) has (m}

1, φ(p1l1))(m2, φ(p2l2)) · · · (mk, φ(pklk))

num-ber 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 7−→ xm _{(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 = {xo, x1. . . xr−1} of

periodic points of period r is said to be a cycle of f is xo = f (xr−1) and

xj = f (xj−1) for 1 ≤ j ≤ r − 1. The length of the cycle is the number of

elements in γ.

Definition 5.2. Let xr= fr(xo). If xr = xo for some positive integer r, then

xo 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 xo and xo an r-periodic point. A

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Definition 5.3. Let q = pk_{, where k ∈ Z}+. Denote q∗(m) by the number which is the largest divisor of φ(q) and relatively prime to m.

M¨obius Inversion Formula: Suppose F and f be an arithmetic functions defined for each n ∈ Z+. Then,

F (n) =X

d|n

f (d),

if and only if,

f (n) =X

d|n

µ(d)F (n|d). See for more details [4].

Theorem 5.1. If q = pk_{, where p > 2 and k ∈ Z}+ then the number r-periodic points of dynamical system f (x) = xm is

Pr(xm, q) =

X

d|r

µ(d)mr/d− 1, φ(q),

and the number of cycles of the length r of f (x) = xm _is

Cr(xm, q) = Pr(xm, q) r = 1 r X d|r µ(d)mr/d− 1, φ(q).

Proof. Suppose f (x) = xm (mod q) monomial dynamical system has r-periodic points. Then,

xmr ≡ x (mod q), xmr− x ≡ 0 (mod q), x(xmr−1− 1) ≡ 0 (mod q).

This above equation has (mr_{− 1, φ(q)) solutions, they are d-periodic points for}

some d|r. So (mr_{− 1, φ(q)) =}P

d|rPd(xm, q). Now we follow M¨obius inversion

formula,

Pr(xm, q) =

X

d|r

µ(d)mr/d− 1, φ(q).

Example 5.1. How to calculate the number of cycle for f (x) = x2 _{(mod 5}2_).

We know, Cr(xm, q) = 1 r X d|r µ(d)mr/d− 1, φ(q). Assume r = 4, C4(x2, q) = 1 4 n µ(1)(24/1− 1, 20) + µ(2)(24/2_{− 1, 20)}o_, C4(x2, q) = 1.

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f (x) = x2 (mod 52) 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) = x2 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) = x2 ₁₉ ₂₁ ₀ ₆ ₁₄ ₂₄ ₁₁ ₀ ₁₆ ₉ ₄ ₁ Table 8: f (x) = x2 _{(mod 5}2₎ Figure 1: f (x) = x2 _{(mod 5}2₎

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Example 5.2. How to calculate the number of cycle (of length r). Let us consider a monomial dynamical system f (x) = x2 (mod 11). We know the formula, Cr(xm, q) = 1 r X d|r µ(d)mr/d− 1, φ(q). Suppose r = 4, C4(x2, q) = 1 4 n µ(1)(24/1− 1, 10) + µ(2)(24/2_{− 1, 10)}o_, C4(x2, q) = 1.

Infact the monomial dynamical system f (x) = x2 _{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) = x2 0 1 4 9 5 3 3 5 9 4 1 Table 9: f (x) = x2 _{(mod 11)}

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Example 5.3. Second way to find the periodic points of f (x) = x2 (mod 11) by using xo= xm

r

o . If xois an r − periodic point of f (x) = xmthen xo= fr(xo).

xmor = xo, xm_or− xo= 0, xo n xm_or−1− 1o= 0, xo n x2_o4−1− 1o= 0, xox15o − 1 = 0. x ≡ 0 (mod 11). (5.1) x15≡ 1 (mod 11). (5.2)

By using the index calculus method we find the following solution. x ≡ 1, 3, 4, 5, 9 (mod 11).

Hence the final solutions are,

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 under-stand 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_b br(m) ≡ 1

(mod q∗(m)) and_br(m) ≤ φ(q∗(m))

Theorem 6.1. Let p > 2 be a fixed prime number and m ≥ 2 be a natural number. If R ≥_br(m). Then,

R

X

r=1

Pr(xm, p) = p∗(m) + 1.

For more detail see [4]

Lemma 6.2. Iff (mr_{− 1)|(m}g_{− 1) then r|g.}

Proof. If r|g then we have to prove (mr_−1)|(mg_{−1). Here r|g, there q ∈ Z such}

that g = rq. Now mg_{−1 = (m}r₎q_{−1 = (m}r_−1)(mr(q−1)_+mr(q−2)_{· · ·+m}r_+1),

so (mr_{− 1)|(m}g_{− 1).}

Conversely, (mr_−1)|(mg_{−1) then we have to prove r|g. Here (m}r_−1)|(mg₋

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(mod mr− 1). By division algorithm g = qr + c, where 0 ≤ c < r. Now mqr+c− 1 ≡ 0 (mod mr_{− 1). Then (m}r₎q_mc _{− 1 ≡ 0 (mod m}r_{− 1) where}

(mr)q ≡ 1 (mod mr_{−1). So m}c_{−1 ≡ 0 (mod m}r_{−1) therefore (m}r_−1)|(mc_−1).

Since c < r so c = 0, when mc− 1 < mr_{− 1. Hence g = q · r and r|g.}

Lemma 6.3. Let q = pk

for some prime p > 2. Then for each r ∈ N, (mr_{− 1, φ(q)) = (m}r_{− 1, q}∗_(m)).

Proof. We know mr_{− 1 ≡ −1 (mod v}

1), where v1|m. Then we can remove the

prime factors from φ(q), which divides m. Because they would not change the value of (mr_{− 1, φ(q)).}

Lemma 6.4. We have (mbr− 1, q∗(m)) = q∗(m)).

Proof. Since (m, q∗(m)) = 1, then from Euler theorem we have mφ(q∗(m)) ≡ 1 (mod q∗(m)). There exist a smallest _br(m) integer such that mbr(m) ≡ 1

(mod q∗(m)) and_br(m) ≤ φ(q∗(m)). Infact_br(m)|φ(q∗(m)).

Theorem 6.5. Let q = pk _{be a prime and m ≥ 2 be a natural number. If}

r|_br(m). Then, R X r=1 Pr(xm, q) = X r|_br(m) Pr(xm, q) = q∗(m) + 1.

Proof. We are finding the r −periodic points of the monomial dynamical system by using the equation,

xmr ≡ x (mod q), x(xmr−1− 1) ≡ 0 (mod q). First we solve,

xmr−1≡ 1 (mod q). (6.1)

The equation (6.1) has (mr_{− 1, q}∗_{(m)) solutions from Lemma 6.3}

Let u be a primitive root of q = pk_{. For all x that are relatively prime to q,}

x = ui _{where i ∈ {0, 1, · · · φ(q)}.}

Now use x = uiin equation (6.1). Then it becomes, ui(mr−1)≡ 10 (mod q), this is equivalent to

i(mr− 1) ≡ 0 (mod φ(q)).

There are (mr_{− 1, φ(q)) solutions to the above equation. From Lemma 6.3 it}

is (mr_{− 1, q}∗_{(m)) solutions. There are q}∗_{(m) possible values of i. When r =}

b r from Lemma 6.4 we have exactly q∗(m) solutions.

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from xmr−1 ≡ 1 (mod q) are either _br − periodic points or r − periodic points for some r|r._b

Assume r >_br and a be an r − periodic point. Then amr−1≡ 1 (mod q). But a must belong to set of solutions of xmbr−1

≡ 1 (mod q∗_{(m)), which is impossible}

because r >_br. If r ≤_br then r|_{r. So, if r -}_b _br then there are no periodic points. Here x ≡ 0, if p|x, then p - xmr−1_{− 1. So we have q}∗_{(m) + 1 periodic points.}

Theorem 6.6. If w = pk and v = ql are relatively prime numbers. Then, P (xm, wv) = P (xm, w)P (xm, 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) = xm (mod n), where n is composite number with n = wv.

We have to prove P (xm, wv) = P (xm, w)P (xm, v). Here P (xm, wv) are the periodic points of f (x) = xm _{(mod n). But P (x}m_{, w), P (x}m_{, v) are the periodic}

points of f (x) = xm _{(mod w), and f (x) = x}m _{(mod v) respectively.}

We consider f (x) = xm _{(mod w). If x is an r − periodic points of f (x) = x}m

(mod w) then x(xmrb−1

− 1) ≡ 0 (mod w). By solving x ≡ a1, a2, a3· · · ak1

(mod w) = R1(total number of periodic points for w). Again, for f (x) =

xm (mod v), we get x ≡ b1, b2, b3· · · bk2 (mod v) = R2(total number of

peri-odic points for v). Now using Theorem 2.1 on R1 and R2 and we get x ≡

c1, c2, c3· · · ck3 (mod wv) = R3(total number of periodic points for wv) then

R3 = R1 · R2. But, for f (x) = xm (mod n), we obtain x ≡ c1, c2, c3· · · ck3

(mod n) = R3(total number of periodic points for n). So we can write R3 =

R1· R2. Then,

P (xm, wv) = P (xm, w)P (xm, v). By using Theorem 6.5 we get,

P (xm, wv) = (1 + w∗(m))(1 + v∗(m)).

Example 6.1. Find the r-periodic point of a monomial dynamical system f (x) = x3 _{(mod 15).}

f (x) = x3 (mod 3 · 5). Now we split it into two parts,

f (x) = x3 (mod 3). (6.2)

f (x) = x3 (mod 5). (6.3)

First solve equation (6.2). According to definition 5.3 f (x) = x3 _{(mod 3) has}

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Now we solve (6.3), by the definition 5.3 f (x) = x3 (mod 5) has v∗(3) = 4 periodic points and by 32 ≡ 1 (mod 4), _br = 2. We use formula to find the r-periodic points for dynamical system f (x) = x3 (mod 15).

P (x3, 3 · 5) = P (x3, 3) · P (x3, 5), P (x3, 3 · 5) = (1 + w∗(3))(1 + v∗(5)). Put the values of w∗_{(3) = 2 and v}∗_{(3) = 4,}

P (x3, 3 · 5) = (1 + 2))(1 + 4), P (x3, 3 · 5) = 15.

Hence monomial dynamical system f (x) = x3 _{(mod 15) has 15 periodic points.}

Example 6.2. Second way to find the r-periodic points of f (x) = x3 _{(mod 15).}

x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

f (x) = x3 ₀ ₁ ₈ ₁₂ ₄ ₅ ₆ ₁₃ ₂ ₉ ₁₀ ₁₁ ₃ ₇ ₁₄

Table 10: f (x) = x3 (mod 5)

Where φ(15) = 8, p∗(3) = 8 and by 32_{≡ 1 (mod 8), we get}

b

r = 2. If x is an r − periodic point of f (x) = x3 _{(mod 15). Then,}

x(x32−1− 1) = 0 (mod 15), x(x8− 1) = 0 (mod 3 · 5). Now, x ≡ 0 (mod 3). (6.4) x8≡ 1 (mod 5). (6.5) x ≡ 0 (mod 5). (6.6) x8≡ 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,

x ≡ 0, 1, 2 . . . , 14 (mod 15). We can also see by Figure 3.

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Figure 3: f (x) = x3 _{(mod 15)}

Example 6.3. Find the periodic point of dynamical system f (x) = x2 _{(mod 2}2_·

32_{). Now we split this equation into two parts,}

f (x) = x2 (mod 22). f (x) = x2 (mod 32).

First we solve equation f (x) = x2 _{(mod 2}2_{). By definition 5.3 f (x) = x}2

(mod 22_{) has w}∗_{(2) = 1 periodic points and}

b

r = 1 by 21_{≡ 1 (mod 1). Now we}

solve the equation f (x) = x2 _{(mod 3}2_{). According to definition (5.3) f (x) = x}2

(mod 32) has v∗(2) = 3 periodic points and r = 2 by 2_b 2 ≡ 1 (mod 3). We use formula to find the r-periodic points for dynamical system f (x) = x2 (mod 22· 32_).

P (x2, 22· 32) = P (x2, 22) · P (x2, 32), P (x2, 22· 32_{) = (1 + w}∗_{(2))(1 + v}∗_(2)).

Put the values of w∗(2) = 1 and v∗(2) = 3 in above formula, P (x2, 22· 32) = (1 + 1))(1 + 3), P (x2, 22· 32_{) = 8.}

So monomial dynamical system f (x) = x2 (mod 22· 32_{) has 8 periodic points,}

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x 0 1 2 3 4 5 6 7 8 9 10 11 f (x) = x2 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) = x2 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) = x2 0 13 28 9 28 13 0 25 16 9 4 1 Table 11: f (x) = x2 _{(mod 2}2_{· 3}2₎ Figure 4: f (x) = x2 _{(mod 2}2_{· 3}2₎

References

[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 Boktryck-eri 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 Num-bers and Their Finite Extensions, Printed by: Intellecta Docusys, G˝oteborg 2005.

[5] Andrei Yu. Khrennikov and Marcus Nilsson, P-adic Deterministic and Ran-dom Dynamics Published by Kluwer Academic Publisher, P.O.Box 17, 3300 AA Dordrecht, The Netherlands 2004.

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[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, dynam-ical systems and bilogdynam-ical models, Kluwer, Dordrecht, 1997.

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NUMBER OF PERIODIC POINTS OF CONGRUENTIAL MONOMIAL DYNAMICAL SYSTEMS

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