TRIPLE SYSTEM and FANO PLANE STRUCTURE in Zn

∗ TRIPLE SYSTEM AND FANO PLANE STRUCTURE IN Zn

Dennis Kinoti Gikunda

A research project submitted in partial fulﬁlment of the requirements for the award of the degree of Masters of Science (Pure Mathematics) in the School of

Pure and Applied Sciences, Kenyatta University

December 2020 ii

DECLARATION

Declaration by the Candidate

This project is my original work and has not been published or presented for a degree award in any other university or any other award.

Signature......

Date......

Dennis Kinoti Gikunda

I56/37638/2017

Declaration by the supervisor

This project has been submitted for examination with my approval as the University supervisor.

Signature......

Date......

Dr.Benard Kivunge

Mathematics Department,

Kenyatta University, Kenya. iii

DEDICATION

This work is dedicated to the community of researchers, in the ﬁeld of modern number theory, combinatorics and computer science. It’s my hope that this work will be put into use, in building secure algorithms for cryptography. iv

ACKNOWLEDGEMENT

I sincerely acknowledge with thanks, my family members, for their unwavering support in the entire period of this study. No words can express my gratitude to you for your precious support.

You’re my rock – I am blessed to have you. Thanks a lot.

Church family, your prayers and encouragement have born fruits. The Gospel is indeed bearing much fruit of its own inherent power.

To my friends, I’m grateful for cheering me on, through this course of life. I sincerely thank you for always giving me an extra push – through words of encouragement and applauding my eﬀorts after every successful step completed.

To my colleagues Waweru, Maina, Tanui, Kioko, Ogeto and Kinyanjui, I am grateful that you made the struggles bearable. I now understand why they say, “on the road to success, there is always we, not me.” We were a rare unit of support. In my next level, I hope to do it again with you, or better still, colleagues like you.

In a world where academic supervisors are largely blamed for delayed research, I wish to sincerely appreciate my supervisor, Dr. Bernard Kivunge for valued support through out my study. You are a true leader. I proudly applause you as my academic mentor. When I start to practise, I deﬁnitely want to be like you. This appreciation extends to the other lecturers from Mathematics

Department, Kenyatta University, for rightly equipping us.

Finally and most importantly, all glory to the Almighty God for the opportunity, strength and provision of all that I needed for this journey. Indeed, You are a Faithful God. v

ABSTRACT

A triple system is an absolutely fascinating concept in projective geometry. This project is an extension of previously done work on triple systems, speciﬁcally the triples that ﬁt into a Fano plane and the (i, j, k) triples of the quaternion group. Here, we have explored and determined

∗ m the existence of triple systems in Zn for n = p, n = pq, n = 2 p and n = pqr with m ∈ N,

∗ p, q, r ∈ P, and p > q > r. A triple system in Zn has been denoted by (k1, k2, k3) where there

2 exists ki > 1, i = 1, 2, 3, such that ki ≡ 1(mod n) with k1k2 ≡ k3(mod n), k1k3 ≡ k2 (mod

∗ n) and k2k3 ≡ k1 (mod n). We have successfully proved that there exists no triples in Zn, for

∗ n = p and n = 2p, p ∈ P. Further, we have established the existence of triples in Zn, for

m n = pq, n = 2 p and n = pqr, where m ∈ N, p, q, r are odd primes and p > q > r. Finally,

∗ m we have ﬁtted the triples of Zn, n = 2 p and n = pqr into Fano Planes. TABLE OF CONTENTS

DECLARATION ...... ii

DEDICATION ...... iii

ACKNOWLEDGEMENT ...... iv

ABSTRACT ...... v

TABLE OF CONTENTS ...... vi

LIST OF TABLES ...... viii

LIST OF FIGURES ...... ix

NOTATIONS ...... x

Chapter 1. INTRODUCTION 1

1.1 Background Information ...... 1

1.2 Deﬁnition of Terms ...... 3

1.3 Statement of the Problem ...... 7

1.4 Objectives of the Study ...... 7

1.4.1 Main Objective ...... 7

1.4.2 Speciﬁc Objectives ...... 7

1.5 Signiﬁcance of the Study ...... 8

Chapter 2. LITERATURE REVIEW 9

2.1 Introduction ...... 9

2.2 Triple Systems ...... 9

2.3 Steiner Triple Systems and Fano Planes ...... 11

2.4 Euler’s Totient Function ...... 12

Chapter 3. RESEARCH METHODOLOGY 14

3.1 Introduction ...... 14

3.2 Research Design ...... 14

vi vii

3.3 Methodology ...... 14

Chapter 4. MAIN RESULTS 16

∗ 4.1 Triple System and Fano Plane Structure in Zn, for n = p, p ∈ P...... 16

∗ m 4.2 Triple System and Fano Plane Structure in Zn, for n = 2 p, p ∈ P, m ∈ N . . . 18

∗ 4.3 Triple System and Fano Plane Structure in Zn, for n = pq, p, q ∈ P, with p > q > 2 30

∗ 4.4 Triple System and Fano Plane Structure in Zn, for n = pqr, p, q, r ∈ P, with p > q > r ...... 45

Chapter 5. CONCLUSIONS AND RECOMMENDATIONS 58

5.1 Conclusion ...... 58

5.2 Recommendations ...... 59

REFERENCES ...... 60 List of Tables

4.1 Multiplication table for k2(mod p) ...... 17

4.2 Multiplication table for k2(mod 2p)...... 19

4.3 Triple systems in n = 22p ...... 21

4.4 Triple systems in n = 2mp ...... 23

4.5 Triple systems in n = 3p ...... 33

4.6 Triple systems in n = 5p ...... 35

4.7 Triple systems in n = 7p ...... 37

4.8 Triple systems in n = 11p ...... 39

4.9 Triple systems in n = 13p ...... 42

4.10 Triple systems in n = pqr ...... 47

viii List of Figures

∗ 4.1 Fano Plane Structure for Z24 ...... 25

∗ 4.2 Fano Plane Structure for Z80 ...... 26

∗ 4.3 Fano Plane Structure for Z224 ...... 28

∗ 4.4 Fano Plane Structure for Z704 ...... 29

∗ 4.5 Fano Plane Structure for Z105 ...... 49

∗ 4.6 Fano Plane Structure for Z165 ...... 51

∗ 4.7 Fano Plane Structure for Z195 ...... 53

∗ 4.8 Fano Plane Structure for Z455 ...... 55

∗ 4.9 Fano Plane Structure for Z1001 ...... 57

ix x

NOTATIONS

P - Set of prime numbers N - Set of natural numbers Z - Set of integers R - Set of real numbers C - Set of complex numbers R - A ring R∗ - The set of units in a ring R H - The quaternion group Zn - The set of integers modulo n ∗ Zn - The set of units in Zn also known as the multiplicative group of Zn ϕ - Eulers phi function. Chapter 1. INTRODUCTION

1.1 Background Information

Any natural number n can be expressed uniquely as a product of primes, for n 6= 0, 1. This implies that prime factors are the ‘building blocks’ for any n ∈ N. Mathematicians ﬁnd delight in understanding the compositional properties of the primes and try to ﬁgure out the structure of their sequence (if any exists). The nature of their existence makes them useful in puzzles, cryptography and generation of security codes. Most of the common private and public key cryptographic systems rely on prime numbers and their role in decomposing other numbers into prime components by factorisation. Considering a case study of RSA (Rivest-Shamir-Adleman), which is an algorithm used to encrypt and decrypt messages by modern computers, a pair of same size prime numbers is needed for every key pair. RSA system is popularly used, an example is SSL (Secure Socket Layer) which is used in https (secure http). While using “secure sites” on the internet, the computer generates both private and public keys for the user. The private key ensures that all personal details like passwords and credit card details remain a secret while on the sites. Examples of such sites include emails and online-pay services.

Thanks to Euclid’s proof that there exist inﬁnitely many prime numbers, in cryptography there are many choice numbers to use as keys or use in generating of keys. This makes it hard for a third party to accurately guess by generating all keys especially those equal to or above

512 − bit, hence guaranteeing secure algorithms.

The prime number theorem has an interesting corollary: Let x be a random “large” number.

The possibility of x being a prime number is 1/ln x. Taking a random string of size n − bit and setting 1 as the least signiﬁcant bit so that it’s always an odd number, the probability of having n as a prime is 1/n (ln 2). From this formula, if you use a 1024 − bit prime, the odds of getting the right code is 1 in 700. With larger primes, the odds get lower. This makes it

1 2 suitable and more secure to use prime numbers as keys or components for key generation, provided there is an eﬀective way of testing whether a number is prime. Galbreath (2003) discusses Miller-Rabin test and the APR test as the common probabilistic tests for primality.

In recent years, mathematicians have made a considerably great progress in a sub-branch of mathematics that concern finite geometries. Herein, we find the concept of triple systems referring to a vector space V over a field K together with K- trilinear mapping

V ⊗ V ⊗ V → V . Girard Desargues (1591 − 1661) discovered the projective geometry derived from Euclidean geometry, which involved 3- dimension ﬁnite geometry. An Italian mathematician, Gino Fano (1871 − 1952) later discussed the 3-dimension ﬁnite geometry with

3 points in each line and 7 points on each plane. The total number of points was 15, with 35 lines and 15 planes. The 3 points on each line form unique triple systems.

Another interesting triple is the (i, j, k) triples of the quaternion group, denoted by

H = {±1, ±i, ±j, ±k}. The discovery of H is quite a famous story in mathematics. William Rowen Hamilton, an Irish mathematician spent much of his life seeking a 3-dimension number system. On 16th Oct 1843, he discovered the fundamental formula for quaternion (action on

3D);

i2 = j2 = k2 = ijk = −1, where i, j, k are imaginary points and 1 is the identity element for multiplication in the set R. The equation above, satisfactorily linked the imaginary part to the commonly known real part.

A very useful relationship of the i, j, k triples in H is also given by:

ij = k = −ji, jk = i = −kj, ki = j = −ik.

∗ The main aspect of this research is the question, “how does the ring Zn, n = p , n = pq and n = pqr, p > q > r for p, q, r ∈ P, connect to the concept of the triple systems in projective geometry, speciﬁcally the Fano plane?” 3

1.2 Deﬁnition of Terms

Definition 1.2.1. A geometry can be defined by a set G = (P,I), where P is the set of points and lines and I the incidence relation that is both reflexive and symmetric. We say a point is incident to the line it lies on and two lines are incident only when they have a point in common.

A relation R on a set A is called reﬂexive if (a, a) ∈ R holds for every element a ∈ A .i.e. if set A = {a, b} then R = (a, a), (b, b) is reﬂexive relation while symmetric relation refers to transformations mapping the object to itself i.e. a relation R on a set A is called symmetric if

(b, a) ∈ R holds when (a, b) ∈ R.

Deﬁnition 1.2.2. Suppose G = (Π,I) is a geometry. Then a ﬂag of G is a set of elements of

Π which are mutually incident. If there is no element outside of the ﬂag, F, which can be added and also be a ﬂag, then F is called maximal.

Deﬁnition 1.2.3. A geometry G = (Π,I) is of rank n if it can be dismembered into sets

Π1, ...... Πn such that every maximal mutually incident set contains precisely one element of each given set.

Deﬁnition 1.2.4. The study of geometric properties that are invariant with respect to projective transformations deﬁne projective geometry. Let a triple G = (P, L, I) be a rank 2 geometry with P = set of points and L = set of lines. Any geometry satisfying the following axioms is a type of projective geometry.

G1 : Any 2 distinct points are incident to a unique line.

G2 : Any 2 lines on the plane meet.

G3 : Any line is incident with atleast 3 points.

G4 : There are atleast 2 lines. 4

Deﬁnition 1.2.5. A projective space is a projection of a 2 - dimensional space to a 3 - dimensional space by adding a point at inﬁnity so that there exists no parallel lines. A projective space with at least 2 lines, such that any 2 distinct lines are incident to a unique point is called a projective plane.

Definition 1.2.6. The order of a finite projective space is given by the number of points that are incident to each line, minus one. Any finite projective plane of order n contains n2 + n + 1 points. A Fano plane is the smallest finite projective plane. It is of order n = 2. the total number of points is 7.

Deﬁnition 1.2.7. A Steiner system, denoted by S(t, k, v), is a set X of v points, and a collection of subsets of X of size k (called blocks), such that any t points of X are in exactly one of the blocks. The special case t = 2 and k = 3 corresponds to a so-called Steiner triple system.

Deﬁnition 1.2.8. Let n be any positive whole number, if 1 and n 6= 1 are the only factors of n, n is said to be a prime number, denoted by p. We will denote the set of prime numbers by

P. If n has more than 2 distinct factors, it is called a composite number.

Deﬁnition 1.2.9. x ≡ y (mod n) implies that n|(x-y) and is read as ‘x is congruent to y modulo n’. In other words, x and y have the same remainders when divided by n.

Deﬁnition 1.2.10. A ring is a non-empty set R with 2 binary operations + (additon) and ·

(multiplication) such that the following axioms are satisﬁed.

R1 : (R, +) is an abelian group. i.e it satisﬁes the following axioms (G1 to G4).

G1 : closure; ∀ a, b ∈ R : a + b = b + a ∈ R.

G2 : associativity; ∀ a, b, c ∈ R :(a + b) + c = a + (b + c). 5

G3 : identity; ∃ 0R ∈ R : 0R + a = a = a + 0R.

G4 : inverse; ∀ a ∈ R, ∃ − a ∈ R : a + (−a) = 0R = (−a) + a

R2 : multiplication is associative i.e. (a · b) · c = a · (b · c) ∀ a, b, c ∈ R.

R3 : multiplication is distributive over addition i.e ∀ a, b, c ∈ R;

a · (b + c) = a · b + a · c → left distributive law and

(a + b) · c = a · c + b · c → right distributive law.

Deﬁnition 1.2.11. Let (R, +, ·) be a ring. Then the set of units for this ring is given by R∗; where the units are elements in R with multiplicative inverse.

2 Deﬁnition 1.2.12. An element k ∈ Zn is idempotent if k ≡ 1(modn).

+ Deﬁnition 1.2.13. Let k ∈ Z . Then Eulers phi function ϕ(k) denotes the number of positive integers ≤ k and relatively prime to k.

Theorem 1.2.1. Euler’s Theorem

ϕ(n) Given k ∈ N with (k, n) = 1, k ≡ 1 (mod n).

Theorem 1.2.2. (Chinese Remainder Theorem)

Let gcd (p, q) = 1. Given a, b ∈ Z, the system of equations x ≡ a(mod p) and x ≡ b(mod q) has a unique solution for x(mod pq)

Corollary 1.2.1. Let n1, n2, ...... , nk be pairwise co-prime positive integers and x1, x2, ...... , xk be arbitrary integers. The system of simultaneous congruence 6

a ≡ x1(mod n1) a ≡ x2(mod n2) . . a ≡ xk(mod nk) has a unique solution mod n, where n = n1n2 . . . nk

k1 k2 kr Theorem 1.2.3. If n is an odd number with n = p1 p2 ...... pr where p1, p2, ...... , pr

2 are distinct odd primes and ki > 0 for 1 ≤ i ≤ r, then the equation x ≡ 1 (mod n) has exactly 2r distinct solutions (mod n)

2 k1 k2 kr k1 k2 kr 2 Proof. Suppose x ≡ 1(mod p1 p2 ...... pr ), then p1 p2 ...... pr | (x − 1). But since

0 k1 k2 kr 2 ki 2 pis are distinct primes, p1 p2 ...... pr | (x − 1) only happens iﬀ pi | (x − 1) for all i, 1 ≤ i ≤ r.

ki But each of the congruences only has two solutions i.e. x ≡ 1 ± (mod pi ).

For each i, 1 ≤ i ≤ r choose yi = ±1 and utilize the linear congruences system.

k1 x ≡ y1(mod p1 ) k2 x ≡ y2(mod p2 ) . . kr x ≡ yr(mod pr )

k1 k2 kr By Chinese Remainder Theorem, the above system has a unique solution (mod n) = (mod p1 p2 ...... pr ).

Since we have 2 choices for each yi (namely ±1), and we have r congruences, then, the possible

r choices for y1 ... , yr are 2

r Assuming that the 2 choices of x are not distinct (mod n), i.e x1 ≡ x2(mod n), then x1 ≡

ki ki x2(mod pi ) for all i. However, any 2 values of x are not congruent pi for at least one i. Therefore, the above system of linear congruences has 2r distinct solutions x(mod n). Any of

r 2 ki r the 2 choices satisﬁes x ≡ (mod pi ) for 1 ≤ i ≤ r. Hence, there are 2 distinct solutions to x ≡ 1(mod n). 7

1.3 Statement of the Problem

A good number of mathematicians have done research on the concept of triples. The previous studies have majorly involved the relationship between points and lines on a 3D - projective plane using both the Euclidean and non-Euclidean approaches.

∗ Equivalently, Zn, the multiplicative group of Zn, has been of interest to mathematicians especially in the ﬁeld of number theory . One property of Zn as a ring, is existence of idempotent elements under multiplication modulo n, n ∈ N.

∗ Here, we determine the number of idempotent elements in Zn, the number of triple systems, Fano planes and higher projective geometry.

1.4 Objectives of the Study

1.4.1 Main Objective

∗ To explore and determine the existence of triple systems in Zn for n = p , n = pq and n = pqr with p, q, r ∈ P, p > q > r and give their projective geometry structure.

1.4.2 Speciﬁc Objectives

∗ i. To determine the number of idempotent elements in Zn

∗ ii. To test existence of triple system in Zn for n = p (prime).

∗ iii. To examine existence of triple systems in Zn for n = pq where q = 2, q = 3, q = 5, q = 7, q = 11 and q = 13 and determine the number of triple systems (where they

exist)and further, derive the general form of the triple systems.

∗ m iv. To establish the existence of triple system and Fano plane structure in Zn, for n = 2 p,

p ∈ P, m ∈ N.

∗ v. To compute triple systems and Fano plane structure in Zn, for n = pqr, p, q, r ∈ P, with p > q > r. 8

1.5 Signiﬁcance of the Study

This research study is an interesting endeavour of the rich relationship between projective geometry and number theory. Of great interest is the connection between the set of units in integers modulo n, the triples and Fano planes. These are valuable ﬁndings to researchers of projective geometry and those in the ﬁeld of number theory.

Consequently, this geometric intuition is important to computer scientists who build on abstract concepts to develop cryptography for secure algorithms. The modulus operations establish periodic functions that physicists may use in working with waves restricted to a certain neighbourhood. Thus, those very basic items in the properties of Zn have very important application. Chapter 2. LITERATURE REVIEW

2.1 Introduction

Coxeter (2003) discusses the genesis of projective geometry as an extension of Euclids work in geometry. He says that Euclidean geometry is so well known to describe the sides of objects in terms of lengths, intersecting lines and angles between the intersecting lines. Additionally, two parallel lines lying on the same plane meet at no point. Considering other geometrical perspec- tives like a cameras imaging process, the insuﬃciency of Euclidean geometry is revealed. Here, transformations increase to include perspective projections, which alter lengths and angles and often a time distort parallelism hence the development of projective geometry.

2.2 Triple Systems

In their book on triple systems, Colbourn and Rosa (1999) argue that among the combinatorial designs, the triple systems are the simplest. They present the triple systems as not limited to geometry but also cutting across algebra, ﬁnite ﬁelds and group theory. Triple system have found a clear application in cryptography and coding theory, among other computer application.

From the deﬁnition of a Steiner triple system as a ﬁnite linear space, (X, A) that is non-empty and all lines of size 3, Doyen and Wilson (1973) discuss the embedding of Steiner triple system.

In their paper, they heavily rely on modulo operations to argue their case.

Still on the concept of modulo operations, Hung and Mendelsohn (1973) show that directed triple systems can be categorized under the examples of block designs while dealing with directed graphs. From a set S of say, r elements, they made a collection of k-subsets of the set S and assigned elements of each k-subset to the vertices of the directed graph. The collection of the subsets was shown to be a block design on the directed graph.

9 10

Ball and Coxeter (1937) deﬁned the Kirkman triple system of order v = 6n + 3 to refer to a triple system with parallelism. Here, parallelism in any aﬃne plane referes to two lines that are either disjoint or equal. Stinson (1987) developed the idea of introducing frames to the Kirkman triple system.

Johnnson and Wellet (2001) constructed a class of parity-check matrices with weight 3 columns to present an analytic method for constructing low-density parity-check (LDPC) codes. The weight 3 columns were a concept of the Kirkman triple systems.

Using Zn, the set of integers modulon, Lu (1983) let S(v) represent the maximum parallelism in the Steiner triple system and proved that if S(2 + n) = n, for p ∈ P, p ≡ 7(mod8) or p ∈ {5, 17, 19, 29} and if (p, n) 6= (5, 1), then S(2 + pn) = pn.

While discussing Steiner triple systems that are non–isomorphic, Wilson (1974) deﬁnes the concept of isomorphism in Steiner Triple systems. He argues that 2 Steiner triple systems, say

(X1, B1) and (X2, B2) are said to be isomorphic provided there is a 1–1 and onto mapping

θ : X1 → X2 such that B ∈ B1 iﬀ θ(A) ∈ B2.

Skolem (1959) showed easy ways of constructing Steiner triple systems for sets containing (6n+1) or (6n + 3) elements. He showed that the sets containing (6n + 1) elements had triples of the form; (x, 2n + x, 4n + x), (n + x, 2n + x, 6n), (3n + x, 4n + x, 6n), (5n + x, x, 6n) and

(x, y, z) for x = 0, 1, . . . , n − 1 with x + y ≡ 2z (mod 2n). The sets containing (6n + 3) elements had triples of the form; (x, 2n + x + 1, 4n + x + 2) and (x, y, z) for x = 0, 1,..., 2n with x + y ≡ 2z (mod 2n + 1). Later, he was able to construct triple systems given a set whose number of elements is a product of primes, of the form (6n + 1). One of his examples was for a set containing (6k + 1) elements. His method yielded 2k distinct Steiner systems. 11

2.3 Steiner Triple Systems and Fano Planes

Steiner triple systems require a set with at least 3 elements and is equivalent to 1 or 3 mod6.

Such sets have a cardinality of 3, 7, 9, 13, 15, 21, 25, 27, and so on. Cremona (1960) argues that the Steiner triple system of cardinality 7 is of special interest as it happens to be the smallest ﬁnite projective plane of order 2, containing 7 points and 7 lines. This order 2 ﬁnite plane is the Fano plane. A Fano plane is therefore a 2-dimensional projective plane of order 2 and is a good example of a Steiner triple system that involves the set of integers and the modulus operations.

An application of Fano plane concept was applied by Lehmer and Lehmer (1974) and successfully factored integers by quadratic form. They used the triples to form lines of the projective geometry on seven points. The planes formed were Fano conﬁgurations that corresponded to pairs of residue classes mod 24. They represented the information in tabular form of a 7 by 7 matrix. The method can be used to factor an arbitrary N by representing N by one of at most three quadratic forms: λN = x2 −Dy2, λ = 1, −1, 2 and D = −1, ±2, ±3, ±6. These three forms appropriate to N, together with inequalities for y, are given for all N prime to 6. This method is practically eﬀective and economical while dealing with numbers with digits as many as 20 to 25.

Using multiplicative properties of the Quaternion set, H = {±1, ±i, ±j, ±k}, Ramo (2011) ex- plains the multiplication of Octonions, denoted by O = {1, i0, i1, i2, i3, i4, i5, i6} and ﬁts them into a Fano plane with directed lines. Vector 1 ∈ O is the multiplicative identity and raising

2 2 2 2 the other vector basis to power 2, they are all equal to −1. I.e. i0 = i1 = i2 = i3 =

2 2 2 i4 = i5 = i6 = −1. Every line has vectors ir, is and it where iris = ±it. The associativity property does not hold in this set, hence the directed Fano plane. (Hung & Mendelsohn 1973).

In mathematical combinatorics, we get the Transylvanian lottery, which is a lottery where the player picks three numbers between 1 and 14 for a single ticket, with the three numbers chosen 12 randomly. If two of the numbers on a given ticket are among the numbers selected in random, the player wins. Using the the Fano plane, one can tabulate the number of tickets a player must buy in order to be certain of winning the lottery. (Mazur 2010, p.280 problem 15).

A player that buys a total of 14 tickets, in two sets of seven, is sure to win. A single set of seven represents all the 7 lines of a Fano plane with the numbers 1−7 and the other with numbers 8−14, i.e.: (1, 2, 5), (1, 3, 6), (1, 4, 7), (2, 3, 7), (2, 4, 6), (3, 4, 5), (5, 6, 7), (8, 9, 12), (8, 10, 13), (8, 11, 14),

(9, 10, 14), (9, 11, 13), (10, 11, 12), (12, 13, 14). The probability of getting all the three numbers in one ticket is 1/26, the probability of having a ticket with two matched numbers is 21/26 and the probability of having three diﬀerent tickets with two of the mentioned numbers is 4/26.

2.4 Euler’s Totient Function

Euler is well known to researchers in the ﬁeld of Number Theory for investigating properties of numbers, especially the distribution of prime numbers. Koshy(2002) views Euler’s phi function, denoted by ϕ, as one of the important functions Euler deﬁned for measuring “breakability” of a number. Considering any number n, ϕ(n) is an output for the count of integers ≤ n, sharing no common factor with n. Example, to compute ϕ(6), we examine all values from 1 to 6, then count the integers with which 6 does not share a common factor > 1. The numbers

2, 3 and 4 are excluded, leaving 5 and 1. Hence ϕ(6) = 2. By Euler’s theorem, we can verify that 52 ≡ 1 (mod 6) and 12 ≡ 1 (mod 6).

It may be hard and tiresome calculating the phi function for a “large” n, except in the case where n is a prime number. Since prime numbers have no factors other than the integer 1 and the number itself, the phi of any p, a prime number, is simply p − 1 i.e. ϕ(p) = p − 1

Koshy(2002). Consider a prime number 11. We have that ϕ(11) = 10, since all integers in the range 1 to 10 share no factor with 11. Consequently, asked to ﬁnd phi of the ten thousandth 13 prime, you’d only need to subtract one to get the solution. I.e. ϕ(104729) = 104728. Hence, given any prime, it’s easy to compute its phi function.

Another interesting property of the phi function is the fact that it’s also multiplicative. I.e.

ϕ(A · B) = ϕ(A) · ϕ(B). Considering a number n, which is a product of two primes, say p and q, then ϕ(n) = ϕ(p · q) = ϕ(p) · ϕ(q) ⇒ ϕ(n) = (p − 1)(q − 1). Consequently, ϕ(pn) = pn − pn−1. Chapter 3. RESEARCH METHODOLOGY

3.1 Introduction

Here is the outline of the research designs and methodology that was used in achieving the objectives mentioned.

3.2 Research Design

This project is an assembly of deﬁnitions, propositions, theorems and proofs drawn from both the projective geometry and number theory for presenting the qualitative results. We also apply

∗ quantitative approach in computation of Zn triples. To compute the triples, we use the Euler’s phi function to assert their accurate count. Chinese remainder theorem and its corollary are key in proving some of the results. The Fano planes are generated by idempotent elements that total seven and have triples. Using various examples, we illustrate workability of the results obtained.

3.3 Methodology

Using the Euler’s phi function, also known as Euler Totient function, we determined the number

∗ of integers k, that are relatively prime to n in Zn, for the range 1 < n < 200. We developed and used algorithms in Ms. Excel that assisted in computation, veriﬁcation and listing of these prime numbers. This included an algorithm that lists the numbers co-prime to n, followed by algorithms that discriminate the idempotent elements.

We also constructed another algorithm in Ms. Excel program for generating the triple systems for integer multiplication modulo n. The program listed the solutions to k2 ≡ 1(mod n), for n = p , n = pq and n = pqr, p > q > r with p, q, r ∈ P. The general results were obtained by deduction and conﬁrmed to be true using the Chinese Remainder Theorem.

Some of the triple solutions were seen to ﬁt into the Fano Planes and conclusions made on the

14 15

∗ projective geometry structure in Zn.

The computational tests are in three categories:

(i) For n = p we investigated if there exists triple system for p = 2, 3.

(ii) For n = pq, p > q with p, q ∈ P, we found out the existing triple systems by computing for the range 1 < p < 200, 2 < q < 13. By deductive reasoning, we drew conclusions,

from which we have the general form for any n = pq, p > q with p, q ∈ P.

(iii) For n = pqr, p > q > r with p, q, r ∈ P, we computed the triples existing for the range p, q, r ∈ {3, 5, 7, 11, 13}. The resulting triples were ﬁtted into Fano plane structures.

Using deductive reasoning, we have made conclusions, from which we have the general

form for any n = pqr, p > q > r with p, q, r ∈ P.

We have also employed the idea of the Steiner Triple system to construct sets of triples which

0 2 are subsets of the set containing kis satisfying k ≡ 1(mod n),. The Steiner Triple Systems are best explained by a collection of elements, say in set S, denoted by points from the set of positive integers. From set S, subsets of 3 elements are constructed and their collection is said to belong to the set T, with the projective geometry property that every pair of unique elements in the set S, appear in exactly one triple. When the set S has 7 elements, then T, the set of triples, perfectly ﬁt into a Fano plane.

∗ For purpose of illustration, we have picked distinct examples for which triples in Zn ﬁt into a Fano plane. The examples presented are a guide on how to ﬁt the triples into Fano Planes. Chapter 4. MAIN RESULTS

∗ 4.1 Triple System and Fano Plane Structure in Zn, for n = p, p ∈ P.

Proposition 4.1.1. If p is prime, 1 and p − 1 are the only integers satisfying k2 ≡ 1(mod p)

∗ in the set Zp

Proof. BWOC, assume that 1 and p − 1 are not the only integers satisfying k2 ≡ 1(mod p) Con-

∗ ∗ sider Zp with Zp = {1, 2, .., p − 1}. Say ∃ another distinct element m with m ∈ {2, 3, ....., p − 2} such that m2 ≡ 1(mod p).

Now, m2 = ap+1, (where a is a natural number). m2 −1 = ap ⇒ p|(m2 −1) ⇒ p|(m−1)(m+1).

Since p is a prime, p|(m − 1) or p|(m + 1). But 1 ≤ m − 1 ≤ p − 3 and 3 ≤ m + 1 ≤ p − 1. Hence p does not divide (m − 1) or (m + 1). Therefore, other than 1 and p − 1, @ integer satisfying k2 ≡ 1(mod p).

16 17

Table 4.1: Multiplication table for k2(mod p) k k2 (mod 5) k2 (mod 7) k2 (mod 11) k2 (mod 13) k2 (mod 17) k2 (mod 19) k2 (mod 23) 1 1 1 1 1 1 1 1 2 4 4 4 4 4 4 4 3 4 2 9 9 9 9 9 4 1 2 5 3 16 16 16 5 4 3 12 8 6 2 6 1 3 10 2 17 13 7 5 10 15 11 3 8 9 12 13 7 18 9 4 3 13 5 12 10 1 9 15 5 8 11 4 2 7 6 12 1 8 11 6 13 16 17 8 14 9 6 12 15 4 16 18 16 1 9 3 17 4 13 18 1 2 19 16 20 9 21 4 22 1 18

∗ m 4.2 Triple System and Fano Plane Structure in Zn, for n = 2 p, p ∈ P, m ∈ N

Proposition 4.2.1. Let n = 2p, p (prime), 1 and 2p − 1 are the only integers satisfying

2 ∗ k ≡ 1(mod 2p) in the set Zn.

Proof. BWOC, assume that 1 and 2p − 1 are not the only integers satisfying k2 ≡ 1(mod 2p)

∗ 2 in Z2p = 1, 2, .., 2p − 1. Suppose ∃ m ∈ 2, 3, ....., 2p − 2 such that m ≡ 1(mod 2p) ⇒ m2 = (2p − 1)2 = 4p2 − 4p + 1 ≡ 1(mod 2p). We have, 2p|(4p2 − 4p) resulting to 2p − 2. Now, m2 = a(2p) + 1, where a is a natural number.

m2 − 1 = a(2p) ⇒ 2p|(m2 − 1) ⇒ 2p|(m − 1)(m + 1)

Since p is prime, p must divide (m − 1) or (m + 1), which means that m ≡ ±1(mod p).

Which is a contradiction, since, other than 1 and p − 1, there exists no other number with m ≡ ±1(mod p). 19

Table 4.2: Multiplication table for k2(mod 2p) k k2 (mod 10) k2 (mod 14) k2 (mod 22) k2 (mod 26) k2 (mod 34) k2 (mod 38) k2 (mod 46) 1 1 1 1 1 1 1 1 2 4 4 4 4 4 4 4 3 9 9 9 9 9 9 9 4 6 2 16 16 16 16 16 5 5 11 3 25 25 25 25 6 6 8 14 10 2 36 36 7 9 7 5 23 15 11 3 8 4 8 20 16 30 26 18 9 1 11 15 3 13 5 35 10 2 12 22 32 24 8 11 9 11 17 19 7 29 12 4 12 14 8 30 6 13 1 15 13 33 17 31 14 20 14 26 6 12 15 5 17 21 35 41 16 14 22 18 28 26 17 3 3 17 23 13 18 16 16 18 20 2 19 9 23 21 19 39 20 4 10 26 20 32 21 1 25 33 23 27 22 16 8 23 24 23 9 19 35 23 24 4 32 7 24 25 1 13 17 27 26 30 30 32 27 15 7 39 28 2 24 2 29 25 5 13 30 16 26 26 31 9 11 41 32 4 36 12 33 1 25 31 34 16 6 35 9 29 36 4 8 37 1 35 38 18 39 3 40 36 41 25 42 16 43 9 44 4 45 1 20

Proposition 4.2.2. Let n = 22p = 4p, p (prime). The equation x2 ≡ 1(mod 22p) has 4

∗ solutions in the set Zn

Proof. x2 ≡ 1(mod 22p) ⇒ x2 ≡ 1(mod 22) and x2 ≡ 1(mod p)

By Chinese Remainder Theorem, x2 ≡ 1(mod 22) has solutions x ≡ ±1(mod 22) while x2 ≡ 1(mod p) has solutions x ≡ ±1(mod p).

Hence, x2 ≡ 1(mod 22p) has 4 solutions.

Remark 4.2.1. By computational analysis, we get similar results as above. The 4 solutions

2 2 ∗ 2 to x ≡ 1(mod 2 p) in the set Zn are 1, 2p − 1, 2p + 1 and 2 p − 1 = 4p − 1. The 3 non-unit solutions form triple systems for all odd p ∈ P. The results are well captured in the table that follows. 21

Table 4.3: Triple systems in n = 22p

2 p n = 2 p k1 k2 k3 2p − 1 2p + 1 22p − 1 3 12 5 7 11 5 20 9 11 19 7 28 13 15 27 11 44 21 23 43 13 52 25 27 51 17 68 33 35 67 19 76 37 39 75 23 92 45 47 91 29 116 57 59 115 31 124 61 63 123 37 148 73 75 147 41 164 81 83 163 43 172 85 87 171 47 188 93 95 187 53 212 105 107 211 59 236 117 119 235 61 244 121 123 243 67 268 133 135 267 71 284 141 143 283 73 292 145 147 291 79 316 157 159 315 83 332 165 167 331 89 356 177 179 355 97 388 193 195 387 101 404 201 203 403 103 412 205 207 411 107 428 213 215 427 109 436 217 219 435 113 452 225 227 451 127 508 253 255 507 131 524 261 263 523 137 548 273 275 547 139 556 277 279 555 149 596 297 299 595 22

∗ m Theorem 4.2.1. Consider the set Zn for n = 2 p, where p is an odd prime and m ∈ N with m ≥ 3. The equation x2 ≡ 1(mod n) has 8 distinct solutions.

Proof. x2 ≡ 1(mod 2mp) ⇒ x2 ≡ 1(mod 2m) and x2 ≡ 1(mod p)

We note that:

i. x2 ≡ 1(mod 23) has 4 solutions, namely x ≡ ±1, ±3(mod 23).

ii. x2 ≡ 1(mod 24) has 4 solutions, namely x ≡ ±1, ±7(mod 24).

iii. x2 ≡ 1(mod 25) has 4 solutions, namely x ≡ ±1, ±15(mod 25).

iv. x2 ≡ 1(mod 26) has 4 solutions, namely x ≡ ±1, ±31(mod 26).

By deductive reasoning, if m ≥ 3 then, x2 ≡ 1(mod 2m) has exactly 22 = 4 solutions i.e. x ≡ ±1, ±(2m−1 − 1)(mod 2m).

By Theorem 1.2.3, there are 21 distinct solutions to the equation x2 ≡ 1(mod n), for n = p,

2 m with p ∈ P where p ≥ 3. Combining with the results above, x ≡ 1(mod 2 p) has exactly 22+1 = 8 solutions 23

Table 4.4: Triple systems in n = 2mp 2m p n = 2mp values of x satisfying x2 ≡ 1(mod n) 23 3 24 5 7 11 13 17 19 23 23 5 40 9 11 19 21 29 31 39 23 7 56 13 15 27 29 41 43 55 23 11 88 21 23 43 45 65 67 87 23 13 104 25 27 51 53 77 79 103

24 3 48 7 17 23 25 31 41 47 24 5 80 9 31 39 41 49 71 79 24 7 112 15 41 55 57 71 97 111 24 11 176 23 65 87 89 111 153 175 24 13 208 25 79 103 105 129 183 207

25 3 96 17 31 47 49 65 79 95 25 5 160 31 49 79 81 111 129 159 25 7 224 15 97 111 113 127 209 223 25 11 352 65 111 175 177 241 287 351 25 13 416 79 129 207 209 287 337 415

26 3 192 31 65 95 97 127 161 191 26 5 320 31 129 159 161 191 289 319 26 7 448 97 127 223 225 321 351 447 26 11 704 65 287 351 353 417 639 703 26 13 832 129 287 415 417 545 703 831 24

From the table above, we pick 4 examples, from which we illustrate computational analysis

m for the triples of n = 2 p for an odd prime p and m ∈ N where m ≥ 3 and ﬁt the triples into Fano Planes.

Example 4.2.1. From the 7 non-unit solutions in Z24, the triples are given by:

i. 5 ∗ 7 ≡ 11(mod 24), 5 ∗ 11 ≡ 7(mod 24), 7 ∗ 11 ≡ 5(mod 24)

⇒ The triple is given by (5, 7, 11)

ii. 5 ∗ 13 ≡ 17(mod 24), 5 ∗ 17 ≡ 13(mod 24), 13 ∗ 17 ≡ 5(mod 24)

⇒ The triple is given by (5, 13, 17)

iii. 5 ∗ 19 ≡ 23(mod 24), 5 ∗ 23 ≡ 19(mod 24), 19 ∗ 23 ≡ 5(mod 24)

⇒ The triple is given by (5, 19, 23)

iv. 7 ∗ 13 ≡ 19(mod 24), 7 ∗ 19 ≡ 13(mod 24), 13 ∗ 19 ≡ 7(mod 24)

⇒ The triple is given by (7, 13, 19)

v. 7 ∗ 17 ≡ 23(mod 24), 7 ∗ 23 ≡ 17(mod 24), 17 ∗ 23 ≡ 7(mod 24)

⇒ The triple is given by (7, 17, 23)

vi. 11 ∗ 13 ≡ 23(mod 24), 11 ∗ 23 ≡ 13(mod 24), 13 ∗ 23 ≡ 11(mod 24)

⇒ The triple is given by (11, 13, 23)

vii. 11 ∗ 17 ≡ 19(mod 24), 11 ∗ 19 ≡ 17(mod 24), 17 ∗ 19 ≡ 11(mod 24)

⇒ The triple is given by (11, 17, 19) 25

Fitting the triples into a Fano plane we have: 23

19 17

5 11 7

∗ Figure 4.1: Fano Plane Structure for Z24

Example 4.2.2. From the 7 non-unit solutions in Z80, the triples are given by:

i. 9 ∗ 31 ≡ 39(mod 80), 9 ∗ 39 ≡ 31(mod 80), 31 ∗ 39 ≡ 9(mod 80)

⇒ The triple is given by (9, 31, 39)

ii. 9 ∗ 41 ≡ 49(mod 80), 9 ∗ 49 ≡ 41(mod 80), 41 ∗ 49 ≡ 9(mod 80)

⇒ The triple is given by (9, 41, 49)

iii. 9 ∗ 71 ≡ 79(mod 80), 9 ∗ 79 ≡ 71(mod 80), 71 ∗ 79 ≡ 9(mod 80)

⇒ The triple is given by (9, 71, 79)

iv. 31 ∗ 41 ≡ 71(mod 80), 31 ∗ 71 ≡ 41(mod 80), 41 ∗ 71 ≡ 31(mod 80)

⇒ The triple is given by (31, 41, 71) 26

v. 31 ∗ 49 ≡ 79(mod 80), 31 ∗ 79 ≡ 49(mod 80), 49 ∗ 79 ≡ 31(mod 80)

⇒ The triple is given by (31, 49, 79)

vi. 39 ∗ 41 ≡ 79(mod 80), 39 ∗ 79 ≡ 41(mod 80), 41 ∗ 79 ≡ 39(mod 80)

⇒ The triple is given by (39, 41, 79)

vii. 39 ∗ 49 ≡ 71(mod 80), 39 ∗ 71 ≡ 49(mod 80), 49 ∗ 71 ≡ 39(mod 80)

⇒ The triple is given by (39, 49, 71)

Fitting the triples into a Fano plane we have: 79

71 49

9 31 39

∗ Figure 4.2: Fano Plane Structure for Z80 27

Example 4.2.3. From the 7 non-unit solutions in Z224, the triples are given by:

i. 15 ∗ 97 ≡ 111(mod 224), 15 ∗ 111 ≡ 97(mod 224), 97 ∗ 111 ≡ 15(mod 224)

⇒ The triple is given by (15, 97, 111)

ii. 15 ∗ 113 ≡ 127(mod 224), 15 ∗ 127 ≡ 113(mod 224), 113 ∗ 127 ≡ 15(mod 224)

⇒ The triple is given by (15, 113, 127)

iii. 15 ∗ 209 ≡ 223(mod 224), 15 ∗ 223 ≡ 209(mod 224), 209 ∗ 223 ≡ 15(mod 224)

⇒ The triple is given by (15, 209, 223)

iv. 97 ∗ 113 ≡ 209(mod 224), 97 ∗ 209 ≡ 113(mod 224), 113 ∗ 209 ≡ 97(mod 224)

⇒ The triple is given by (97, 113, 209)

v. 97 ∗ 127 ≡ 223(mod 224), 97 ∗ 223 ≡ 127(mod 224), 127 ∗ 223 ≡ 97(mod 224)

⇒ The triple is given by (97, 127, 223)

vi. 11 ∗ 13 ≡ 223(mod 224), 111 ∗ 223 ≡ 113(mod 224), 113 ∗ 223 ≡ 111(mod 224)

⇒ The triple is given by (111, 113, 223)

vii. 111 ∗ 127 ≡ 209(mod 224), 111 ∗ 209 ≡ 127(mod 224), 127 ∗ 209 ≡ 111(mod 224)

⇒ The triple is given by (111, 127, 209) 28

Fitting the triples into a Fano plane we have: 223

209 127

113

15 111 97

∗ Figure 4.3: Fano Plane Structure for Z224

Example 4.2.4. From the 7 non-unit solutions in Z704, the triples are given by:

i. 65 ∗ 287 ≡ 351(mod 704), 65 ∗ 351 ≡ 287(mod 704), 287 ∗ 351 ≡ 65(mod 704)

⇒ The triple is given by (65, 287, 351)

ii. 65 ∗ 353 ≡ 417(mod 704), 65 ∗ 417 ≡ 353(mod 704), 353 ∗ 417 ≡ 65(mod 704)

⇒ The triple is given by (65, 353, 417)

iii. 65 ∗ 639 ≡ 703(mod 704), 65 ∗ 703 ≡ 639(mod 704), 639 ∗ 703 ≡ 65(mod 704)

⇒ The triple is given by (65, 639, 703)

iv. 287 ∗ 353 ≡ 639(mod 704), 287 ∗ 639 ≡ 353(mod 704), 353 ∗ 639 ≡ 287(mod 704)

⇒ The triple is given by (287, 353, 639) 29

v. 287 ∗ 417 ≡ 703(mod 704), 287 ∗ 703 ≡ 417(mod 704), 417 ∗ 703 ≡ 287(mod 704)

⇒ The triple is given by (287, 417, 703)

vi. 287 ∗ 417 ≡ 703(mod 704), 287 ∗ 703 ≡ 417(mod 704), 417 ∗ 703 ≡ 287(mod 704)

⇒ The triple is given by (287, 417, 703)

vii. 351 ∗ 417 ≡ 639(mod 704), 351 ∗ 639 ≡ 417(mod 704), 417 ∗ 639 ≡ 351(mod 704)

⇒ The triple is given by (351, 417, 639)

Fitting the triples into a Fano plane we have: 703

639 417

353

65 351 287

∗ Figure 4.4: Fano Plane Structure for Z704 30

∗ 4.3 Triple System and Fano Plane Structure in Zn, for n = pq, p, q ∈ P, with p > q > 2

∗ Proposition 4.3.1. If n = 3p, p (prime) with p > 3, Zn has 2 possible cases of triple system;

Case 1 If p = 3k + 1, the triple system is given by (p + 1, 2p − 1, 3p − 1)

Proof. First, we show that (p + 1)(2p − 1) ≡ (3p − 1)(mod n)

(p + 1)(2p − 1) ≡ (3k + 2)(6k + 1) (mod n)

= 18k2 + 15k + 2 (mod 3p) since n = 3p

= 18k2 + 15k + 2 (mod 9k + 3) since p = 3k + 1 but 18k2 + 15k + 2 = 2k + 1 remainder (−1) 9k + 3 i.e

18k2 + 15k + 2 ≡ −1 (mod 3p)

= 3p − 1

Now we show that (p + 1)(3p − 1) ≡ (2p − 1)(mod n)

We already have:

(3p − 1) ≡ −1 (mod 3p) hence,

(p + 1)(3p − 1) ≡ (p + 1)(−1) (mod 3p)

= (−p − 1) (mod 3p)

= 3p − p − 1

= 2p − 1 31

Finally, we show that (2p − 1)(3p − 1) ≡ (p + 1)(mod n).

We have (3p − 1) ≡ −1 (mod 3p) hence,

(2p − 1)(3p − 1) ≡ (2p − 1)(−1) (mod 3p)

= (−2p + 1) (mod 3p)

= 3p − 2p + 1

= p + 1

Example 4.3.1. For p = 7, n = 3p = 21 and the triple is given by (8, 13, 20).

8 ∗ 13 = 104 ≡ 20 mod (21)

8 ∗ 20 = 160 ≡ 13 mod (21)

13 ∗ 20 = 260 ≡ 8 mod (21)

82 ≡ 132 ≡ 202 ≡ 1(mod 21)

Case 2 If p = 3k + 2, the triples are given by (p − 1, 2p + 1, 3p − 1)

Proof. First, we show that (p − 1)(2p + 1) = (3p − 1)(mod n).

(p − 1)(2p = 1) ≡ (3k + 1)(6k + 5) (mod n)

= 18k2 + 21k + 5 (mod 3p) since n = 3p

= 18k2 + 21k + 5 (mod 9k + 6) since p = 3k + 1 but 18k2 + 21k + 5 = 2k + 1 remainder (−1) 9k + 3 32 i.e

18k2 + 21k + 5 ≡ −1 (mod 3p)

= 3p − 1

Now we show that (p − 1)(3p − 1) = (2p + 1)(mod n)

We already have (3p − 1) ≡ −1 (mod 3p) hence,

(p − 1)(3p − 1) ≡ (p − 1)(−1) (mod 3p)

= (−p + 1) (mod 3p)

= 3p − p + 1

= 2p + 1

Finally, we show that (2p + 1)(3p − 1) = (p − 1)(mod n)

We have (3p − 1) ≡ −1 (mod 3p) hence,

(2p + 1)(3p − 1) ≡ (2p + 1)(−1) (mod 3p)

= (−2p − 1) (mod 3p)

= 3p − 2p − 1

= p − 1

Example 4.3.2. For p = 5, n = pq = 5 ∗ 3 = 15 and the triple is given by (4, 11, 14)

4 ∗ 11 = 44 ≡ 14 mod (15)

4 ∗ 14 = 56 ≡ 11 mod (15)

11 ∗ 14 = 154 ≡ 4 mod (15)

42 ≡ 112 ≡ 142 ≡ 1(mod 15) 33

Table 4.5: Triple systems in n = 3p

p n=3p p(mod 3) k1 k2 k3 1 2p-1 p+1 3p-1 7 21 1 13 8 20 13 39 1 25 14 38 19 57 1 37 20 56 31 93 1 61 32 92 37 111 1 73 38 110 43 129 1 85 44 128 61 183 1 121 62 182 67 201 1 133 68 200 73 219 1 145 74 218 79 237 1 157 80 236 97 291 1 193 98 290 103 309 1 205 104 308 109 327 1 217 110 326 127 381 1 253 128 380 139 417 1 277 140 416 151 453 1 301 152 452 157 471 1 313 158 470 163 489 1 325 164 488 181 543 1 361 182 542 193 579 1 385 194 578 199 597 1 397 200 596 2 p-1 2p+1 3p-1 5 15 2 4 11 14 11 33 2 10 23 32 17 51 2 16 35 50 23 69 2 22 47 68 29 87 2 28 59 86 41 123 2 40 83 122 47 141 2 46 95 140 53 159 2 52 107 158 59 177 2 58 119 176 71 213 2 70 143 212 83 249 2 82 167 248 89 267 2 88 179 266 101 303 2 100 203 302 107 321 2 106 215 320 113 339 2 112 227 338 131 393 2 130 263 392 137 411 2 136 275 410 149 447 2 148 299 446 167 501 2 166 335 500 173 519 2 172 347 518 179 537 2 178 359 536 191 573 2 190 383 572 197 591 2 196 395 590 34

∗ Proposition 4.3.2. If n = 5p, p (prime) with p > 5, for some k ∈ N, Zn has 4 possible cases of triple system;

Case 1. if p = 5k + 1 the triple is (2p − 1, 3p + 1, 5p − 1) e.g. for p = 11, the triple is given by

(21, 34, 54)

Case 2. if p = 5k + 2 the triple is (p − 1, 4p + 1, 5p − 1) e.g. for p = 7, the triple is given by

(6, 29, 34)

Case 3. if p = 5k + 3 the triple is (p + 1, 4p − 1, 5p − 1) e.g. for p = 13, the triple is given by

(14, 51, 64)

Case 4. if p = 5k + 4 the triple is (2p + 1, 3p − 1, 5p − 1) e.g. for p = 19, the triple is given by

(39, 56, 94)

Proof. Similar to the proof of Proposition 4.3.1 35

Table 4.6: Triple systems in n = 5p p n=5p p(mod 5) k1 k2 k3 1 2p-1 3p+1 5p-1 11 55 1 21 34 54 31 155 1 61 94 154 41 205 1 81 124 204 61 305 1 121 184 304 71 355 1 141 214 354 101 505 1 201 304 504 131 655 1 261 394 654 151 755 1 301 454 754 181 905 1 361 544 904 191 955 1 381 574 954 2 p-1 4p+1 5p-1 7 35 2 6 29 34 17 85 2 16 69 84 37 185 2 36 149 184 47 235 2 46 189 234 67 335 2 66 269 334 97 485 2 96 389 484 107 535 2 106 429 534 127 635 2 126 509 634 137 685 2 136 549 684 157 785 2 156 629 784 167 835 2 166 669 834 197 985 2 196 789 984 3 4p-1 p+1 5p-1 13 65 3 51 14 64 23 115 3 91 24 114 43 215 3 171 44 214 53 265 3 211 54 264 73 365 3 291 74 364 83 415 3 331 84 414 103 515 3 411 104 514 113 565 3 451 114 564 163 815 3 651 164 814 173 865 3 691 174 864 193 965 3 771 194 964 4 3p-1 2p+1 5p-1 19 95 4 56 39 94 29 145 4 86 59 144 59 295 4 176 119 294 79 395 4 236 159 394 89 445 4 266 179 444 109 545 4 326 219 544 139 695 4 416 279 694 149 745 4 446 299 744 179 895 4 536 359 894 199 995 4 596 399 994 36

∗ Proposition 4.3.3. If n = 7p, p(prime) with p > 7 for some k ∈ N, Zn has 6 possible cases of triple system;

Case 1. if p = 7k + 1 the triple is (2p − 1, 5p + 1, 7p − 1) e.g. for p = 29, the triple is given by

(57, 146, 202)

Case 2. if p = 7k + 2 the triple is (p − 1, 6p + 1, 7p − 1) e.g. for p = 23, the triple is given by

(22, 139, 160)

Case 3. if p = 7k + 3 the triple is (3p − 1, 4p + 1, 7p − 1) e.g. for p = 17, the triple is given by

(50, 69, 118)

Case 4. if p = 7k + 4 the triple is (3p + 1, 4p − 1, 7p − 1) e.g. for p = 11, the triple is given by

(34, 43, 76)

Case 5. if p = 7k + 5 the triple is (p + 1, 6p − 1, 7p − 1) e.g. for p = 19, the triple is given by

(20, 113, 132)

Case 6. if p = 7k + 6 the triple is (2p + 1, 5p − 1, 7p − 1) e.g. for p = 13, the triple is given by

(27, 64, 90)

Proof. Similar to the proof of Proposition 4.3.1 37

Table 4.7: Triple systems in n = 7p

p n=7p p(mod 7) k1 k2 k3 1 2p-1 5p+1 7p-1 29 203 1 57 146 202 43 301 1 85 216 300 71 497 1 141 356 496 113 791 1 225 566 790 127 889 1 253 636 888 197 1379 1 393 986 1378 2 p-1 6p+1 7p-1 23 161 2 22 139 160 37 259 2 36 223 258 79 553 2 78 475 552 107 749 2 106 643 748 149 1043 2 148 895 1042 163 1141 2 162 979 1140 191 1337 2 190 1147 1336 3 3p-1 4p+1 7p-1 17 119 3 50 69 118 31 217 3 92 125 216 59 413 3 176 237 412 73 511 3 218 293 510 101 707 3 302 405 706 157 1099 3 470 629 1098 199 1393 3 596 797 1392 4 4p-1 3p+1 7p-1 11 77 4 43 34 76 53 371 4 211 160 370 67 469 4 267 202 468 109 763 4 435 328 762 137 959 4 547 412 958 151 1057 4 603 454 1056 179 1253 4 715 538 1252 193 1351 4 771 580 1350 5 6p-1 p+1 7p-1 19 133 5 113 20 132 47 329 5 281 48 328 61 427 5 365 62 426 89 623 5 533 90 622 103 721 5 617 104 720 131 917 5 785 132 916 173 1211 5 1037 174 1210 6 5p-1 2p+1 7p-1 13 91 6 64 27 90 41 287 6 204 83 286 83 581 6 414 167 580 97 679 6 484 195 678 139 973 6 694 279 972 167 1169 6 834 335 1168 181 1267 6 904 363 1266 38

∗ Proposition 4.3.4. If n = 11p, p(prime) with p > 11 for some k ∈ N, Zn has 10 possible cases of triple system;

Case 1. if p = 11k + 1 the triple is (2p − 1, 9p + 1, 11k − 1) e.g. for p = 23, the triple is given

by (45, 208, 252)

Case 2. if p = 11k + 2 the triple is (p − 1, 10p + 1, 11k − 1) e.g. for p = 13, the triple is given

by (12, 131, 142)

Case 3. if p = 11k + 3 the triple is (3p + 1, 8p − 1, 11k − 1) e.g. for p = 47, the triple is given

by (142, 375, 516)

Case 4. if p = 11k + 4 the triple is (5p + 1, 6p − 1, 11k − 1) e.g. for p = 37, the triple is given

by (186, 221, 406)

Case 5. if p = 11k + 5 the triple is (4p + 1, 7p − 1, 11k − 1) e.g. for p = 71, the triple is given

by (285, 496, 780)

Case 6. if p = 11k + 6 the triple is (4p − 1, 7p + 1, 11k − 1) e.g. for p = 17, the triple is given

by (67, 120, 186)

Case 7. if p = 11k + 7 the triple is (5p − 1, 6p + 1, 11k − 1) e.g. for p = 29, the triple is given

by (144, 175, 318)

Case 8. if p = 11k + 8 the triple is (3p − 1, 8p + 1, 11k − 1) e.g. for p = 19, the triple is given

by (56, 153, 208)

Case 9. if p = 11k + 9 the triple is (p + 1, 10p − 1, 11k − 1) e.g. for p = 31, the triple is given

by (32, 309, 340)

Case 10. if p = 11k + 10 the triple is (2p + 1, 9p − 1, 11k − 1) e.g. for p = 43, the triple is given

by (87, 386, 472)

Proof. Similar to the proof of Proposition 4.3.1 39

Table 4.8: Triple systems in n = 11p

p n=11p p(mod11) k1 k2 k3 1 2p-1 9p+1 11p-1 23 253 1 45 208 252 67 737 1 133 604 736 89 979 1 177 802 978 199 2189 1 397 1792 2188 2 p-1 10p+1 11p-1 13 143 2 12 131 142 79 869 2 78 791 868 101 1111 2 100 1011 1110 167 1837 2 166 1671 1836 3 8p-1 3p+1 11p-1 47 517 3 375 142 516 113 1243 3 903 340 1242 157 1727 3 1255 472 1726 179 1969 3 1431 538 1968 4 6p-1 5p+1 11p-1 37 407 4 221 186 406 59 649 4 353 296 648 103 1133 4 617 516 1132 191 2101 4 1145 956 2100 5 7p-1 4p+1 11p-1 71 781 5 496 285 780 137 1507 5 958 549 1506 181 1991 5 1266 725 1990 6 4p-1 7p+1 11p-1 17 187 6 67 120 186 61 671 6 243 428 670 83 913 6 331 582 912 127 1397 6 507 890 1396 149 1639 6 595 1044 1638 193 2123 6 771 1352 2122 7 5p-1 6p+1 11p-1 29 319 7 144 175 318 73 803 7 364 439 802 139 1529 7 694 835 1528 8 3p-1 8p+1 11p-1 19 209 8 56 153 208 41 451 8 122 329 450 107 1177 8 320 857 1176 151 1661 8 452 1209 1660 173 1903 8 518 1385 1902 9 10p-1 p+1 11p-1 31 341 9 309 32 340 53 583 9 529 54 582 97 1067 9 969 98 1066 163 1793 9 1629 164 1792 10 9p-1 2p+1 11p-1 43 473 10 386 87 472 109 1199 10 980 219 1198 131 1441 10 1178 263 1440 197 2167 10 1772 395 2166 40

∗ Proposition 4.3.5. If n = 13p, p(prime) with p > 13 for some k ∈ N, Zn has 12 possible cases of triple system;

Case 1. if p = 13k + 1 the triple is (2p − 1, 11p + 1, 13k − 1) e.g. for p = 53, the triple is given

by (105, 584, 688)

Case 2. if p = 13k + 2 the triple is (p − 1, 12p + 1, 13k − 1) e.g. for p = 41, the triple is given

by (40, 493, 532)

Case 3. if p = 13k + 3 the triple is (5p − 1, 8p + 1, 13k − 1) e.g. for p = 29, the triple is given

by (144, 233, 376)

Case 4. if p = 13k + 4 the triple is (6p + 1, 7p − 1, 13k − 1) e.g. for p = 17, the triple is given

by (103, 118, 220)

Case 5. if p = 13k + 5 the triple is (3p − 1, 10p + 1, 13k − 1) e.g. for p = 31, the triple is given

by (92, 311, 402)

Case 6. if p = 13k + 6 the triple is (4p + 1, 9p − 1, 13k − 1) e.g. for p = 19, the triple is given

by (77, 170, 246)

Case 7. if p = 13k + 7 the triple is (4p − 1, 9p + 1, 13k − 1) e.g. for p = 59, the triple is given

by (235, 532, 766)

Case 8. if p = 13k + 8 the triple is (3p + 1, 10p − 1, 13k − 1) e.g. for p = 47, the triple is given

by (142, 469, 610)

Case 9. if p = 13k + 9 the triple is (6p − 1, 7p + 1, 13k − 1) e.g. for p = 61, the triple is given

by (365, 428, 792)

Case 10. if p = 13k + 10 the triple is (5p + 1, 8p − 1, 13k − 1) e.g. for p = 23, the triple is given

by (116, 183, 298)

Case 11. if p = 13k + 11 the triple is (p + 1, 12p − 1, 13k − 1) e.g. for p = 37, the triple is given

by (38, 443, 480) 41

Case 12. if p = 13k + 12 the triple is (2p + 1, 10p − 1, 13k − 1) e.g. for p = 103, the triple is given

by (207, 1132, 1338)

Proof. Similar to the proof of Proposition 4.3.1 42

Table 4.9: Triple systems in n = 13p

p n=13p p(mod 13) k1 k2 k3 1 2p-1 11p+1 13p-1 53 689 1 105 584 688 79 1027 1 157 870 1026 131 1703 1 261 1442 1702 157 2041 1 313 1728 2040 2 p-1 12p+1 13p-1 41 533 2 40 493 532 67 871 2 66 805 870 197 2561 2 196 2365 2560 3 5p-1 8p+1 13p-1 29 377 3 144 233 376 107 1391 3 534 857 1390 4 7p-1 6p+1 13p-1 17 221 4 118 103 220 43 559 4 300 259 558 173 2249 4 1210 1039 2248 199 2587 4 1392 1195 2586 5 3p-1 10p+1 13p-1 31 403 5 92 311 402 83 1079 5 248 831 1078 109 1417 5 326 1091 1416 6 9p-1 4p+1 13p-1 19 247 6 170 77 246 71 923 6 638 285 922 97 1261 6 872 389 1260 149 1937 6 1340 597 1936 7 4p-1 9p+1 13p-1 59 767 7 235 532 766 137 1781 7 547 1234 1780 163 2119 7 651 1468 2118 8 10p-1 3p+1 13p-1 47 611 8 469 142 610 73 949 8 729 220 948 151 1963 8 1509 454 1962 9 6p-1 7p+1 13p-1 61 793 9 365 428 792 113 1469 9 677 792 1468 139 1807 9 833 974 1806 191 2483 9 1145 1338 2482 10 8p-1 5p+1 13p-1 23 299 10 183 116 298 101 1313 10 807 506 1312 127 1651 10 1015 636 1650 179 2327 10 1431 896 2326 11 12p-1 p+1 13p-1 37 481 11 443 38 480 89 1157 11 1067 90 1156 167 2171 11 2003 168 2170 193 2509 11 2315 194 2508 12 11p-1 2p+1 13p-1 103 1339 12 1132 207 1338 181 2353 12 1990 363 2352 43

∗ Proposition 4.3.6. In the ring Zn for n = pq, where p, q ∈ P, with p > q, the unit

∗ elements in Zn form q − 1 triple systems for every q ≥ 3. The triple system is of the form (sp + 1, [q − s]p − 1, qp − 1), with 1 ≤ s ≤ q − 1

The general form of the triples is given by;

(p + 1, (q − 1)p − 1, pq − 1)

(2p + 1, (q − 2)p − 1, pq − 1)

(3p + 1, (q − 3)p − 1, pq − 1)

. .

(q − 1)p + 1, p − 1, pq − 1)

∗ Theorem 4.3.1. Consider the set Zn, for n = pq, p, q ∈ P, with p > q > 2. We have 4 integer solutions satisfying x2 ≡ 1(mod n), with 3 of the solutions forming a triple for each

case of q.

Proof. By Theorem 1.2.3, there are 22 = 4 distinct solutions to the equation x2 ≡ 1(mod n),

for n = pq, with p, q ∈ P where p > q > 2.

By Chinese Remainder Theorem, the solutions are of the form:

Case A: We have two trivial solutions that correspond to items (i) and (ii) i. x ≡ 1(mod p) ≡ 1(mod q) ⇒ x ≡ 1(mod pq)

{(i) represents the unit solution. The remaining 3 are non-unit solutions, which form the triple

system } 44 ii. x ≡ −1(mod p) ≡ −1(mod q) ⇒ x ≡ pq − 1(mod pq)

Case B: The non-trivial solutions correspond to items (iii) to (viii) iii. x ≡ −1(mod p) ≡ 1(mod q) iv. x ≡ 1(mod p) ≡ −1(mod q)

⇒ There is exactly one triple system for each case of q. 45

∗ 4.4 Triple System and Fano Plane Structure in Zn, for n = pqr, p, q, r ∈ P, with p > q > r

Lemma 4.4.1. (xy)2 ≡ 1(mod n) if x2 ≡ 1(mod n) and y2 ≡ 1(mod n)

Proof.

(xy)2 = x2y2 ≡ 1 ∗ 1(mod n) = 1(mod n)

Hence, (xy)2 ≡ 1(mod n) if x2 ≡ 1(mod n) and y2 ≡ 1(mod n)

Lemma 4.4.2. Given :

x(xy) = x2y = 1 ∗ y = y(mod n)

y(xy) = xy2 = x ∗ 1 = x(mod n)

Then, xy ≡ ±1(mod n)

Proof. From x2 ≡ 1(mod n) and y2 ≡ 1(mod n) we have, x ≡ ±1(mod n) and y ≡ ±1(mod n).

Hence, xy ≡ (±1)(±1)(mod n) = (±1)(mod n)

Lemma 4.4.3. Suppose,

x ≡ 1(mod p) ≡ 1(mod q) ≡ −1(mod r)

y ≡ 1(mod p) ≡ −1(mod q) ≡ 1(mod r)

z ≡ −1(mod p) ≡ 1(mod q) ≡ 1(mod r)

Then

xy, xz, yz, xyz ≡ ±1(mod p) ≡ ±1(mod q) ≡ ±1(mod r).

Proof.

1. xy ≡ 1 ∗ 1(mod p) ≡ 1 ∗ (−1)(mod q) ≡ (−1) ∗ 1(mod r)

⇒ xy ≡ 1(mod p) ≡ −1(mod q) ≡ −1(mod r) 46

2. xz ≡ 1 ∗ (−1)(mod p) ≡ 1 ∗ 1(mod q) ≡ (−1) ∗ 1(mod r)

⇒ xz ≡ −1(mod p) ≡ 1(mod q) ≡ −1(mod r)

3. yz ≡ 1 ∗ (−1)(mod p) ≡ (−1) ∗ 1(mod q) ≡ 1 ∗ 1(mod r)

⇒ yz ≡ −1(mod p) ≡ −1(mod q) ≡ 1(mod r)

4. xyz ≡ 1 ∗ 1 ∗ (−1)(mod p) ≡ 1 ∗ (−1) ∗ 1(mod q) ≡ (−1) ∗ 1 ∗ 1(mod r)

⇒ xyz ≡ −1(mod p) ≡ −1(mod q) ≡ −1(mod r)

∗ Theorem 4.4.1. Consider the set Zn for n = pqr, where p, q, r are distinct odd primes. The equation x2 ≡ 1(mod n) has 8 distinct solutions.

Proof. By Theorem 1.2.3, there are 23 = 8 distinct solutions to the equation x2 ≡ 1(mod n),

for n = pqr, with p, q, r ∈ P where p > q > r > 2.

By Chinese Remainder Theorem, the solutions are of the form:

Case A: We have two trivial solutions that correspond to items (i) and (ii) i. x ≡ 1(mod p) ≡ 1(mod q) ≡ 1(mod r)

{ Item (i) represents the unit solution. The remaining 7 are non-unit solutions } ii. x ≡ −1(mod p) ≡ −1(mod q) ≡ −1(mod r)

Case B: The non-trivial solutions correspond to items (iii) to (viii) iii. x ≡ 1(mod p) ≡ 1(mod q) ≡ −1(mod r) iv. x ≡ 1(mod p) ≡ −1(mod q) ≡ 1(mod r) v. x ≡ −1(mod p) ≡ 1(mod q) ≡ 1(mod r) vi. x ≡ −1(mod p) ≡ −1(mod q) ≡ 1(mod r) vii. x ≡ −1(mod p) ≡ 1(mod q) ≡ −1(mod r) viii. x ≡ 1(mod p) ≡ −1(mod q) ≡ −1(mod r) 47

Table 4.10: Triple systems in n = pqr p q r n = pqr values of x satisfying x2 ≡ 1(mod n) 3 5 7 105 29 34 41 64 71 76 104 3 5 11 165 34 56 76 89 109 131 164 3 5 13 195 14 64 79 116 131 181 194 3 7 11 231 34 43 76 155 188 197 230 3 7 13 273 64 92 118 155 181 209 272 3 11 13 429 131 142 155 274 287 298 428 5 7 11 385 34 76 111 274 309 351 384 5 7 13 455 64 181 209 246 274 391 454 7 11 13 1001 155 274 428 573 727 846 1000

From the table above, we pick ﬁve examples, from which we illustrate computational analysis for the triples of n = pqr and ﬁt the triples into Fano Planes.

Example 4.4.1. From the 7 non-unit solutions in Z105, the triples are given by:

i. 29∗34 ≡ 986(mod 105) ≡ 41(mod 105), 29∗41 ≡ 1189(mod 105) ≡ 34(mod 105), 34∗41 ≡

1394(mod 105) ≡ 29(mod 105)

⇒ The triple is given by (29, 34, 41)

ii. 29∗64 ≡ 1856(mod 105) ≡ 71(mod 105), 29∗71 ≡ 2059(mod 105) ≡ 64(mod 105), 64∗71 ≡

4544(mod 105) ≡ 29(mod 105)

⇒ The triple is given by (29, 64, 71)

iii. 29 ∗ 76 ≡ 2204(mod 105) ≡ 104(mod 105), 29 ∗ 104 ≡ 3016(mod 105) ≡ 76(mod 105), 76 ∗

104 ≡ 7904(mod 105) ≡ 29(mod 105)

⇒ The triple is given by (29, 76, 104) 48

iv. 34∗64 ≡ 2176(mod 105) ≡ 76(mod 105), 34∗76 ≡ 2584(mod 105) ≡ 64(mod 105), 64∗76 ≡

4864(mod 105) ≡ 34(mod 105)

⇒ The triple is given by (34, 64, 76)

v. 34 ∗ 71 ≡ 2414(mod 105) ≡ 104(mod 105), 34 ∗ 104 ≡ 3536(mod 105) ≡ 71(mod 105), 71 ∗

104 ≡ 7384(mod 105) ≡ 34(mod 105)

⇒ The triple is given by (34, 71, 104)

vi. 41 ∗ 64 ≡ 2624(mod 105) ≡ 104(mod 105), 41 ∗ 104 ≡ 4264(mod 105) ≡ 64(mod 105), 64 ∗

104 ≡ 6656(mod 105) ≡ 41(mod 105)

⇒ The triple is given by (41, 64, 104)

vii. 41∗71 ≡ 2911(mod 105) ≡ 76(mod 105), 41∗76 ≡ 3116(mod 105) ≡ 71(mod 105), 71∗76 ≡

5396(mod 105) ≡ 41(mod 105)

⇒ The triple is given by (41, 71, 76)

Observe that:

29 ≡ 2(mod 3) ≡ 4(mod 5) ≡ 1(mod 7)

34 ≡ 1(mod 3) ≡ 4(mod 5) ≡ 6(mod 7)

41 ≡ 2(mod 3) ≡ 1(mod 5) ≡ 6(mod 7)

64 ≡ 1(mod 3) ≡ 4(mod 5) ≡ 1(mod 7)

71 ≡ 2(mod 3) ≡ 1(mod 5) ≡ 1(mod 7)

76 ≡ 1(mod 3) ≡ 1(mod 5) ≡ 6(mod 7)

104 ≡ 2(mod 3) ≡ 4(mod 5) ≡ 6(mod 7) 49

Further,

29 ≡ −1(mod 3) ≡ −1(mod 5) ≡ 1(mod 7)

34 ≡ 1(mod 3) ≡ −1(mod 5) ≡ −1(mod 7)

41 ≡ −1(mod 3) ≡ 1(mod 5) ≡ −1(mod 7)

64 ≡ 1(mod 3) ≡ −1(mod 5) ≡ 1(mod 7)

71 ≡ −1(mod 3) ≡ 1(mod 5) ≡ 1(mod 7)

76 ≡ 1(mod 3) ≡ 1(mod 5) ≡ −1(mod 7)

104 ≡ −1(mod 3) ≡ −1(mod 5) ≡ −1(mod 7)

Fitting the triples into a Fano plane we have:

76 64

104

41 34 29

∗ Figure 4.5: Fano Plane Structure for Z105 50

Example 4.4.2. From the 7 non-unit solutions in Z165, the triples are given by:

i. 34 ∗ 56 ≡ 89(mod 165), 34 ∗ 89 ≡ 56(mod 165), 56 ∗ 89 ≡ 34(mod 165)

⇒ The triple is given by (34, 56, 89)

ii. 34 ∗ 76 ≡ 109(mod 165), 34 ∗ 109 ≡ 76(mod 165), 76 ∗ 109 ≡ 34(mod 165)

⇒ The triple is given by (34, 76, 109)

iii. 34 ∗ 131 ≡ 164(mod 165), 34 ∗ 164 ≡ 131(mod 165), 131 ∗ 164 ≡ 34(mod 165)

⇒ The triple is given by (34, 131, 164)

iv. 56 ∗ 76 ≡ 131(mod 165), 56 ∗ 131 ≡ 76(mod 165), 76 ∗ 131 ≡ 56(mod 165)

⇒ The triple is given by (56, 76, 131)

v. 56 ∗ 109 ≡ 164(mod 165), 56 ∗ 164 ≡ 109(mod 165), 109 ∗ 164 ≡ 56(mod 165)

⇒ The triple is given by (56, 109, 164)

vi. 76 ∗ 89 ≡ 164(mod 165), 76 ∗ 164 ≡ 89(mod 165), 89 ∗ 164 ≡ 76(mod 165)

⇒ The triple is given by (76, 89, 164)

vii. 89 ∗ 109 ≡ 131(mod 165), 89 ∗ 131 ≡ 109(mod 165), 109 ∗ 131 ≡ 89(mod 165)

⇒ The triple is given by (89, 109, 131)

Observe that:

34 ≡ 1(mod 3) ≡ −1(mod 5) ≡ 1(mod 11)

56 ≡ −1(mod 3) ≡ 1(mod 5) ≡ 1(mod 11) 51

76 ≡ 1(mod 3) ≡ 1(mod 5) ≡ −1(mod 11)

89 ≡ −1(mod 3) ≡ −1(mod 5) ≡ 1(mod 11)

109 ≡ 1(mod 3) ≡ −1(mod 5) ≡ −1(mod 11)

131 ≡ 1(mod 3) ≡ 1(mod 5) ≡ −1(mod 11)

164 ≡ −1(mod 3) ≡ −1(mod 5) ≡ −1(mod 11)

Fitting the triples into a Fano plane we have: 164

131 109

34 89 56

∗ Figure 4.6: Fano Plane Structure for Z165 52

Example 4.4.3. From the 7 non-unit solutions in Z195, the triples are given by:

i. 16 ∗ 64 ≡ 116(mod 195), 16 ∗ 116 ≡ 64(mod 195), 64 ∗ 116 ≡ 16(mod 195)

⇒ The triple is given by (16, 64, 116)

ii. 14 ∗ 79 ≡ 131(mod 195), 14 ∗ 131 ≡ 79(mod 195), 79 ∗ 131 ≡ 14(mod 195)

⇒ The triple is given by (14, 79, 131)

iii. 14 ∗ 181 ≡ 194(mod 195), 14 ∗ 194 ≡ 181(mod 195), 181 ∗ 194 ≡ 14(mod 195)

⇒ The triple is given by (14, 181, 194)

iv. 64 ∗ 79 ≡ 181(mod 195), 64 ∗ 181 ≡ 79(mod 195), 79 ∗ 181 ≡ 64(mod 195)

⇒ The triple is given by (64, 79, 181)

v. 64 ∗ 131 ≡ 194(mod 195), 64 ∗ 194 ≡ 131(mod 195), 131 ∗ 194 ≡ 64(mod 195)

⇒ The triple is given by (64, 131, 194)

vi. 79 ∗ 116 ≡ 194(mod 195), 79 ∗ 194 ≡ 116(mod 195), 116 ∗ 194 ≡ 79(mod 195)

⇒ The triple is given by (79, 116, 194)

vii. 116 ∗ 131 ≡ 181(mod 195), 116 ∗ 181 ≡ 131(mod 195), 131 ∗ 181 ≡ 116(mod 195)

⇒ The triple is given by (116, 131, 181)

Observe that:

14 ≡ −1(mod 3) ≡ −1(mod 5) ≡ 1(mod 13)

64 ≡ 1(mod 3) ≡ −1(mod 5) ≡ −1(mod 13) 53

79 ≡ 1(mod 3) ≡ −1(mod 5) ≡ 1(mod 13)

116 ≡ −1(mod 3) ≡ 1(mod 5) ≡ −1(mod 13)

131 ≡ 1(mod 3) ≡ 1(mod 5) ≡ 1(mod 13)

181 ≡ 1(mod 3) ≡ 1(mod 5) ≡ −1(mod 13)

194 ≡ −1(mod 3) ≡ −1(mod 5) ≡ −1(mod 13)

Fitting the triples into a Fano plane we have: 194

181 131

14 116 64

∗ Figure 4.7: Fano Plane Structure for Z195 54

Example 4.4.4. From the 7 non-unit solutions in Z455, the triples are given by:

i. 64 ∗ 181 ≡ 209(mod 455), 64 ∗ 209 ≡ 181(mod 455), 181 ∗ 209 ≡ 64(mod 455)

⇒ The triple is given by (64, 181, 209)

ii. 64 ∗ 246 ≡ 274(mod 455), 64 ∗ 274 ≡ 246(mod 455), 246 ∗ 274 ≡ 64(mod 455)

⇒ The triple is given by (64, 246, 274)

iii. 64 ∗ 391 ≡ 454(mod 455), 64 ∗ 454 ≡ 391(mod 455), 391 ∗ 454 ≡ 64(mod 455)

⇒ The triple is given by (64, 391, 454)

iv. 181 ∗ 246 ≡ 391(mod 455), 181 ∗ 391 ≡ 246(mod 455), 246 ∗ 391 ≡ 181(mod 455)

⇒ The triple is given by (181, 246, 391)

v. 181 ∗ 274 ≡ 454(mod 455), 181 ∗ 454 ≡ 274(mod 455), 274 ∗ 454 ≡ 181(mod 455)

⇒ The triple is given by (181, 274, 454)

vi. 209 ∗ 246 ≡ 454(mod 455), 209 ∗ 454 ≡ 246(mod 455), 246 ∗ 454 ≡ 209(mod 455)

⇒ The triple is given by (209, 246, 454)

vii. 209 ∗ 274 ≡ 391(mod 455), 209 ∗ 391 ≡ 274(mod 455), 274 ∗ 391 ≡ 209(mod 455)

⇒ The triple is given by (209, 274, 391)

Observe that:

64 ≡ −1(mod 5) ≡ 1(mod 7) ≡ −1(mod 13)

181 ≡ 1(mod 5) ≡ −1(mod 7) ≡ −1(mod 13) 55

209 ≡ −1(mod 5) ≡ −1(mod 7) ≡ 1(mod 13)

246 ≡ 1(mod 5) ≡ 1(mod 7) ≡ −1(mod 13)

274 ≡ −1(mod 5) ≡ 1(mod 7) ≡ 1(mod 13)

391 ≡ 1(mod 5) ≡ −1(mod 7) ≡ 1(mod 13)

454 ≡ −1(mod 5) ≡ −1(mod 7) ≡ −1(mod 13)

Fitting the triples into a Fano plane we have: 181

209 274

246

64 391 454

∗ Figure 4.8: Fano Plane Structure for Z455 56

Example 4.4.5. From the 7 non-unit solutions in Z1001, the triples are given by:

i. 155 ∗ 274 ≡ 428(mod 1001), 155 ∗ 428 ≡ 274(mod 1001), 274 ∗ 428 ≡ 155(mod 1001)

⇒ The triple is given by (155, 274, 428)

ii. 155 ∗ 573 ≡ 727(mod 1001), 155 ∗ 727 ≡ 573(mod 1001), 573 ∗ 727 ≡ 155(mod 1001)

⇒ The triple is given by (155, 573, 727)

iii. 155 ∗ 846 ≡ 1000(mod 1001), 155 ∗ 1000 ≡ 846(mod 1001), 846 ∗ 1000 ≡ 155(mod 1001)

⇒ The triple is given by (155, 846, 1000)

iv. 274 ∗ 573 ≡ 846(mod 1001), 274 ∗ 846 ≡ 573(mod 1001), 573 ∗ 846 ≡ 274(mod 1001)

⇒ The triple is given by (274, 573, 846)

v. 274 ∗ 727 ≡ 1000(mod 1001), 274 ∗ 1000 ≡ 727(mod 1001), 727 ∗ 1000 ≡ 274(mod 1001)

⇒ The triple is given by (274, 727, 1000)

vi. 428 ∗ 573 ≡ 1000(mod 1001), 428 ∗ 1000 ≡ 573(mod 1001), 573 ∗ 1000 ≡ 428(mod 1001)

⇒ The triple is given by (428, 573, 1000)

vii. 428 ∗ 727 ≡ 846(mod 1001), 428 ∗ 846 ≡ 727(mod 1001), 727 ∗ 846 ≡ 428(mod 1001)

⇒ The triple is given by (428, 727, 846)

Observe that:

155 ≡ 1(mod 7) ≡ 1(mod 11) ≡ −1(mod 13)

274 ≡ 1(mod 7) ≡ −1(mod 11) ≡ 1(mod 13) 57

428 ≡ 1(mod 7) ≡ −1(mod 11) ≡ −1(mod 13)

573 ≡ −1(mod 7) ≡ 1(mod 11) ≡ 1(mod 13)

727 ≡ −1(mod 7) ≡ 1(mod 11) ≡ −1(mod 13)

846 ≡ −1(mod 7) ≡ −1(mod 11) ≡ 1(mod 13)

1000 ≡ −1(mod 7) ≡ −1(mod 11) ≡ −1(mod 13)

Fitting the triples into a Fano plane we have: 428

274 727

573

155 1000 846

∗ Figure 4.9: Fano Plane Structure for Z1001 Chapter 5. CONCLUSIONS AND RECOMMENDATIONS

5.1 Conclusion

We have obtained the following results in this project:

In section 4.1, we coﬁrmed that if p is prime, 1 and p − 1 are the only integers satisfying

2 ∗ k ≡ 1(mod p) in the set Zp.

In section 4.2, we obtained three results:

(i) We proved that for n = 2p, p (prime), 1 and 2p − 1 are the only integers satisfying

2 ∗ k ≡ 1(mod 2p) in the set Zn.

2 2 ∗ (ii) We proved that the equation x ≡ 1(mod 2 p) has 4 solutions in the set Zn for n =

2 2 2 ∗ 2 p = 4p, p (prime). The 4 solutions to x ≡ 1(mod 2 p) in the set Zn are 1, 2p − 1, 2p + 1 and 22p − 1 = 4p − 1. The 3 non-unit solutions form triple systems for all odd

p ∈ P.

2 ∗ (iii) We proved that the equation x ≡ 1(mod n) has 8 distinct solutions in the set Zn for

m n = 2 p, where p is an odd prime and m ∈ N with m ≥ 3. We proved this theorem using Theorem 1.2.3 and the results under Remark ??. The seven non-unit solutions were

ﬁtted into Fano planes. We gave four examples for purpose of illustration.

∗ In section 4.3, by induction, we were able to prove that set Zn, for n = pq, p, q ∈ P, with p > q > 2 has 4 integer solutions satisfying x2 ≡ 1(mod n), with 3 of the solutions forming a triple for each case of q.

In section 4.4, using Theorem 1.2.3 we proved that the equation x2 ≡ 1(mod n) has 8 distinct

∗ solutions in the set Zn for n = pqr, where p, q, r are distinct odd primes. The seven non-unit

58 59 solutions were ﬁtted into Fano planes. We gave examples for purpose of illustration.

All the results were captured in tabular form and the triples ﬁtted in Fano planes in cases where

∗ we obtained seven idempotent elements in the set Zn.

5.2 Recommendations

∗ m Having computed the idempotent elements for Zn where n = p, n = 2p, n = 2 q n = pq and n = pqr for p, q, r ∈ P and m ∈ N, this research can be extended to various structures

∗ of n, a composition of four or more prime numbers i.e. considering a ﬁnite set Zn for n =

k1 k2 k3 km p1 p2 p3 ...... pm , where pi0s ∈ P for all i ∈ N. Compute the number of triples, Fano Planes and projective geometry structures formed for various values of n. 60

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