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DIGITAL TEXT BOOK
- 1. A textbook of NUMBERS BINTU GEORGE JACOB MOUNT TABOR TRAINING COLLEGE PATHANAPURAM. MOC PUBLICATIONS, KOLLAM 2014
- 2. ii Preface The 1990s saw a wave of calculus reform whose aim was to teach students to think for themselves and to solve substantial problems, rather than merely memorizing formulas and performing rote algebraic manipulations. This book has a similar, albeit somewhat more ambitious, goal: to lead you to think mathematically and to experience the thrill of independent intellectual discovery. Our chosen subject, Number Theory, is particularly well suited for this purpose. The natural numbers 1, 2, 3, ... satisfy a multitude of beautiful patterns and relationships, many of which can be discerned at a glance; others are so subtle that one marvels they were noticed at all. Experimentation requires nothing more than paper and pencil, but many false alleys beckon to those who make conjectures on too scanty evidence. It is only by rigorous demonstration that one is finally convinced that the numerical evidence reflects a universal truth. This book will lead you through the groves wherein lurk some of the brightest flowers of Number Theory, as it simultaneously encourages you to investigate, analyze, conjecture, and ultimately prove your own beautiful number theoretic results. This book was originally written to serve as a text for Math 42, a course created by Jeff Hoffstein at Brown University in the early 1990s. Math 42 was designed to attract nonscience majors, those with little interest in pursuing the standard calculus sequence, and to convince them to study some college mathematics. The intent was to create a course similar to one on, say, βThe Music of Mozartβ or βElizabethan Drama,β wherein an audience is introduced to the overall themes and methodology of an entire discipline through the detailed study of a particular facet of the subject. Math 42 has been extremely successful, attracting both its intended audience and also scientifically oriented undergraduates interested in a change of pace from their large-lecture, cookbook-style courses. The prerequisites for reading this book are few. Some facility with high school algebra is required, and those who know how to program a computer will have fun generating reams of data and implementing assorted algorithms, but in truth the reader needs nothing more than a simple calculator. Concepts from calculus are mentioned in passing, but are not used in an essential way. However, and the reader is hereby forewarned, it is not possible to truly appreciate Number Theory without an eager and questioning mind and a spirit that is not afraid to experiment, to make mistakes and profit from them, to accept frustration and persevere to the ultimate triumph. Readers who are able to cultivate these qualities will find themselves richly rewarded, both in their study of Number Theory and their appreciation of all that life has to offer.
- 3. CONTENTS CHAPTER 1 : PRELIMINARIES 1 CHAPTER 2 : MULTIPLICATIVE FUNCTIONS 5 CHAPTER 3 : PERFECT NUMBERS 25 BIBLIOGRAPHY 35 iii
- 4. Chapter 1 PRELIMINARIES In this chapter we give some definitions and theorems which we will use in the subsequent chapters. We start with Peano axioms for Natural numbers. 1 Peano Axioms. ν΅ is a set with the following properties. (1) ν΅ has a distinguished element which we call β²1β². (2) There exists a set map ν: ν΅ β ν΅. (3) ν is one-to-one (injective). (4) There does not exist an element ν β ν΅ such that ν(ν) = 1. (5) (Principle of Induction) Let ν β ν΅ such that a) 1 β ν and b) if ν β ν, then ν(ν) β ν. Then ν = ν΅. We call such a set ν΅ to be the set of natural numbers and elements of this said to be natural numbers. The principle of mathematical induction is a tool which can be used to prove a wide variety of mathematical statements. The set of integers is denoted by ν: ν = {β¦ , β3, β2, β1, 0, 1, 2, 3, β¦ }.
- 5. 2 Theorem 1.1 (The Principle of Mathematical Induction) Let ν(ν) be a statement satisfying the following conditions, where ν β ν: 1. ν(ν0 ) is true for some integer ν0. 2. If ν(ν) is truefor an arbitrary integer ν β₯ ν0, then ν(ν + 1) is true. Then ν(ν) is true for every integer ν β₯ ν0 . The following is a strong version of the principle of induction. Theorem 1.2 (The Second Principle of Mathematical Induction) Let ν(ν) be a statement satisfying the following conditions, where ν β ν: (1') ν(ν0 ) is true for some integer ν0. (2') If ν is an arbitrary integer greater than or equal to ν0 such that ν(ν0), ν(ν0 + 1), β¦, and ν(ν) are true, then ν(ν + 1) is also true. Then ν(ν) is true for every integer ν β₯ ν0. Definition 1.3 (Prime Numbers)
- 6. Any integer greater than one which has no factors other than one and itself is 3 called a prime number. Prime numbers are the building blocks of integers. Euclid proved the fundamental theorem of arithmetic and presented the Euclid algorithm for finding the greatest common divisor of two numbers. Theorem 1.4 (Unique Factorisation) Any natural number greater than 1, factors into a product of primes. Definition 1.5 (Common Divisor) Let ν, ν be the integers, let ν0, we say that ν divides ν or ν is a divisor of ν if ν = νν for some integer ν. When ν divides ν we write ν|ν. Definition 1.6 (Greatest Common Divisor) A number ν is the greatest common divisor (gcd) of two non zero numbers ν and ν, if ν is the common divisor of ν and ν and no common divisor of ν and ν is larger than ν. gcd of ν and ν is denoted by (ν, ν). Definition 1.7 Two positive integers ν and ν are relatively prime if their gcd is . Theorem 1.8 (Euclidean Algorithm)
- 7. Let ν and ν are any positive integers, and ν be the reminder, when ν is 4 divided by ν. Then (ν, ν) = (ν, ν). Theorem 1.9 (Fundamental Theorem of Arithmetic) Any natural number ν β₯ 2 is either a prime or can be expressed as a product of primes. The factorization into primes is unique except for the order of factors . Definition 1.10 (Congruence Modulo ν) Let ν be a positive integer. Then an integer ν is congruent to an integer ν modulo ν if ν|(ν β ν). We write ν β‘ ν(mod ν) and ν is called the modulus of the congruence relation. Definition 1.11 (Eulerβs Phi Function) Let ν be a positive integer. Then Eulerβs phi function ν(ν) denotes the number of positive integers ν and relatively prime to ν.
- 8. Chapter 2 MULTIPLICATIVE FUNCTIONS Eulerβs phi function is one of the most important numberβtheoretic functions (also known as arithmetic functions). Arithmetic functions are defined for all positive integers. Eulerβs phi function belongs to a large class of arithmetic 5 functions called multiplicative functions. Definition 2.1 A number-theoretic function ν is multiplicative if ν(νν) = ν(ν)ν(ν), whenever ν and ν are relatively prime. Example 2.2 The constant function ν(ν) = 1 is multiplicative, since ν(νν) = 1 = 1 β 1 = ν(ν) β ν(ν). The functions ν(ν) = νν , ν being a fixed integer is multiplicative, since ν(νν) = (νν)ν = νν νν = ν(ν)ν(ν). οΏ Theorem 2.3 Let ν be a multiplicative function and ν be a positive integer with canonical ν1 ν2 decomposition ν = ν1 ν2 β¦ νν νν . Then
- 9. ν1 , then ν(ν) = ν(ν1 ν1 )ν(ν2 6 ν1 )ν(ν2 ν(ν) = ν(ν1 ν2 )β¦ ν(νν νν ). Proof. If ν = 1, that is, if ν = ν1 ν1 ), so the theorem is trivially true. Assume it is true for any integer with canonical decomposition consisting of ν distinct primes. That is ν(ν) = ν(ν1 ν2 )β¦ ν(νν νν ). Let ν be any integer with ν + 1 distinct primes in its canonical ν1 ν2 decomposition, say, ν = ν1 ν2 β¦ νν νν . ν1 ν2 Since (ν1 ν2 β¦ νν νν , νν+1 νν+1 ) = 1 and ν is multiplicative, ν1 ν2 ν(ν1 ν2 β¦ νν νν νν+1 νν+1 ) = ν(ν1 ν1 ν2 ν2 β¦ νν νν )ν(νν+1 νν+1 ) ν1 )ν(ν2 = ν(ν1 ν2 ) β¦ ν(νν νν )ν(νν+1 νν+1 ), by induction. Therefore by induction the result is true for any positive integer ν. οΏ Theorem 2.4 Let ν be a prime and ν any positive integer. Then ν(νν) = νν β ννβ1. Proof. ν(νν) = number of positive integers less than or equal to ννand relatively prime to it.
- 10. = (number of positive integers less than or equal to νν)-(number of positive integers less than or equal to νν and not relatively prime 7 to it) The positive integers less than or equal to νν and not relatively prime to it are the various multiples of ν, namely, ν, 2ν, 3ν, β¦ β¦ β¦ β¦ (ννβ1 )ν , and they are ννβ1 in number. Thus Ο(pe) = pe-pe-1 . οΏ Lemma 2.5 Let ν and ν be relatively prime positive integers. Then the integers ν, ν + ν, 2ν + ν, β¦ , (ν β 1)ν + ν are congruent modulo ν to 0, 1 , 2, β¦ , (ν β 1) in some order. Example 2.6 ν(8) = ν(23 ) = 23 β 22 = 8 β 4 = 4. Thus there are four positive integers less than or equal to 8 and relatively prime to it; 1, 3, 5 and 7. ν(81) = ν(34) = 34 β 33 = 54. ν(15625) = ν(56) = 56 β 55 = 12500. οΏ Theorem 2.7 The function Ο is multiplicative.
- 11. Proof. Let ν and ν be positive integers such that (ν, ν) = 1. We have to prove that ν(νν) = ν(ν)ν(ν). Arrange the integers 1 through νν in ν rows of ν each: 1. ν + 1 2ν + 1 β¦ . (ν β 1)ν + 1 2. ν + 2 2ν + 2 β¦ . (ν β 1)ν + 2 3. ν + 3 2ν + 3 β¦ . (ν β 1)ν + 3 8 . ν. ν + ν 2ν + ν β¦ (ν β 1)ν + ν . . . ν. 2ν 3ν . . . νν Let ν be a positive integer less than or equal to ν such that (ν, ν) > 1. We will show that no element of the νth row is relatively prime to νν.
- 12. Let ν = (ν, ν). Then ν|ν and ν|ν, so ν|νν + ν for any integer ν; that is, ν is a factor of every element in the νth row. Thus no element in the νth row is relatively prime to ν, and hence to νν if (ν, ν) > 1. That is the elements in the array are relatively prime to νν from the νth row only if (ν, ν) = 1. By definition, there are ν(ν) such integers ν and hence ν(ν) such rows. 9 Now, in νth row, where (ν, ν) = 1: ν , ν + ν , 2ν + ν , β¦ , (ν β 1)ν + ν By Lemma 2.5, their least residues modulo ν are a permutation of 0 ,1 ,2 , β¦ , (ν β 1) of which ν(ν) are relatively prime to ν. Therefore exactly ν(ν) elements in νth row are relatively prime to ν and hence to νν. Thus there are Ο(m) rows containing positive integers relatively prime to νν, and each row contains ν(ν) elements relatively prime to it. So the array contains ν(ν)ν(ν) positive integers less than or equal to νν and relatively prime to νν. Therefore ν(νν) = ν(ν)ν(ν). οΏ Example 2.8 1. ν(221) = ν(13 β 17) = ν(13) β ν(17) = 12 β 16 = 192. 2. ν(625) = ν(53 β 72 ) = ν(53 ) β ν(72)
- 13. = (53 β 52)(72 β 7) = 4200. οΏ ν1 )ν(ν2 ν1 (1 β 1 10 Theorem 2.9 ν1 ν2 Let ν = ν1 ν2 β¦ νν νν be the canonical decomposition of a positive integer ν. Then ν(ν) = ν (1 β 1 ν1 ) (1 β 1 ν2 )β¦ (1 β 1 νν ). Proof. Since ν is multiplicative, by theorem 2.1 ν(ν) = ν(ν1 ν2 ) β¦ ν(νν νν ) = ν1 ν1 ν2 (1 β 1 ) ν2 ν2 νν (1 β 1 )β¦ νν νν ) ν1 ν2 = ν1 νν (1 β 1 ν2 β¦ νν ν1 ) (1 β 1 ν2 )β¦ (1 β 1 νν ) = ν (1 β 1 ν1 ) (1 β 1 ν2 ) β¦ (1 β 1 νν ). οΏ Example 2.10 1. 666 = 2 β 32 β 37 Thus ν(666) = 666 (1 β 1 2 ) (1 β 1 3 ) (1 β 1 37 ) = 216. 2. 1976 = 23 β 13 β 19 Thus ν(1976) = 1976(1 β 1 2 ) (1 β 1 13 ) (1 β 1 ) 19
- 14. = 864. οΏ 11 Theorem 2.11 If ν β₯ 3, then ν(ν) is even. Proof. If ν is even obviously ν(ν) is even. If ν is odd then for each prime factor ν of ν, the numerator of (1 β 1 ν ) is even. Thus ν(ν) is even for ν β₯ 3. οΏ Theorem 2.12 Let ν be a positive integer, then Ξ£ν|ν ν(ν) = ν. Here we turn to illustrate its proof. Example 2.13 The positive divisors of 18 are 1, 2, 3, 6, 9 and 18. So Ξ£ν|18 ν(18) = ν(1)+ ν(2)+ ν(3)+ ν(6)+ ν(9)+ ν(18) = 1 + 1 + 2 + 2 + 6 + 6 = 18. οΏ The Tau and Sigma Functions Definition 2.14 (The Tau Function) Let ν be a positive integer. Then ν(ν) denotes the number of positive divisors of ν; that is ν(ν) = Ξ£ν|ν 1.
- 15. 12 Example 2.15 ο· The positive divisors of 18 are 1, 2, 3, 6, 9 and 18, So ν(18) = 6. ο· 23, being a prime, has exactly two positive divisors, So ν(23) = 2. οΏ Remark. If ν is prime, then ν(ν) = 2; if ν(ν) = 2, then ν is prime. Definition 2.16 (The Sigma Function) Let ν be positive integer. Then ν(ν) denotes the sum of the positive divisors of ν; that is ν(ν) = Ξ£ν|ν ν. Example 2.17 ο· The positive divisors of 12 are 1, 2, 3, 4, 5, 6 and 12; So ν(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28. ο· The positive divisors of 28 are 1, 2, 4, 7, 14 and 28; So ν(28) = 1 + 2 + 4 + 7 + 14 + 28 = 56. οΏ
- 16. Let ν be a multiplicative function. Define a new function νΉ by νΉ(ν) = Ξ£ν|ν ν(ν). The following example shows that the new function νΉ can be a 13 multiplicative function. Example 2.18 We have (4,7) = 1. Then νΉ(4 β 7) = Ξ£ν|28ν(28) = ν(1) + ν(2) + ν(4) + ν(7) + ν(14) + ν(28) = ν(1 β 1) + ν(1 β 2) + ν(1 β 4) + ν(1 β 7) + ν(2 β 7) + ν(4 β 7) = [ν(1) + ν(2) + ν(4)]ν(1) + [ν(1) + ν(2) + ν(4)]ν(7) = [ν(1) + ν(2) + ν(4)][ν(1) + ν(7)] = Ξ£ν(ν) ν|4 β Ξ£ν(ν) ν|7 = νΉ(4)νΉ(7). οΏ Theorem 2.19 If ν is a multiplicative function, then νΉ(ν) = Ξ£ν|ν ν(ν) is also multiplicative.
- 17. Proof. Let ν and ν be relatively prime positive integers. We want to show that 14 νΉ(νν) = νΉ(ν)νΉ(ν). By definition, νΉ(νν) = Ξ£ν|νν ν(ν). Since (ν, ν) = 1, every positive divisor of νν, is the product of a unique pair of positive divisors ν1 of ν and ν2 of ν, where, (ν, ν) = 1. Therefore νΉ(νν) = Ξ£ν1|ν,ν2|ν ν(ν1ν2). Since ν is multiplicative, ν(ν1ν2) = ν(ν1 )ν(ν2 ). Therefore νΉ(νν) = Ξ£ν1|ν,ν2|νν(ν1ν2) = Ξ£ νΉ(ν)ν(ν2 ) ν2 |ν = νΉ(ν)Ξ£ ν(ν2 ) ν2 |ν = νΉ(ν)νΉ(ν). Thus νΉ is multiplicative. οΏ Corollary 2.20 The Tau and Sigma functions are multiplicative. Proof. By theorem 2.6, the functions Ξ£ν|ν ν(ν) = Ξ£ν|ν 1 = ν(ν) and Ξ£ν|ν ν(ν) = Ξ£ν|ν ν = ν(ν) are multiplicative.
- 18. (Since we have constant function ν(ν) = 1 and the identity function ν(ν) = 15 ν are multiplicative.) That is, if ν(ν, ν) = 1, then ν(νν) = ν(ν)ν(ν) and ν(νν) = ν(ν)ν(ν). Example 2.21 36 = (4)(9). Also (4, 9) = 1. Thus ν(36) = ν(4)ν(9) = 3 β 3 = 9. And ν(36) = ν(4)ν(9) = (1 + 2 + 4)(1 + 3 + 9) = 91. οΏ Theorem 2.22 Let ν be any prime and ν any positive integer. Then ν(νν) = ν + 1 and ν(νν) = νν+1 β1 νβ1 . Proof. The positive factors of νν are of the form νν , where 0 β€ ν β€ ν. There are ν + 1 such factors. So ν(νν) = ν + 1. ν ν(νν) = Ξ£νν = 0 νν+1 β 1 ν β 1 . The following theorem gives two formulas. They are a consequence of Corollary 2.20 and Theorem 2.2. οΏ Theorem 2.23
- 19. Let ν be a positive integer with canonical decomposition ν = ν1 16 ν1 ν2 ν2 β¦ νν νν. ν1 β1 ν1 β1 Then ν(ν) = (ν1 + 1)(ν2 + 2) β¦ (νν + ν) and ν(ν) = ν1 ν2 β1 ν2 β1 β ν2 β β¦ β νν β1 ννβ1 ν ν . Proof. Since ν is multiplicative, ν1 ) β ν(ν2 ν(ν) = ν(ν1 ν2 ) β¦ ν(νν νν ) = (ν1 + 1)(ν2 + 1) β¦ (νν + 1). Since ν is multiplicative, ν1 ) β ν(ν2 ν(ν) = ν(ν1 νν ) = ν1 ν2 )β¦ ν(νν ν1 β1 ν1 β1 ν2 β1 ν2 β1 β ν2 β β¦ β νν β1 ννβ1 ν ν . οΏ Example 2.24 The canonical decomposition of 6120 = 23 β 32 β 5 β 17 Therefore ν(6120) = (3 + 1)(2 + 1)(1 + 1)(1 + 1) = 48 and ν(6120) = 23+1β1 2β1 β 32+1β1 3 β1 β 51+1β1 5β1 β 171+1β1 17β1 = 15 β 14 β 6 β 18 = 22680. οΏ The MΣ§bius Function The MΣ§bius function ν is an important number-theoretic function which was discovered by the German mathematician August Ferdinand MΣ§bius. It plays an important role in the study of the distribution of primes. Definition 2.25 (The MΣ§bius function ν)
- 20. 1 if ν = 1 0 if ν2 |ν for some prime ν (β1)ν if ν = ν1 ν2 β¦ νν 17 Let ν be a positive integer. Then ν(ν) = { For example ν(2) = β1, ν(3) = β1, ν(4) = 0, ν(12) = 0, ν(35) = ν(5.7) = (β1)2 and ν(672) = ν(25 β 3 β 7) = 0. Thus ν assigns β1,0 or to each positive integer; ν(ν) = 0 if the canonical decomposition of ν contains a perfect square, that is, if ν is not square-free; it is (β1)ν if it consists of ν distinct prime factors. Theorem 2.26 The function ν is multiplicative. Proof. Let ν and ν be relatively prime positive integers. If ν = 1 or ν = 1, then clearly ν(νν) = ν(ν)ν(ν). Suppose ν or ν is divisible by ν2 for some prime ν. Then ν(ν)ν(ν) = 0. If ν2 |ν or ν2 |ν, then ν2 |νν, so ν(νν) = 0. Thus ν(νν) = ν(ν)ν(ν). Finally, suppose both ν and ν are square-free, let ν = ν1ν2 β¦ νν and ν = ν1 ν2 β¦ νν , where the νν βs and ννβs are distinct primes, since (ν, ν) = 1. ν(ν) = (β1)ν and ν(ν) = (β1)5. Then νν = ν1 ν2 β¦ ννν1ν2 β¦ νν , a product of distinct primes.
- 21. Therefore ν(νν) = (β1)ν+ν = (β1)ν β (β1)ν = ν(ν)ν(ν). Thus, in every case, ν(νν) = ν(ν)ν(ν), so ν is multiplicative. οΏ 18 Example 2.27 Let us determine ν(420). Since ν is multiplicative, ν(420) = ν(15)ν(28) Now 15 = 3 β 5. So by definition, ν(15) = 1. 28 = 22 β 7. So by definition, ν(28) = 0. 420 = 22 β 3 β 5 β 7. Since 420 is not squareβfree, ν(420) = 0. Thus ν(420) = 0 = 1 β 0 = ν(15)ν(28). οΏ Next we develop a formula for Ξ£ν|ν ν(ν). When n = 1, Ξ£ν|1 ν(ν) = ν(1) = 1. If ν > 1, we can compute the sum using the canonical decomposition of ν and Theorem 8.7 provided we know the sum when ν is a prime-power νν. For this we need to introduce a new function, as the following lemma shows. Lemma 2.28 Let νΉ(ν) = Ξ£ν|ν ν(ν). Then νΉ(νν) = 0, where ν > 1. Proof. νΉ(νν) = Ξ£ν|νν ν(ν) = Ξ£ ν(νν) νν =0
- 22. = ν(1) + ν(ν) + ν(ν2 ) + β― + ν(νν) = 1 + (β1) = 0 + β― + 0 = 0. οΏ (ν|6) . 19 Theorem 2.29 1 if ν = 1 0 othewise Let ν be a positive integer. Then Ξ£ν|ν ν(ν) = { . ν1 ν2 Proof. If ν = 1, Ξ£ν|1 ν(ν) = ν(1) = 1. So let ν > 1 and let ν = ν1 ν2 β¦ νν νν be the canonical decomposition of ν. Let, νΉ(ν) = Ξ£ν|ν ν(ν). Since ν is multiplicative, νΉ is also multiplicative, by Theorem 8.7. Therefore νΉ(ν) = Ξ νΉ(ννν) νν =1 νΉ(ν) = Ξ 0 νν =1 = 0. οΏ Example 2.30 Ξ£ ν(ν) = ν(1) + ν(2) + ν(3) + ν(6) + ν(9) + ν(18) ν|18 = 1 + (β1) + (β1) + (β1)2 + 0 + 0 = 0. Problem. Let ν be a number-theoretic function. Show that Ξ£ Ξ£ ν(ν)ν(νβ² ) = Ξ£ Ξ£ ν(νβ²)ν(ν) (ν|(6 ν )) (νβ² |6) (νβ²|(6 ν ))
- 23. (νβ² |6) . οΏ 20 Solution. Ξ£ Ξ£ ν(ν)ν(νβ²) (νβ²|(6 ν )) ν|6 = Ξ£ ν(1)ν(νβ² ) νβ² |6 + Ξ£ ν(2)ν(νβ²) νβ² |3 + Ξ£ ν(3)ν(νβ² ) νβ² |2 +Ξ£ ν(6)ν(νβ²) νβ² |1 = ν(1)Ξ£ ν(νβ²) νβ² |6 + ν(2)Ξ£ ν(νβ² ) + ν(3)Ξ£ ν(νβ² ) + ν(6)Ξ£ ν(νβ²) νβ² |3 νβ² |2 νβ² |1 = ν(1)[ν(1) + ν(2) + ν(3) + ν(6)] + ν(2)[ν(1) + ν(3)] + ν(3)[ν(1) + ν(2)] + ν(6)ν(1) = ν(1)[ν(1) + ν(2) + ν(3) + ν(6)] + ν(2)[ν(1) + ν(3)] + ν(3)[ν(1) + ν(2)] + ν(6)ν(1) = ν(1) Ξ£ ν(ν) ν|6 + ν(2) Ξ£ ν(ν) ν |3 + ν(3) Ξ£ ν(ν) ν|2 + ν(6)Ξ£ν|1 ν(ν) = Ξ£νβ² |6 ν(νβ²)Ξ£ ν(ν) ν |( 6 νβ²) = Ξ£ Ξ£ ν(νβ²)ν(ν) (ν|(6 ν )) Theorem 2.31 (MΣ§bius Inversion Formula) Let ν be a number-theoretic function and let νΉ(ν) = Ξ£ν|ν ν(ν). Then ν(ν) = Ξ£ ν(ν)νΉ (ν ν|ν ). ν Proof. νΉ(ν) = Ξ£ν|ν ν(ν). Then νΉ (ν ν ) = Ξ£ ν(νβ²) νβ² |(ν ν ) .
- 24. ν|ν ν|ν ) . νβ²ν|ν νβ² |ν ) νβ²) ν|ν ] νβ²) equals if ν|ν ) = ν(νβ²) β 1, where νβ² = ν. ν|ν ) can also be written asν(ν) = 21 So ν(ν)νΉ (ν ν ) = ν(ν) Ξ£ ν(νβ² ) νβ² |(ν ν ) = Ξ£ ν(ν)ν(νβ² ) νβ² |(ν ν ) Therefore Ξ£ ν(ν)νΉ (ν ν ) = Ξ£ Ξ£ ν(ν)ν(νβ² ) νβ² |(ν ν As ν runs over the positive divisors of ν, so does νβ²; also ννβ² runs over the positive factors of ν; that is, the sum of all its values of ν(ν)ν(νβ²) as ννβ² runs over the positive factors of ν; that is, the sum of all its values for all pairs ν and νβ² such that νβ² ν and ν| ( ν νβ² ). That is, Ξ£ Ξ£ ν(ν) νβ² |(ν ν ) ν(νβ² ) = Ξ£ Ξ£ ν(ν)ν(νβ² ) ν|( ν Thus Ξ£ ν(ν)νΉ (ν ν ) = Ξ£νβ² |ν ν(νβ²) [Ξ£ ν(ν) ν|( ν But by Theorem 8.17, Ξ£ ν(ν) ν|( ν ν νβ² = 1; that is, if ν = νβ², and otherwise. Thus Ξ£ ν(ν)νΉ (ν ν In other words, ν(ν) = Ξ£ ν(ν)νΉ ( ν νβ² ν|ν ). οΏ Remark. As ν runs over the positive factors of ν, so does ν ν . Therefore, the inversion formula ν(ν) = Ξ£ ν(ν)νΉ (ν ν Ξ£ ν (ν ) νΉ(ν) ν|ν . ν
- 25. Notice that the definition ν(ν) = Ξ£ν|ν ν(ν) expresses νΉ in terms of ν, whereas the inversion formula ν(ν) = Ξ£ ν(ν)νΉ (ν ν|ν ) expresses ν in terms of νΉ. 22 ν To illustrate the inversion formula, recall that ν(ν) = Ξ£ν|ν 1 and ν(ν) = Ξ£ν|ν ν because both the constant function ν(ν) = 1 and the identity function ν(ν) = ν are multiplicative, it follows by Theorem 2.32 that ν ν 1 = Ξ£ν(ν)ν ( ν ν ) = Ξ£ν ( ) ν(ν) ν|ν ν|ν and ν ν ν = Ξ£ν(ν)ν ( ν ν ) = Ξ£ν ( ) ν(ν) ν|ν ν|ν We can illustrate these results by the following example. Example 2.32 6 ν Ξ£ν(ν)ν ( ) = ν(1)ν(6) + ν(2)ν(3) + ν(3)ν(2) + ν(6)ν(1) ν|6 = 1 β 4 + (β1) β 2 + (β1) β 2 + (β1)2 β 1 = 1 6 ν Ξ£ν(ν)ν ( ) = ν(1)ν(6) + ν(3)ν(2) + ν(6)ν(1) ν|6
- 26. = 1 β 12 + (β1) β 4 + (β1) β 3 + (β1)2 β 1 = 6. οΏ ν|ν ) =Ξ£ ν(ν) ν|ν ν . οΏ 23 Theorem 2.33 ν(ν) = ν Ξ£ ν(ν) ν|ν ν . Proof. By Theorem 2.12, ν = Ξ£ν|ν ν(ν). Let ν denote the identity function given by ν(ν) = ν for all ν. Therefore ν(ν) = ν = Ξ£ν|ν ν(ν). Then by MΣ§bius inversion formula, ν(ν) = Ξ£ ν(ν)ν (ν ν Problem. Verify the formula in Theorem 2.35 for n=12. Solution. By Theorem 2.9, ν(12) = ν(22 β 3) = 4. Now, 12 Ξ£ ν(ν) ν 12 ν = Ξ£ ( ) ν(ν) ν|12 ν|12 = 12ν(1) + 6ν(2) + 4ν(3) + 3ν(4) + 2ν(6) + 1ν(12)
- 27. = 12 β 1 + 6 β (β1) + 4 β (β1) + 3 β 0 + 2 β (β1)2 + 1 β 0 ν|ν ). Then νΉ(ν) = Ξ£ν|ν ν(ν). ) ν|ν ]. 24 = 4 = ν(12). Theorem 2.34 Let νΉ and ν be number-theoretic functions such that ν(ν) = Ξ£ ν(ν)νΉ (ν ν Proof. By the definition of ν, ν(ν) = Ξ£ ν(νβ² )νΉ ( ν νβ² νβ² |ν ). Therefore Ξ£ ν(ν) = Ξ£ Ξ£ ν(νβ² )νΉ ( ν νβ² ν|ν ν|ν νβ² |ν ). Let ν νβ² = ν and ν ν = ν,then ν = ννβ² and ν = νν = νννβ².So this equation yields Ξ£ν|ν ν(ν) = Ξ£ν|ν Ξ£ννβ²=ν ν(νβ²)νΉ(ν) = Ξ£ννβ² |ν ν(νβ²)νΉ(ν) = Ξ£ νΉ(ν) [Ξ£ ν(νβ²) νβ² |(ν ν By Theorem 8.17, Ξ£ ν(νβ² ) νβ² |(ν ν ) equals if ν = ν and otherwise. So the equation becomes Ξ£ ν(ν) = νΉ(ν) β (1) (ν|ν) , where ν = ν. That is, νΉ(ν) = Ξ£ν|ν ν(ν). οΏ
- 28. Chapter 3 PERFECT NUMBERS We can use the sigma function to study a marvelous class of numbers called perfect numbers. The term perfect numbers was coined by the Pythagoreans. The ancient Greeks thought these numbers had mystical power and held them to be βgoodβ numbers. They were also studied by the early Hebrews; Rabbi Josef ben Jehuda in the 12th century recommended their study in his book, βHealing of 25 soulsβ. Historically, some biblical scholars considered 6 a perfect number, because they believed god created the world in six days and godβs work is perfect. St. Augustine, on the other hand, believed Godβs work to be perfect because 6 is a perfect number. He writes, βSix is a number perfect in itself, and not because God created all things in six days; rather the inverse is true; God created all things in six days because this number is perfect. And it would remain perfect even if the work of the six days did not exist.β
- 29. The Pythagoreans regarded 6 as the symbol of βmarriage and health and beauty on account of the integrity of its parts and the agreement existing in it.β What is mystical about ? The Pythagoreans observed that 6 equals the sum of its proper divisors: 6 = 1 + 2 + 3. The next two perfect numbers are 28 and 496: 26 28 = 1 + 2 + 4 + 7 + 14 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 Their discovery is sometimes attributed to the Greek mathematician Nichomachus (ca. A.D. 100). We can now formalise the definition of a perfect number. Definition 3.1 (Perfect number) A positive integer ν is a perfect number if the sum of its proper divisiors equals ν. Thus ν is perfect if ν(ν) β ν = ν, that is, if ν(ν) = 2ν. The first eight perfect numbers are: 6 = 2(22 β 1) 28 = 22 (23 β 1) 496 = 24 (25 β 1) 8,128 = 26 (27 β 1) 33,550,336 = 212(213 β 1)
- 30. 8,589,869,056 = 216 (217 β 1) 137,438,691,328 = 218(219 β 1) 2,305,843,008,139,952,128 = 230 (231 β 1). Of which only the first four were known to the ancient Greeks; they are listed in Nichomachusβs Introductio Arithmeticae. The next perfect numbers was discovered by the Greek mathematician Hudalrichus Regius around 1536. The Italian mathematician Pietro Antonic Cataldi (1548-1626) discovered the next two in 1588. Euler discovered the eighth perfect numbers in 1750. Interestingly, a medieval German nun, Hrotsvit, a Benedictine in the Abbey of Gandersheim in Saxony and the first woman German poet, listed the first four perfect numbers in her 10th century play, Sapientia. Mathematicians of the middle ages, basing their assumptions on the first four perfect numbers, conjectured that: ο· There is a perfect numbers between any two consecutive of 10; that is there is a perfect numbers of n digits for every positive integer n; and ο· Perfect numbers end alternatively in 6 and 8. Unfortunately, both conjectures are false. There are no perfect numbers that are five digits long. Even perfect numbers do end in 6 or 8, but not alternately; for instance, the fifth and sixth even perfect number end in 6 or 8, but not alternately 27
- 31. for instance, the fifth and sixth even perfect numbers end in 6; the next four end in 28 8. Unfortunately, both conjectures are false. There are no perfect numbers that are five digits long. Even perfect numbers do end in 6 or 8, but not alternatively; for instance, the fifth and sixth even perfect numbers end in 6; the next four end in 8. Unfortunately, both conjectures are false. There are no perfect numbers do end in 6 or 8, but not alternately; for instance, the fifth and sixth even perfect numbers end in 6; the next four end in 8. Notice that every perfect number in the preceding list is even and is of the form 2νβ1 (2ν β 1), where ν and 2νβ1 are primes. We should be doubly impressed that Euclid proved that every such number is a perfect number, as the following theorem confirms. Theorem 3.2 (Euclid) If ν is an intiger greater than or equal to such that 2ν β 1 is a prime, then ν = 2νβ1 (2ν β 1) is a perfect number. Proof. Since 2ν β 1 is a prime ν(2ν β 1) = 1 + (2ν β 1) = 2ν . Because ν is multiplicative, ν(ν) = ν(2νβ1 )ν(2ν β 1) = (2ν β 1)(2ν) = 2ν(2ν β 1) = 2ν.
- 32. Thus ν is a perfect number. οΏ About 2000 years after Euclidβs discovery, Euler proved that the converse of this theorem is also true; that is, every even perfect number is of the form 2ν β 1 is a prime. Theorem 3.2 and 3.3 categorically characterize even perfect numbers. 29 Theorem 3.3 (Euler) Every even perfect numbers ν is of the form ν = 2νβ1 (2ν β 1), where 2ν β 1 is a prime. Proof. Let ν be of the form 2νν , where ν is odd and ν β₯ 1. Since ν is perfect, ν(ν) = 2ν = 2ν+1 ν . Clearly, (2ν, ν ) = 1. So ν(ν) = ν(2νν ) = ν(2ν)ν(ν ). That is 2ν+1 ν = (2ν+1 β 1)ν(ν ). (1) Since (2ν+1 , 2ν+1 β 1) = 1, it follows by corollaryβ¦β¦. that 2ν+1 |ν(ν ). Therefore ν(ν ) = 2ν+1 ν‘ for some positive integer ν‘. Substituting for ν(ν ) in equation (1), 2ν+1 ν = (2ν+1 β 1)2ν+1 ν‘. (2) Therefore ν = (2ν+1 β 1)ν‘. (3) This implies ν‘|ν and ν‘ < ν , since ν‘ = ν implies ν = 0, a contradiction.
- 33. We will now show that ν‘ = 1. For that equation (2) can be rewritten as ν + 30 ν‘ = 2ν+1 ν‘. That is, ν + ν‘ = ν(ν ). (4) This shows that ν‘ is the sum of the proper divisors of ν ; but, by equation (3), ν‘ itself is a proper divisor of ν . So, for the relationship (4) to hold, ν‘ must be . Thus ν + 1 = ν(ν ), so ν has exactly two positive factors and ν . Consequently, ν = 2ν+1 β 1 must be a prime. Thus ν = 2ν(2ν+1 β 1), where 2ν+1 β 1 is a prime. οΏ Problem. Let us find all perfect numbers of the form νν + 1. Solution. Take ν = νν + 1 Case 1 Let ν be odd. Then ν is an even perfect number, so ν must be of the form ν = 2νβ1 (2ν β 1), where 2ν β 1 is a prime. Clearly ν can be factored as ν = νν + 1 = (ν + 1)ν, where ν = ννβ1 β ννβ2 + β― β ν + 1. We claim that (ν + 1, ν) = 1. To show this, notice that since ν is odd, ν is odd and ν + 1 is even. Let ν + 1 = 2ν , where ν‘ is an odd integer greater than or equal to .Then ν = 2ν ν‘ν, where both ν‘ and ν are odd. Since ν is an even perfect number this is possible only if ν‘ =
- 34. 1; so ν + 1 = 2ν and hence (ν + 1, ν) = 1. If ν = 1, then ν = νν + 1; so ν = 1. Then ν = 2, which is not a perfect number. Since ν = 2νβ1 (2ν β 1) = (ν + 1)ν = 2ν ν, where 2ν β 1 is an odd prime and ν is odd, 2ν = 2νβ1 = ν + 1 and ν = 2ν β 1 = 2(ν + 1) β 1 = 2ν + 31 1. Therefore ν = νν + 1 = (ν + 1)(2ν + 1) = 2ν2 + 3ν + 1. This gives νν = 2ν2 + 3ν. Thus ννβ1 = 2ν + 3. Since ν is an integer, this equation has a unique solution . (See the graph in figureβ¦..) Then ν = 33 + 1 = 28. Thus 28 is the only even perfect number of the desired form.
- 35. Case 2 Let ν be even, say, ν = 2ν. Then ν is odd, νν is a square, and νν β‘ β1(mod ν). We claim that 3 <β€ ν. Suppose on the contrary that 3|ν. Then νν β‘ 32 β1(mod ν); that is, (2ν)2ν β‘ β1(mod 3) 4ν β ν2ν β‘ 2(mod 3) 1 β ν2ν β‘ 2(mod 3) ν2ν β‘ 2(mod 3) Cleraly, ν β’ 0 or modulo . If ν β‘ 2(mod 3), then the above congruence gives 2ν β‘ 2(mod 3)
- 36. 4ν β‘ 2(mod 3) 1 β‘ 2(mod 3), a contradiction So ν cannot be congruent to 0,1 or , which is absurd. Thus 3 β€ ν. By Touchardβs theorem, ν = 12ν + 1 or 36ν + 9 for some integer ν. If ν = 36ν + 9, then 3|ν, a contradiction. So ν = 12ν + 1; that is, νν = 12ν. Since 3|12ν, 3|νν, so 3|ν. Thus 2|ν and 3|ν, so 6|ν. 33 Let ν = ν6 + 1,where ν = ν ν 6 > 1. Then ν can be factored as ν = (ν2 + 1)(ν4 β ν2 + 1). (5) We will now see that these factors of ν are relatively prime. To this end, let ν be a common prime factor of the two factors ν2 + 1 and ν4 β ν2 + 1. Since ν4 β ν2 + 1 = (ν4 + 2ν2 + 1) β 3ν2 = (ν2 + 1)2 β 3ν2 = (ν2 + 1)2 β 3(ν2 + 1) + 3, ν|3. That is, ν = 3. This implies 3|ν, a contradiction; so the factors ν2 + 1 and ν4 β ν2 + 1 are relatively prime. Besides, since ν is odd, both factors are also odd. Since ν is perfect and ν is multiplicative, equation (5) yields ν(ν) = ν(ν2 + 1) β ν(ν4 β ν2 + 1). That is, 2ν = ν(ν2 + 1) β ν(ν4 β ν2 + 1).
- 37. Since ν is odd, one of the factors on the right-hand side must be odd. But, if ν and ν(ν) are both odd, then ν is a square. This implies that either ν2 + 1 or ν4 β ν2 + 1 is a square. But ν2 < ν2 + 1 < (ν + 1)2 and (ν2 β 1)2 < ν4 β ν2 + 1 < (ν2 )2. Thus neither can be a square, a contradiction. Consequently, there are no odd perfect numbers of the form νν + 1. Thus 28 is the only perfect number of the desired form. οΏ 34 Odd Perfect Numbers The question remains unanswered as to whether or not there are any odd perfect numbers. Although a host of conditions such a number ν must satisfy have been established, no one has been successful in finding one, in spite of large computer searches with modern supercomputers. For example, ν must be congruent to modulo 12 or to modulo 36; it must have at least eight different prime factors; and in 1991 R. P. Brent, G. L. Cohen, and H. J. J. te Riele showed that it must be greater than 10300. In 1998, G. L Cohen of the University of Technology, Sydney, and P. Hagis, Jr. of Temple University proved that the largest prime factor of an odd perfect number exceed 106; and three years earlier, D. E. Iannucci showed that the second prime factor exceed 104 and the third prime factor exceed 100. In 2000, Paul A. Weiner of St. Maryβs University of Minnesota established that if 3ν(ν) = 5ν for some integer ν, then 5ν is an odd perfect
- 38. number. There is however, a strong belief in the mathematical community that there may not be any odd perfect numbers. 35 BIBLIOGRAPHY [1] W. Edwin Clark, Elementary Number Theory, Department of Mathematics, University of Florida (2002). [2] G. H. Hardy and E. M. Wright, An introduction to the Theory of Numbers, Fifth Edition, Oxford University Press (1979). [3] S. B. Malik, S. K. Maheshwari, S. P. Misra, Basic Number Theory, Second Edition, Vikas Publishing House (1998). [4] Thomas Koshy, Elementary Number Theory with Applications, Academic Press (2002). [5] Tom M. Apostol, Introduction to Analytic Number Theory, Narosa (2000).