A textbook of

NUMBERS

BINTU GEORGE JACOB

MOUNT TABOR TRAINING COLLEGE

PATHANAPURAM.

MOC PUBLICATIONS, KOLLAM

2014

PrefaceThe 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 memorizingformulas and performing rote algebraic manipulations. This book has a similar,albeit somewhat more ambitious, goal: to lead you to think mathematically andto experience the thrill of independent intellectual discovery. Our chosen subject,Number Theory, is particularly well suited for this purpose. The natural numbers1, 2, 3, ... satisfy a multitude of beautiful patterns and relationships, many ofwhich can be discerned at a glance; others are so subtle that one marvels they werenoticed at all. Experimentation requires nothing more than paper and pencil, butmany false alleys beckon to those who make conjectures on too scanty evidence. Itis only by rigorous demonstration that one is finally convinced that the numericalevidence reflects a universal truth. This book will lead you through the groveswherein lurk some of the brightest flowers of Number Theory, as it simultaneouslyencourages you to investigate, analyze, conjecture, and ultimately prove your ownbeautiful number theoretic results.This book was originally written to serve as a text for Math 42, a course createdby Jeff Hoffstein at Brown University in the early 1990s. Math 42 was designed toattract nonscience majors, those with little interest in pursuing the standard calculussequence, and to convince them to study some college mathematics. The intent wasto create a course similar to one on, say, “The Music of Mozart” or “ElizabethanDrama,” wherein an audience is introduced to the overall themes and methodologyof 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 andalso scientifically oriented undergraduates interested in a change of pace from theirlarge-lecture, cookbook-style courses.The prerequisites for reading this book are few. Some facility with high schoolalgebra is required, and those who know how to program a computer will have fungenerating reams of data and implementing assorted algorithms, but in truth thereader needs nothing more than a simple calculator. Concepts from calculus arementioned in passing, but are not used in an essential way. However, and the readeris hereby forewarned, it is not possible to truly appreciate Number Theory withoutan eager and questioning mind and a spirit that is not afraid to experiment, to makemistakes and profit from them, to accept frustration and persevere to the ultimatetriumph. Readers who are able to cultivate these qualities will find themselvesrichly rewarded, both in their study of Number Theory and their appreciation of allthat life has to offer.

ii

CONTENTS

CHAPTER 1 : PRELIMINARIES 1

CHAPTER 2 : MULTIPLICATIVE

FUNCTIONS 5

CHAPTER 3 : PERFECT NUMBERS 25

BIBLIOGRAPHY 35

iii

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.

Peano Axioms.

N is a set with the following properties.

(1) N has a distinguished element which we call ' 1 '.

(2) There exists a set map P : N → N .

(3) P is one-to-one (injective).

(4) There does not exist an element n∈N such that P(n)=1.

(5) (Principle of Induction) Let S⊂N such that a) 1∈S and b)

if n∈S, then P(n)∈S . Then S=N .

We call such a set N 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 Z:

Z={… ,−3 ,−2 ,−1,0 , 1 ,2 , 3 ,…}.

1

Theorem 1.1

(The Principle of Mathematical Induction) Let p(n) be a statement

satisfying the following conditions, where n∈Z:

1. p(n0) is true for some integer n0.

2. If p(k) is truefor an arbitrary integer k ≥ n0, then p(k+1) is true.

Then p(n) is true for every integer n ≥ n0.

The following is a strong version of the principle of induction.

Theorem 1.2

(The Second Principle of Mathematical Induction) Let p(n) be a

statement satisfying the following conditions, where n∈Z:

(1') p(n0) is true for some integer n0.

(2') If k is an arbitrary integer greater than or equal to n0 such that

p (n0 ) , p (n0+1 ) ,…, and p(k) are true, then p(k+1) is also true.

Then p(n) is true for every integer n ≥ n0.

Definition 1.3 (Prime Numbers)

Any integer greater than one which has no factors other than one and itself is

called a prime number.

2

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 a ,b be the integers, let a0, we say that a divides b or a is a divisor of b if

b=aq for some integer q. When a divides b we write a∨b.

Definition 1.6 (Greatest Common Divisor)

A number d is the greatest common divisor (gcd) of two non zero numbers a

and b, if d is the common divisor of a and b and no common divisor of a and b is

larger than d. gcd of a and b is denoted by (a ,b).

Definition 1.7

Two positive integers a and b are relatively prime if their gcd is .

Theorem 1.8 (Euclidean Algorithm)

Let a and b are any positive integers, and r be the reminder, when a is

divided by b. Then (a ,b)=(b , r ).

3

Theorem 1.9 (Fundamental Theorem of Arithmetic)

Any natural number n ≥ 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 m)

Let m be a positive integer. Then an integer a is congruent to an integer b

modulo m if m∨(a−b). We write a ≡ b(mod m) and m is called the modulus of the

congruence relation.

Definition 1.11 (Euler’s Phi Function)

Let m be a positive integer. Then Euler’s phi function φ (m) denotes the

number of positive integers m and relatively prime to m.

Chapter 2

MULTIPLICATIVE FUNCTIONS

4

Euler’s phi function is one of the most important number–theoretic functions

(also known as arithmetic functions). Arithmetic functions are defined for all

positive integers. Euler’s phi function belongs to a large class of arithmetic

functions called multiplicative functions.

Definition 2.1

A number-theoretic function f is multiplicative if f (mn)=f (m) f (n), whenever

m and n are relatively prime.

Example 2.2

The constant function f (n)=1 is multiplicative, since

f (mn)=1=1 ∙1=f (m) ∙ f (n).

The functions g(n)=nk, k being a fixed integer is multiplicative, since

g (mn )=(mn)k=mk nk=g (m ) g(n).

Theorem 2.3

Let f be a multiplicative function and n be a positive integer with canonical

decomposition n=p1e1 p2

e2 … pkek . Then

f ( n )=f ( p1e1) f ( p2

e2 )…f ( pkek ).

Proof. If k=1, that is, if n=p1e1, then f (n)=f ( p1

e1), so the theorem is trivially true.

5

Assume it is true for any integer with canonical decomposition consisting of

k distinct primes. That is f ( n )=f ( p1e1) f ( p2

e2 )…f ( pkek ) .

Let n be any integer with k+1 distinct primes in its canonical decomposition,

say, n=p1e1 p2

e2 … pkek .

Since ( p1e1 p2

e2… pkek , pk+1

ek+1 )=1 and f is multiplicative,

f ( p1e1 p2

e2… pkek pk+1

ek+1 )=f ¿

¿ f ¿,

by induction. Therefore by induction the result is true for any positive integer n.

Theorem 2.4

Let p be a prime and e any positive integer. Then φ ( pe)=pe−pe−1.

Proof. φ ( pe)=¿ number of positive integers less than or equal to peand relatively

prime to it.

= (number of positive integers less than or equal to pe)-(number

of positive integers less than or equal to pe and not relatively prime

to it)

6

The positive integers less than or equal to pe and not relatively prime to it are

the various multiples of p, namely, p ,2 p , 3 p ,…… ……( pe−1) p, and they are pe−1 in

number. Thus φ ( pe)=pe−pe−1 . Lemma 2.5

Let m and n be relatively prime positive integers.

Then the integers r , m+r ,2m+r , …,(n−1)m+r are congruent modulo n to

0 ,1 ,2 , …,(n−1) in some order.

Example 2.6

φ (8 )=φ (23 )=23−22=8−4=4.

Thus there are four positive integers less than or equal to 8 and relatively prime to

it; 1, 3, 5 and 7.

φ (81 )=φ (34 )=34−33=54.

φ (15625 )=φ (56 )=56−55=12500.

Theorem 2.7

The function φ is multiplicative.

Proof. Let m and n be positive integers such that (m , n)=1.

We have to prove that φ (mn )=φ (m ) φ (n ) .

Arrange the integers 1 through mn in m rows of n each:

1.m+12 m+1 ….(n−1)m+1

7

2.m+22 m+2 ….(n−1)m+2

3.m+32m+3 ….(n−1)m+3

.

r . m+r2 m+r …(n−1)m+r

.

.

.

m .2m 3m ... nm

Let r be a positive integer less than or equal to m such that (r , m)>1.

We will show that no element of the r th row is relatively prime to mn .

Let d=(r , m). Then d∨r and d∨m, so d∨km+r for any integer k; that is, d is a

factor of every element in the r th row. Thus no element in the r th row is relatively

prime to m, and hence to mn if (r , m)>1. That is the elements in the array are

relatively prime to mn from the r th row only if (r , m)=1. By definition, there are

φ (m) such integers r and hence φ (m) such rows.

Now, in r th row, where (r , m)=1:

8

r , m+r ,2m+r , …,(n−1)m+r

By Lemma 2.5, their least residues modulo n are a permutation of

0 ,1 ,2 , …,(n−1) of which φ (n) are relatively prime to n. Therefore exactly φ (n)

elements in r th row are relatively prime to n and hence to mn.

Thus there are φ (m) rows containing positive integers relatively prime to mn,

and each row contains φ (n) elements relatively prime to it. So the array contains

φ (m)φ(n) positive integers less than or equal to mn and relatively prime to mn.

Therefore φ (mn )=φ (m ) φ (n ) .

Example 2.8

1. φ (221 )=φ (13 ∙17 )=φ (13 ) ∙ φ (17 )

¿12 ∙16=192.

2. φ (625 )=φ (53 ∙ 72 )=φ (53 ) ∙φ (72 )

¿ (53−52 ) ( 72−7 )=4200.

Theorem 2.9

Let n=p1e1 p2

e2 … pkek be the canonical decomposition of a positive integer n.

Then φ (n )=n(1−1p1

)(1− 1p2

)…(1− 1pk

).

Proof. Since φ is multiplicative, by theorem 2.1

9

φ (n )=φ ( p1e1 ) φ ( p2

e2) …φ ( pkek )

¿ p1e1(1− 1

p1) p2

e2(1−1p2

)… pkek(1− 1

pk)

¿ p1e1 p2

e2 … pkek(1−

1p1

)(1− 1p2

)…(1−1pk

)

¿n(1−1p1

)(1− 1p2

)…(1−1pk

).

Example 2.10

1. 666=2∙32 ∙37

Thus φ (666 )=666(1−12 )(1−1

3 )(1− 137 )

¿216.

2. 1976=23 ∙13 ∙19

Thus φ (1976 )=1976(1−12 )(1− 1

13 )(1− 119 )

¿864.

Theorem 2.11

If n ≥ 3 , then φ (n) is even.

Proof. If n is even obviously φ (n) is even. If n is odd then for each prime factor p

of n, the numerator of (1−1p ) is even. Thus φ (n) is even for n ≥ 3.

10

Theorem 2.12

Let n be a positive integer, then ∑d∨n

φ (d )=n.

Here we turn to illustrate its proof.

Example 2.13

The positive divisors of 18 are 1 ,2, 3 , 6 , 9 and 18.

So ∑d∨18

φ (18 )=φ (1 )+φ (2 )+φ (3 )+φ (6 )+φ (9 )+φ (18)

¿1+1+2+2+6+6=18.

The Tau and Sigma Functions

Definition 2.14 (The Tau Function)

Let n be a positive integer. Then τ (n) denotes the number of positive divisors

of n; that is τ (n )=∑d∨n

1.

Example 2.15

The positive divisors of 18 are 1 ,2, 3 , 6 , 9 and 18,

So τ (18 )=6.

23, being a prime, has exactly two positive divisors,

11

So τ (23 )=2.

Remark. If n is prime, then τ (n )=2; if τ (n )=2, then n is prime.

Definition 2.16 (The Sigma Function)

Let n be positive integer. Then σ (n) denotes the sum of the positive divisors

of n; that is σ (n )=∑d∨n

d.

Example 2.17

The positive divisors of 12 are 1 ,2, 3 , 4 ,5 , 6 and 12;

So σ (12 )=1+2+3+4+6+12=28.

The positive divisors of 28 are 1 ,2, 4 , 7 ,14 and 28;

So σ (28 )=1+2+4+7+14+28=56.

Let f be a multiplicative function. Define a new function F by F (n )=∑d∨n

f (d).

The following example shows that the new function F can be a multiplicative

function.

Example 2.18

We have (4,7 )=1.

Then F ( 4 ∙7 )=∑d∨28

f (28)

12

¿ f (1 )+ f (2 )+ f (4 )+ f (7 )+ f (14 )+ f (28)

¿ f (1 ∙1 )+ f (1∙ 2 )+ f (1 ∙ 4 )+ f (1∙7 )+ f (2∙7 )+ f (4 ∙7)

¿ [ f (1 )+ f (2 )+ f (4 ) ] f (1 )+[ f (1 )+ f (2 )+ f (4 ) ] f (7)

¿ [ f (1 )+ f (2 )+ f (4 ) ] [ f (1 )+ f (7 )]

¿∑d∨4

f (d )∙∑d∨7

f (d )

¿ F ( 4 ) F(7).

Theorem 2.19

If f is a multiplicative function, then F (n )=∑d∨n

f (d) is also multiplicative.

Proof. Let m and n be relatively prime positive integers. We want to show that

F (mn)=F (m)F (n).

By definition, F (mn )= ∑d∨mn

f (d ). Since (m , n)=1, every positive divisor of mn, is

the product of a unique pair of positive divisors d1 of m and d2 of n, where, (m , n)=1.

Therefore F (mn )= ∑d1∨m,d 2∨m

f (d1 d2) .

Since f is multiplicative, f ( d1d2 )=f (d1 ) f (d2).

Therefore F (mn )= ∑d1∨m,d 2∨m

f (d1 d2)

13

¿∑d2∨n

F (m ) f (d2)

¿ F (m)∑d2∨n

f (d2)

¿ F (m ) F(n).

Thus F is multiplicative.

Corollary 2.20

The Tau and Sigma functions are multiplicative.

Proof. By theorem 2.6, the functions ∑d∨n

f (d )=∑d∨n

1=τ (n) and ∑d∨n

g (d )=∑d∨n

d=σ (n) are

multiplicative.

(Since we have constant function f (n)=1 and the identity function g(n)=n are

multiplicative.)

That is, if f (m ,n)=1, then τ (mn )=τ (m) τ (n) and σ (mn )=σ (m )σ (n).

Example 2.21

36 = (4)(9). Also (4, 9) = 1.

Thus τ (36 )=τ (4 ) τ (9 )=3 ∙ 3=9.

And σ (36 )=σ ( 4 ) σ (9 )=(1+2+4 ) (1+3+9 )=91.

Theorem 2.22

14

Let p be any prime and e any positive integer. Then τ ( pe )=e+1 and

σ ( pe)= pe+ 1−1p−1

.

Proof. The positive factors of pe are of the form pi, where 0 ≤ i≤ e. There are e+1

such factors. So τ ( pe )=e+1.

σ ( pe)=∑0

e

p i=¿ pe+1−1p−1

.¿

The following theorem gives two formulas. They are a consequence of

Corollary 2.20 and Theorem 2.2.

Theorem 2.23

Let n be a positive integer with canonical decomposition n=p1e1 p2

e2 … pkek. Then

τ (n )=(e1+1 ) ( e2+2 ) … (ek+k ) and σ (n )=p1

e1−1p1−1

∙p2

e2−1p2−1

∙ …∙pk

e k−1pk−1

.

Proof. Since τ is multiplicative,

τ (n )=τ ( p1e1 ) ∙ τ ( p2

e2 )…τ ( pke k )=(e1+1 ) ( e2+1 ) …(ek+1).

Since σ is multiplicative,

σ (n )=σ ( p1e1 ) ∙ σ ( p2

e2 )…σ ( pkek )= p1

e1−1p1−1

∙p2

e2−1p2−1

∙…∙pk

ek−1pk−1

.

Example 2.24

The canonical decomposition of 6120=23 ∙32 ∙5 ∙17

15

Therefore τ (6120 )=(3+1 ) (2+1 ) (1+1 ) (1+1 )=48

and σ (6120 )=23+1−12−1

∙32+1−1

3−1∙

51+1−15−1

∙171+1−1

17−1=15 ∙ 14 ∙ 6 ∙18=22680.

The Mӧbius Function

The Mӧbius function μ is an important number-theoretic function which was

discovered by the German mathematician August Ferdinand Mӧbius. It plays an

important role in the study of the distribution of primes.

Definition 2.25 (The Mӧbius function μ)

Let n be a positive integer. Then

μ (n )={ 1 if n=10 if p2∨n for some prime p

(−1)k if n=p1 p2 … pk

For example μ (2 )=−1 , μ (3 )=−1 , μ ( 4 )=0 , μ (12 )=0 , μ (35 )=μ (5.7 )=(−1)2 and

μ (672 )=μ (25 ∙3 ∙ 7 )=0.

Thus μ assigns −1,0 or to each positive integer; μ (n )=0 if the canonical

decomposition of n contains a perfect square, that is, if n is not square-free; it is

(−1)k if it consists of k distinct prime factors.

Theorem 2.26

16

The function μ is multiplicative.

Proof. Let m and n be relatively prime positive integers. If m=1 or n=1, then clearly

μ (mn )=μ (m ) μ(n).

Suppose m or n is divisible by p2 for some prime p. Then μ (m ) μ (n )=0. If p2∨m

or p2∨n, then p2∨mn, so μ (mn )=0. Thus μ (mn )=μ (m ) μ(n).

Finally, suppose both m and n are square-free, let m=p1 p2 … pr and n=q1 q2…qs

, where the pi’s and q j’s are distinct primes, since (m , n)=1. μ (m )=(−1)r and

μ (n )=(−1)5.

Then mn=p1 p2 … pr q1 q2 …qs, a product of distinct primes.

Therefore μ (mn )=(−1)r+s=(−1)r ∙(−1)s=μ (m ) μ (n). Thus, in every case, μ (mn )=μ (m ) μ(n)

, so μ is multiplicative.

Example 2.27

Let us determine μ (420 ).

Since μ is multiplicative, μ (420 )=μ (15 ) μ (28)

Now 15=3∙ 5. So by definition, μ (15 )=1.

28=22 ∙ 7. So by definition, μ (28 )=0.

420=22 ∙3 ∙ 5∙ 7. Since 420 is not square–free, μ (420 )=0. Thus

μ (420 )=0=1∙ 0=μ (15 ) μ(28).

17

Next we develop a formula for ∑d∨n

μ (d ). When n = 1, ∑d∨1

μ ( d )=μ (1 )=1. If n>1,

we can compute the sum using the canonical decomposition of n and Theorem 8.7

provided we know the sum when n is a prime-power pe. For this we need to

introduce a new function, as the following lemma shows.

Lemma 2.28

Let F (n )=∑d∨n

μ (d ). Then F ( pe)=0, where e>1.

Proof. F ( pe)=∑d∨pe

μ(d)

¿∑i=0

n

μ( pi)

¿ μ (1 )+μ ( p )+μ ( p2 )+…+μ( pe )

¿1+(−1 )=0+…+0=0.

Theorem 2.29

Let n be a positive integer. Then ∑d∨n

μ (d )={ 1 if n=10 othewise.

Proof. If n=1, ∑d∨1

μ ( d )=μ (1 )=1. So let n>1 and let n=p1e1 p2

e2 … pkek be the canonical

decomposition of n. Let, F (n )=∑d∨n

μ (d ). Since μ is multiplicative, F is also

multiplicative, by Theorem 8.7.

18

Therefore F (n )=∏i=1

k

F ( pei)

F (n )=∏i=1

k

0=0.

Example 2.30

∑d∨18

μ (d )=μ (1 )+μ (2 )+μ (3 )+μ (6 )+μ (9 )+μ (18 )

¿1+(−1 )+(−1 )+(−1)2+0+0=0.

Problem.

Let f be a number-theoretic function. Show that

∑(d∨6)

∑(d '|( 6

d ))μ (d ) f ( d' )= ∑

(d¿¿ '∨6) ∑(d|( 6

d ))f (d ' ) μ (d)

¿¿¿.

Solution. ∑d∨6

∑(d '|( 6

d ))μ (d ) f ( d ' )

¿∑d'∨6

μ (1 ) f (d ' )+∑d'∨3

μ (2 ) f (d ' )+∑d'∨2

μ (3 ) f (d ' )+∑d'∨1

μ (6 ) f ( d' )

¿ μ (1 )∑d '∨6

f (d ')+μ(2)∑d '∨3

f (d ' )+μ(3)∑d'∨2

f (d ' )+μ (6 )∑d'∨1

f ( d' )

¿ μ (1 ) [ f (1 )+ f (2 )+ f (3 )+ f (6 ) ]+μ (2 ) [ f (1 )+f (3 ) ]+μ (3 ) [ f (1 )+ f (2 ) ]+μ (6 ) f (1)

¿ f (1 ) [ μ (1 )+μ (2 )+μ (3 )+μ (6 ) ]+ f (2 ) [μ (1 )+μ (3 ) ]+ f (3 ) [ μ (1 )+μ (2 ) ]+ f (6 ) μ(1)

¿ f (1 )∑d∨6

μ ( d )+ f (2 )∑d∨3

μ (d )+f (3 )∑d∨2

μ (d )+ f (6)∑d∨1

μ(d)

19

¿∑

d'∨6

f (d ' ) ∑d∨( 6

d' )

μ(d)

¿ ∑

(d¿¿ '∨6) ∑(d|( 6

d ))f ( d' ) μ (d)¿

¿.

Theorem 2.31 (Mӧbius Inversion Formula)

Let f be a number-theoretic function and let F (n )=∑d∨n

f (d). Then

f ( n )=∑d∨n

μ (d ) F ( nd ).

Proof.F (n )=∑d∨n

f (d). Then F ( nd )= ∑

d '∨( nd )

f (d' ).

So μ (d ) F ( nd )=μ (d ) ∑

d '∨( nd )

f (d ' )= ∑d '∨( n

d )μ (d ) f (d ' )

Therefore ∑d∨n

μ (d ) F ( nd )=∑

d∨n∑

d'∨( nd )

μ (d ) f (d ').

As d runs over the positive divisors of n, so does d '; also dd ' runs over the

positive factors of n; that is, the sum of all its values of μ (d ) f (d ') as dd ' runs over

the positive factors of n; that is, the sum of all its values for all pairs d and d ' such

that d '

n and d∨( n

d ' ).

That is, ∑d∨n

∑d'∨( n

d )μ (d ) f (d ' )=∑

d '∨n∑

d∨( nd ' )

μ (d ) f (d ')

20

Thus ∑d∨n

μ (d ) F ( nd )=∑

d '∨n

f (d' )[ ∑d∨( nd' )

μ (d )]But by Theorem 8.17,

∑d∨( n

d' )μ (d )

equals if n

d '=1; that is, if n=d ', and

otherwise. Thus ∑d∨n

μ (d ) F ( nd )=f (d ' )∙ 1, where d '=n.

In other words, f ( n )=∑d∨n

μ (d ) F ( n

d ' ).

Remark. As d runs over the positive factors of n, so does nd . Therefore, the

inversion formula f ( n )=∑d∨n

μ (d ) F ( nd ) can also be written asf ( n )=∑

d∨n

μ ( nd )F (d ).

Notice that the definition f ( n )=∑d∨n

f (d) expresses F in terms of f , whereas the

inversion formula f ( n )=∑d∨n

μ (d ) F ( nd ) expresses f in terms of F.

To illustrate the inversion formula, recall that τ (n )=∑d∨n

1 and σ (n )=∑d∨n

d

because both the constant function f ( n )=1 and the identity function g(n)=n are

multiplicative, it follows by Theorem 2.32 that

1=∑d∨n

μ (d ) τ ( nd )=∑

d∨n

μ( nd )τ (d)

and

21

n=∑d∨n

μ (d ) σ ( nd )=∑

d∨n

μ ( nd )σ (d )

We can illustrate these results by the following example.

Example 2.32

∑d∨6

μ (d ) τ ( 6d )=μ (1 ) τ (6 )+μ (2 ) τ (3 )+μ (3 ) τ (2 )+μ (6 ) τ (1 )

¿1 ∙4+(−1 ) ∙ 2+ (−1 ) ∙2+(−1)2 ∙1

¿1

∑d∨6

μ (d ) σ ( 6d )=μ (1 ) σ (6 )+μ (3 ) σ (2 )+μ (6 ) σ (1)

¿1 ∙12+(−1 ) ∙ 4+ (−1 ) ∙ 3+(−1)2 ∙1

¿6.

Theorem 2.33

φ (n )=n∑d∨n

μ(d)d

.

Proof. By Theorem 2.12, n=∑d∨n

φ(d ). Let g denote the identity function given by

g (n )=n for all n. Therefore g (n )=n=∑d∨n

φ(d). Then by Mӧbius inversion formula,

φ (n )=∑d∨n

μ ( d ) g( nd )=¿∑

d∨n

μ (d )d

¿.

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Problem.

Verify the formula in Theorem 2.35 for n=12.

Solution.

By Theorem 2.9, φ (12 )=φ(2¿¿2 ∙ 3)=4.¿

Now,

12 ∑d∨12

μ (d)d

=∑d∨12

( 12d )μ (d )

¿12 μ (1 )+6 μ (2 )+4 μ (3 )+3μ ( 4 )+2 μ (6 )+1 μ(12)

¿12 ∙1+6 ∙ (−1 )+4 ∙ (−1 )+3 ∙ 0+2 ∙(−1)2+1 ∙ 0

¿4=φ (12 ) .

Theorem 2.34

Let F andf be number-theoretic functions such that f ( n )=∑d∨n

μ (d ) F ( nd ). Then

F (n )=∑d∨n

f (d).

Proof. By the definition of f , f ( d )=∑d'∨d

μ ( d ' ) F ( d

d ' ).

Therefore ∑d∨n

f (d )=∑d∨n

∑d'∨d

μ ( d ' ) F ( d

d ' ).

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Let d

d '=k and

nd=l,then d=kd ' and n=ld=kld '.So this equation yields

∑d∨n

f (d )=∑d∨n

∑k d'=d

μ ( d ' ) F (k )

¿ ∑k d'∨n

μ (d ' ) F(k )

¿∑

k∨n

F (k )[ ∑d'∨( nk )

μ(d ' )].

By Theorem 8.17, ∑

d'∨( nk )

μ(d ' ) equals if n=k and otherwise. So the equation

becomes ∑(d|n)

f (d )=F (k ) ∙(1), where n=k.

That is, F (n )=∑d∨n

f (d).

Chapter 3

PERFECT NUMBERS

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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

“good” numbers. They were also studied by the early Hebrews; Rabbi Josef ben

Jehuda in the 12th century recommended their study in his book, “Healing of

souls”.

Historically, some biblical scholars considered 6 a perfect number, because

they believed god created the world in six days and god’s work is perfect. St.

Augustine, on the other hand, believed God’s work to be perfect because 6 is a

perfect number. He writes, “Six 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.”

The Pythagoreans regarded 6 as the symbol of “marriage and health and

beauty on account of the integrity of its parts and the agreement existing in it.”

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:

28=1+2+4+7+14

496=1+2+4+8+16+31+62+124+248

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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 n is a perfect number if the sum of its proper divisiors

equals n. Thus n is perfect if σ (n )−n=n, that is, if σ (n )=2n.

The first eight perfect numbers are:

6=2(22−1)

28=22(23−1)

496=24(25−1)

8,128=26(27−1)

33,550,336=212(213−1)

8,589,869,056=216(217−1)

137,438,691,328=218(219−1)

2,305,843,008,139,952,128=230 (231−1 ) .

Of which only the first four were known to the ancient Greeks; they are

listed in Nichomachus’s Introductio Arithmeticae. The next perfect numbers was

discovered by the Greek mathematician Hudalrichus Regius around 1536. The

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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:

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

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

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 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.

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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 2p−1(2p−1), where p and 2p−1 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 n is an intiger greater than or equal to such that 2n−1 is a prime, then

N=2n−1(2n−1) is a perfect number.

Proof. Since 2n−1 is a prime σ (2n−1 )=1+(2n−1 )=2n. Because σ is multiplicative,

σ ( N )=σ (2n−1 ) σ (2n−1)

¿ (2n−1 ) ( 2n )=2n (2n−1 )=2N .

Thus N is a perfect number.

About 2000 years after Euclid’s discovery, Euler proved that the converse of

this theorem is also true; that is, every even perfect number is of the form 2n−1 is a

prime. Theorem 3.2 and 3.3 categorically characterize even perfect numbers.

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Theorem 3.3 (Euler)

Every even perfect numbers N is of the form N=2n−1(2n−1), where 2n−1 is a

prime.

Proof. Let N be of the form 2e s, where s is odd and e≥ 1. Since N is perfect,

σ ( N )=2 N=2e+1 s. Clearly, (2e , s )=1. So σ ( N )=σ (2e s )=σ (2e) σ (s).

That is 2e+1 s=(2e+1−1 ) σ (s). (1)

Since (2e+1 , 2e+1−1 )=1, it follows by corollary……. that 2e+1∨σ (s). Therefore

σ ( s)=2e+1t for some positive integer t. Substituting for σ (s ) in equation (1),

2e+1 s=(2e+1−1)2e+1t . (2)

Therefore s=(2e+1−1) t. (3)

This implies t∨s and t <s, since t=s implies e=0, a contradiction.

We will now show that t=1. For that equation (2) can be rewritten as

s+t=2e+1 t.

That is, s+t=σ (s ). (4)

This shows that t is the sum of the proper divisors of s; but, by equation (3),

t itself is a proper divisor of s. So, for the relationship (4) to hold, t must be . Thus

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s+1=σ (s), so s has exactly two positive factors and s. Consequently, s=2e+1−1

must be a prime.

Thus N=2e(2e+1−1), where 2e+1−1 is a prime.

Problem.

Let us find all perfect numbers of the form nn+1.

Solution. Take N=nn+1

Case 1 Let n be odd. Then N is an even perfect number, so N must be of the form

N=2m−1(2m−1), where 2m−1 is a prime.

Clearly N can be factored as N=nn+1=(n+1 )r, where r=nn−1−nn−2+…−n+1.

We claim that (n+1 , r )=1.

To show this, notice that since n is odd, r is odd and n+1 is even. Let n+1=2s,

where t is an odd integer greater than or equal to .Then N=2s tr, where both t and r

are odd. Since N is an even perfect number this is possible only if t=1; so n+1=2s

and hence (n+1 , r )=1. If r=1, then N=nn+1; so n=1. Then N=2, which is not a

perfect number.

Since N=2m−1 (2m−1 )=( n+1 )r=2s r, where 2m−1 is an odd prime and r is odd,

2s=2m−1=n+1 and r=2m−1=2 (n+1 )−1=2n+1.

Therefore N=nn+1=(n+1 ) (2n+1 )=2n2+3n+1.

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This gives nn=2 n2+3 n. Thus nn−1=2n+3.

Since n is an integer, this equation has a unique solution . (See the graph in

figure…..) Then N=33+1=28. Thus 28 is the only even perfect number of the

desired form.

Case 2 Let n be even, say, n=2 k. Then N is odd, nn is a square, and nn≡−1(mod N ).

We claim that 3<∤N . Suppose on the contrary that 3∨N . Then nn≡−1(mod N ); that

is,

(2k )2 k ≡−1(mod 3)

4k ∙ k2 k ≡ 2(mod 3)

31

1 ∙ k2 k ≡ 2(mod 3)

k 2k ≡2(mod 3)

Cleraly, k ≢ 0 or modulo . If k ≡ 2(mod 3), then the above congruence gives

2k ≡2(mod3)

4k ≡2(mod3)

1 ≡2(mod3), a contradiction

So k cannot be congruent to 0,1 or , which is absurd. Thus 3∤N .

By Touchard’s theorem, N=12 m+1 or 36 m+9 for some integer m. If

N=36 m+9, then 3∨N , a contradiction. So N=12 m+1; that is, nn=12 m. Since 3∨12m,

3∨nn, so 3∨n. Thus 2∨n and 3∨n, so 6∨n.

Let N=a6+1,where a=nn6 >1. Then N can be factored as N= (a2+1 ) (a4−a2+1 ).

(5)

We will now see that these factors of N are relatively prime. To this end, let

p be a common prime factor of the two factors a2+1 and a4−a2+1. Since

a4−a2+1=( a4+2a2+1 )−3 a2

¿(a2+1)2−3a2

¿(a2+1)2−3 (a2+1 )+3, p∨3.

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That is, p=3. This implies 3∨N , a contradiction; so the factors a2+1 and a4−a2+1

are relatively prime. Besides, since N is odd, both factors are also odd. Since N is

perfect and σ is multiplicative, equation (5) yields σ ( N )=σ (a2+1 ) ∙ σ (a4−a2+1).

That is, 2 N=σ ( a2+1 ) ∙ σ (a4−a2+1).

Since N is odd, one of the factors on the right-hand side must be odd. But, if

m and σ (m) are both odd, then m is a square. This implies that either a2+1 or

a4−a2+1 is a square. But a2<a2+1<(a+1)2 and (a2−1)2<a4−a2+1<(a2)2. Thus neither

can be a square, a contradiction.

Consequently, there are no odd perfect numbers of the form nn+1. Thus 28 is

the only perfect number of the desired form.

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 N must satisfy have

been established, no one has been successful in finding one, in spite of large

computer searches with modern supercomputers. For example, N 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

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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. Mary’s University of Minnesota

established that if 3 σ (n )=5 n for some integer n, then 5 n is an odd perfect number.

There is however, a strong belief in the mathematical community that there may

not be any odd perfect numbers.

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).

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[4] Thomas Koshy, Elementary Number Theory with Applications, Academic

Press (2002).

[5] Tom M. Apostol, Introduction to Analytic Number Theory, Narosa (2000).

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