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, “The Music of Mozart” or “Elizabethan
Drama,” 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.
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.
푵 is a set with the following properties.
(1) 푵 has a distinguished element which we call ′1′.
(2) There exists a set map 푃: 푵 → 푵.
(3) 푃 is one-to-one (injective).
(4) There does not exist an element 푛 ∈ 푵 such that 푃(푛) = 1.
(5) (Principle of Induction) Let 푆 ⊂ 푵 such that a) 1 ∈ 푆 and b)
if 푛 ∈ 푆, then 푃(푛) ∈ 푆. Then 푆 = 푵.
We call such a set 푵 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 풁:
풁 = {… , −3, −2, −1, 0, 1, 2, 3, … }.
5. 2
Theorem 1.1
(The Principle of Mathematical Induction) Let 푝(푛) be a statement
satisfying the following conditions, where 푛 ∈ 풁:
1. 푝(푛0 ) is true for some integer 푛0.
2. If 푝(푘) is truefor an arbitrary integer 푘 ≥ 푛0, then 푝(푘 + 1) is true.
Then 푝(푛) is true for every integer 푛 ≥ 푛0 .
The following is a strong version of the principle of induction.
Theorem 1.2
(The Second Principle of Mathematical Induction) Let 푝(푛) be a
statement satisfying the following conditions, where 푛 ∈ 풁:
(1') 푝(푛0 ) is true for some integer 푛0.
(2') If 푘 is an arbitrary integer greater than or equal to 푛0 such that
푝(푛0), 푝(푛0 + 1), …, and 푝(푘) are true, then 푝(푘 + 1) is also true.
Then 푝(푛) is true for every integer 푛 ≥ 푛0.
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 푎, 푏 be the integers, let 푎0, we say that 푎 divides 푏 or 푎 is a divisor of 푏
if 푏 = 푎푞 for some integer 푞. When 푎 divides 푏 we write 푎|푏.
Definition 1.6 (Greatest Common Divisor)
A number 푑 is the greatest common divisor (gcd) of two non zero numbers 푎
and 푏, if 푑 is the common divisor of 푎 and 푏 and no common divisor of 푎 and 푏 is
larger than 푑. gcd of 푎 and 푏 is denoted by (푎, 푏).
Definition 1.7
Two positive integers 푎 and 푏 are relatively prime if their gcd is .
Theorem 1.8 (Euclidean Algorithm)
7. Let 푎 and 푏 are any positive integers, and 푟 be the reminder, when 푎 is
4
divided by 푏. Then (푎, 푏) = (푏, 푟).
Theorem 1.9 (Fundamental Theorem of Arithmetic)
Any natural number 푛 ≥ 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 풎)
Let 푚 be a positive integer. Then an integer 푎 is congruent to an integer 푏
modulo 푚 if 푚|(푎 − 푏). We write 푎 ≡ 푏(mod 푚) and 푚 is called the modulus of
the congruence relation.
Definition 1.11 (Euler’s Phi Function)
Let 푚 be a positive integer. Then Euler’s phi function 휑(푚) denotes the
number of positive integers 푚 and relatively prime to 푚.
8. Chapter 2
MULTIPLICATIVE FUNCTIONS
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
5
functions called multiplicative functions.
Definition 2.1
A number-theoretic function 푓 is multiplicative if 푓(푚푛) = 푓(푚)푓(푛),
whenever 푚 and 푛 are relatively prime.
Example 2.2
The constant function 푓(푛) = 1 is multiplicative, since 푓(푚푛) = 1 = 1 ∙
1 = 푓(푚) ∙ 푓(푛).
The functions 푔(푛) = 푛푘 , 푘 being a fixed integer is multiplicative, since
푔(푚푛) = (푚푛)푘 = 푚푘 푛푘 = 푔(푚)푔(푛).
Theorem 2.3
Let 푓 be a multiplicative function and 푛 be a positive integer with canonical
푒1 푝2
decomposition 푛 = 푝1
푒2 … 푝푘
푒푘 . Then
9. 푒1 , then 푓(푛) = 푓(푝1
푒1 )푓(푝2
6
푒1 )푓(푝2
푓(푛) = 푓(푝1
푒2 )… 푓(푝푘
푒푘 ).
Proof. If 푘 = 1, that is, if 푛 = 푝1
푒1 ), so the theorem is trivially
true.
Assume it is true for any integer with canonical decomposition consisting of
푘 distinct primes. That is 푓(푛) = 푓(푝1
푒2 )… 푓(푝푘
푒푘 ).
Let 푛 be any integer with 푘 + 1 distinct primes in its canonical
푒1 푝2
decomposition, say, 푛 = 푝1
푒2 … 푝푘
푒푘 .
푒1 푝2
Since (푝1
푒2 … 푝푘
푒푘 , 푝푘+1
푒푘+1 ) = 1 and 푓 is multiplicative,
푒1 푝2
푓(푝1
푒2 … 푝푘
푒푘 푝푘+1
푒푘+1 ) = 푓(푝1
푒1 푝2
푒2 … 푝푘
푒푘 )푓(푝푘+1
푒푘+1 )
푒1 )푓(푝2
= 푓(푝1
푒2 ) … 푓(푝푘
푒푘 )푓(푝푘+1
푒푘+1 ),
by induction. Therefore by induction the result is true for any positive integer 푛.
Theorem 2.4
Let 푝 be a prime and 푒 any positive integer. Then 휑(푝푒) = 푝푒 − 푝푒−1.
Proof. 휑(푝푒) = number of positive integers less than or equal to 푝푒and relatively
prime to it.
10. = (number of positive integers less than or equal to 푝푒)-(number
of positive integers less than or equal to 푝푒 and not relatively prime
7
to it)
The positive integers less than or equal to 푝푒 and not relatively prime to it
are the various multiples of 푝, namely, 푝, 2푝, 3푝, … … … … (푝푒−1 )푝 , and they are
푝푒−1 in number. Thus φ(pe) = pe-pe-1 . Lemma
2.5
Let 푚 and 푛 be relatively prime positive integers.
Then the integers 푟, 푚 + 푟, 2푚 + 푟, … , (푛 − 1)푚 + 푟 are congruent modulo 푛 to
0, 1 , 2, … , (푛 − 1) in some order.
Example 2.6
휑(8) = 휑(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.
휑(81) = 휑(34) = 34 − 33 = 54.
휑(15625) = 휑(56) = 56 − 55 = 12500.
Theorem 2.7
The function φ is multiplicative.
11. Proof. Let 푚 and 푛 be positive integers such that (푚, 푛) = 1.
We have to prove that 휑(푚푛) = 휑(푚)휑(푛).
Arrange the integers 1 through 푚푛 in 푚 rows of 푛 each:
1. 푚 + 1 2푚 + 1 … . (푛 − 1)푚 + 1
2. 푚 + 2 2푚 + 2 … . (푛 − 1)푚 + 2
3. 푚 + 3 2푚 + 3 … . (푛 − 1)푚 + 3
8
.
푟. 푚 + 푟 2푚 + 푟 … (푛 − 1)푚 + 푟
.
.
.
푚. 2푚 3푚 . . . 푛푚
Let 푟 be a positive integer less than or equal to 푚 such that (푟, 푚) > 1.
We will show that no element of the 푟th row is relatively prime to 푚푛.
12. Let 푑 = (푟, 푚). Then 푑|푟 and 푑|푚, so 푑|푘푚 + 푟 for any integer 푘; that is, 푑
is a factor of every element in the 푟th row. Thus no element in the 푟th row is
relatively prime to 푚, and hence to 푚푛 if (푟, 푚) > 1. That is the elements in the
array are relatively prime to 푚푛 from the 푟th row only if (푟, 푚) = 1. By definition,
there are 휑(푚) such integers 푟 and hence 휑(푚) such rows.
9
Now, in 푟th row, where (푟, 푚) = 1:
푟 , 푚 + 푟 , 2푚 + 푟 , … , (푛 − 1)푚 + 푟
By Lemma 2.5, their least residues modulo 푛 are a permutation of
0 ,1 ,2 , … , (푛 − 1) of which 휑(푛) are relatively prime to 푛. Therefore exactly
휑(푛) elements in 푟th row are relatively prime to 푛 and hence to 푚푛.
Thus there are φ(m) rows containing positive integers relatively prime to
푚푛, and each row contains 휑(푛) elements relatively prime to it. So the array
contains 휑(푚)휑(푛) positive integers less than or equal to 푚푛 and relatively prime
to 푚푛. Therefore 휑(푚푛) = 휑(푚)휑(푛).
Example 2.8
1. 휑(221) = 휑(13 ∙ 17) = 휑(13) ∙ 휑(17)
= 12 ∙ 16 = 192.
2. 휑(625) = 휑(53 ∙ 72 ) = 휑(53 ) ∙ 휑(72)
14. = 864.
11
Theorem 2.11
If 푛 ≥ 3, then 휑(푛) is even.
Proof. If 푛 is even obviously 휑(푛) is even. If 푛 is odd then for each prime factor 푝
of 푛, the numerator of (1 − 1
푝
) is even. Thus 휑(푛) is even for 푛 ≥ 3.
Theorem 2.12
Let 푛 be a positive integer, then Σ푑|푛 휑(푑) = 푛.
Here we turn to illustrate its proof.
Example 2.13
The positive divisors of 18 are 1, 2, 3, 6, 9 and 18.
So Σ푑|18 휑(18) = 휑(1)+ 휑(2)+ 휑(3)+ 휑(6)+ 휑(9)+ 휑(18)
= 1 + 1 + 2 + 2 + 6 + 6 = 18.
The Tau and Sigma Functions
Definition 2.14 (The Tau Function)
Let 푛 be a positive integer. Then 휏(푛) denotes the number of positive
divisors of 푛; that is 휏(푛) = Σ푑|푛 1.
15. 12
Example 2.15
The positive divisors of 18 are 1, 2, 3, 6, 9 and 18,
So 휏(18) = 6.
23, being a prime, has exactly two positive divisors,
So 휏(23) = 2.
Remark. If 푛 is prime, then 휏(푛) = 2; if 휏(푛) = 2, then 푛 is prime.
Definition 2.16 (The Sigma Function)
Let 푛 be positive integer. Then 휎(푛) denotes the sum of the positive divisors
of 푛; that is 휎(푛) = Σ푑|푛 푑.
Example 2.17
The positive divisors of 12 are 1, 2, 3, 4, 5, 6 and 12;
So 휎(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28.
The positive divisors of 28 are 1, 2, 4, 7, 14 and 28;
So 휎(28) = 1 + 2 + 4 + 7 + 14 + 28 = 56.
16. Let 푓 be a multiplicative function. Define a new function 퐹 by 퐹(푛) =
Σ푑|푛 푓(푑). The following example shows that the new function 퐹 can be a
13
multiplicative function.
Example 2.18
We have (4,7) = 1.
Then 퐹(4 ∙ 7) = Σ푑|28푓(28)
= 푓(1) + 푓(2) + 푓(4) + 푓(7) + 푓(14) + 푓(28)
= 푓(1 ∙ 1) + 푓(1 ∙ 2) + 푓(1 ∙ 4) + 푓(1 ∙ 7) + 푓(2 ∙ 7) + 푓(4 ∙ 7)
= [푓(1) + 푓(2) + 푓(4)]푓(1) + [푓(1) + 푓(2) + 푓(4)]푓(7)
= [푓(1) + 푓(2) + 푓(4)][푓(1) + 푓(7)]
= Σ푓(푑)
푑|4
∙ Σ푓(푑)
푑|7
= 퐹(4)퐹(7).
Theorem 2.19
If 푓 is a multiplicative function, then 퐹(푛) = Σ푑|푛 푓(푑) is also
multiplicative.
17. Proof. Let 푚 and 푛 be relatively prime positive integers. We want to show that
14
퐹(푚푛) = 퐹(푚)퐹(푛).
By definition, 퐹(푚푛) = Σ푑|푚푛 푓(푑). Since (푚, 푛) = 1, every positive
divisor of 푚푛, is the product of a unique pair of positive divisors 푑1 of 푚 and 푑2
of 푛, where, (푚, 푛) = 1.
Therefore 퐹(푚푛) = Σ푑1|푚,푑2|푚 푓(푑1푑2).
Since 푓 is multiplicative, 푓(푑1푑2) = 푓(푑1 )푓(푑2 ).
Therefore 퐹(푚푛) = Σ푑1|푚,푑2|푚푓(푑1푑2)
= Σ 퐹(푚)푓(푑2 )
푑2 |푛
= 퐹(푚)Σ 푓(푑2 )
푑2 |푛
= 퐹(푚)퐹(푛).
Thus 퐹 is multiplicative.
Corollary 2.20
The Tau and Sigma functions are multiplicative.
Proof. By theorem 2.6, the functions Σ푑|푛 푓(푑) = Σ푑|푛 1 = 휏(푛) and Σ푑|푛 푔(푑) =
Σ푑|푛 푑 = 휎(푛) are multiplicative.
18. (Since we have constant function 푓(푛) = 1 and the identity function 푔(푛) =
15
푛 are multiplicative.)
That is, if 푓(푚, 푛) = 1, then 휏(푚푛) = 휏(푚)휏(푛) and 휎(푚푛) = 휎(푚)휎(푛).
Example 2.21
36 = (4)(9). Also (4, 9) = 1.
Thus 휏(36) = 휏(4)휏(9) = 3 ∙ 3 = 9.
And 휎(36) = 휎(4)휎(9) = (1 + 2 + 4)(1 + 3 + 9) = 91.
Theorem 2.22
Let 푝 be any prime and 푒 any positive integer. Then 휏(푝푒) = 푒 + 1 and
휎(푝푒) = 푝푒+1 −1
푝−1
.
Proof. The positive factors of 푝푒 are of the form 푝푖 , where 0 ≤ 푖 ≤ 푒. There are
푒 + 1 such factors. So 휏(푝푒) = 푒 + 1.
푒
휎(푝푒) = Σ푝푖 =
0
푝푒+1 − 1
푝 − 1
.
The following theorem gives two formulas. They are a consequence of
Corollary 2.20 and Theorem 2.2.
Theorem 2.23
19. Let 푛 be a positive integer with canonical decomposition 푛 = 푝1
16
푒1 푝2
푒2 … 푝푘
푒푘.
푒1 −1
푝1 −1
Then 휏(푛) = (푒1 + 1)(푒2 + 2) … (푒푘 + 푘) and 휎(푛) = 푝1
푒2 −1
푝2 −1
∙ 푝2
∙ … ∙
푒푘 −1
푝푘−1
푝
푘
.
Proof. Since 휏 is multiplicative,
푒1 ) ∙ 휏(푝2
휏(푛) = 휏(푝1
푒2 ) … 휏(푝푘
푒푘 ) = (푒1 + 1)(푒2 + 1) … (푒푘 + 1).
Since 휎 is multiplicative,
푒1 ) ∙ 휎(푝2
휎(푛) = 휎(푝1
푒푘 ) = 푝1
푒2 )… 휎(푝푘
푒1 −1
푝1 −1
푒2 −1
푝2 −1
∙ 푝2
∙ … ∙
푒푘 −1
푝푘−1
푝
푘
.
Example 2.24
The canonical decomposition of 6120 = 23 ∙ 32 ∙ 5 ∙ 17
Therefore 휏(6120) = (3 + 1)(2 + 1)(1 + 1)(1 + 1) = 48
and 휎(6120) = 23+1−1
2−1
∙ 32+1−1
3 −1
∙ 51+1−1
5−1
∙ 171+1−1
17−1
= 15 ∙ 14 ∙ 6 ∙ 18 = 22680.
The Mӧbius Function
The Mӧbius function 휇 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 흁)
20. 1 if 푛 = 1
0 if 푝2 |푛 for some prime 푝
(−1)푘 if 푛 = 푝1 푝2 … 푝푘
17
Let 푛 be a positive integer. Then
휇(푛) = {
For example 휇(2) = −1, 휇(3) = −1, 휇(4) = 0, 휇(12) = 0, 휇(35) =
휇(5.7) = (−1)2 and 휇(672) = 휇(25 ∙ 3 ∙ 7) = 0.
Thus 휇 assigns −1,0 or to each positive integer; 휇(푛) = 0 if the canonical
decomposition of 푛 contains a perfect square, that is, if 푛 is not square-free; it is
(−1)푘 if it consists of 푘 distinct prime factors.
Theorem 2.26
The function 휇 is multiplicative.
Proof. Let 푚 and 푛 be relatively prime positive integers. If 푚 = 1 or 푛 = 1, then
clearly 휇(푚푛) = 휇(푚)휇(푛).
Suppose 푚 or 푛 is divisible by 푝2 for some prime 푝. Then 휇(푚)휇(푛) = 0. If
푝2 |푚 or 푝2 |푛, then 푝2 |푚푛, so 휇(푚푛) = 0. Thus 휇(푚푛) = 휇(푚)휇(푛).
Finally, suppose both 푚 and 푛 are square-free, let 푚 = 푝1푝2 … 푝푟 and 푛 =
푞1 푞2 … 푞푠, where the 푝푖 ’s and 푞푗’s are distinct primes, since (푚, 푛) = 1. 휇(푚) =
(−1)푟 and 휇(푛) = (−1)5.
Then 푚푛 = 푝1 푝2 … 푝푟푞1푞2 … 푞푠, a product of distinct primes.
21. Therefore 휇(푚푛) = (−1)푟+푠 = (−1)푟 ∙ (−1)푠 = 휇(푚)휇(푛). Thus, in every case,
휇(푚푛) = 휇(푚)휇(푛), so 휇 is multiplicative.
18
Example 2.27
Let us determine 휇(420).
Since 휇 is multiplicative, 휇(420) = 휇(15)휇(28)
Now 15 = 3 ∙ 5. So by definition, 휇(15) = 1.
28 = 22 ∙ 7. So by definition, 휇(28) = 0.
420 = 22 ∙ 3 ∙ 5 ∙ 7. Since 420 is not square–free, 휇(420) = 0. Thus 휇(420) =
0 = 1 ∙ 0 = 휇(15)휇(28).
Next we develop a formula for Σ푑|푛 휇(푑). When n = 1, Σ푑|1 휇(푑) = 휇(1) =
1. If 푛 > 1, we can compute the sum using the canonical decomposition of 푛 and
Theorem 8.7 provided we know the sum when 푛 is a prime-power 푝푒. For this we
need to introduce a new function, as the following lemma shows.
Lemma 2.28
Let 퐹(푛) = Σ푑|푛 휇(푑). Then 퐹(푝푒) = 0, where 푒 > 1.
Proof. 퐹(푝푒) = Σ푑|푝푒 휇(푑)
= Σ 휇(푝푖) 푛푖
=0
22. = 휇(1) + 휇(푝) + 휇(푝2 ) + ⋯ + 휇(푝푒)
= 1 + (−1) = 0 + ⋯ + 0 = 0.
(푑|6) .
19
Theorem 2.29
1 if 푛 = 1
0 othewise
Let 푛 be a positive integer. Then Σ푑|푛 휇(푑) = {
.
푒1 푝2
Proof. If 푛 = 1, Σ푑|1 휇(푑) = 휇(1) = 1. So let 푛 > 1 and let 푛 = 푝1
푒2 … 푝푘
푒푘 be
the canonical decomposition of 푛. Let, 퐹(푛) = Σ푑|푛 휇(푑). Since 휇 is
multiplicative, 퐹 is also multiplicative, by Theorem 8.7.
Therefore 퐹(푛) = Π 퐹(푝푒푖) 푘푖
=1
퐹(푛) = Π 0 푘푖
=1 = 0.
Example 2.30
Σ 휇(푑) = 휇(1) + 휇(2) + 휇(3) + 휇(6) + 휇(9) + 휇(18) 푑|18
= 1 + (−1) + (−1) + (−1)2 + 0 + 0 = 0.
Problem.
Let 푓 be a number-theoretic function. Show that
Σ Σ 휇(푑)푓(푑′ ) = Σ Σ 푓(푑′)휇(푑)
(푑|(6
푑
))
(푑′ |6)
(푑′|(6
푑
))
24. 푑|푛 푑|푛 ) .
푑′푑|푛 푑′ |푛 )
푑′) 푑|푛 ]
푑′) equals if
푑|푛 ) = 푓(푑′) ∙ 1, where 푑′ = 푛.
푑|푛 ) can also be written as푓(푛) =
21
So 휇(푑)퐹 (푛
푑
) = 휇(푑) Σ 푓(푑′ )
푑′ |(푛
푑
) = Σ 휇(푑)푓(푑′ )
푑′ |(푛
푑
)
Therefore Σ 휇(푑)퐹 (푛
푑
) = Σ Σ 휇(푑)푓(푑′ )
푑′ |(푛
푑
As 푑 runs over the positive divisors of 푛, so does 푑′; also 푑푑′ runs over the
positive factors of 푛; that is, the sum of all its values of 휇(푑)푓(푑′) as 푑푑′ runs
over the positive factors of 푛; that is, the sum of all its values for all pairs 푑 and
푑′ such that
푑′
푛
and 푑| ( 푛
푑′ ).
That is, Σ Σ 휇(푑)
푑′ |(푛
푑
) 푓(푑′ ) = Σ Σ 휇(푑)푓(푑′ )
푑|( 푛
Thus Σ 휇(푑)퐹 (푛
푑
) = Σ푑′ |푛 푓(푑′) [Σ 휇(푑)
푑|( 푛
But by Theorem 8.17, Σ 휇(푑)
푑|( 푛
푛
푑′ = 1; that is, if 푛 = 푑′, and
otherwise. Thus Σ 휇(푑)퐹 (푛
푑
In other words, 푓(푛) = Σ 휇(푑)퐹 ( 푛
푑′ 푑|푛 ).
Remark. As 푑 runs over the positive factors of 푛, so does
푛
푑
. Therefore, the
inversion formula 푓(푛) = Σ 휇(푑)퐹 (푛
푑
Σ 휇 (푛
) 퐹(푑) 푑|푛 .
푑
25. Notice that the definition 푓(푛) = Σ푑|푛 푓(푑) expresses 퐹 in terms of 푓,
whereas the inversion formula 푓(푛) = Σ 휇(푑)퐹 (푛
푑|푛 ) expresses 푓 in terms of 퐹.
22
푑
To illustrate the inversion formula, recall that 휏(푛) = Σ푑|푛 1 and 휎(푛) =
Σ푑|푛 푑 because both the constant function 푓(푛) = 1 and the identity function
푔(푛) = 푛 are multiplicative, it follows by Theorem 2.32 that
푛
푑
1 = Σ휇(푑)휏 (
푛
푑
) = Σ휇 (
) 휏(푑)
푑|푛 푑|푛
and
푛
푑
푛 = Σ휇(푑)휎 (
푛
푑
) = Σ휇 (
) 휎(푑)
푑|푛 푑|푛
We can illustrate these results by the following example.
Example 2.32
6
푑
Σ휇(푑)휏 (
) = 휇(1)휏(6) + 휇(2)휏(3) + 휇(3)휏(2) + 휇(6)휏(1)
푑|6
= 1 ∙ 4 + (−1) ∙ 2 + (−1) ∙ 2 + (−1)2 ∙ 1
= 1
6
푑
Σ휇(푑)휎 (
) = 휇(1)휎(6) + 휇(3)휎(2) + 휇(6)휎(1)
푑|6
26. = 1 ∙ 12 + (−1) ∙ 4 + (−1) ∙ 3 + (−1)2 ∙ 1
= 6.
푑|푛 ) =Σ 휇(푑)
푑|푛 푑 .
23
Theorem 2.33
휑(푛) = 푛 Σ 휇(푑)
푑|푛 푑 .
Proof. By Theorem 2.12, 푛 = Σ푑|푛 휑(푑). Let 푔 denote the identity function given
by 푔(푛) = 푛 for all 푛. Therefore 푔(푛) = 푛 = Σ푑|푛 휑(푑). Then by Mӧbius
inversion formula, 휑(푛) = Σ 휇(푑)푔 (푛
푑
Problem.
Verify the formula in Theorem 2.35 for n=12.
Solution.
By Theorem 2.9, 휑(12) = 휑(22 ∙ 3) = 4.
Now,
12 Σ
휇(푑)
푑
12
푑
= Σ (
) 휇(푑)
푑|12 푑|12
= 12휇(1) + 6휇(2) + 4휇(3) + 3휇(4) + 2휇(6) + 1휇(12)
27. = 12 ∙ 1 + 6 ∙ (−1) + 4 ∙ (−1) + 3 ∙ 0 + 2 ∙ (−1)2 + 1 ∙ 0
푑|푛 ). Then 퐹(푛) = Σ푑|푛 푓(푑).
) 푘|푛 ].
24
= 4 = 휑(12).
Theorem 2.34
Let 퐹 and 푓 be number-theoretic functions such that 푓(푛) =
Σ 휇(푑)퐹 (푛
푑
Proof. By the definition of 푓, 푓(푑) = Σ 휇(푑′ )퐹 ( 푑
푑′ 푑′ |푑 ).
Therefore Σ 푓(푑) = Σ Σ 휇(푑′ )퐹 ( 푑
푑′ 푑|푛 푑|푛 푑′ |푑 ).
Let
푑
푑′ = 푘 and
푛
푑
= 푙,then 푑 = 푘푑′ and 푛 = 푙푑 = 푘푙푑′.So this equation
yields Σ푑|푛 푓(푑) = Σ푑|푛 Σ푘푑′=푑 휇(푑′)퐹(푘)
= Σ푘푑′ |푛 휇(푑′)퐹(푘)
= Σ 퐹(푘) [Σ 휇(푑′)
푑′ |(푛
푘
By Theorem 8.17, Σ 휇(푑′ )
푑′ |(푛
푘
) equals if 푛 = 푘 and otherwise. So the
equation becomes Σ 푓(푑) = 퐹(푘) ∙ (1) (푑|푛) , where 푛 = 푘.
That is, 퐹(푛) = Σ푑|푛 푓(푑).
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
“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
25
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.”
29. 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:
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 푛 is a perfect number if the sum of its proper divisiors
equals 푛. Thus 푛 is perfect if 휎(푛) − 푛 = 푛, that is, if 휎(푛) = 2푛.
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)
30. 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
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
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 2푝−1 (2푝 − 1), where 푝 and 2푝−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 푛 is an intiger greater than or equal to such that 2푛 − 1 is a prime, then
푁 = 2푛−1 (2푛 − 1) is a perfect number.
Proof. Since 2푛 − 1 is a prime 휎(2푛 − 1) = 1 + (2푛 − 1) = 2푛 . Because 휎 is
multiplicative, 휎(푁) = 휎(2푛−1 )휎(2푛 − 1)
= (2푛 − 1)(2푛) = 2푛(2푛 − 1) = 2푁.
32. Thus 푁 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 2푛 − 1 is
a prime. Theorem 3.2 and 3.3 categorically characterize even perfect numbers.
29
Theorem 3.3 (Euler)
Every even perfect numbers 푁 is of the form 푁 = 2푛−1 (2푛 − 1), where
2푛 − 1 is a prime.
Proof. Let 푁 be of the form 2푒푠, where 푠 is odd and 푒 ≥ 1. Since 푁 is perfect,
휎(푁) = 2푁 = 2푒+1 푠. Clearly, (2푒, 푠) = 1. So 휎(푁) = 휎(2푒푠) = 휎(2푒)휎(푠).
That is 2푒+1 푠 = (2푒+1 − 1)휎(푠). (1)
Since (2푒+1 , 2푒+1 − 1) = 1, it follows by corollary……. that 2푒+1 |휎(푠).
Therefore 휎(푠) = 2푒+1 푡 for some positive integer 푡. Substituting for 휎(푠) in
equation (1), 2푒+1 푠 = (2푒+1 − 1)2푒+1 푡. (2)
Therefore 푠 = (2푒+1 − 1)푡. (3)
This implies 푡|푠 and 푡 < 푠, since 푡 = 푠 implies 푒 = 0, a contradiction.
33. We will now show that 푡 = 1. For that equation (2) can be rewritten as 푠 +
30
푡 = 2푒+1 푡.
That is, 푠 + 푡 = 휎(푠). (4)
This shows that 푡 is the sum of the proper divisors of 푠; but, by equation (3),
푡 itself is a proper divisor of 푠. So, for the relationship (4) to hold, 푡 must be .
Thus 푠 + 1 = 휎(푠), so 푠 has exactly two positive factors and 푠. Consequently,
푠 = 2푒+1 − 1 must be a prime.
Thus 푁 = 2푒(2푒+1 − 1), where 2푒+1 − 1 is a prime.
Problem.
Let us find all perfect numbers of the form 푛푛 + 1.
Solution. Take 푁 = 푛푛 + 1
Case 1 Let 푛 be odd. Then 푁 is an even perfect number, so 푁 must be of the form
푁 = 2푚−1 (2푚 − 1), where 2푚 − 1 is a prime.
Clearly 푁 can be factored as 푁 = 푛푛 + 1 = (푛 + 1)푟, where 푟 = 푛푛−1 −
푛푛−2 + ⋯ − 푛 + 1. We claim that (푛 + 1, 푟) = 1.
To show this, notice that since 푛 is odd, 푟 is odd and 푛 + 1 is even. Let 푛 +
1 = 2푠, where 푡 is an odd integer greater than or equal to .Then 푁 = 2푠 푡푟, where
both 푡 and 푟 are odd. Since 푁 is an even perfect number this is possible only if 푡 =
34. 1; so 푛 + 1 = 2푠 and hence (푛 + 1, 푟) = 1. If 푟 = 1, then 푁 = 푛푛 + 1; so 푛 = 1.
Then 푁 = 2, which is not a perfect number.
Since 푁 = 2푚−1 (2푚 − 1) = (푛 + 1)푟 = 2푠푟, where 2푚 − 1 is an odd
prime and 푟 is odd, 2푠 = 2푚−1 = 푛 + 1 and 푟 = 2푚 − 1 = 2(푛 + 1) − 1 = 2푛 +
31
1.
Therefore 푁 = 푛푛 + 1 = (푛 + 1)(2푛 + 1) = 2푛2 + 3푛 + 1.
This gives 푛푛 = 2푛2 + 3푛. Thus 푛푛−1 = 2푛 + 3.
Since 푛 is an integer, this equation has a unique solution . (See the graph in
figure…..) Then 푁 = 33 + 1 = 28. Thus 28 is the only even perfect number of the
desired form.
35. Case 2 Let 푛 be even, say, 푛 = 2푘. Then 푁 is odd, 푛푛 is a square, and 푛푛 ≡
−1(mod 푁). We claim that 3 <∤ 푁. Suppose on the contrary that 3|푁. Then 푛푛 ≡
32
−1(mod 푁); that is,
(2푘)2푘 ≡ −1(mod 3)
4푘 ∙ 푘2푘 ≡ 2(mod 3)
1 ∙ 푘2푘 ≡ 2(mod 3)
푘2푘 ≡ 2(mod 3)
Cleraly, 푘 ≢ 0 or modulo . If 푘 ≡ 2(mod 3), then the above congruence gives
2푘 ≡ 2(mod 3)
36. 4푘 ≡ 2(mod 3)
1 ≡ 2(mod 3), a contradiction
So 푘 cannot be congruent to 0,1 or , which is absurd. Thus 3 ∤ 푁.
By Touchard’s theorem, 푁 = 12푚 + 1 or 36푚 + 9 for some integer 푚. If
푁 = 36푚 + 9, then 3|푁, a contradiction. So 푁 = 12푚 + 1; that is, 푛푛 = 12푚.
Since 3|12푚, 3|푛푛, so 3|푛. Thus 2|푛 and 3|푛, so 6|푛.
33
Let 푁 = 푎6 + 1,where 푎 = 푛
푛
6 > 1. Then 푁 can be factored as 푁 =
(푎2 + 1)(푎4 − 푎2 + 1). (5)
We will now see that these factors of 푁 are relatively prime. To this end, let
푝 be a common prime factor of the two factors 푎2 + 1 and 푎4 − 푎2 + 1. Since
푎4 − 푎2 + 1 = (푎4 + 2푎2 + 1) − 3푎2
= (푎2 + 1)2 − 3푎2
= (푎2 + 1)2 − 3(푎2 + 1) + 3, 푝|3.
That is, 푝 = 3. This implies 3|푁, a contradiction; so the factors 푎2 + 1 and 푎4 −
푎2 + 1 are relatively prime. Besides, since 푁 is odd, both factors are also odd.
Since 푁 is perfect and 휎 is multiplicative, equation (5) yields 휎(푁) = 휎(푎2 + 1) ∙
휎(푎4 − 푎2 + 1).
That is, 2푁 = 휎(푎2 + 1) ∙ 휎(푎4 − 푎2 + 1).
37. Since 푁 is odd, one of the factors on the right-hand side must be odd. But, if
푚 and 휎(푚) are both odd, then 푚 is a square. This implies that either 푎2 + 1 or
푎4 − 푎2 + 1 is a square. But 푎2 < 푎2 + 1 < (푎 + 1)2 and (푎2 − 1)2 < 푎4 − 푎2 +
1 < (푎2 )2. Thus neither can be a square, a contradiction.
Consequently, there are no odd perfect numbers of the form 푛푛 + 1. Thus
28 is the only perfect number of the desired form.
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 푁 must satisfy have
been established, no one has been successful in finding one, in spite of large
computer searches with modern supercomputers. For example, 푁 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. Mary’s University of Minnesota
established that if 3휎(푛) = 5푛 for some integer 푛, then 5푛 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).
