An-Introductory-in-Elementary-Number-Theory

Algebraic Operations With Integers

The set Z of all integers, which this book is all about, consists of all positive and negative integers as well as 0 Thus Z is the set given by

Z = { , −4, −3, −2, −1, 0, 1, 2, 3, 4, } (1.1) While the set of all positive integers, denoted by N, is defined by

On Z, there are two basic binary operations, namely addition (denoted by +) and multiplication (denoted by ã), that satisfy some basic properties from which every other property for Z emerges.

1 The Commutativity property for addition and multiplication a + b = b + a a ã b = b ã a

2 Associativity property for addition and multiplication

3 The distributivity property of multiplication over addition a ã (b + c) = a ã b + a ã c.

The Well Ordering Principle and Mathematical Induction

The Well Ordering Principle

The Well Ordering Principle: A least element exist in any non empty set of positive integers.

The principle that any set of positive integers is well-ordered, while the set of all integers is not, serves as a fundamental axiom in number theory This distinction is crucial for proving numerous theorems related to integers.

The Pigeonhole Principle

The Pigeonhole Principle: If s objects are placed in k boxes for s > k, then at least one box contains more than one object.

In this proof, we assume that each box contains at most one object, which implies a maximum of k objects However, this assumption contradicts the existence of s objects when s is greater than k.

The Principle of Mathematical Induction

We now present a valuable tool for proving results about integers This tool is the principle of mathematical induction

Theorem 1 The First Principle of Mathematical Induction: If a set of positive integers has the property that, if it contains the integer k, then it also contains n

1.2 THE WELL ORDERING PRINCIPLE AND MATHEMATICAL

Induction states that if a property holds for the positive integer 1 and is also true for n + 1 whenever it is true for n, then this property must be true for all positive integers Thus, if a set includes the number 1, it encompasses all positive integers.

We use the well ordering principle to prove the first principle of mathematical induction

The set S consists of positive integers starting with 1 and including k + 1 whenever it contains k Assuming S does not encompass all positive integers, there exists at least one integer not in S, which must have a least element α according to the well-ordering principle Since 1 is in S, α cannot be less than 1, making α equal to 1 However, since α - 1 is also in S, it follows that α must be in S as well Consequently, S must include all positive integers.

We now present some examples in which we use the principle of induction.

Example 1 Use mathematical induction to show that ∀n ∈ N

X j = 1 = 1 ã 2 j=1 2 and thus the the statement is true for n = 1 For the remaining inductive step, suppose that the formula holds for n, that is P n j = n(n+1) We show that

2 j=1 to complete the proof by induction Indeed n+1 n

Example 2 Use mathematical induction to prove that n! ≤ n n for all positive integers n.

Note that 1! = 1 ≤ 1 1 = 1 We now present the inductive step Suppose that n! ≤ n n for some n, we prove that (n + 1)! ≤ (n + 1) n+1 Note that

Theorem 2 The Second Principle of Mathematical Induction: A set of positive integers that has the property that for every integer k, if it contains all the integers

In mathematical terms, if a property holds for the positive integer 1 and remains true for all integers up to n + 1 whenever it is valid for all integers up to n, then that property is universally applicable to all positive integers This principle underscores the foundational nature of positive integers in mathematics.

The second principle of induction is also known as the principle of strong induction Also, the first principle of induction is known as the principle of weak induction.

To prove the second principle of induction, we use the first principle of induction.

Let T be a set of integers that includes 1 and satisfies the condition that for any positive integer k, if it contains all integers from 1 to k, it must also contain k + 1 Consequently, we define S as the set of all positive integers k for which every positive integer less than or equal to k is included in T.

The set S includes the element 1 and also contains k + 1, indicating that S encompasses all positive integers Consequently, since S is a subset of T, it follows that T must also consist of all positive integers.

Divisibility and the Division Algorithm

Integer Divisibility

Definition 1 If a and b are integers such that a = 0, then we say ”a divides b” if there exists an integer k such that b = ka.

If a divides b, we also say ”a is a factor of b” or ”b is a multiple of a” and we write a | b If a doesn’t divide b, we write a - b For example 2 | 4 and 7 | 63, while 5 - 26.

Every even integer can be expressed as 2k, where k is an integer, while any odd integer takes the form 2k + 1 Consequently, an integer n is even if it is divisible by 2 (2|n), and it is odd if it is not divisible by 2 (2 - n).

14 CHAPTER 1 INTRODUCTION b) ∀a ∈ Z one has that a | 0. c) If b ∈ Z is such that |b| < a, and b = 0, then a - b.

Theorem 3 If a, b and c are integers such that a | b and b | c, then a | c.

Proof Since a | b and b | c, then there exist integers k 1 and k 2 such that b = k 1 a and c = k 2 b As a result, we have c = k 1 k 2 a and hence a | c.

The following theorem states that if an integer divides two other integers then it divides any linear combination of these integers.

Theorem 4 If a, b, c, m and n are integers, and if c | a and c | b, then c | (ma + nb).

Proof Since c | a and c | b, then by definition there exists k1 and k2 such that a = k 1 c and b = k 2 c Thus ma + nb = mk 1 c + nk 2 c = c(mk 1 + nk 2 ), and hence c | (ma + nb).

Theorem 4 can be generalized to any finite linear combination as follows If a | b1 , a | b2 , , a | bn then n a | X k j b j (1.6) j=1 for any set of integers k1, ã ã ã , kn ∈ Z It would be a nice exercise to prove the generalization by induction.

1.3 DIVISIBILITY AND THE DIVISION ALGORITHM 15

The Division Algorithm

The following theorem states somewhat an elementary but very useful result.

Theorem 5 The Division Algorithm If a and b are integers such that b > 0, then there exist unique integers q and r such that a = bq + r where 0 ≤ r < b.

In the proof, we define the set A = {a − bk ≥ 0 | k ∈ Z}, which is nonempty because for k < a/b, a − bk is positive According to the well-ordering principle, A has a least element denoted as r = a − bq for some integer q, ensuring that r is nonnegative Furthermore, if r is greater than or equal to b, it follows that r is also greater than a − b(q + 1), reinforcing the relationship between r, a, and b.

This leads to a contradiction since r is assumed to be the least positive integer of the form r = a − bq As a result we have 0 ≤ r < b.

We will show that q and r are unique Suppose that a = bq 1 + r 1 and a bq 2 + r 2 with 0 ≤ r 1 < b and 0 ≤ r 2 < b Then we have b(q 1 − q2 ) + (r 1 − r2) 0.

And since − max(r1 , r2) ≤ |r2 − r1| ≤ max(r1, r2 ), and b > max(r1, r2 ), then r 2 − r1 must be 0, i.e r 2 = r 1 And since bq 1 + r 1 = bq 2 + r 2 , we also get that q1 = q2 This proves uniqueness.

Example 5 If a = 71 and b = 6, then 71 = 6 ã 11 + 5 Here q = 11 and r = 5. Exercises

2 Use the division algorithm to find the quotient and the remainder when 76 is divided by 13.

3 Use the division algorithm to find the quotient and the remainder when -100 is divided by 13.

4 Show that if a, b, c and d are integers with a and c nonzero, such that a | b and c | d, then ac | bd.

5 Show that if a and b are positive integers and a | b, then a ≤ b.

6 Prove that the sum of two even integers is even, the sum of two odd integers is even and the sum of an even integer and an odd integer is odd.

7 Show that the product of two even integers is even, the product of two odd integers is odd and the product of an even integer and an odd integer is even.

8 Show that if m is an integer then 3 divides m 3 − m.

9 Show that the square of every odd integer is of the form 8m + 1.

10 Show that the square of any integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

11 Show that if ac | bc, then a | b.

12 Show that if a | b and b | a then a = ±b.

Representations of Integers in Different Bases

This section demonstrates that every positive integer can be uniquely expressed in any positive base integer expansion While decimal notation is commonly used to represent integers, we will illustrate the process of converting an integer from decimal to any other positive base and vice versa.

1.4 REPRESENTATIONS OF INTEGERS IN DIFFERENT BASES 17 decimal notation in daily life is simply better because we have ten fingers which facilitates all the mathematical operations.

Notation An integer a written in base b expansion is denoted by (a) b

Theorem 6 Let b be a positive integer with b > 1 Then any positive integer m can be written uniquely as m = al b l + a l 1 − b l−1 + + a b + a ,1 0 where l is a positive integer, 0 ≤ aj < b for j = 0, 1, , l and a l = 0.

Proof We start by dividing m by b and we get m = bq0 + a0, 0 ≤ a0 < b.

If q0 = 0 then we continue to divide q0 by b and we get q0 = bq1 + a1, 0 ≤ a1 < b.

We continue this process and hence we get q 1 bq2 + a2, 0 ≤ a2 < b,

. ql−2 ql−1 bql−1 + al−1, 0 ≤ al−1 < b, b ã 0 + a l , 0 ≤ a l < b.

Note that the sequence q 0 , q 1 , is a decreasing sequence of positive integers with a last term q l that must be 0.

Now substituting the equation q0 = bq1 + a1 in m = bq0 + a0, we get m = b(bq1 + a1 ) + a0 = b 2 q1 + a1 b + a0,

Successively substituting the equations in m, we get m = b 3 q 2 + a 2 b 2 + a 1 b + a 0 ,

What remains to prove is that the representation is unique Suppose now that m = a l b l + a l 1 b l−1 + + a 1 b + a 0

= c l b l + c l−1b l−1 + + c 1 b + c 0 where if the number of terms is different in one expansion, we add zero coefficients to make the number of terms agree Subtracting the two expansions, we get

If the two expansions are different, then there exists 0 ≤ j ≤ l such that cj = aj

As a result, we get b j ((al − cl )b l−j + + (aj+1 − cj+1 )b + (aj − cj )) 0 and since b = 0, we get

The equation aj − cj can be expressed as a sum involving the differences (al − cl)bl−j and (aj+1 − cj+1)b, leading to the conclusion that b divides (aj − cj) Given that both aj and cj are constrained within the range of 0 to b, it follows that aj must equal cj This contradiction demonstrates that the expansion is unique.

1.4 REPRESENTATIONS OF INTEGERS IN DIFFERENT BASES 19

Binary representation, known as base 2 representation, is essential in computing While arithmetic operations can be performed on integers in various positive integer bases, this article focuses on converting decimal integers to other base representations and vice versa For instance, to expand the number 214 in base 3, specific conversion methods are applied.

As a result, to obtain a base 3 expansion of 214, we take the remainders of divisions and we get that (214) 10 = (21221) 3

Example 7 To find the base 10 expansion, i.e the decimal expansion, of (364)7:

In base b expansions where b exceeds 10, additional characters are introduced to represent values greater than 9 Typically, the alphabetic letters are utilized to denote integers beyond 9 in these base b representations.

To convert from one base to the other, the simplest way is to go through base

10 and then convert to the other base There are methods that simplify conversion from one base to the other but it will not be addressed in this book.Exercises

4 Convert (AB6C 7D)16 to decimal notation.

The Greatest Common Divisor

In this section, we define the greatest common divisor (gcd) of two integers and explore its key properties Additionally, we demonstrate that the gcd can be expressed as a linear combination of the two integers.

Two integers, a and b, which are not both zero, possess a finite number of divisors, leading to a limited quantity of common divisors This section focuses on determining the greatest common divisor (GCD) of a and b It is important to consider the divisors of both integers in this context.

Definition 2 The greatest common divisor of two integers a and b is the greatest integer that divides both a and b.

We denote the greatest common divisor of two integers a and b by (a, b) We also define (0, 0) = 0.

Example 8 Note that the greatest common divisor of 24 and 18 is 6 In other words (24, 18) = 6.

There are couples of integers (e.g 3 and 4, etc ) whose greatest common divisor is 1 so we call such integers relatively prime integers.

Definition 3 Two integers a and b are relatively prime if (a, b) = 1.

Example 9 The greatest common divisor of 9 and 16 is 1, thus they are relatively prime.

Every integer possesses both positive and negative divisors If a is a positive divisor of m, then its negative counterpart, −a, is also a divisor of m Consequently, according to the definition of the greatest common divisor, we can conclude that (a, b) is equivalent to (|a|, |b|).

This theorem addresses the greatest common divisor (GCD) of two integers, asserting that when both integers are divided by their GCD, the resulting integers are relatively prime.

Proof We will show that a/d and b/d have no common positive divisors other than 1 Assume that k is a positive common divisor such that k | a/d and k | b/d.

As a result, there are two positive integers m and n such that a/d = km and b/d = kn

Thus we get that a = kmd and b = knd.

Hence kd is a common divisor of both a and b Also, kd ≥ d However, d is the greatest common divisor of a and b As a result, we get that k = 1.

The next theorem shows that the greatest common divisor of two integers does not change when we add a multiple of one of the two integers to the other.

Theorem 8 Let a, b and c be integers Then (a, b) = (a + cb, b).

In this article, we demonstrate that every divisor of integers $ a $ and $ b $ is also a divisor of both $ a + cb $ and $ b $, and vice versa, establishing that $ a $ and $ b $ share identical divisors Consequently, the greatest common divisor (gcd) of $ a $ and $ b $ is also the gcd of $ a + cb $ and $ b $ Let $ k $ represent a common divisor of $ a $ and $ b $; according to Theorem 4, $ k $ retains its properties across these expressions.

22 CHAPTER 1 INTRODUCTION and hence k is a divisor of a + cb Now assume that l is a common divisor of a + cb and b Also by Theorem 4 we have , l | ((a + cb) − cb) = a.

As a result, l is a common divisor of a and b and the result follows.

We now present a theorem which proves that the greatest common divisor of two integers can be written as a linear combination of the two integers.

Theorem 9 The greatest common divisor of two integers a and b, not both 0 is the least positive integer such that ma + nb = d for some integers m and n.

To prove the statement, we start by assuming that a and b are positive integers We examine the set of all positive integer linear combinations of a and b, which is non-empty since both a and b can be expressed in this form According to the well-ordering principle, this set contains a least element, denoted as d This element can be represented as d = ma + nb for some integers m and n Our goal is to demonstrate that d divides both a and b, establishing it as the greatest common divisor of the two integers.

By the division algorithm, we have a = dq + r, 0 ≤ r < d.

Thus we have r = a − dq = a − q(ma + nb) = (1 − qm)a − qnb.

In this discussion, we establish that r, a linear combination of a and b, satisfies the condition 0 ≤ r < d, where d is the smallest positive integer that can be expressed as a linear combination of a and b Consequently, we conclude that r equals 0, leading to the result that a is divisible by d The same reasoning applies to b, confirming that d divides both a and b Furthermore, by Theorem 4, any common divisor c of a and b also divides any linear combination of a and b, implying that c must divide d This establishes that any common divisor of a and b is less than or equal to d, thereby proving that d is the greatest common divisor of a and b.

As a result, we conclude that if (a, b) = 1 then there exist integers m and n such that ma + nb = 1.

The greatest common divisor (GCD) of a set of integers a1, a2, , an, where not all integers are zero, is defined as the largest integer that can evenly divide each of the integers in the set This is represented mathematically as (a1, a2, , an).

Definition 5 The integers a 1 , a 2 , , a n are said to be mutually relatively prime if

Example 11 The integers 3, 6, 7 are mutually relatively prime since (3, 6, 7) 1 although (3, 6) = 3.

Definition 6 The integers a 1 , a 2 , , a n are called pairwise prime if for each i j, we have (ai , aj ) = 1.

Example 12 The integers 3, 14, 25 are pairwise relatively prime Notice also that these integers are mutually relatively prime.

Notice that if a 1 , a 2 , , a n are pairwise relatively prime then they are mutually relatively prime.

1 Find the greatest common divisor of 15 and 35.

2 Find the greatest common divisor of 100 and 104.

3 Find the greatest common divisor of -30 and 95.

4 Let m be a positive integer Find the greatest common divisor of m and m + 1.

5 Let m be a positive integer, find the greatest common divisor of m and m + 2.

6 Show that if m and n are integers such that (m, n) = 1, then (m+n,m-n)=1 or 2.

7 Show that if m is a positive integer, then 3m + 2 and 5m + 3 are relatively prime.

8 Show that if a and b are relatively prime integers, then (a + 2b, 2a + b) 1or

9 Show that if a 1 , a 2 , , a n are integers that are not all 0 and c is a positive integer, then (ca1 , ca2, , can ) = c(a1 , a2, an ).

The Euclidean Algorithm

In this section we describe a systematic method that determines the greatest common divisor of two integers This method is called the Euclidean algorithm.

Lemma 1 If a and b are two integers and a = bq + r where also q and r are integers, then (a, b) = (r, b).

Proof Note that by theorem 8, we have (bq + r, b) = (b, r).

The Euclidean algorithm, in its general form, establishes that the greatest common divisor (GCD) of two integers can be determined by identifying the last non-zero remainder from a series of divisions.

Theorem 10 Let a = r0 and b = r1 be two positive integers where a ≥ b If we apply the division algorithm successively to obtain that r j = r j+1 q j+1 + r j+2 where 0 ≤ r j+2 < r j+1

1.6 THE EUCLIDEAN ALGORITHM 25 for all j = 0, 1, , n − 2 and r n+1 = 0.

Proof By applying the division algorithm, we see that r0 r1

It is important to note that we will eventually reach a remainder of 0, as all remainders are integers and each subsequent remainder is smaller than the one before it This observation aligns with Lemma 1.

Example 13 We will find the greatest common divisor of 4147 and 10672:

The Euclidean algorithm allows us to express the greatest common divisor (GCD) of two integers as a linear combination of those integers In this article, we will illustrate the process through a specific example, which will help determine the variables m and n as outlined in Theorem 9 Although the algorithm can be generalized, we will focus on a straightforward example to clearly demonstrate the steps involved in finding the GCD of two integers as a linear combination.

Example 14 Express 29 as a linear combination of 4147 and 10672:

As a result, we see that 29 = 175 ã 4147 − 68 ã 10672.

1 Use the Euclidean algorithm to find the greatest common divisor of 412 and

32 and express it in terms of the two integers.

2 Use the Euclidean algorithm to find the greatest common divisor of 780 and

150 and express it in terms of the two integers.

3 Find the greatest common divisor of 70, 98, 108.

4 Let a and b be two positive even integers Prove that (a, b) = 2(a/2, b/2).

5 Show that if a and b are positive integers where a is even and b is odd, then(a, b) = (a/2, b).

Lame’s Theorem

This section provides an estimate of the steps required to determine the greatest common divisor (GCD) of two integers using the Euclidean algorithm To facilitate this, we introduce Fibonacci numbers to establish a lemma that estimates their growth within the Fibonacci sequence This lemma will play a crucial role in the proof of Lame’s theorem.

Definition 7 The Fibonacci sequence is defined recursively by f1 = 1, f2 = 1, and f n = f n−1 + f n−2 for n ≥

The terms in the sequence are called Fibonacci numbers.

In the following lemma, we give a lower bound on the growth of Fibonacci numbers We will show that Fibonacci numbers grow faster than a geometric series with common ratio α = (1 +

Lemma 2 For n ≥ 3, we have fn > α n−2 where α = (1 +

Proof We use the second principle of mathematical induction to prove our result It is easy to see that this is true for n = 3 and n = 4 Assume that α k−2

< f k for all integers k where k ≤ n Now since α is a solution of the polynomial x 2 − x − 1 = 0, we have α 2 = α + 1 Hence α n−1 = α 2 α n−3 = (α + 1).α n−3 = α n−2 + α n−3

By the inductive hypothesis, we have α n−2 < f n , α n−3 < f n−1 After adding the two inequalities, we get α n−1 < fn + f n−1 = fn+1

We now present Lame’s theorem.

Theorem 11 states that when applying the Euclidean algorithm to determine the greatest common divisor (GCD) of two positive integers, the number of divisions required will not exceed five times the number of decimal digits in the smaller of the two integers.

Proof Let a and b be two positive integers where a > b Applying the Euclidean algorithm to find the greatest common divisor of two integers with a = r 0 and b = r 1 , we get r 0 r 1

Notice that each of the quotients q 1 , q 2 , , q n−1 are all greater than 1 and q n ≥ 2 and this is because rn < rn−1 Thus we have r n r n−1 r n−2 r n−3

≥ r3 + r4 ≥ fn−1 + fn−2 = fn , r1 ≥ r2 + r3 ≥ fn + fn−1 = fn+1

Thus notice that b ≥ fn+1 By Lemma 2, we have f n+1 > α n−1 for n > 2 As a result, we have b > α n−1 Now notice since we see that

Now let b has k decimal digits As a result, we have b < 10 k and thus log 10 b < k Hence we conclude that n − 1 < 5k Since k is an integer, we conclude that n ≤ 5k.

1 Find an upper bound for the number of steps in the Euclidean algorithm that is used to find the greatest common divisor of 38472 and 957748838.

To determine the upper bound for the number of steps in the Euclidean algorithm used to find the greatest common divisor (GCD) of 15 and 75, we can analyze the process The Euclidean algorithm involves a series of divisions, where the larger number is divided by the smaller one, and the remainder is used in the next step In this case, the GCD of 15 and 75 can be found quickly, as 75 divided by 15 equals 5 with a remainder of 0, indicating that 15 is the GCD Therefore, the upper bound for the number of steps is 1, confirming that the algorithm efficiently finds the GCD in just one iteration.

Prime numbers serve as the fundamental components of integers and have been the focus of extensive study throughout history The unique representation of an integer as a product of prime numbers underpins number theory and leads to fascinating results within this field Numerous theorems, applications, and conjectures have emerged, highlighting the significance of prime numbers and their properties.

In this chapter, we explore techniques for identifying prime and composite numbers through the ancient Greek method developed by Eratosthenes We demonstrate that prime numbers are infinite and illustrate that every integer can be expressed uniquely as a product of prime numbers.

We introduce as well the concept of diophantine equations where integer solutions from given equations are determined using the greatest common divisor.

We then mention the Prime Number theorem without giving a proof of course in addition to other conjectures and major results related to prime numbers.

The Sieve of Eratosthenes

Definition 8 A prime is an integer greater than 1 that is only divisible by 1 and itself.

Example 15 The integers 2, 3, 5, 7, 11 are prime integers.

Note that any integer greater than 1 that is not prime is said to be a composite number.

The Sieve of Eratosthenes is an ancient algorithm developed by the Greek mathematician Eratosthenes for identifying prime numbers up to a given integer While there are various methods to ascertain if a number is prime or composite, this sieve remains one of the most efficient To understand its application, we will first introduce a lemma essential for proving several related theorems.

Lemma 3 Every integer greater than one has a prime divisor.

We prove this Lemma by contradiction by assuming there exists an integer greater than one without any prime divisors Given that the set of integers greater than one without prime divisors is nonempty, the well-ordering principle indicates there is a least positive integer n greater than one with this property Consequently, n must be composite, as it divides itself, leading to the conclusion that n can be expressed as the product of two integers, a and b, where both 1 < a < n and 1 < b < n.

Notice that a < n and as a result since n is minimal, a must have a prime divisor which will also be a divisor of n.

Theorem 12 If n is a composite integer, then n has a prime factor not exceeding

Proof Since n is composite, then n = ab, where a and b are integers with 1 < a ≤ b < n Suppose now that a > √ n, then

Therefore a ≤ √ n Also, by Lemma 3, a must have a prime divisor a 1 which is also a prime divisor of n and thus this divisor is less than a 1 ≤ a ≤ √ n.

We now present the algorithm of the Sieve of Eratosthenes that is used to determine prime numbers up to a given integer.

The Algorithm of the Sieve of Eratosthenes

1 Write a list of numbers from 2 to the largest number n you want to test.

Note that every composite integer less than n must have a prime factor less than √ n Hence you need to strike off the multiples of the primes that are less than √ n

2 Strike off all multiples of 2 greater than 2 from the list The first remaining number in the list is a prime number.

3 Strike off all multiples of this number from the list.

4 Repeat the above steps until no more multiples are found of the prime integers that are less than √ n Exercises

1 Use the Sieve of Eratosthenes to find all primes less than 100.

2 Use the Sieve of Eratosthenes to find all primes less than 200.

3 Show that no integer of the form a 3 + 1 is a prime except for 2 = 1 3 + 1.

4 Show that if 2 n − 1 is prime, then n is prime.

Hint: Use the identity (a kl − 1) = (a k − 1)(a k(l−1) + a k(l−2) + + a k + 1).

The infinitude of Primes

We demonstrate that there are infinitely many prime numbers, a fact that can be proven through various methods An alternative proof is available as an exercise, while the proof we will discuss is attributed to Euclid in his renowned work, "Elements."

Theorem 13 There are infinitely many primes.

Proof We present the proof by contradiction Suppose there are finitely many primes p 1 , p 2 , , p n , where n is a positive integer Consider the integer Q such that

By Lemma 3, Q has at least a prime divisor, say q If we prove that q is not one of the primes listed then we obtain a contradiction Suppose now that q = pi for

1 ≤ i ≤ n Thus q divides p 1 p 2 p n and as a result q divides Q − p 1 p 2 p n Therefore q divides 1 But this is impossible since there is no prime that divides

1 and as a result q is not one of the primes listed.

The theorem addresses the existence of significant gaps between prime numbers, asserting that these gaps can be arbitrarily large It highlights the irregular spacing of prime numbers within the series, emphasizing the unpredictable nature of their distribution.

Theorem 14 Given any positive integer n, there exists n consecutive composite integers.

Proof Consider the sequence of integers

+ 1Notice that every integer in the above sequence is composite because k divides(n + 1)! + k if 2 ≤ k ≤ n + 1 by 4.

The Fundamental Theorem of Arithmetic

To prove the fundamental theorem of arithmetic, we need to prove some lemmas about divisibility.

Lemma 4 If a,b,c are positive integers such that (a, b) = 1 and a | bc, then a | c.

Proof Since (a, b) = 1, then there exists integers x, y such that ax + by = 1 As a result, cax + cby = c Notice that since a | bc, then by Theorem 4, a divides cax + cby and hence a divides c.

We can generalize the lemma by stating that if (a, n_i) = 1 for each i from 1 to n, and if a divides the product n_1, n_2, , n_(k+1), then it follows that a also divides n_(k+1) We will demonstrate a specific case of this generalization, which will aid in proving the fundamental theorem of arithmetic.

Lemma 5 If p divides n1 n2 n3 nk , where p is a prime and ni > 0 for all 1 ≤ i ≤ k, then there is an integer j with 1 ≤ j ≤ k such that p | n j

We demonstrate the proof of this result through mathematical induction For the base case of k = 1, the result is evident Assuming the result holds for k, we examine the case for n1, n2, , nk+1, which is divisible by p It is important to note that either

Now if (p, n1 n2 nk ) = 1 then by Lemma 4, p | nk+1 Now if p | n1n2 nk , then by the induction hypothesis, there exists an integer i such that p | n i

We now state the fundamental theorem of arithmetic and present the proof using Lemma 5.

Theorem 15 The Fundamental Theorem of Arithmetic Every positive integer different from 1 can be written uniquely as a product of primes. i

2.3 THE FUNDAMENTAL THEOREM OF ARITHMETIC 37

If $ n $ is a prime integer, it is represented as a product of primes with only one factor In the case of composite integers, we apply proof by contradiction Assume there exists a positive integer that cannot be expressed as a product of primes, and let $ n $ be the smallest such integer Therefore, we can express $ n $ as $ n = ab $, where $ 1 < a < n $ and $ 1 < b < n $.

Every integer can be expressed as a product of prime numbers, establishing that all integers are products of primes To demonstrate the uniqueness of this representation for positive integers, we consider an integer $ n $ with two distinct factorizations: $ n = p_1 p_2 \ldots p_s = q_1 q_2 \ldots q_r $, where $ p_1, p_2, \ldots, p_s $ and $ q_1, q_2, \ldots, q_r $ are primes arranged in non-decreasing order By canceling out the common primes from both factorizations, we arrive at the equation $ p_{j_1} p_{j_2} \ldots p_{j_u} = q_{i_1} q_{i_2} \ldots q_{i_v} $, leading to a contradiction and proving that the factorization of a positive integer into primes is unique.

All primes on the left side are distinct from those on the right side Given that any prime $ p_j^l $ (where $ l = 1, 2, , n $) divides the product $ p_j^1 p_j^2 \ldots p_j^u $, it follows that $ p_j^l $ must also divide the product $ q_i^1 q_i^2 \ldots q_i^v $.

, and hence by Lemma 5, pj 1 must divide qj k for some 1 ≤ k ≤ v which is impossible Hence the representation is unique.

Remark 1 The unique representation of a positive integer n as a product of primes can be written in several ways We will present the most common representations For example, n = p1p2p3 pk where pi for 1 ≤ i ≤ k are not necessarily distinct Another example would be n = p a 1 1 p2 a 2 p3 a 3 pj a j (2.1) where all the p i are distinct for 1 ≤ i ≤ j One can also write a formal product n = Y all primes p i p α i , (2.2) i n

38 CHAPTER 2 PRIME NUMBERS where all but finitely many of the α 0 s are 0.

Example 17 The prime factorization of 120 is given by 120 = 2ã2ã2ã3ã5 = 2 3 ã3ã5 Notice that 120 is written in the two ways described in 1.

Prime factorization is a valuable method for determining the greatest common divisor (GCD) of two integers For integers $ a $ and $ b $, we can express them in terms of their prime factors: $ a = p_1^{a_1} p_2^{a_2} \ldots p_n^{a_n} $ and $ b = p_1^{b_1} p_2^{b_2} \ldots p_n^{b_n} $, where primes with a power of 0 in both integers are excluded In this representation, if a prime $ p_i $ is present in $ a $ but not in $ b $, then $ a_i = 0 $ while $ b_i $ remains non-zero This approach enables us to easily identify the common prime factors and their respective minimum powers, which are essential for calculating the GCD.

= 0, and vise versa Then the greatest common divisor is given by

(a, b) = p 1 min(a 1 ,b 2 ) min(a p 2 2 ,b 2 ) p min(a n ,b n ) where min(n, m) is the minimum of m and n.

The following lemma is a consequence of the Fundamental Theorem of Arith- metic.

Lemma 6 Let a and b be relatively prime positive integers Then if d divides ab, there exists d 1 and d 2 such that d = d 1 d 2 where d 1 is a divisor of a and d 2 is a divisor of b Conversely, if d 1 and d 2 are positive divisors of a and b, respectively, then d = d 1 d 2 is a positive divisor of ab.

To prove the relationship between d, d1, and d2, we define d1 as (a, d) and d2 as (b, d) Given that (a, b) = 1 and considering the prime decomposition of a and b, we conclude that d = d1 d2 and that d1 and d2 are coprime It is important to note that each prime power in the factorization of d must be represented in either d1 or d2 Specifically, the prime powers from the factorization of d that divide a will be included in d1, while those that divide b will be included in d2.

2.3 THE FUNDAMENTAL THEOREM OF ARITHMETIC 39

Now conversely, let d 1 and d 2 be positive divisors of a and b, respectively Then is a divisor of ab. d = d 1 d 2

More on the Infinitude of Primes

There are also other theorems that discuss the infinitude of primes in a given arithmetic progression The most famous theorem about primes in arithmetic progression is Dirichlet’s theorem

Theorem 16 Dirichlet’s Theorem Given an arithmetic progression of terms an + b , for n = 1, 2, ,the series contains an infinite number of primes if a and b are relatively prime,

Dirichlet's theorem, initially conjectured by Gauss, was first proven using complex analysis, although the proof is notably complex In this article, we will focus on a special case of this theorem, demonstrating that there are infinitely many prime numbers within a specific arithmetic progression Before presenting this theorem, we will establish a lemma that is essential for the subsequent proof.

Lemma 7 If a and b are integers both of the form 4n + 1, then their product ab is of the form 4n +

Proof Let a = 4n1 + 1 and b = 4n2 + 1, then ab = 16n 1 n 2 + 4n 1 + 4n 2 + 1 = 4(4n 1 n 2 + n 1 + n 2 ) + 1 = 4n 3 +

Theorem 17 There are infinitely many primes of the form 4n + 3, where n is a positive integer.

Proof Suppose that there are finitely many primes of the form 4n + 3, say p 0 3, p 1 , p 2 , , p n Let

Every odd prime can be expressed as either 4n + 1 or 4n + 3 In the prime factorization of N, there must be at least one prime of the form 4n + 3; otherwise, according to Lemma 7, N would take the form 4n + 1 Our goal is to demonstrate that this prime factor of N is not among the primes p0 = 3, p1, p2, , pn.

3 | 4p1p2 pn which is impossible since p i = 3 for every i Hence 3 doesn’t divide N Also, the other primes p1 , p2, , pn don’t divide N because if pi | N , then p i | (N − 4p 1 p 2 p n ) = 3.

Hence none of the primes p 0 , p 1 , p 2 , , p n divides N Thus there are infinitely many primes of the form 4n + 3.

1 Find the prime factorization of 32, of 800 and of 289.

2 Find the prime factorization of 221122 and of 9!.

3 Show that all the powers of in the prime factorization of an integer a are even if and only if a is a perfect square.

4 Show that there are infinitely many primes of the form 6n + 5.

Least Common Multiple

We can use prime factorization to find the smallest common multiple of two positive integers.

Definition 9 The least common multiple (l.c.m.) of two positive integers is the smallest positive integer that is a multiple of both.

We denote the least common multiple of two positive integers a an b by ha, bi. Example 18 h2, 8i = 8, h5, 8i = 40

We can figure out ha, bi once we have the prime factorization of a and b

To establish the relationship between the least common multiple (LCM) and the greatest common divisor (GCD) of two positive integers, let a and b be expressed in their prime factorization forms, excluding any prime raised to the power of zero in both The LCM can be determined by taking the maximum power of each prime factor present in either a or b Specifically, the LCM is calculated as the product of the primes raised to their respective maximum powers, which can be denoted as max(a₁, b₁) max(a₂, b₂) max(aₙ, bₙ) This theorem, which connects LCM and GCD, is often defined in mathematical literature as the foundational definition of the least common multiple.

To prove the theorem we present a lemma

Lemma 8 If a and b are two real numbers, then min(a, b) + max(a, b) = a + b Proof Assume without loss of generality that a ≥ b Then max(a, b) = a and min(a, b) = b, and the result follows.

Theorem 18 Let a and b be two positive integers Then

3 If a | m and b | m, then ha, bi | m

Proof The proof of part 1 follows from the definition As for part 2, let a = p a 1 p a 2 p a n and b = p b 1 p b 2 p b n Notice that since

(a, b) = p 1 min(a 1 ,b 2 ) min(a p 2 2 ,b 2 ) p min(a n ,b n ) and max(a 1 ,b 1 ) max(a 2 ,b 2 ) then ha, bi = p 1 p 2 p max(a n ,b n ) , max(a 1 ,b 1 ) max(a 2 ,b 2 ) min(a 1 ,b 2 ) min(a 2 ,b 2 ) ha, bi(a, b) = p 1 p 2 p max(a n ,b n ) p 1 p 2 p min(a n ,b n )

= p max(a 1 1 ,b 1 )+min(a 1 ,b 1 ) max(a p 2 2 ,b 2 )+min(a 2 ,b 2 p ) max(a n ,b n )+min(a n ,b n )

Note also that we used Lemma 8 in the above equations For part 3, it would be a nice exercise to show that ab/(a, b) | m (Exercise 6) Thus ha, bi | m.

1 Find the least common multiple of 14 and 15.

2 Find the least common multiple of 240 and 610.

3 Find the least common multiple and the greatest common divisor of 2 5 5 6

4 Show that every common multiple of two positive integers a and b is divisible by the least common multiple of a and b.

Linear Diophantine Equations

If $ a $ and $ b $ are positive integers, then their greatest common divisor (GCD) divides their least common multiple (LCM) This relationship can be expressed mathematically as $ \text{GCD}(a, b) \times \text{LCM}(a, b) = a \times b $ The GCD and LCM of two numbers are equal when both numbers are the same, as in the case where $ a = b $.

This section focuses on diophantine equations, which involve two variables and require integer solutions The objective is to identify the set of integer-coordinate points that satisfy these equations Geometrically, diophantine equations represent straight lines, and our task is to locate the integer points that lie on these lines.

Definition 10 A linear equation of the form ax + by = c where a, b and c are integers is known as a linear diophantine equation.

A linear diophantine equation of the form ax + by = c has integer solutions if and only if the greatest common divisor d of a and b divides c When one solution (x0, y0) is found, it leads to infinitely many solutions expressed as x = x0 + (b/d)t and y = y0 - (a/d)t, where t is any integer.

Proof Suppose that the equation ax + by = c has integer solution x and y. Thus since d | a and d | b, then d | (ax + by) = c.

Now we have to prove that if d | c, then the equation has integral solution Assume that d | c By theorem 9, there exist integers m and n such that d = am + bn.

And also there exists integer k such that c = dk Now since c = ax + by, we have c = dk = (ma + nb)k = a(km) + b(nk).

Hence a solution for the equation ax + by = c is x 0 = km and y 0 = kn.

What is left to prove is that we have infinitely many solutions Let x = x0 + (b/d)t and y = y0 − (a/d)t.

We have to prove now that x and y are solutions for all integers t Notice that ax + by = a(x0 + (b/d)t) + b(y0 − (a/d)t) = ax0 + by0 c.

We now show that every solution for the equation ax + by = c is of the form x = x 0 + (b/d)tand y = y 0 − (a/d)t.

Notice that since ax 0 + by 0 = c, we have a(x − x 0 ) + b(y − y 0 ) = 0.

Dividing both sides by d, we get a/d(x − x 0 ) = b/d(y − y 0

Notice that (a/d, b/d) = 1 and thus we get by Lemma 4 that a/d | y − y0 As a result, there exists an integer t such that y = y0 − (a/d)t Now substituting y − y0 in the equation

Example 19 The equation 3x+6y = 7 has no integer solution because (3, 6) 3 does not divide 7.

The equation 4x + 6y = 8 has infinitely many integer solutions because the greatest common divisor of 4 and 6 is 2, which divides 8 By applying the Euclidean algorithm, we find integers m and n such that 4m + 6n = 2; specifically, 4(-1) + 6(1) = 2 This leads to a particular solution of x₀ = -4 and y₀ = 4 Consequently, the general solutions can be expressed for all integers t.

1 Either find all solutions or prove that there are no solutions for the diophantine equation 21x + 7y = 147.

4 A grocer orders apples and bananas at a total cost of $8.4 If the apples cost

25 cents each and the bananas 5 cents each, how many of each type of fruit did he order.

The function [x] , the symbols ”O”, ”o” and ”∼”

The Function [x]

Definition 11 The function [x] represents the largest integer not exceeding x In other words, for real x, [x] is the unique integer such that x − 1 < [x] ≤ x < [x] +

We also define ((x)) to be the fractional part of x In other words ((x)) x − [x].

We now list some properties of [x] that will be used in later or in more advanced courses in number theory.

3 [x] + [−x] is 0 if x is an integer and -1 otherwise.

4 The number of integers m for which x < m ≤ y is [y] − [x].

2.6 THE FUNCTION [X ] , THE SYMBOLS ”O”, ”O” AND ”∼” 47

5 The number of multiples of m which do not exceed x is [x/m].

Using the definition of [x], it will be easy to see that the above properties are direct consequences of the definition.

In this section, we introduce specific symbols that will aid in estimating the growth of number theoretic functions While these symbols may not be emphasized within the context of this book, they play a significant role in numerous analytic proofs.

The ”O” and ”o” Symbols

In mathematical analysis, if f(x) is a positive function and g(x) is any function, then O(f(x)), pronounced "big-oh" of f(x), represents the set of functions g(x) that grow at a rate constrained by f(x) The conventional way to express that g(x) is part of this set is by using the notation g(x) = O(f(x)).

This means that for sufficiently large x,

|f (x)| where M is some positive number.

Example 21 sin(x) = O(x), and also sin(x) = O(1).

Now, the relation g(x) = o(f (x)), pronounced ”small-oh” of f (x), is used to indicate that f (x) grows much faster than g(x) It formally says that lim g(x)

= 0 (2.4) x→∞ f (x) More generally, g(x) = o(f (x)) at a point b if lim g(x)

Example 22 sin(x) = o(x) at ∞, and x k = o(e x ) also at ∞ for every constant k.

The notation that f (x) is asymptotically equal to g(x) is denoted by ∼ For- mally speaking, we say that f (x) ∼ g(x) if lim f (x)

The introduction of these symbols aims to simplify complex mathematical expressions by breaking them down into a principal part and a remainder term The remainder can be easily represented using the specified notations, facilitating the combination of multiple expressions We will outline some properties of these symbols, which are straightforward to prove based on their definitions.

There are some other properties that we did not mention here, properties that are rarely used in number theoretic proofs.

1 Prove the five properties of the [x]

2 Prove the five properties of the O and o notations in Example 24. p

Theorems and Conjectures involving prime numbers

2.7 Theorems and Conjectures involving prime numbers

The existence of infinitely many primes and the presence of arbitrarily large gaps between them lead to the important question of estimating the number of primes less than a given number This is addressed by the prime number theorem, which defines π(x) as the count of primes less than a positive number x After extensive conjecture and research by mathematicians, Chebyshev established the estimate as x/logx The prime number theorem was ultimately proved in 1896 by Hadamard and Poussin through independent methods Before delving into the theorem, we will present and prove a lemma related to primes that will be relevant in subsequent chapters.

Lemma 9 Let p be a prime and let m ∈ Z + Then the highest power of p dividing m! is

The highest power of a prime $ p $ that divides $ m! $ can be determined by counting the integers divisible by $ p $ among the integers from 1 to $ m $ Specifically, there are $ h(m, p) $ integers divisible by $ p $, represented as $ p, 2p, \ldots, h(m, p)p $ Similarly, the count of integers divisible by $ p^i $ is given by $ h(m, p^i) $ Thus, the total highest power of $ p $ dividing $ m! $ is the sum of these counts across all relevant powers of $ p $.

Theorem 20 The Prime Number Theorem Let x > 0 then π(x) ∼ x/logx

The theorem indicates that to estimate the number of prime numbers less than a given value x, it is unnecessary to identify all the primes below that threshold Instead, one can simply calculate x/log(x) for sufficiently large values of x to obtain an accurate approximation of the prime count.

Numerous theorems have been established regarding prime numbers, with many renowned mathematicians tackling various problems associated with them Despite significant progress, several open problems remain in the field of prime number theory that continue to intrigue researchers.

Conjecture 1 Twin Prime Conjecture There are infinitely many pairs primes p and p + 2.

Conjecture 2 Goldbach’s Conjecture Every even positive integer greater than 2 can be written as the sum of two primes.

Conjecture 3 The n 2 + 1 Conjecture There are infinitely many primes of the form n 2 + 1, where n is a positive integer.

Conjecture 4 Polignac Conjecture For every even number 2n are there infinitely many pairs of consecutive primes which differ by 2n.

Conjecture 5 Opperman Conjecture Is there always a prime between n 2 and(n + 1) 2 ?

A congruence refers to a statement concerning divisibility, a concept introduced by mathematician Carl Friedrich Gauss Gauss significantly advanced the foundational principles of congruences and demonstrated various theorems associated with this mathematical theory.

We start by introducing congruences and their properties We proceed to prove theorems about the residue system in connection with the Euler φ-function.

This article explores solutions to linear congruences, laying the groundwork for understanding the Chinese remainder theorem Additionally, it highlights significant congruence theorems established by notable mathematicians such as Wilson, Fermat, and Euler.

Introduction to congruences

As we mentioned in the introduction, the theory of congruences was developed by

Gauss at the beginning of the nineteenth century.

Definition 12 Let m be a positive integer We say that a is congruent to b modulo m if m | (a − b) where a and b are integers, i.e if a = b + km where k

If a is congruent to b modulo m, we write a ≡ b(mod m).

Example 24 19 ≡ 5(mod 7) Similarly 2k + 1 ≡ 1(mod 2) which means every odd number is congruent to 1 modulo 2.

There are many common properties between equations and congruences Some properties are listed in the following theorem.

Theorem 21 Let a, b, c and d denote integers Let m be a positive integers Then:

2 If a ≡ b(mod m) and b ≡ c(mod m), then a ≡ c(mod m).

5 If a ≡ b(mod m), then ac ≡ bc(mod m).

6 If a ≡ b(mod m), then ac ≡ bc(mod mc), for c > 0.

7 If a ≡ b(mod m) and c ≡ d(mod m) then a + c ≡ (b + d)(mod m).

8 If a ≡ b(mod m) and c ≡ d(mod m) then a − c ≡ (b − d)(mod m).

9 If a ≡ b(mod m) and c ≡ d(mod m) then ac ≡ bd(mod m).

Proof 1 If a ≡ b(mod m), then m | (a − b) Thus there exists integer k such that a − b = mk, this implies b − a = m(−k) and thus m | (b − a) Consequently b ≡ a(mod m).

2 Since a ≡ b(mod m), then m | (a − b) Also, b ≡ c(mod m), then m | (b − c) As a result, there exit two integers k and l such that a = b + mk and b = c+ml, which imply that a = c+m(k+l) giving that a c(mod m).

3 Since a ≡ b(mod m), then m | (a − b) So if we add and subtract c we get m | ((a + c) − (b + c)) and as a result a + c ≡ b + c(mod m).

4 Since a ≡ b(mod m), then m | (a − b) so we can subtract and add c and we get and as a result m | ((a − c) − (b − c)) a − c ≡ b − c(mod m).

5 If a ≡ b(mod m), then m | (a − b) Thus there exists integer k such that a − b = mk and as a result ac − bc = m(kc) Thus m | (ac − bc) and hence ac ≡ bc(mod m).

6 If a ≡ b(mod m), then m | (a − b) Thus there exists integer k such that a − b = mk and as a result ac − bc = mc(k).

Thus and hence mc | (ac − bc) ac ≡ bc(mod mc).

7 Since a ≡ b(mod m), then m | (a − b) Also, c ≡ d(mod m), then m | (c −d) As a result, there exits two integers k and l such that a −b mk and c − d = ml Note that

8 If a = b + mk and c = d + ml where k and l are integers, then

9 There exit two integers k and l such that a − b = mk and c − d = ml and thus ca − cb = m(ck) and bc − bd = m(bl) Note that

(ca − cb) + (bc − bd) = ac − bd = m(kc − lb).

As a result, hence m | (ac − bd), ac ≡ bd(mod m).

3.1 INTRODUCTION TO CONGRUENCES 55 Examples 1 1 Because 14 ≡ 8(mod 6) then 8 ≡ 14(mod 6).

2 Because 22 ≡ 10(mod 6) and 10 ≡ 4(mod 6) Notice that 22 ≡ 4(mod 6).

7 Because 19 ≡ 3(mod 8) and 17 ≡ 9(mod 8), then 19 + 17 = 36 ≡ 3 + 9 12(mod 8).

8 Because 19 ≡ 3(mod 8) and 17 ≡ 9(mod 8), then 19 − 17 = 2 ≡ 3 − 9 −6(mod 8).

9 Because 19 ≡ 3(mod 8) and 17 ≡ 9(mod 8), then 19(17) = 323 ≡ 3(9) 27(mod 8).

This theorem highlights a key distinction between equations and congruences: when both sides of an equation are divided by a non-zero number, the equality remains valid However, this principle does not apply to congruences, as dividing both sides by the same integer does not guarantee the preservation of the congruence.

Theorem 22 1 If a, b, c and m are integers such that m > 0, d = (m, c) and ac ≡ bc(mod m), then a ≡ b(mod m/d).

2 If (m, c) = 1 then a = b(mod m) if ac ≡ bc(mod m).

Proof Part 2 follows immediately from Part 1 For Part 1, if ac ≡ bc(mod m), then m | (ac − bc) = c(a − b). i=1ai ≡

Hence there exists k such that c(a − b) = mk Dividing both sides by d, we get (c/d)(a − b) = k(m/d) Since (m/d, c/d) = 1, it follows that m/d | (a − b). Hence a ≡ b(mod m/d).

Example 25 38 ≡ 10(mod 7) Since (2, 7) = 1 then 19 ≡ 5(mod 7).

The following theorem combines several congruences of two numbers with different moduli.

Theorem 23 If a ≡ b(mod m 1 ), a ≡ b(mod m 2 ), , a ≡ b(mod m t ) where a, b, m 1 , m 2 , , m t are integers and m 1 , m 2 , , m t are positive, then a ≡ b(mod hm 1 , m 2 , m t i)

Proof Since a ≡ b(mod m i ) for all 1 ≤ i ≤ t Thus mi | (a − b) As a result, hm1, m 2 , , m t i | (a − b)

(prove this as an exercise) Thus a ≡ b(mod hm 1 , m 2 , m t i).

1 Determine whether 3 and 99 are congruent modulo 7 or not.

2 Show that if x is an odd integer, then x 2 ≡ 1(mod 8)

3 Show that if a, b, m and n are integers such that m and n are positive, n | m and a ≡ b(mod m), then a ≡ b(mod n).

4 Show that if a i ≡ bi (mod m) for i = 1, 2, , n, where m is a positive integer and ai , bi are integers for j = 1, 2, , n, then P n P n i=1bi (mod m)

5 For which n does the expression 1 + 2 + + (n − 1) ≡ 0(mod n) holds.

Residue Systems and Euler’s φ-Function

Residue Systems

In the context of positive integers, when dividing two integers a and b, the division algorithm reveals that a can be expressed as a = bm + r, where r is the least non-negative residue of a modulo m, satisfying the condition 0 ≤ r < m Consequently, any integer is congruent to one of the integers in the range of 0 to m − 1 when considered modulo m.

Definition 13 A complete residue system modulo m is a set of integers such that every integer is congruent modulo m to exactly one integer of the set.

The easiest complete residue system modulo m is the set of integers 0, 1, 2, , m−

1 Every integer is congruent to one of these integers modulo m.

Example 26 The set of integers {0, 1, 2, 3, 4} form a complete residue system modulo 5 Another complete residue system modulo 5 could be 6, 7, 8, 9, 10.

Definition 14 A reduced residue system modulo m is a set of integers r i such that

(ri , m) = 1 for all i and ri = rj (mod m) if i j.

Notice that, a reduced residue system modulo m can be obtained by deleting all the elements of the complete residue system set that are not relatively prime to m.

Example 27 The set of integers {1, 5} is a reduced residue system modulo

The following lemma will help determine a complete residue system modulo any positive integer m.

Lemma 10 A set of m incongruent integers modulo m forms a complete residue system modulo m.

We will prove this lemma by contradiction If the set of m integers does not form a complete residue system modulo m, there exists at least one integer a that is not congruent to any element in the set Consequently, none of the set's elements are congruent to the remainder of a when divided by m, resulting in at most m − 1 possible remainders According to the pigeonhole principle, this implies that at least two integers in the set share the same remainder modulo m This creates a contradiction, as the set consists of m integers that are incongruent modulo m.

Theorem 24 If a1 , a2, , am is a complete residue system modulo m, and if k is a positive integer with (k, m) = 1, then ka 1 + b, ka 2 + b, , ka m + b is another complete residue system modulo m for any integer b.

Proof Let us prove first that no two elements of the set {ka1 +b, ka 2 +b, , ka m + b} are congruent modulo m Suppose there exists i and j such that kai + b ≡ kaj + b(mod m).

Thus we get that kai ≡ kaj (mod m).

Now since (k, m) = 1, we get ai ≡ aj (mod m)

But for i = j, a i is inequivalent to a j modulo m Thus i = j Now notice that there are m inequivalent integers modulo m and thus by Lemma 10, the set form a complete residue system modulo m.

Euler’s φ-Function

In this chapter, we introduce the Euler φ-function, which counts the positive integers less than a specified integer that are relatively prime to it We will explore the properties of the Euler φ-function in detail in Chapter 5, but for now, it is important to familiarize ourselves with its notation.

Definition 15 The Euler φ-function of a positive integer n, denoted by φ(n) counts the number of positive integers less than n that are relatively prime to n.

Example 28 Since 1 and 3 are the only two integers that are relatively prime to

4 and less than 4, then φ(4) = 2 Also, 1,2, ,6 are the integers that are relatively prime to 7 that are less than 7, thus φ(7) = 6.

Now we can say that the number of elements in a reduced residue system modulo n is φ(n).

Theorem 25 If a 1 , a 2 , , a φ(n) is a reduced residue system modulo n and (k, n)

1, then ka 1 , ka 2 , , ka φ(n) is a reduced residue system modulo n.

Proof The proof proceeds exactly in the same way as that of Theorem 24.

1 Give a reduced residue system modulo 12.

2 Give a complete residue system modulo 13 consisting only of odd integers.

Linear Congruences

In this chapter, we introduce the Euler φ-function, which counts the positive integers less than a specified integer that are relatively prime to it While a detailed discussion of the properties of the Euler φ-function will be provided in chapter 5, we will focus on the notation and its relevance for our current purposes.

Definition 15 The Euler φ-function of a positive integer n, denoted by φ(n) counts the number of positive integers less than n that are relatively prime to n.

Example 28 Since 1 and 3 are the only two integers that are relatively prime to

4 and less than 4, then φ(4) = 2 Also, 1,2, ,6 are the integers that are relatively prime to 7 that are less than 7, thus φ(7) = 6.

Now we can say that the number of elements in a reduced residue system modulo n is φ(n).

Theorem 25 If a 1 , a 2 , , a φ(n) is a reduced residue system modulo n and (k, n)

1, then ka 1 , ka 2 , , ka φ(n) is a reduced residue system modulo n.

Proof The proof proceeds exactly in the same way as that of Theorem 24.

1 Give a reduced residue system modulo 12.

2 Give a complete residue system modulo 13 consisting only of odd integers.

Linear congruences, similar to equations, prompt inquiries about solutions to linear equations This section focuses on one-variable linear congruences and their solutions, beginning with a clear definition of linear congruences.

Definition 16 A congruence of the form ax ≡ b(mod m) where x is an unknown integer is called a linear congruence in one variable.

Understanding linear congruences is crucial, as any integer $ x_i $ that satisfies the condition $ x_i \equiv x_0 \, (\text{mod} \, m) $ is also a solution if $ x_0 $ is a known solution Additionally, the equation $ ax \equiv b \, (\text{mod} \, m) $ can be transformed into a linear Diophantine equation, indicating that there exists an integer $ y $ such that $ ax - my = b $ This forms the basis for proving theorems related to the solutions of linear congruences.

Theorem 26 states that for integers a, b, and m (with m > 0), if c is the greatest common divisor of a and m, and c does not divide b, then the congruence ax ≡ b (mod m) has no solutions Conversely, if c divides b, the congruence ax ≡ b (mod m) has exactly c incongruent solutions modulo m.

Proof As we mentioned earlier, ax ≡ b(mod m) is equivalent to ax − my = b.

According to Theorem 19 on Diophantine equations, the equation ax - my = b has no solutions if c does not divide b Conversely, if c divides b, the equation has infinitely many solutions, with the variable x given by x = x0 + (m/c)t, where x0 is a particular solution and t is an integer parameter.

The values of x identified are solutions to the congruence ax ≡ b (mod m) To find the number of incongruent solutions, we consider the case where two solutions are congruent, expressed as x₀ + (m/c)t₁ ≡ x₀ + (m/c)t₂ (mod m).

Now notice that (m, m/c) = m/c and thus t1 ≡ t2(mod c).

Thus we get a set of incongruent solutions given by x = x 0 + (m/c)t, where t is taken modulo c.

Remark 2 Notice that if c = (a, m) = 1, then there is a unique solution modulo m for the equation ax ≡ b(mod m).

To solve the congruence 3x ≡ 12 (mod 6), we first observe that the greatest common divisor (gcd) of 3 and 6 is 3, which divides 12 This indicates that there are three incongruent solutions modulo 6 By applying the Euclidean algorithm to the equation 3x − 6y = 12, we find a particular solution, x₀ = 6 Consequently, the three incongruent solutions can be expressed as x₁ = 6 (mod 6), x₂ = 2 (mod 6), and x₃ = 4 (mod 6).

As we mentioned earlier in Remark 2, the congruence ax ≡ b(mod m) has a unique solution if (a, m) = 1 This will allow us to talk about modular inverses.

Definition 17 A solution for the congruence ax ≡ 1(mod m) for (a, m) = 1 is called the modular inverse of a modulo m We denote such a solution by a¯.

Example 30 The modular inverse of 7 modulo 48 is 7 Notice that a solution for

1 Find all solutions of 3x ≡ 6(mod 9).

2 Find all solutions of 3x ≡ 2(mod 7).

3 Find an inverse modulo 13 of 2 and of 11.

4 Show that if a¯ is the inverse of a modulo m and ¯b is the inverse of b modulo m, then a¯¯b is the inverse of ab modulo m.

The Chinese Remainder Theorem

This section addresses the solution of systems of congruences with varying moduli For example, we aim to find a number that gives a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 3 when divided by 5 Such problems can be expressed using congruences Consequently, this chapter introduces a systematic approach to solving these systems of congruences.

Theorem 27 The system of congruences x ≡ b1(mod n1), x ≡ b2 (mod n2),

. x ≡ b t (mod n t ), has a unique solution modulo N = n 1 n 2 n t if n 1 , n 2 , , n t are pairwise relatively prime positive integers.

Proof Let N k = N/n k Since (n i , n j ) = 1 for all i = j, then (N k , n k ) = 1. Hence by Theorem 26 , we can find an inverse y k of N k modulo n k such that

Since thus we see that t x = X bi Ni yi i=1

Nj ≡ 0(mod nk ) for all j = k, x ≡ b k N k y k (mod n k ).

Also notice that N k y k ≡ 1(mod nk ) Hence x is a solution to the system of t congruences We have to show now that any two solutions are congruent modulo

N Suppose now that you have two solutions x0, x1 to the system of congruences Then x 0 ≡ x 1 (mod n k ) for all 1 ≤ k ≤ t Thus by Theorem 23, we see that x 0 ≡ x1 (mod N ).

Thus the solution of the system is unique modulo N

We now present an example that will show how the Chinese remainder theorem is used to determine the solution of a given system of congruences.

Example 31 Solve the system x ≡ 1(mod 2) x ≡ 2(mod 3) x ≡ 3(mod 5).

So we have to solve now 15y1 ≡ 1(mod 2) Thus y1 ≡ 1(mod 2).

In the same way, we find that y2 ≡ 1(mod 3)and y3 ≡ 1(mod 5).

As a result, we get x ≡ 1.15.1 + 2.10.1 + 3.6.1 ≡ 53 ≡ 23(mod

1 Find an integer that leaves a remainder of 2 when divided by either 3 or 5, but that is divisible by 4.

2 Find all integers that leave a remainder of 4 when divided by 11 and leaves a remainder of 3 when divided by 17.

3 Find all integers that leave a remainder of 1 when divided by 2, a remainder of 2 when divided by 3 and a remainder of 3 when divided by 5.

Theorems of Fermat, Euler, and Wilson

This section explores three significant applications of congruences in number theory First, we discuss Wilson’s theorem, which asserts that for any prime number p, the expression (p − 1)! + 1 is divisible by p Next, we introduce Fermat’s little theorem, which states that if p is a prime and a is an integer, then a raised to the power of p minus a is congruent to the same remainder when divided by p Lastly, we examine Euler’s theorem, a generalization of Fermat’s theorem, which claims that for any positive integer m that is relatively prime to an integer a, the equation a raised to the power of φ(m) is congruent to 1 modulo m, where φ represents Euler’s φ-function We begin with a proof concerning the inverse of integers modulo prime numbers.

Theorem 28 Let p be a prime A positive integer m is its own inverse modulo p if and only if p divides m + 1 or p divides m −

Proof Suppose that m is its own inverse Thus

Hence p | m 2 − 1 As a result, m.m ≡ 1(mod p). p | (m − 1)or p | (m + 1).

3.5 THEOREMS OF FERMAT, EULER, AND WILSON 65

We get that m ≡ 1(mod p) or m ≡ −1(mod p).

Conversely, suppose that m ≡ 1(mod p)or m ≡ −1(mod p).

Theorem 29 Wilson’s Theorem If p is a prime number, then p divides (p−1)!+1.

Proof When p = 2, the congruence holds Now let p > 2 Using Theorem

For every integer $ m $ within the range $ 1 \leq m \leq p $, there exists an inverse $ \bar{m} $ such that $ m \bar{m} \equiv 1 \mod p $ According to Theorem 28, the only integers that possess their own inverses are 1 and $ p - 1 $ This leads to a coupling of the integers from 1 to $ p $.

2 to p − 2 each with its inverse, we get

As a result, we have (p − 1)! ≡ −1(mod p).

Note also that the converse of Wilson’s theorem also holds The converse tells us whether an integer is prime or not.

Theorem 30 If m is a positive integer with m ≥ 2 such that

Proof Suppose that m has a proper divisor c 1 and that

That is m = c 1 c 2 where 1 < c 1 < m and 1 < c 2 < m Thus c 1 is a divisor of (m − 1)! Also, since m | ((m − 1)! + 1), we get c 1 | ((m − 1)! + 1).

As a result, by Theorem 4, we get that c 1 | ((m − 1)! + 1 − (m − 1)!), which gives that c 1 | 1 This is a contradiction and hence m is prime.

Fermat’s Little Theorem asserts that for a prime number $ p $ that does not divide $ a $, the remainder of $ a^{p-1} $ when divided by $ p $ is 1 Additionally, Euler’s Theorem states that if $ a $ is relatively prime to a positive integer $ m $, then the remainder of $ a^{\phi(m)} $ when divided by $ m $ is also 1.

1 We prove Euler’s Theorem only because Fermat’s Theorem is nothing but a special case of Euler’s Theorem This is due to the fact that for a prime number p, φ(p) = p − 1.

Theorem 31 Euler’s Theorem If m is a positive integer and a is an integer such that (a, m) = 1, then a φ(m) ≡ 1(mod m)

Example 32 Note that 3 4 = 81 ≡ 1(mod 5) Also, 2 φ(9) = 2 6 = 64 ≡ 1(mod

We now present the proof of Euler’s theorem.

3.5 THEOREMS OF FERMAT, EULER, AND WILSON 67

Proof Let k 1 , k 2 , , k φ(m) be a reduced residue system modulo m By Theorem

{ak 1 , ak 2 , , ak φ(m) } also forms a reduced residue system modulo m Thus ak1 ak2 akφ(m) = a φ(m) k1k2 kφ(m) ≡ k1k2 kφ(m) (mod m).

Now since (k i , m) = 1 for all 1 ≤ i ≤ φ(m), we have (k 1 k 2 k φ(m) , m) = 1. Hence by Theorem 22 we can cancel the product of k’s on both sides and we get a φ(m) ≡ 1(mod m).

An immediate consequence of Euler’s Theorem is:

Corollary 1 Fermat’s Theorem If p is a prime and a is a positive integer with p - a, then a p−1 ≡ 1(mod p).

We introduce two theorems derived from Fermat's theorem The first reformulates Fermat's theorem, asserting that the remainder of a number $ p $ when divided by $ p $ is identical to the remainder of another number $ a $ when divided by $ p $ The second theorem provides a method for finding the inverse of an integer $ a $ modulo $ p $ where $ p $ is greater than $ a $.

Theorem 32 If p is a prime number and a is a positive integer, then a p ≡ a(mod p).

Proof If p - a, by Fermat’s theorem we know that a p−1 ≡ 1(mod p).

Now if p | a, we have a p ≡ a ≡ 0(mod p).

Theorem 33 If p is a prime number and a is an integer such that p - a, then a p−2 is the inverse of a modulo p.

Proof If p - a, then Fermat’s theorem says that a p−1 ≡ 1(mod p).

As a result, a p−2 is the inverse of a modulo p.

1 Show that 10!+1 is divisible by 11.

2 What is the remainder when 5!25! is divided by 31?

3 What is the remainder when 5 100 is divided by 7?

4 Show that if p is an odd prime, then 2(p − 3)! ≡ −1(mod p).

5 Find a reduced residue system modulo 2 m , where m is a positive integer.

6 Show that if a 1 , a 2 , , a φ(m) is a reduced residue system modulo m, where m is a positive integer with m = 2, then a1 + a2 + + aφ(m) ≡ 0(mod m).

7 Show that if a is an integer such that a is not divisible by 3 or such that a is divisible by 9, then a 7 ≡ a(mod 63).

In this chapter, we explore multiplicative functions defined on integers, which possess the unique property that the function's value at the product of two relatively prime integers equals the product of the function's values at those integers We begin by establishing several key theorems regarding multiplicative functions for future reference Additionally, we investigate special functions, demonstrating that the Euler φ-function is indeed multiplicative Furthermore, we introduce and define the sum of divisors and the number of divisors functions.

The Mobius function is defined to analyze integers based on their prime factorization Its summatory function calculates the sum of the function's values at the divisors of a specific integer $ n $ We explore the Mobius inversion of this function, expressing the values of $ f $ in relation to its summatory function The chapter concludes by examining integers with unique properties and demonstrating some of their characteristics.

70 CHAPTER 4 MULTIPLICATIVE NUMBER THEORETIC FUNCTIONS

Definitions and Properties

Definition 18 An arithmetic function is a function whose domain of definition is the set N of positive integers.

Definition 19 An arithmetic function f is called multiplicative if f (ab) = f (a)f (b) for all a, b ∈ N such that (a, b) = 1.

Definition 20 An arithmetic function f is called completely multiplicative if f (ab) = f (a)f (b) (4.1) for all positive integers a, b.

Example 33 The function f (a) = 1 where k is a completely multiplicative function since f (ab) = 1 = f (a)f

Notice also that a completely multiplicative function is a multiplicative function but not otherwise.

We now prove a theorem about multiplicative functions We will be interested in studying the properties of multiplicative functions rather than the completely multiplicative ones.

Theorem 34 Given a multiplicative function f Let n = Q s p a k be the prime factorization of n Then s f (n) = Y f (p a k ). k=1

Proof We prove this theorem by induction on the number of primes in the factorization of n Suppose that n = p a 1 Thus the result follow easily.

4.1 DEFINITIONS AND PROPERTIES 71 we have s f (n) = Y f (p a k ).

So we have to prove that if k=1 s+1 n = Y p a k , then k=1 s+1 f (n) = Y f (p a k ).

Notice that for k=1 s+1 n = Y p a k , we have (Q s k=1 p a k , p a s+1 ) = 1 Thus we have get k=1 k s+1 s+1 s f (n) = f (Y p a k ) = f (Y p a k )f (p a s+1 ) k k=1 which by the inductive step gives k k=1 s+1 s+1 s+1 f (Y p a k ) = f (n) = Y f (p a k ). k k=1 k k=1

To evaluate a multiplicative function at an integer, it is sufficient to know the function's values at the primes found in the integer's prime factorization.

We now define summatory functions which represents the sum of the values of a given function at the divisors of a given number.

Definition 21 Let f be an arithmetic function Define

Then F is called the summatory function of f

This function determines the sum of the values of the arithmetic function at the divisors of a given integer.

Example 34 If f (n) is an arithmetic function, then

Theorem 35 If f is a multiplicative function, then the summatory function of f denoted by F (n) = P d n f (d) is also multiplicative.

Proof We have to prove that F (mn) = F (m)F (n) whenever (m, n) = 1 We have

According to Lemma 6, every divisor of the product mn can be uniquely expressed as a product of relatively prime divisors d1 of m and d2 of n Furthermore, the multiplication of any two divisors from m and n results in a divisor of mn.

Notice that since f is multiplicative, we have f (d1d2)

1 Determine whether the arithmetic functions f (n) = n! and g(n) = n/2 are completely multiplicative or not.

Multiplicative Number Theoretic Functions

The Euler φ-Function

The Euler φ-function quantifies the count of integers that are less than a specified integer and are relatively prime to it Initially, we will evaluate the phi-function for prime numbers and their powers.

Theorem 36 If p is prime, then φ(p) = p − 1 Conversely, if p is an integer such that φ(p) = p − 1, then p is prime.

To prove that a number p is prime, we start by noting that all positive integers less than p are relatively prime to p If p is not prime, it must either be 1 or a composite number In the case of p being 1, we find that φ(p) = p - 1 For composite numbers, p has a positive divisor, leading to the conclusion that φ(p) = p - 1 as well This creates a contradiction, confirming that p must indeed be a prime number.

We now find the value of φ at prime powers.

Theorem 37 Let p be a prime and m a positive integer, then φ(p m ) = p m − p m−1 Proof Note that all integers that are relatively prime to p m and that are less than p m are those that are not multiple of p Those integers are p, 2p, 3p, , p m−1 p.

There are p m−1 of those integers that are not relatively prime to p m and that are less than p m Thus φ(p m ) = p m − p m−1 j j j j p p p

We now prove that φ is a multiplicative function.

Theorem 38 Let m and n be two relatively prime positive integers Then φ(mn) φ(m)φ(n).

Let φ(m) be represented as s, with k₁, k₂, , kₛ forming a reduced residue system modulo m Similarly, let φ(n) be denoted as t, with k₀₁, k₂₀, , kₜ₀ as a reduced residue system modulo n It is important to note that if x is part of a reduced residue system modulo mn, then certain properties regarding the residues hold true.

The congruences x ≡ ki (mod m) and x ≡ k0 (mod n) imply that x belongs to a reduced residue system modulo mn, given that (x, mn) = 1 To derive a reduced residue system modulo mn, one must identify all x values congruent to ki and k0 under moduli m and n, respectively According to the Chinese Remainder Theorem, the equations x ≡ ki (mod m) and x ≡ k0 (mod n) possess a unique solution, meaning that varying i and j will produce distinct results Consequently, it follows that φ(mn) = st.

We now derive a formula for φ(n).

Theorem 39 Let n = p 1 a 1 p 2 a 2 p s a s be the prime factorization of n Then

Proof By Theorem 37, we can see that for all 1 ≤ i ≤ k φ(p a i ) = p a i − p a i −1 = p a i 1 − 1

5 Theorem 40 Let n be a positive integer greater than 2 Then φ(n) is even.

Proof Let n = p 1 2 a 1 p a 2 p k a k Since φ is multiplicative, then k φ(n) = Y φ(p a j ). j=1

Thus by Theorem 39, we have φ(p a j j ) = pj a j −1−1 (pj − 1).

We see then φ(p a j )is even if p is an odd prime Notice also that if pj = 2, then it follows that φ(p a j ) is even Hence φ(n) is even.

Theorem 41 Let n be a positive integer Then

To analyze the integers from 1 to n, we categorize them into classes based on their greatest common divisor (gcd) with n An integer m belongs to class C_d if its gcd with n equals d The count of integers in class C_d corresponds to the number of positive integers up to n/d that are relatively prime to n/d, represented by φ(n/d) Consequently, the total number of integers can be expressed as the sum of φ(n/d) for all divisors d of n, leading to the conclusion that n equals the sum of φ(n/d) over all divisors d of n.

As d runs over all divisors of n, so does n/d Hence n = X φ(n/d) = X φ(d). d|n d|n

The Sum-of-Divisors Function

The sum of divisors function, denoted by σ(n), is the sum of all positive divisors of n.

Note that we can express σ(n) as σ(n) = P d n d.

We now prove that σ(n) is a multiplicative function.

Theorem 42 The sum of divisors function σ(n) is multiplicative.

Proof We have proved in Theorem 35 that the summatory function is multiplicative once f is multiplicative Thus let f (n) = n and notice that f (n) is multiplicative As a result, σ(n) is multiplicative.

Once we found out that σ(n) is multiplicative, it remains to evaluate σ(n) at powers of primes and hence we can derive a formula for its values at any positive integer.

Theorem 43 Let p be a prime and let n = p 1 2 a 1 p a 2 p t a t be a positive integer Then σ(p a ) = p a+1 − 1

Proof Notice that the divisors of p a are 1, p, p 2 , , p a Thus σ(p a ) = 1 + p + p 2 + + p a p a+1− 1

. p − 1 where the above sum is the sum of the terms of a geometric progression.

Now since σ(n) is multiplicative, we have σ(n) = σ(p a 1 )σ(p a 2 ) σ(p a t ) p a 1 +1 a 2 +1 a t +1

The Number-of-Divisors Function

The number of divisors function, denoted by τ (n), is the sum of all positive divisors of n.

We can also express τ (n) as τ (n) = P d n 1.

We can also prove that τ (n) is a multiplicative function.

Theorem 44 The number of divisors function τ (n) is multiplicative Proof By Theorem 36, with f (n) = 1, τ (n) is multiplicative.

We also find a formula that evaluates τ (n) for any integer n.

Theorem 45 Let p be a prime and let n = p a 1 p a 2 p a t be a positive integer

Proof The divisors of p a as mentioned before are 1, p, p 2 , , p a Thus τ (p a ) = a + 1 Now since τ (n) is multiplicative, we have τ (n) = τ (p a 1 )τ (p a 2 ) τ (p a t )

The Mobius Function and the Mobius Inversion Formula

3 Find all positive integers n such that φ(n) = 6.

4 Show that if n is a positive integer, then φ(2n) = φ(n) if n is odd.

5 Show that if n is a positive integer, then φ(2n) = 2φ(n) if n is even.

6 Show that if n is an odd integer, then φ(4n) = 2φ(n).

7 Find the sum of positive integer divisors and the number of positive integer divisors of 35

8 Find the sum of positive integer divisors and the number of positive integer divisors of 2 5 3 4 5 3 7 3 13.

9 Which positive integers have an odd number of positive divisors.

10 Which positive integers have exactly two positive divisors.

4.3 The Mobius Function and the Mobius Inversion

The Mobius function is defined to analyze integers based on their prime factorization Subsequently, we explore the Mobius inversion formula, which allows us to express the values of a function f at a specific integer through its summatory function.

(−1) t if n = p p p where the p are distinct primes;1 2 t i

Note that if n is divisible by a power of a prime higher than one then à(n) 0 In connection with the above definition, we have the following

Definition 23 An integer n is said to be square-free, if no square divides it, i.e. if there does not exist an integer k such that k 2 | n.

It is immediate (prove as exercise) that the prime-number factorization of a square-free integer contains only distinct primes.

We now prove that à(n) is a multiplicative function.

Theorem 46 The Mobius function à(n) is multiplicative.

Proof Let m and n be two relatively prime integers We have to prove that à(mn) = à(m)à(n).

When both m and n equal 1, the equality is satisfied Furthermore, if m is greater than 1, the equality remains evident In cases where either m or n is divisible by a prime power greater than 1, the result is that à(mn) equals 0, which can be expressed as à(m) multiplied by à(n).

To demonstrate that m and n are square-free integers, we express m as the product of distinct primes $ m = p_1 p_2 \ldots p_s $ and n as $ n = q_1 q_2 \ldots q_t $ Since the greatest common divisor (gcd) of m and n is 1, it follows that m and n share no common prime factors Consequently, the values of the Möbius function are given by $ \mu(m) = (-1)^s $, $ \mu(n) = (-1)^t $, and $ \mu(mn) = (-1)^{s+t} $.

In the following theorem, we prove that the summatory function of the Mobius function takes only the values 0 or 1.

4.3 THE MOBIUS FUNCTION AND THE MOBIUS INVERSION FORMULA81

Proof For n = 1, we have F (1) = à(1) = 1 Let us now find à(p k ) for any integer k > 0 Notice that

Thus by Theorem 36, for any integer n = p 1 2 a 1 p a 2 p t a t > 1 we have,

We now define the Mobius inversion formula The Mobius inversion formula expresses the values of f in terms of its summatory function of f

Theorem 48 Suppose that f is an arithmetic function and suppose that F is its summatory function, then for all positive integers n we have f (n) = X à(d)F (n/d). d|n

Notice that P d (n/e) à(d) = 0 unless n/e = 1 and thus e = n Consequently we get X f (e) X à(d) = f (n).1 = f (n). e|n d|(n/d)

Example 42 A good example of a Mobius inversion formula would be the inversion of σ(n) and τ (n) These two functions are the summatory functions of f (n) = n and f (n) = 1 respectively Thus we get n = X à(n/d)σ(d) d|n and

2 Find the value of à(n) for each integer n with 100 ≤ n ≤ 110.

3 Use the Mobius inversion formula and the identity n = P d n φ(n/d) to show that φ(p t ) = p t − p t−1 where p is a prime and t is a positive integer.

Perfect, Mersenne, and Fermat Numbers

Integers with certain properties were studied extensively over the centuries We present some examples of such integers and prove theorems related to these integers and their properties.

We start by defining perfect numbers.

Definition 24 A positive integer n is called a perfect number if σ(n) = 2n.

4.4 PERFECT, MERSENNE, AND FERMAT NUMBERS 83

In other words, a perfect number is a positive integer which is the sum of its proper divisors.

Example 43 The first perfect number is 6, since σ(6) = 12 You can also view this as 6 = 1 + 2 + 3 The second perfect number is 28, since σ(28) = 56 or

The following theorem tells us which even positive integers are perfect.

Theorem 49 The positive integer n is an even perfect number if and only if n = 2 l−1 (2 l − 1), where l is an integer such that l ≥ 2 and 2 l − 1 is prime.

Proof We show first that if n = 2 l−1 (2 l − 1) where l is an integer such that l

≥ 2 and 2 l − 1 is prime then n is perfect Notice that 2 l − 1 is odd and thus

(2 l−1 , 2 l − 1) = 1 Also, notice that σ is a multiplicative function and thus σ(n) = σ(2 l−1 )σ(2 l − 1).

Notice that σ(2 l−1 ) = 2 l − 1 and since 2 l − 1 is prime we get σ(2 l − 1) = 2 l Thus σ(n) = 2n.

We now prove the converse Suppose that n is a perfect number Let n = 2 r s, where r and s are positive integers and s is odd Since (2 r , s) = 1, we get σ(n) = σ(2 r )σ(s) = (2 r+1 − 1)σ(s).

Since n is perfect, we get

Notice now that (2 r+1 − 1, 2 r+1 ) = 1 and thus 2 r+1 | σ(s) Therefore there exists an integer q such that σ(s) = 2 r+1 q As a result, we have

84 CHAPTER 4 MULTIPLICATIVE NUMBER THEORETIC FUNCTIONS and thus we get

So we get that q | s We add q to both sides of the above equation and we get s + q = (2 r+1 − 1)q + q = 2 r+1 q = σ(s).

We have to show now that q = 1 Notice that if q = 1, then s will have three divisors and thus σ(s) ≥ 1 + s + q Hence q = 1 and as a result s = 2 r+1 −

1 Also notice that σ(s) = s + 1 This shows that s is prime since the only divisors of s are 1 and s As a result, where (2 r+1 − 1) is prime. n = 2 r (2 r+1 − 1),

Theorem 50 reveals that identifying even perfect numbers requires finding primes of the form 2^l - 1 The existence of odd perfect numbers remains an open question in mathematics.

Theorem 50 If 2 l − 1 is prime where l is a positive integer, then l must be prime Proof Suppose that l is composite, that is l = rs where 1 < r < m and 1

< s < m Thus after factoring, we get that

Notice that the two factors above are both greater than 1 Thus 2 m −1 is not prime This is a contradiction.

The above theorem motivates the definition of interesting numbers called Mersenne numbers.

Definition 25 Let l be a positive integer An integer of the form M l = 2 l − 1 is called the lth Mersenne number; if l is prime then M l = 2 l − 1 is called the lth Mersenne prime. n

Example 44 M 3 = 2 3 − 1 = 7 is the third Mersenne prime.

We prove a theorem that help decide whether Mersenne numbers are prime.

Theorem 51 Divisors of Mp = 2 p − 1 for prime p is of the form 2mp + 1, where m is a positive integer.

Proof Let p 1 be a prime dividing M p = 2 p − 1 By Fermat’s theorem, we know that p 1 | (2 p 1 −1 − 1) Also, it is easy to see that

Since $ p $ is a common divisor of $ 2p - 1 $ and $ 2p_1 - 1 $, they are not relatively prime, leading to the conclusion that $ (p, p_1 - 1) = p $ Therefore, $ p $ divides $ (p_1 - 1) $, which implies the existence of a positive integer $ k $ such that $ p_1 - 1 = kp $ Given that $ p_1 $ is odd, $ k $ must be even, allowing us to express $ k $ as $ 2m $ Consequently, we can represent $ p_1 $ as $ p_1 = kp + 1 = 2mp + 1 $.

Because any divisor of Mp is a product of prime divisors of Mp , each prime divisor of M p is of the form 2mp + 1 and the result follows.

Example 45 M 23 = 2 23 − 1 is divisible by 47 = 46k + 1 We know this by trial and error and thus looking at all primes of the form 46k + 1 that are less than√

We now define Fermat numbers and prove some theorems about the properties of these numbers.

Definition 26 Integers of the form F n = 2 2 + 1 are called Fermat numbers.

Fermat initially conjectured that certain integers, known as Fermat numbers, are prime; however, this was later disproven For instance, the first few Fermat numbers are F0 = 3, F1 = 5, F2 = 17, F3 = 257, and F4 = 65,537, but F5 is a composite number divisible by 641 This article will explore several theorems related to the properties of Fermat numbers.

Theorem 52 For all positive integers n, we have

Proof We will prove this theorem by induction For n = 1, the above identity is true Suppose now that holds We claim that

Using Theorem 53, we prove that Fermat numbers are relatively prime. Theorem 53 Let s = t be nonnegative integers Then (F s , F t ) = 1.

Proof Assume without loss of generality that s < t Thus by Theorem 52, we have

Assume now that there is a common divisor d of Fs and Ft thus we see that d divides

Thus d = 1 or d = 2 But since Ft is odd for all t We have d = 1 Thus Fs and

1 Find the six smallest even perfect numbers.

2 Find the eighth perfect number.

4 We say n is abundant if σ(n) > 2n Prove that if n = 2 m−1 (2 m − 1) where m is a positive integer such that 2 m − 1 is composite, then n is abundant.

5 Show that there are infinitely many even abundant numbers.

6 Show that there are infinitely many odd abundant numbers.

9 Find all primes of the form 2 2 n + 5 where n is a nonnegative integer.

In this chapter, we discuss the multiplicative structure of the integers modulo n.

This article explores the order of integers modulo n, examining its properties in depth We define primitive roots modulo n and outline the method for identifying whether a given integer is primitive in this context Additionally, we identify all positive integers that possess primitive roots and provide proofs for related findings.

This article explores the concept of quadratic residues, outlining their fundamental properties It introduces the Legendre symbol and examines its essential characteristics, followed by a discussion on the law of quadratic reciprocity Furthermore, the article extends the Legendre symbol to the Jacobi symbol and addresses the associated law of reciprocity for the Jacobi symbol.

The order of Integers and Primitive Roots

In this section, we study the order of an integer modulo n, where n is positive.

We also define primitive roots and related results Euler’s theorem in Chapter 4 states that if a positive integer a is relatively prime to n, then a φ(n) ≡ 1(mod n). Thus

90 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC RESIDUES by the well ordering principle, there is a least positive integer x that satisfies this congruence a x ≡ 1(mod n).

Definition 1 Let (a, b) = 1 The smallest positive integer x such that a x ≡ 1(mod b) is called the order of a modulo b We denote the order of a modulo b by ord b a.

Example 46 ord 7 2 = 3 since 2 3 ≡ 1(mod 7) while 2 1 ≡ 2(mod 7) and 2 2 ≡ 4(mod 7).

To find all integers x such that a x ≡ 1(mod b), we need the following theorem.

Theorem 54 If (a, b) = 1 with b > 0, then the positive integer x is a solution of the congruence a x ≡ 1(mod b) if and only if ordb a | x.

Proof Having ord b a | x, then we have that x = k.ord b a for some positive integer k Thus a x = a kord b a = (a ord b a ) k ≡ 1(mod b).

Now if a x ≡ 1(mod b), we use the division algorithm to write x = qordb a + r, 0 ≤ r < ordb a.

Thus we see that a x ≡ a qord b a+r ≡ (a ord b a ) q a r ≡ a r (mod b).

Now since a x ≡ 1(mod b),we have a r ≡ 1(mod b) Since ordb a, we get r = 0. Thus x = q.ord b a and hence ord b a | x.

Example 47 Since ord 7 2 = 3, then 2 15 ≡ 1(mod 7) while 10 is not a solution for 2 x ≡ 1(mod 7).

Theorem 55 If (a, b) = 1 with b > 0, then a i ≡ a j (mod b)

5.1 THE ORDER OF INTEGERS AND PRIMITIVE ROOTS 91 where i and j are nonnegative integers, if and only if i ≡ j(mod ordb a)

Proof Suppose that i ≡ j(mod ord b a) and 0 ≤ j ≤ i.

Then we have i − j = k.ordb a, where k is a positive integer Hence a i = a j+k.ord b a = a j (a ord b a ) k ≡ a j (mod b).

Assume now that a i ≡ a j (mod b) with i ≥ j Thus we have a i ≡ a j a i−j ≡ a j (mod b)

Since (a, b) = 1, we have (a j , b) = 1 and thus by Theorem 22, we get a i−j ≡ 1(mod b).

By theorem 54, we get that ord b a | (i − j) and hence i ≡ j(mod b).

This article introduces primitive roots and explores their properties, focusing on integers with an order modulo another integer equal to φ(b) It includes an exercise demonstrating that if integers a and b are relatively prime, then the order of a modulo b divides φ(b).

Definition 2 If (r, m) = 1 with m > 0 and if ord m r = φ(m) then r is called a primitive root modulo m.

Example 48 Notice that φ(7) = 6 hence 2 is not a primitive root modulo 7 While ord73 = 6 and thus 3 is a primitive root modulo 7.

Theorem 56 If (r, m) = 1 with m > 0 and if r is a primitive root modulo n, then the integers {r 1 , r 2 , r φ(m) } form a reduced residue set modulo m.

92 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC RESIDUES

To demonstrate that the set {r₁, r₂, , rₓ(m)} constitutes a reduced residue system modulo m, we must establish that every pair of elements is relatively prime and that no two elements are congruent modulo m Given that (r, m) = 1, it follows that (rⁿ, m) = 1 for all positive integers n, indicating that all powers of r are relatively prime to m To confirm that no two powers in this set are equivalent modulo m, we start by assuming that rᵢ ≡ rⱼ (mod m).

By Theorem 55, we see that i ≡ j(mod ordm φ(m)).

Theorem 57 If ordm a = t and if u is a positive integer, then ord m (a u ) = t/(t, u).

Because t 1 = t/(t, u), we want to show that ord m (a u ) = t 1 To do this, we will show that (a u ) t 1 ≡ 1(mod m) and that if (a u ) v ≡ 1(mod m), then t1 | v. First note that

Hence by Theorem 54, we have v | t1 Now on the other hand, since

5.1 THE ORDER OF INTEGERS AND PRIMITIVE ROOTS 93

Example 49 We see that ord 7 3 4 = 6/(6, 4) since ord 7 3 = 6.

Corollary 2 Let r be a primitive root modulo m, where m is a positive integer, m > 1 Then r u is a primitive root modulo m if and only if (u, φ(m)) 1 Proof By Theorem 57, we see that ord m r u = ord m r/(u, ord m r) = φ(m)/(u, φ(m)).

Thus ordm r u = φ(m) and r u is a primitive root if and only if (u, φ(m)) = 1.

The above corollary leads to the following theorem

Theorem 58 If the positive integer m has a primitive root, then it has a total of φ(φ(m)) incongruent primitive roots.

In this proof, we consider a primitive root $ r $ modulo $ m $ According to Theorem 56, the set $ \{ r^1, r^2, \ldots, r^{\phi(m)} \} $ constitutes a reduced residue system modulo $ n $ Furthermore, as established by Corollary 1, $ r^u $ is a primitive root modulo $ m $ if and only if $ (u, \phi(m)) = 1 $ Consequently, there are precisely $ \phi(\phi(m)) $ integers $ u $ that are relatively prime to $ \phi(m) $, leading to the conclusion that there are exactly $ \phi(\phi(m)) $ primitive roots modulo $ m $.

3 Show that 5 is a primitive root of 6.

4 Show that if a¯ is an inverse of a modulo n, then ordn a = ordn a¯.

5 Show that if n is a positive integer, and a and b are integers relatively prime to n such that (ord n a, ord n b) = 1, then ord n (ab) = ord n a.ord n b.

6 Show that if a is an integer relatively prime to the positive integer m and ordm a = st, then ordm a t = s.

7 Show that if a and n are relatively prime with n > 0, then ord n a | φ(n).

Primitive Roots for Primes

In this section, we show that every integer has a primitive root To do this we need to introduce polynomial congruence.

Let f (x) be a polynomial with integer coefficients We say that an integer a is a root of f (x) modulo m if f (a) ≡ 0(mod m).

Example 50 Notice that x ≡ 3(mod 11) is a root for f (x) = 2x 2 + x + 1 since f (3) = 22 ≡ 0(mod 11).

Lagrange's theorem for primes, which operates under modulo p, is a key component of the fundamental theorem of algebra This theorem plays a crucial role in demonstrating that every prime number possesses a primitive root.

Theorem 59 Lagrange’s Theorem Let m(x) = b n x n + bn

1x n−1 + + b 1 x + b 0 be a polynomial of degree n, n ≥ 1 with integer coefficients and with leading coefficient b n not divisible by a prime p Then m(x) has at most n distinct incongruent roots modulo p.

Proof Using induction, notice that if n = 1, then we have m(x) = b 1 x + b 0 and p - b 1

A root of m(x) is a solution for b 1 x+b 0 (mod p) Since p - b 1 , then this congruence has exactly one solution by Theorem 26.

Assuming the theorem holds for polynomials of degree n − 1, consider m(x), a polynomial of degree n with integer coefficients whose leading coefficient is not divisible by p If m(x) possesses n + 1 incongruent roots modulo p, denoted as x₀, x₁, , xₙ, it follows that m(xₖ) ≡ 0 (mod p) for each k.

5.2 PRIMITIVE ROOTS FOR PRIMES 95 for 0 ≤ k ≤ n Thus we have m(x) − m(x0) = b n (x n − x n ) + b n

+ x x 0 + + xx 0 + x 0 ) + + b 1 (x − c 0 ) where f (x) is a polynomial of degree n − 1 with leading coefficient bn Notice that since m(x k ) ≡ m(x0 )(mod p), we have m(x k ) − m(x0) = (x k − x0)f (x k ) ≡ 0(mod p).

Thus f (xk ) ≡ 0(mod p) for all 1 ≤ k ≤ n and thus x1 , x2, , xn are roots of f (x) This is a contradiction since we a have a polynomial of degree n − 1 that has n distinct roots.

We now use Lagrange’s Theorem to prove the following result.

Theorem 60 Consider the prime p and let p − 1 = kn for some integer k Then x n − 1 has exactly n incongruent roots modulo p Proof Since p − 1 = kn, we have x p−1 − 1 = (x n − 1)(x n(k−1) + x n(k−2) + + x n +

According to Fermat’s little theorem, the expression $x^{p-1} - 1$ has $p - 1$ incongruent roots modulo $p$ These roots are also associated with the polynomial $f(x)$ or the equation $x^n - 1$ Lagrange’s Theorem indicates that $f(x)$ can have at most $p - n - 1$ roots modulo $p$, which implies that the equation $x^n - 1$ must have at least $n$ roots modulo $p$.

Theorem, since we have that x n − 1 has at most n roots, thus we get that x n − 1 has exactly n incongruent roots modulo p.

We now prove a lemma that gives us how many incongruent integers can have a given order modulo p.

Lemma 11 Let p be a prime and let m be a positive integer such that p − 1 mk for some integer k.

Proof For each positive integer m dividing p −

Notice that if S(m) = 0, then S(m) ≤ φ(m) If S(m) > 0, then there is an integer a of order m modulo p Since ordp a = m, then a, a 2 , a m are incongruent modulo p Also each power of a is a root of x m − 1 modulo p because

The equation $ (a^k)^m \equiv (a^m)^k \equiv 1 \mod p $ holds for all positive integers $ k $ According to Theorem 60, the polynomial $ x^m - 1 $ has exactly $ m $ incongruent roots modulo $ p $, meaning each root corresponds to a power of $ a $ Additionally, Theorem 57 indicates that the powers of $ a^k $ are relevant when $ (k, m) $ is considered.

= 1 have order m There are exactly φ(m) such integers with 1 ≤ k ≤ m and thus if there is one element of order m modulo p, there must be exactly φ(m) such positive integers less than p Hence S(m) ≤ φ(m).

In the following theorem, we determine how many incongruent integers can have a given order modulo p We actually show the existence of primitive roots for prime numbers.

Theorem 61 Every prime number has a primitive root.

To prove the statement, let p represent a prime number and m denote a positive integer where p - 1 can be expressed as mk for some integer k Define F(m) as the count of positive integers with order m modulo p that are less than p Since the order of an integer not divisible by p divides p - 1, it follows that p - 1 can be factored as a sum of terms where each term corresponds to a divisor m of p - 1.

By Theorem 42, we see that p − 1 = X m|p−1 φ(m).

By Lemma 1, F (m) ≤ φ(m) when m | (p − 1) Together with

F (m) = X m|p−1 φ(m) we see that F (m) = φ(m) for each positive divisor m of p − 1 Thus we conclude that F (m) = φ(m) As a result, we see that there are p − 1 incongruent integers of order p − 1 modulo p Thus p has φ(p − 1) primitive roots.

1 Find the incongruent roots modulo 11 of x 2 + 2.

2 Find the incongruent roots modulo 11 of x 4 + x 2 + 1.

3 Find the incongruent roots modulo 13 of x 3 + 12.

4 Find the number of primitive roots of 13 and of 47.

5 Find a complete set of incongruent primitive roots of 13.

8 Let r be a primitive root of p with p ≡ 1(mod 4) Show that −r is also a primitive root.

9 Show that if p is a prime and p ≡ 1(mod 4), then there is an integer x such that x 2 ≡ −1(mod p).

98 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATICRESIDUES

The Existence of Primitive Roots

This section explores the integers that possess primitive roots, beginning with the assertion that every power of an odd prime is guaranteed to have a primitive root We establish this by first demonstrating that every square of an odd prime also has a primitive root.

Theorem 62 If p is an odd prime with primitive root r, then one can have either r or r + p as a primitive root modulo p 2

Proof Notice that since r is a primitive root modulo p, then ord p r = φ(p) = p − 1.

By Theorem 54, we have r m ≡ 1(mod p 2 ). r m ≡ 1(mod p). p − 1 | m.

By Exercise 7 of section 6.1, we also have that m | φ(p 2 ).

Also, φ(p 2 ) = p(p − 1) and thus m either divides p or p − 1 And since p − 1 | m then we have m = p − 1 or m = p(p − 1).

If m = p(p − 1) and ord p 2 r = φ(p 2 ) then r is a primitive root modulo p 2 Other- wise, we have m = p − 1 and thus r p−1 ≡ 1(mod p 2 ).

5.3 THE EXISTENCE OF PRIMITIVE ROOTS 99

Let s = r + p Then s is also a primitive root modulo p Hence, ord p 2 s equals either p−1 or p(p−1) We will show that ordp 2 s = p−1 so that ordp 2 s p(p−1) Note that s p−1 = (r + p) p−1 = r p−1 + (p − 1)r p−2 p + + p p−1

Note also that if then

Thus we have p 2 | s p−1 − (1 − pr p−2 p 2 | (s p−1 − 1), p 2 | pr p−2 p | r p−2 which is impossible because p - r Because ord p 2 s = p − 1, we can conclude that ord p 2 s = p(p − 1) = φ(p 2 ).

Thus, s = r + p is a primitive root of p 2

Example 51 Notice that 7 has 3 as a primitive root Either ord 49 3 = 6 or ord 49 3 = 42 But since 3 6 6≡ 1(mod 49) Hence ord 49 3 = 42 Hence 3 is a primitive root of 49.

We now show that any power of an odd prime has a primitive root.

Theorem 63 Let p be an odd prime Then any power of p is a primitive root. Moreover, if r is a primitive root modulo p 2 , then r is a primitive root modulo p m for all positive integers m.

Proof By Theorem 62, we know that any prime p has a primitive root r which is also a primitive root modulo p 2 , thus p 2 - (r p−1 − 1) (5.1)

We will demonstrate by induction that $ p^m - (r p^{m-2} (p-1) - 1) $ holds true for all integers $ m \geq 2 $ After establishing this congruence, we will also confirm that $ r $ serves as a primitive root modulo $ p^m $ Let $ n = \text{ord}_{p^m} r $; according to Theorem 54, $ n $ divides $ \phi(p^m) $, where $ \phi(p^m) = p^m (p - 1) $ Therefore, $ n $ must also divide $ p^m (p - 1) $ Additionally, since $ p^m $ divides $ (r^n - 1) $, it follows that $ p $ divides $ (r^n - 1) $.

Since φ(p) = p − 1, we see that by Theorem 54, we have n = l(p − 1) also n | p m−1 (p − 1), we have that n = p s (p − 1), where 0 ≤ s ≤ m − 1 If n = p s (p − 1) with s ≤ m − 2, then p k | r p m−2 (p−1) − 1, which is a contradiction Hence ordp m r = φ(p m ).

We prove now (7.5) by induction Assume that our assertion is true for all m ≥ 2 Then p m - (r p m−2 (p−1) − 1).

Because (r, p) = 1, we see that (r, p m−1 ) = 1 We also know from Euler’s theorem that p m−1 | (r p m−2 (p−1) − 1).

Thus there exists an integer k such that r p m−2 (p−1) = 1 + kp m−1 where p - k because r p m−2 (p−1) 6≡ 1(mod p m ) Thus we have now r p m−1 (p−1) = (1 + kp m−1 ) p

Example 52 Since 3 is a primitive root of 7, then 3 is a primitive root for 7 k for all positive integers k.

The theorem demonstrates that only the powers of 2, specifically 2 and 4, possess a primitive root This conclusion arises from the fact that for any odd integer m, the order of k modulo m is equal to φ(2^k), which is derived from the condition that 2^k divides (aφ(2)/2 - 1).

Theorem 64 If m is an odd integer, and if k ≥ 3 is an integer, then m 2 k−2 ≡ 1(mod 2 k ).

Proof We prove the result by induction If m is an odd integer, then m = 2n + 1 for some integer n Hence, m 2 = 4n 2 + 4n + 1 = 4n(n + 1) +

102 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC RESIDUES Then there is an integer q such that m 2 k−2 = 1 + q.2 k Thus squaring both sides, we get m 2 k−1 = 1 + q.2 k+1 + q 2 2 2k

Note now that 2 and 4 have primitive roots 1 and 3 respectively We now list the set of integers that do not have primitive roots.

Theorem 65 If m is not p a or 2p a , then m does not have a primitive root.

In this proof, let $ m = p^{s_1} p^{s_2} \ldots p^{s_i} $ If $ m $ possesses a primitive root $ r $, then $ r $ and $ m $ are relatively prime, and the order of $ r $ modulo $ m $ is equal to $ \phi(m) $ Additionally, it follows that $ (r, p^s) = 1 $, where $ p^s $ represents the prime factors of $ m $ According to Euler's theorem, it can be concluded that $ p^s $ divides $ (r^{\phi(p)} - 1) $.

1 2 i r L ≡ 1(mod p s k ) for all 1 ≤ k ≤ m Thus using the Chinese Remainder Theorem, we get m | (r L − 1), which leads to ord m r = φ(m) ≤ L Now because φ(m) = φ(p s 1 1 )φ(p s 2 2 ) φ(p s n n ) ≤ [φ(p 1 s 1 ), φ(p 2 s 2 ), , φ(p n s n )]. s s s s

The inequality is valid only when φ(p s 1 1 ), φ(p s 2 2 ), , φ(p n s n ) are relatively prime According to Theorem 41, these values are not relatively prime unless m equals p s or m equals 2p s, where p represents an odd prime and t is any positive integer.

We now show that all integers of the form m = 2p s have primitive roots.

Theorem 66 states that for a prime number p = 2 and a positive integer s, the expression 2p^s possesses a primitive root Specifically, if r is an odd primitive root modulo p^s, it will also serve as a primitive root modulo 2p^s Conversely, if r is even, then r + p^s qualifies as a primitive root modulo 2p^s.

Proof If r is a primitive root modulo p s , then p s | (r φ(p ) − 1) and no positive exponent smaller than φ(p s ) has this property Note also that φ(2p s ) = φ(p s ), so that

Thus by Theorem 56, we get p s | (r φ(2p ) − 1).

It is crucial to understand that no smaller power of r can be congruent to 1 modulo 2p^s If such a power were congruent to 1 modulo p^s, it would contradict the fact that r is a primitive root of p^s Therefore, we conclude that r is indeed a primitive root modulo 2p^s.

104 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC RESIDUES While, if r is even, then r + p s is odd Hence

As a result, we see that 2p s | ((r + p s ) φ(2p ) − 1) and since for no smaller power of r + p s is congruent to 1 modulo 2p s , we see that r + p s is a primitive root modulo 2p s

As a result, by Theorem 63, Theorem 65 and Theorem 66, we see that

Theorem 67 The positive integer m has a primitive root if and only if n = 2, 4, p s or 2p s for prime p = 2 and s is a positive integer.

1 Which of the following integers 4, 12, 28, 36, 125 have a primitive root.

3 Find all primitive roots modulo 22.

4 Show that there are the same number of primitive roots modulo 2p s as there are modulo p s , where p is an odd prime and s is a positive integer.

5 Find all primitive roots modulo 25.

6 Show that the integer n has a primitive root if and only if the only solutions of the congruence x 2 ≡ 1(modn) are x ≡ ±1(mod n).

Introduction to Quadratic Residues and Nonresidues

5.4 Introduction to Quadratic Residues and Non- residues

The question that we need to answer in this section is the following If p is an odd prime and a is an integer relatively prime to p Is a a perfect square modulo p.

A positive integer m defines a quadratic residue, where an integer a is considered a quadratic residue of m if it satisfies two conditions: (a, m) = 1 and the congruence x² ≡ a (mod m) is solvable Conversely, if the congruence x² ≡ a (mod m) does not have a solution, then a is classified as a quadratic nonresidue of m.

Example 53 Notice that 1 2 = 6 2 ≡ 1(mod 7), 3 2 = 4 2 ≡ 2(mod 7) and

2 2 = 5 2 ≡ 4(mod 7) Thus 1, 2, 4 are quadratic residues modulo 7 while 3, 5, 6 are quadratic nonresidues modulo 7.

Lemma 12 Let p = 2 be a prime number and a is an integer such that p - a. Then either a is quadratic nonresidue modulo p or x 2 ≡ a(mod p) has exactly two incongruent solutions modulo p.

Proof If x 2 ≡ a(mod p) has a solution, say x = x 0 , then −x 0 is a solution as well Notice that −x 0 6≡ x 0 (mod p) because then p | 2x 0 and hence p - x 0

We now show that there are no more than two incongruent solutions Assume that x = x 0 and x = x 00 are both solutions of x 2 ≡ a(mod p) Then we have

The following theorem determines the number of integers that are quadratic residues modulo an odd prime.

Theorem 68 If p = 2 is a prime, then there are exactly (p − 1)/2 quadratic residues modulo p and (p − 1)/2 quadratic nonresidues modulo p in the set of integers 1, 2 , p − 1.

Proof To find all the quadratic residues of p among all the integers 1, 2, , p −

In the study of quadratic residues modulo a prime number $ p $, we analyze the least positive residues of the squares of integers from 1 to $ p-1 $ Each of the $ p-1 $ congruences can yield either no solution or two distinct solutions, leading to the conclusion that there are precisely $ (p-1)/2 $ quadratic residues among the integers 1 through $ p-1 $ Consequently, the remaining integers in this range represent $ (p-1)/2 $ quadratic nonresidues of $ p $.

1 Find all the quadratic residues of 3.

2 Find all the quadratic residues of 13.

3 find all the quadratic residues of 18.

4 Show that if p is prime and p ≥ 7, then there are always two consecutive quadratic residues of p Hint: Show that at least one of 2, 5 or 10 is a quadratic residue of p.

5 Show that if p is prime and p ≥ 7, then there are always two quadratic residues of p that differ by 3.

Legendre Symbol

In this section, we define Legendre symbol which is a notation associated to quadratic residues and prove related theorems. p p p a

Definition 4 Let p = 2 be a prime and a be an integer such that p - a The Legendre symbol a is defined by a p (

1 if a is a quadratic residue of p

−1 if a is a quadratic nonresidue of p.

Example 54 Notice that using the previous example, we see that

In the following theorem, we present a way to determine wether an integer is a quadratic residue of a prime.

Theorem 69 Euler’s Criterion Let p = 2 be a prime and let a be a positive integer such that p - a Then a p φ(p)/2

= 1 Then the congruence x 2 ≡ a(mod p) has a solution say x = x 0 According to Fermat’s theorem, we see that a φ(p)/2 = ((x 0 ) 2 ) φ(p)/2 ≡ 1(mod p).

If $ x^2 \equiv a \mod p $ is not solvable when $ a = -1 $, then according to Theorem 26, for every integer $ k $ coprime to $ p $, there exists an integer $ l $ such that $ kl \equiv a \mod p $ Since the equation $ x^2 \equiv a \mod p $ has no solutions, we can conclude that $ i = j $ Consequently, the integers from 1 to $ p - 1 $ can be paired into $ (p - 1)/2 $ pairs, each of which has a product equal to $ a $ By multiplying these pairs together, we derive a significant result.

Using Wilson’s Theorem, we get a = −1 ≡ a (p−1)/2 (mod p).

We now prove some properties of Legendre symbol.

Theorem 70 Let p = 2 be a prime Let a and b be integers such that p - a, p - b and p | (a − b) then a p b p Proof Since p | (a − b), then x 2 ≡ a(mod p) has a solution if and only if x 2 ≡ b(mod p) has a solution Hence a p b p

Theorem 71 Let p = 2 be a prime Let a and b be integers such that p - a, p - b then a b ab and p

By Euler’s criterion, we have a p b p φ(p)/2

We now show when is −1 a quadratic residue of a prime p

Proof By Euler’s criterion, we know that a

If 4 | (p − 1), then p = 4m + 1 for some integer m and thus we get

(−1) φ(p)/2 = (−1) 2m = 1. and if 4 | (p − 3), then p = 4m + 3 for some integer m and we also get

We now determine when 2 is a quadratic residue of a prime p.

Theorem 72 For every odd prime p we have

Proof Consider the following (p − 1)/2 congruences p − 1 ≡ 1(−1) 1 (mod p)

110 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC RESIDUES where r is either p − (p − 1)/2 or (p − 1)/2 Multiplying all these equations we get,

! 6≡ 0(mod p) and thus we get

Note also that by Euler’s criterion, we get

2 φ(p)/2 p (mod p), and since each member is 1 or -1 the two members are equal.

We now present an important lemma that determines whether an integer is a quadratic residue of a prime or not.

Lemma 13 Gauss’s Lemma Let p = 2 be a prime and a a relatively prime integer to p If k counts the number of least positive residues of the integers a, 2a, , ((p − 1)/2)a that are greater than p/2, then a

In the given proof, we identify integers m1, m2, , ms that are greater than p/2 from the least positive residues of the integers a, 2a, , ((p − 1)/2)a, while n1, n2, , nt represent those less than p/2 We demonstrate that the integers p − m1, p − m2, , p − mk, and p − n1, p − n2, , p − nt correspond exactly to the required set of integers.

5.5 LEGENDRE SYMBOL 111 in the same order.

We will demonstrate that no two integers from the set are congruent modulo p, as there are precisely (p − 1)/2 distinct positive integers, all of which are less than or equal to (p − 1)/2 It follows that for all indices i and j, mi is not congruent to mj modulo p, and ni is not congruent to nj modulo p If any congruences were to fail, it would imply that r is congruent to s modulo p, given that ra is congruent to sa modulo p Additionally, any integer of the form p − mi could be congruent to any of the ni's If such a congruence exists, it leads to the conclusion that ra is congruent to p − sa modulo p, which simplifies to ra being congruent to −sa modulo p Since p does not divide a, this scenario implies that r is congruent to −s modulo p, which is a contradiction Therefore, we conclude that k is less than p − 1.

2 !(mod p), Simplifying, we get m 1 m 2 (p − m k )n 1 n 2 n t ≡ a.2a ((p − 1)/2) = a (p−1)/2 ((p − 1)/2)!(mod p).

As a result, we have that a (p−1)/2 ((p − 1)/2)! ≡ ((p − 1)/2)!(mod p) Note that since (p, ((p − 1)/2)!) = 1, we get

Using Euler’s criterion, the result follows.

Example 56 To find 5 using Gauss’s lemma, we calculate

1 Find all quadratic residues of 3

2 Find all quadratic residues of 19.

3 Find the value of Legendre symbol j for j = 1, 2, 3, 4, 5, 6.

4 Evaluate the Legendre symbol 7 by using Euler’s criterion.

5 Let a and b be integers not divisible by p Show that either one or all three of the integers a, b and ab are quadratic residues of p.

6 Let p be a prime and a be a quadratic residue of p Show that if p ≡ 1(mod 4), then −a is also a quadratic residue of p, whereas if p ≡ 3(mod

4), then −a is a quadratic nonresidue of p.

7 Show that if p is an odd prime and a is an integer not divisible by p then a 2

The Law of Quadratic Reciprocity

In this article, we explore the relationship between two odd prime numbers, p and q, focusing on whether p is a quadratic residue of q, given that we know the status of q as a quadratic residue of p We will first introduce Eisenstein's Lemma, which is essential for proving the law of quadratic reciprocity This lemma establishes a connection between the Legendre symbol and the counting of lattice points within a triangle, setting the stage for our discussion on quadratic residues.

5.6 THE LAW OF QUADRATIC RECIPROCITY 113

Lemma 14 If p = 2 is a prime and a is an odd integer such that p - a, then a p

To prove the statement, we analyze the least positive residues of the integers a, 2a, , ((p − 1)/2)a We define two sets of integers: m₁, m₂, , mₛ, where each mᵢ is greater than p/2, and n₁, n₂, , nₜ, where each nᵢ is less than p/2 Applying the division algorithm, we express ia as p[ia/p] + r, with r being one of the residues from either set By summing these (p − 1)/2 equations, we derive the necessary results.

As in the proof of Gauss’s Lemma, we see that p − m 1 , p − m 2 , , p − m s , p − n 1 , p − n 2 , , p − n t are precisely the integers 1, 2, , (p − 1)/2, in the same order Now we obtain

(p − mi ) + X ni = ps − X mi + X ni (5.4) i=1 i=1 i=1 i=1 i=1

We subtract (5.4) from (5.3) to get

Now since we are taking the following as exponents for −1, it suffice to look at them modulo 2 Thus

Using Gauss’s lemma, we get a (p−1)/2

Theorem 73 The Law of Quadratic Reciprocity Let p and q be distinct odd primes Then p q = (−1) q p p− 1 q − 1

Proof We consider now the pairs of integers also known as lattice points (x, y) with

The number of such pairs is p− 1 q − 1 We divide these pairs into two groups

2 2 de- pending on the sizes of qx and py Note that qx = py for all pairs because p and q are distinct primes.

We now count the pairs of integers (x, y) with

Note that these pairs are precisely those where

For each fixed value of x with 1 ≤ x ≤ (p − 1)/2, there are [qx/p] integers satisfying 1 ≤ y ≤ qx/p Consequently, the total number of pairs with are

1 ≤ x ≤ (p − 1)/2, 1 ≤ y ≤ qx/p, and qx > py is (p−1)/2X i=1

5.6 THE LAW OF QUADRATIC RECIPROCITY 115

Consider now the pair of integers (x, y) with

Similarly, we find that the total number of such pairs of integers is

Adding the numbers of pairs in these classes, we see that

[pi/q] = , i=1 i=1 2 2 and hence using Lemma 14, we get that p p p− 1 q − 1 q q = (−1) 2 2

3 Using the law of quadratic reciprocity, show that if p is an odd prime, then

4 Show that if p is an odd prime, then

5 Find a congruence describing all primes for which 5 is a quadratic residue. m b c i i

116 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC

Jacobi Symbol

The Jacobi symbol is introduced as a generalization of the Legendre symbol, which is specifically defined for prime numbers Unlike the Legendre symbol, the Jacobi symbol extends this concept to any odd integers, utilizing the Legendre symbol in its formulation.

Definition 27 Let n be an odd positive integer with prime factorization n p1 a 1 p 2 a 2 p a m m and let a be an integer relatively prime to n, then a

Example 57 Notice that from the prime factorization of 45, we get that

We now prove some properties for Jacobi symbol that are similar to the properties of Legendre symbol.

Theorem 74 Let n be an odd positive integer and let a and b be integers such that(a, n) = 1 and (b, n) = 1 Then

Proof Proof of 1: Note that if p is in the prime factorization of n, then we have that p | (a − b) Hence by Theorem 70, we get that a p b p m m i i

Proof of 2: Note that by Theorem 71, we have ab a b for any prime p p p p appearing in the prime factorization of n As a result, we have ab n Ym i=1 m ab c i p i c i m c i

In the following theorem, we determine

Theorem 75 Let n be an odd positive integer Then

2 n = (−1) (n −1)/8 Proof Proof of 1: If p is in the prime factorization of n, then by Corollary 3, we see that

118 CHAPTER 5 PRIMITIVE ROOTS AND QUADRATIC RESIDUES Notice that since p i − 1 is even, we have i = (1 + (pi − 1)) ≡ 1 + ci (pi − 1)(mod 4) and hence we get m m n = Y p c i ≡ 1 + X ci (pi − 1)(mod 4).

Proof of 2: If p is a prime, then by Theorem 72 we have

2 P m 2 n = (−1) i=1 c i (p i −1)/8 Because 8 | p 2 − 1, we see similarly that

(1 + (p i 2 − 1)) c i ≡ 1 + ci (p i 2 − 1)(mod 64) and thus m n 2 ≡ 1 + X c i (p 2 − 1)(mod 64), which implies that i=1 m

We now show that the reciprocity law holds for Jacobi symbol.

Theorem 76 Let (a, b) = 1 be odd positive integers Then b a

Proof Notice that since a = Q m p c i and b = Q n q d i we get b a j=1 i n m p i=1 q i c j d i

By the law of quadratic reciprocity, we get b a

As in the proof of part 1 of Theorem 75, we see that and

3 For which positive integers n that are relatively prime to 15 does the Jacobi symbol 15 equal 1?

4 Let n be an odd square free positive integer Show that there is an integer a such that (a, n) = 1 and a

This chapter introduces continued fractions, explores their fundamental properties, and demonstrates their application in solving various mathematical problems Continued fractions are a natural mathematical construct that arise in diverse areas of Mathematics, often in surprising contexts Notably, the Dutch mathematician and astronomer Christian Huygens (1629-1695) contributed significantly to their study.

In 1687, the mathematician (1695) made the first practical use of "anthyphaeiretic ratios," known today as continued fractions, by demonstrating how convergents can provide optimal rational approximations for gear ratios, which helped him select gears with ideal tooth counts for his mechanical planetarium This concept piqued the interest of notable mathematicians such as Euler, Jacobi, Cauchy, and Gauss, leading to ongoing research in continued fractions, which remain relevant in contemporary mathematics Among the current contributors to this field, Australian mathematician A.J van der Poorten stands out as a leading figure.

122 CHAPTER 6 INTRODUCTION TO CONTINUED FRACTIONS

Basic Notations

In general, a (simple) continued fraction is an expression of the form a0 + 1 a 1 + a 1 ,

The expression 2 + a0 + a1 + a2 + involves independent variables represented by letters a0, a1, a2, etc., which can be interpreted in various ways, such as real or complex numbers or functions This expression is meaningful when the number of terms is finite; however, it may lack significance when there are infinitely many terms In this section, we will focus on the simplest classical context.

The letters a 1 , a 2 , denote positive integers The letter a 0 denotes an integer.

The following standard notation is very convenient.

[a 0 ; a 1 , a 2 , , a n ] = a 0 + if the number of terms is finite, and a 1 + a2

2 + for an infinite number of terms.

To ensure the formula is meaningful, a certain amount of work is required when dealing with an infinite number of terms Conversely, the formula is valid and applicable when considering a finite number of terms.

Notation 2 For a finite continued fraction [a 0 ; a 1 , a 2 , , a n ] and a positive integer k ≤ n, the k-th remainder is defined as the continued fraction rk = [ak ; ak+1 , ak+2 , , an ].

Similarly, for an infinite continued fraction [a 0 ; a 1 , a 2 , ] and a positive integer k, the k-th remainder is defined as the continued fraction r k = [a k ; a k+1 , a k+2 ,

Thus, at least in the case of a finite continued fraction, α = [a 0 ; a 1 , a 2 , , a n ] = a 0 + 1/(a 1 + 1/(a 2 + +

1/a n )) we have α = a0 + 1/(a1 + 1/(a2 + + 1/(ak−1 + 1/rk ))) = ”[a0 ; a1, a2 , , ak−1 , rk ]”

(6.1) for any positive k ≤ n Quotation signs appear because we consider the expressions of this kind only with integer entries but the quantity r k may be a non- integer.

Expanding a rational number α into a continued fraction is a straightforward process Begin by defining a0 as the greatest integer less than or equal to α, resulting in a difference δ = α − a0, which satisfies 0 ≤ δ < 1 If δ equals 0, the expansion is complete If not, set r1 = 1/δ, calculate a1 as the greatest integer less than or equal to r1, and determine a new δ = α1 − a1, ensuring it remains non-negative and less than 1 This iterative process continues until δ reaches 0.

Example 59 Consider the continued fraction expansion for 42/31 We obtain a 0 = [42/31] = 1, δ = 42/31 − 1 = 11/31 Now r1 = 1/δ = 31/11 and a 1 = [α 1 ] = [31/11] = 2 The new δ = 31/11 − 2 = 9/11 Now r 2 = 1/δ 11/9 and a2 = [α2] = [11/9] = 1 It follows that δ = 11/9 − 1 = 2/9. Now r 3 = 1/δ = 9/2 and a 3 = [α 3 ] = [9/2] = 4 It follows that δ = 9/2 − 4

= 1/2 Now r 4 = 1/δ = 2 and a 4 = [α 4 ] = [2] = 2 It follows that δ = 2 − 2 0 and we are done.

124 CHAPTER 6 INTRODUCTION TO CONTINUED FRACTIONS Thus we have calculated

The example demonstrates that the algorithm concludes after a finite number of steps, highlighting a common occurrence in this context To reinforce our understanding of the introduced notations, we will now prove a straightforward yet significant proposition.

Proposition 1 Any rational number can be represented as a finite continued fraction.

Proof By construction, all remainders are positive rationals For a positive integer k put rk = A/B and let ak = [rk ] Then

In the algorithm, the condition B B with C is less than B is established by the inequality rk - ak < 1 If C equals zero, the algorithm concludes at this stage Assuming C is not zero, it can be derived from equation (6.1) that rk equals ak plus r A comparison between equations (6.2) and (6.3) can provide further insights into the results.

Since C is less than B, the rational number rk+1 has a smaller denominator than the previous remainder rk Consequently, after a finite number of steps, we arrive at an integer rn, which is a rational number with 1 as its denominator, indicating that the procedure concludes at this stage.

There appear several natural questions in the connection with Proposition 1.

Is such a continued fraction representation unique? The immediate answer is

”no” Here are two ”different” continued fraction representations for 1/2:

However, we require that a n > 1, where a n is the last element of a finite continued fraction Then the answer is ”yes”.

Hint Make use of the formulas (6.5) below.

From now on we assume that a n > 1.

Infinite continued fractions are closely related to real numbers, and while the proof of this relationship is complex and not presented here, we will state the result and direct readers to the literature ([12, Theorem 14]) for a comprehensive proof Additionally, we will offer insights into the significance of convergence in this context.

Theorem 77 An infinite continued fraction converges and defines a real number. There is a one-to-one correspondence between

• all (finite and infinite) continued fractions [a0; a1, a2 , ] with an integer a0 and positive integers a k for k > 0 (and the last term a n > 1 in the case of finite continued fractions) and

Note that the algorithm we developed above can be applied to any real number and provides the corresponding continued fraction.

Theorem 77 holds significant theoretical importance, as it offers a constructive method for representing real numbers derived from integers, independent of any specific numerical base, such as decimal or binary This aligns with L Kronecker's famous assertion, "God created the integers; the rest is work of man," highlighting the foundational role of integers in mathematical constructs.

We will discuss some examples later.

1 Prove that under the assumption a n > 1 the continued fraction representation given in Proposition 1 is unique In other words, the correspondence q

• finite continued fractions [a 0 ; a 1 , a 2 , a n ] with an integer a 0 , positive integers a k for k > 0 and a n > 1 and

• rational numbers is one-to- one.

Main Technical Tool

A continued fraction α can be truncated at the k-th position, resulting in the k-th convergent sk = [a0; a1, a2, , ak], where k is less than n in the case of a finite continued fraction The integers pk and qk are defined such that sk can be expressed in reduced form as pk/qk, with the condition that qk is greater than zero.

The following recursive transformation law takes place.

Regardless of whether we are considering finite or infinite continued fractions, the convergents remain finite To prove this, we apply an induction argument on k, establishing that the statement holds true for k = 2.

Now, assume (6.5) for 2 ≤ k < l Let α = [a ; a , a , a ] = pl

6.2 MAIN TECHNICAL TOOL 127 be an arbitrary continued fraction of length l + 1 We denote by p r /q r the r-th convergent α Consider also the continued fraction β = [a 1 ; a 2 , , a l

] and denote by p 0 r/qr 0 its r-th convergent We have α = a0 + 1/β which translates as pl = a0 p 0 0 l−1

(6.6) ql = p 0 Also, by the induction assumption, p 0 l−1 = al p 0 l−2 + p 0 l−3

By combining the equations (6.6) and (6.7), we derive the formulas for $ p_l $ and $ q_l $, which are expressed in terms of previous values $ p_{l-1} $, $ p_{l-2} $, $ q_{l-1} $, and $ q_{l-2} $ This establishes the induction step, confirming that $ p_k $ and $ q_k $ adhere to the recursive definitions provided in (6.5) Furthermore, we need to verify that these quantities align with the conditions specified in (6.4), ensuring that $ q_k > 0 $ and that $ p_k $ and $ q_k $ are relatively prime, which is supported by the results from (6.5).

To demonstrate the assertion for k > 0, we multiply equations (6.5) by q^(k−1) and p^(k−1) and then subtract the results This leads us to the equation p^k * q^(k−1) - q^k * p^(k−1) = -(p^(k−1) * q^(k−2) - q^(k−1) * p^(k−2)) This completes the proof of Theorem 6.5, and from equation (6.5), we derive important implications.

128 CHAPTER 6 INTRODUCTION TO CONTINUED FRACTIONS p k−1 p k ( 1) k

− qk q k q k−2 Since all the numbers q k and a k are positive, the above formulas imply the following.

Proposition 2 The subsequence of convergents pk /qk for even indices k is increasing.

The subsequence of convergents p k /q k for odd indices k is decreasing.

Every convergent with an odd index is bigger than every convergent with an even index.

Proposition 2 demonstrates that both the odd and even indexed subsequences of convergents converge to limits, indicating a common limit for the infinite continued fraction Establishing that these two limits are equal involves a more technical, yet still elementary, proof.

Theorem 79 Let α = [a0 ; a1, a2, , an ] For k < n we have

An additional inequality establishes the minimum distance between the number α and its k-th convergent, which requires a more complex proof To demonstrate this, we will first explore a method of adding fractions that is often favored by students.

Definition 1 The number a + c b + d is called the mediant of the two fractions a/b and c/d (The quantities a, b, c and d are integers.) k

Lemma 15 If a c then b ≤ d a a + c c b ≤ b + d ≤ d Consider now the sequence of fractions pk

= , (6.10) q k q k + q k+1 qk + 2qk+1 q k + a k q k+1 q k+2 where the last equality follows from (6.5).

The sequence (6.10) exhibits an increasing pattern when k is even and a decreasing pattern when k is odd Notably, the fraction p k + p k+1 / q k + q k+1 (6.11) lies between p k /q k and α Consequently, the distance between p k /q k and the fraction (6.11) is less than the distance between p k /q k and α, emphasizing the closeness of these values.

= qk (qk 1+ qk+1 ). The second (right) inequality in Theorem 79 is now proved This finishes the proof of Theorem 79.

1 Check the assertion of Theorem 78 for k = 2.

Hint Introduce formally p −1 = 1 and q −1 = 0, check that then formulas 6.5 are true also for k = 1.

3 Combine the previous exercises with (6.8) to obtain qk pk−1 − pk qk−1 (−1) for k ≥ 1 Derive from this that qk and pk are relatively prime.

5 Combine (6.9) with Proposition 2 to prove the inequality α − p k 1 qk ≤ qk qk+1

7 Use (6.5) to show that the sign of the difference between two consecutive fractions in (6.10) depends only on the parity of k.

Very Good Approximation

Continued fractions offer a generic and canonical representation of numbers that transcends arbitrary base choices This unique form of representation is considered optimal, and in this section, we will explore and quantify this fundamental concept.

Definition 2 A rational number a/b is referred to as a ”good” approximation to a number α if imply c a d b and 0 < d ≤ b

Remarks 1 Our ”good approximation” is ”the best approximation of the second kind” in a more usual terminology.

2 Although we use this definition only for rational α, it may be used for any real α as well Neither the results of this section nor the proofs alter.

3 Naively, this definition means that a/b approximates α better then any other rational number whose denominator does not exceed b There is another, more a a p a

A rational number x/y is considered "the best approximation of the first kind" if, for any rational number c/d with a denominator not exceeding y, the distance |α - c/d| is greater than |α - x/y| This means that x/y is the closest rational number to α among those with a denominator of y or less Our definition introduces a modified measure of approximation that incorporates the denominator, using the formula b|α - a/b| = |bα - a|, rather than solely focusing on the distance.

Theorem 80 Any ”good” approximation is a convergent.

Proof Let a/b be a ”good” approximation to α = [a 0 ; a 1 , a 2 , , a n ]

We have to prove that a/b = pk /qk for some k.

Thus we have a/b > p 1 /q 1 or a/b lies between two consecutive convergents p k−1 /q k−1 and p k+1 /q k+1 for some k Assume the latter.

Then pk−1 1 and b − qk−1 ≥ bqk−1 p k−1 pk pk−1 1

It follows that b − qk−1 qk

At the same time Theorem 79 (it right inequality multiplied by q k ) reads

FRACTIONS and the latter inequality together with (6.12) show that a/b is not a ”good” approximation of α in this case.

This finishes the proof of Theorem 80.

1 Prove that if a/b is a ”good” approximation then a/b ≥ a0.

2 Show that if a/b > p1/q1 then a/b is not a ”good” approximation to α.

An Application

When using a computer to calculate a quantity known to be a rational number, it is common to receive a highly accurate decimal approximation To determine the exact value of this rational number, one must analyze the decimal output and identify its repeating pattern or terminating nature, which can lead to the precise answer.

To be more specific consider an example.

Example 60 Assume that the desired answer is

123456 121169 and the result of computer calculation with a modest error of 10 −15 is α = 123456/121169 + 10 −15 1.018874464590779169333740478175110795665558022266421279370135

99113634675535821868629765038912593155014896549447465936006734395761292 07 with some two hundred digits of accuracy which, of course come short to help in guessing the period and the exact denominator of 121169.

A Formula of Gauss, a Theorem of Kuzmin and Le´vi and a Prob-

Solution Since 123456/121169 is a good (just in a naive sense) approximation to α, it should be among its convergents This is not an exact statement, but it offers a hope! We have α = [1; 52, 1, 53, 2, 4, 1, 2, 1, 68110, 4, 1, 2, 106, 22, 3, 1, 1, 10, 2, 1, 3, 1, 3, 4, 2, 11].

We observe an irregularity among the convergents, particularly with the element 68110, which significantly deviates from the others To clarify this, we apply the left inequality from Theorem 79 alongside formula (6.5) Remarkably, our approximation of α is highly accurate, with the difference |α − pk /qk| being extremely small, approximately 10^−15, and with a relatively modest value of qk This leads to the relationship qk(qk+1 + qk) = qk(ak+1qk + qk1) = q2(α − pk1/qk).

It follows that 1/q 2 (a k+1 + q k−1 /qk k (a k+1 + q k−1 /q k ) ) is small (smaller than 10 −15 ) and therefore, a k+1 should be big This is exactly what we see Of course, our guess is correct:

In conclusion, an unexpectedly large element in a continued fraction enables us to accurately determine the exact rational quantity by cutting the fraction just before this element While quantifying this procedure is possible, I prefer to use it intuitively for immediate estimation of the correct values.

6.5 A Formula of Gauss, a Theorem of Kuzmin and

Le´vi and a Problem of Arnold

In this connection Gauss asked about a probability c k for a number k to appear as an element of a continued fraction Such a probability is defined in a natural way:

In Chapter 6, the discussion on continued fractions highlights their behavior as a limit when N approaches infinity, specifically regarding the frequency of occurrences of a number k within the first N elements Although Gauss offered a solution to this problem, he did not publish the proof Subsequently, independent proofs were developed by R.O Kuzmin in 1928 and P Lévy in 1929, with a detailed exposition of Kuzmin's proof available in the referenced literature.

Theorem 81 For almost every real α the probability for a number k to appear as an element in the continued fraction expansion of α is c k = 1 ln 2 ln 1

Remarks 1 The words ”for almost every α” mean that the measure of the set of exceptions is zero.

2 Even the existence of pk (defined as a limit) is highly non-trivial.

Theorem 81 is best viewed as a result from ergodic theory, rather than number theory, establishing a connection between these two mathematical fields and shedding light on the growing interest in continued fractions among mathematicians studying dynamical systems Specifically, V.I Arnold posed an open problem that involves examining pairs of integers (a, b) where the points represented on a plane lie within a quarter of a circle with radius N, defined by the equation a² + b² ≤ N.

This article explores the expansion of rational numbers p/q into continued fractions and examines the frequencies s_k associated with the occurrence of the integer k within these fractions It raises the question of whether these frequencies converge to specific limits as N approaches infinity and if these limits relate to the probabilities outlined in equation (6.13) The investigation primarily relies on experimental computer analysis, making it an engaging project for students However, it is important to note that discovering and subsequently proving such phenomena through experimentation presents significant challenges.

Of course, one can consider more general kinds of continued fractions In particular, one may ease the assumption that the elements are positive integers

In the study of mathematical identities, significant contributions were made by three renowned mathematicians, including the English mathematician R.J Gauss, who developed a formula that has been widely recognized Additionally, the theorem of Kuzmin and the problem posed by Arnold further enrich this area of research, allowing for the exploration of arbitrary real numbers and the resolution of convergence issues.

Rogers found and proved these identities in 1894, Ramanujan found the identities (without proof) and formulated them in his letter to Hardy from India in

In 1913, I J Schur independently discovered remarkable identities, publishing two distinct proofs in 1917, a significant development made possible by the separation from England due to the war For a comprehensive exploration of these identities, interested readers are encouraged to consult the detailed discussion in reference [2].

1 Prove that c k really define a probability distribution, namely that

The distribution of prime numbers has captivated modern mathematicians, notably leading to the prime number theorem proposed by Gauss and Legendre This theorem asserts that the quantity of prime numbers less than a positive integer x is asymptotically equivalent to x/logx as x approaches infinity The conjecture was ultimately validated by Hadamard and Poisson, paving the way for the development of Analytic Number Theory.

In this chapter we demonstrate elementary theorems on primes and prove elementary properties and results that will lead to the proof of the prime number theorem.

Introduction

The harmonic series $ P = \sum_{n=1}^{\infty} \frac{1}{n} $ is known to diverge, prompting the need to explore asymptotic formulas that describe the growth of $ P_n(x) = \sum_{n=1}^{x} \frac{1}{n} $ To achieve this, we will utilize Euler’s summation formula, which serves as a crucial tool in deriving the asymptotic behavior of the series.

138 CHAPTER 7 INTRODUCTION TO ANALYTIC NUMBER THEORY

This section explores the intriguing question of what happens when we sum over all prime numbers We demonstrate that this sum also diverges, confirming that the total of prime numbers extends infinitely.

We also show that an interesting product will also diverge From the following theorem, we can actually deduce that there are infinitely many primes.

Euler’s Summation Formula If f has a continuous derivative on an interval [a, b] where a > 0, then

{b})− f (a)({a}). where {t} denotes the fractional part of t.

For the proof of Euler’s summation formula see [3, Chapter 3].

Proposition 3 If x ≥ 1, we have that:

Proof We use Euler’s summation formula by taking f (t) = 1/t We then get

Notice now that {t} ≤ t and hence the two improper integrals exist since they are dominated by integrals that converge We therefore have

In this section, we derive the asymptotic formula, highlighting that γ represents Euler's constant Additionally, we note that similar methods can be employed to determine asymptotic formulas for various sums that involve powers of n.

We now proceed to show that if we sum over the primes instead, we still get a divergent series.

As a result, we have that

Choose m > 0 ∈ Z such that 2 m−1 ≤ x ≤ 2 m Observe also that

= 1 + X 1 p p m p m 1 p m 2 p≤x p i ≤x 1 2 where 1 ≤ m i ≤ m As a result, we get every 1 , n ∈ Z + where each prime factor of n is less than or equal to x(Exercise) Thus we have

Taking the limit as x approaches infinity, we conclude that P (x) diverges.

We proceed now to prove that S(x) diverges Notice that if u > 0, then

We now let u = 1/p for each p ≤ x, then

Thus we have log P (x) = X log(1/1 − p). p≤x

2 And thus S(x) diverges as x approaches infinity.

Theorem 83 (Abel’s Summation Formula) For any arithmetic function f (n), we let

A(x) = X f (n) n≤x where A(x) = 0 for x < 1 Assume also that g has a continuous derivative on the interval [y, x], where 0 < y < x Then we have

Chebyshev’s Functions

The proof of this theorem can be found in [3, Chapter 4].

1 Show that one gets every 1 , n ∈ Z + where each prime factor of n is less than or equal to x in the proof of Theorem 1.

2 Write down the proof of Abel’s summation formula in details.

We introduce some number theoretic functions which play important role in the distribution of primes We also prove analytic results related to those functions.

We start by defining the Van-Mangolt function

Definition 5 Ω(n) = logp if n = p m and vanishes otherwise.

We define also the following functions, the last two functions are called Cheby- shev’s functions.

3 ψ(10) = log2 + log2 + log2 + log3 + log3 + log5 + log7

142 CHAPTER 7 INTRODUCTION TO ANALYTIC NUMBER THEORY Remark 3 It is easy to see that ψ(x) = θ(x) + θ(x 1/2 ) + θ(x 1/3 ) + θ(x 1/m

) where m ≤ log 2 x This remark is left as an exercise.

Notice that the above sum will be a finite sum since for some m, we have that x 1/m < 2 and thus θ(x 1/m ) = 0.

We use Abel’s summation formula now to express the two functions π(x) and θ(x) in terms of integrals.

Theorem 84 For x ≥ 2, we have and θ(x) = π(x) log x −Z x π(t)dt

In this article, we define the characteristic function χ(n), which equals 1 for prime numbers and 0 otherwise This allows us to express the prime counting function π(x) and the prime number theorem θ(x) in terms of χ(n) By utilizing Abel’s summation formula, we can analyze θ(x) with f(n) = χ(n) and π(x) with f(n) = χ(n) log n Thus, we establish a foundational relationship for calculating π(x).

In Theorem 84, by setting g(x) = log x with y = 1, we derive the integral representation of θ(x) Additionally, by choosing g(x) = 1/log x with y = 3/2, we achieve the desired result for π(x), noting that θ(t) equals 0 for t < 2.

This article presents a theorem connecting the Chebyshev functions θ(x) and ψ(x) It asserts that if the limit of either function, θ(x)/x or ψ(x)/x, exists, then the limit of the other function also exists, and both limits are equal.

Getting Closer to the Proof of the Prime Number Theorem

Theorem 85 For x > 0, we have ψ(x) θ(x) (log x) 2

2√ x log 2 Proof From Remark 4, it is easy to see that

0 ≤ ψ(x) − θ(x) = θ(x 1/2 ) + θ(x 1/3 ) + θ(x 1/m ) where m ≤ log2 x Moreover, we have that θ(x) ≤ x log x The result will follow after proving the inequality in Exercise 2.

2 Show that 0 ≤ ψ(x) − θ(x) ≤ (log 2 (x))√ x log √ x and thus the result of Theorem 86 follows.

3 Show that the following two relations are equivalent π(x) = x log x

7.3 Getting Closer to the Proof of the Prime Num- ber Theorem

We know prove a theorem that is related to the defined functions above Keep in mind that the prime number theorem is given as follows: π(x)logx x→∞lim x = 1.

We now state equivalent forms of the prime number theorem.

Theorem 86 The following relations are equivalent π(x) log x x→∞lim x→∞lim x→∞lim x θ(x) x ψ(x) x

In Theorem 86, we established the equivalence of equations (7.2) and (7.3) To complete the proof, it suffices to demonstrate that (7.1) and (7.2) are also equivalent By applying the integral representations of the functions outlined in Theorem 85, we derive the relationship θ(x) π(x) log x.

Now to prove that (7.1) implies (7.2), we need to prove that

Notice also that (7.1) implies that π(t) = O

1 for t ≥ 2 and thus we have

Now once you show that (Exercise 1)

2 log t ≤ log 2 log x then (7.1) implies (7.2) will follow We still need to show that (7.2) implies (7.1) and thus we have to show that lim log x Z x θ(t)dt x→∞ x 2 t log 2 t= 0.

7.3 GETTING CLOSER TO THE PROOF OF THE PRIME NUMBER THEOREM145

Notice that θ(x) = O(x) and hence log x Z x θ(t)dt log x Z x

Now once again we show that (Exercise 2)

2 log 2 t ≤ log 2 2 log 2 x then (7.2) implies (7.1) will follow.

Theorem 87 Define l1 = lim inf x→∞ π(x) x/logx θ(x)

, L1 = lim sup x→∞ π(x) , x/logx θ(x) and l2 = lim inf x→∞ x , L 2 = lim sup x→∞ x , l3 = lim inf x→∞ ψ(x) x , L3 = lim sup x→∞ ψ(x)x , then l 1 = l 2 = l 3 and L 1 = L 2 = L 3

Proof Notice that ψ(x) = θ(x) + θ(x 1/2 ) + θ(x 1/3 ) + θ(x 1/m ) ≥ θ(x) where m ≤ log 2 x

As a result, we have θ(x) ψ(x) π(x) x ≤ x ≤ x/ log x and we get that L 2 ≤ L 3 ≤ L 1 We still need to prove that L 1 ≤ L 2

Let α be a real number where 0 < α < 1, we have θ(x) = X log p ≥ X log p p≤x

> x/ log x− αx α−1 log x Since limx→∞ α log x/x

As a result, we get that

Proving that l 1 = l 2 = l 3 is left as an exercise.

We now present an inequality due to Chebyshev about π(x).

Theorem 88 There exist constants a < A such that for sufficiently large x. alog x < π(x) < Ax x log x

Proof Put l = lim inf x→∞ π(x) x/ log x, L = lim sup x→∞ π(x) , x/ log x

It will be sufficient to prove that L ≤ 4 log 2 and l ≥ log 2 Thus by Theorem

2, we have to prove that and

To prove (7.4), notice that lim sup x→∞ lim inf x→∞ θ(x) x ψ(x) x

N = C (2n, n) =(n + 1)(n + 2) (n + n) n! < 2 2n < (2n + 1)N Suppose now that p is a prime such that n < p < 2n and hence p | N As a result, we have N ≥ Q n 2n log 2 − log(2n + 1).

≥ 1 Thus x − 1 < n < x and we get 2n ≤ x So we get

As a result, we get lim inf x→∞ ψ(x) x ≥ log 2.

Hint: For one side of the inequality, write

2 4n The other side of the inequality will follow with similar arithmetic techniques as the first inequality.

Tiêu đề	An Introductory Course in Elementary Number Theory
Tác giả	Wissam Raji
Người hướng dẫn	Professor Pavel Guerzhoy, Professor Ramez Maalouf
Trường học	University of Hawaii
Chuyên ngành	Number Theory
Thể loại	Course Notes

Định dạng
Số trang	188
Dung lượng	3,39 MB

An-Introductory-in-Elementary-Number-Theory

Algebraic Operations With Integers

The Well Ordering Principle and Mathematical Induction

The Well Ordering Principle

The Pigeonhole Principle

The Principle of Mathematical Induction

Divisibility and the Division Algorithm

Integer Divisibility

The Division Algorithm

Representations of Integers in Different Bases

The Greatest Common Divisor

The Euclidean Algorithm

Lame’s Theorem

The Sieve of Eratosthenes

The infinitude of Primes

The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic

More on the Infinitude of Primes

Least Common Multiple

Linear Diophantine Equations

The function [x] , the symbols ”O”, ”o” and ”∼”

The Function [x]

The ”O” and ”o” Symbols

Theorems and Conjectures involving prime numbers

Introduction to congruences

Residue Systems and Euler’s φ-Function

Residue Systems

Euler’s φ-Function

Linear Congruences

The Chinese Remainder Theorem

Theorems of Fermat, Euler, and Wilson

Definitions and Properties

Multiplicative Number Theoretic Functions

The Euler φ-Function

The Sum-of-Divisors Function

The Number-of-Divisors Function

The Mobius Function and the Mobius Inversion Formula

Perfect, Mersenne, and Fermat Numbers

The order of Integers and Primitive Roots

Primitive Roots for Primes

The Existence of Primitive Roots

Introduction to Quadratic Residues and Nonresidues

Legendre Symbol

The Law of Quadratic Reciprocity

Jacobi Symbol

Basic Notations

Main Technical Tool

Very Good Approximation

An Application

A Formula of Gauss, a Theorem of Kuzmin and Le´vi and a Prob-

Introduction

Chebyshev’s Functions

Getting Closer to the Proof of the Prime Number Theorem

Cryptography

Elliptic Curves

The Riemann Zeta Function