Various Problems on Subproducts of Residue Classes ...

Various Problems on Subproducts of Residue Classes

Modulo a Prime

by

Amir Hossein Parvardi

A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS

FOR THE DEGREE OF

MASTER OF SCIENCE

in

THE FACULTY OF GRADUATE AND POSTDOCTORAL STUDIES

(Mathematics)

THE UNIVERSITY OF BRITISH COLUMBIA

(Vancouver)

August 2019

c© Amir Hossein Parvardi, 2019

The following individuals certify that they have read, and recommend to the Faculty of Graduate and Postdoctoral Studies for acceptance, a thesis/dissertation entitled:

Various problems on subproducts of residue classes modulo a prime

submitted by Amir Hossein Parvardi in partial fulfillment of the requirements for

the degree of Master of Science

in Mathematics

Examining Committee:

Greg Martin

Supervisor

Greg Martin

Supervisory Committee Member

Mike Bennett


Additional Examiner

Additional Supervisory Committee Members:



ii

Abstract

Let p be a prime number. In [1], Booker and Pomerance find an integer y with 1 < y ≤ p such that all

non-zero residue classes modulo p can be written as a square-free product of positive integers up to y. Let

us denote by y(p) the smallest such y. Booker and Pomerance show in their paper that except for p = 5 and

7, we have y(p) ≤ y and some better upper bounds were conjectured.

Later, Munsch and Shparlinski [7] proved those conjectures with even better localization. Their work was

done as the same time as ours, but with fairly more complicated methods in the proof. We were seeking to

find a solution for the problem using Polya-Vinogradov inequality or at most its improvement, the Burgess

bound on character sums. That being said, we removed the condition in the problem that the product has

to be square-free. We proved that for m > p1/(4√e)+o(1), each residue class b of (Z/pZ)× can be written

as a product of elements of the set {1, 2, . . . ,m} modulo p. In fact, we showed that the number of such

sub-products (congruent to b (mod p)) is asymptotic to 2m/(p− 1).

The proofs are based on an identity involving sums of Dirichlet characters modulo p as well as the Burgess

inequality on partial character sums. Basically, we use proof by contradiction to state that if the error term

(for number of sub-product) is large, then there should be many χ values close to 1, which would result in

the character sum being large, thereby contradicting the Burgess inequality (which essentially says bounds

the character sums).

iii

Lay Summary

Let p be a prime number and m < p a positive integer. Consider the set A = {1, 2, . . . ,m}. Multiplying

some elements in this set and taking modulo p gives us an integer in the set {1, 2, . . . , p − 1}. One might

ask if it is possible to hit all these p− 1 residue classes by multiplying elements of A. Of course, the answer

to this question is negative if m is too small. For instance, if m is too small, it is possible that the largest

possible product, m!, is less than p − 1 and this means we cannot hit p − 1. Hence, it makes sense to ask

this modified question: what is the smallest m for which we can hit all the residue classes? This problem,

as well as several similar problems, have been studied deeply in the past.

We can make the problem more difficult by requiring the products to have disitinct elements of A. Again,

we must note that this question is worthy only when m satisfies a lower bound. Using cutting edge results

in number theory such as Burgess inequality, we can expect a possible lower bound for this problem. In this

project, our goal is to prove that the expected bound indeed holds, and it is the best we can get considering

the tools we have in hand. We show that for m > p1/(4√e)+ε (for small ε), every element in {1, 2, . . . , p− 1}

will be hit approximately 2m/(p− 1) times by products (mod p) of numbers from {1, 2, . . . ,m} where each

product has distinct elemets.

One can follow this path further and ask what is the smallest m for which it is possible to hit all non-zero

resiude classes mod p as a product of elements from {1, 2, . . . ,m}, if we require the products not only to have

distinct elements, but to be squarefree as well. Let’s denote by y(p) the smallest such m. This turns out to

be a more complicated question, which was posed by Booker and Pomerance [1]. In fact, we were motivated

by their results (e.g., that y(p) ≤ p for all primes p > 7) to find a better bound for y(p). However, we

soon realized that our usual techniques will not work on this problem. Munsch and Shparlinski [7] happened

to be working on the same problem coincidentally with us, and their proofs using heavy and complicated

computational partial character sums convinced us that the problem is far more difficult to be solved with

our approach. That being said, we decided to work on the easier version of the problem, which was mentioned

in the second paragraph.

iv

Preface

This thesis is an original, unpublished work by Amir Hossein Parvardi, under the supervision of Greg Martin

from the mathematics department at The University of British Columbia.

v

Table of Contents

Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

Lay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi

List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 The Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Proof Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 The General Case:

Subproducts Congruent to b (mod p) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1 The Case m > p1/4+ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 The Case m > p1/(4√e)+ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

A Proof of Proposition 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

B Proof of Propositions 14 and 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

vi

List of Figures

1.1 Proof strategy: all χ values lie in the 2θ arc. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1 Angle θ and the arc MCN in the unit circle in Lemma 11. . . . . . . . . . . . . . . . . . . . 7

2.2 Angles κj in the proof of Theorem 13. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

vii

Acknowledgements

Thanks to Professor Greg Martin for his guidance and helpful comments on my thesis. I’m thankful to my

lovely wife, Nadia Ghobadipasha, for her full support during my master’s education.

viii

Chapter 1

Introduction

1.1 The Problems

In this thesis, I will focus on three main problems. The first one is relatively the “easiest” among the three.

Problem 1. Let p ≥ 3 be a prime and m < p a positive integer. Does the set {1, 2, . . . ,m} generate all

non-zero residue classes modulo p under multiplication?

This problem is rather well-known in the context. We first need to notice that in order to get a positive

answer for this problem, m cannot be too small. Let us denote by G(p) the smallest m which gives a positive

answer to Problem 1 for a given p. Also, let n2(p) denote the smallest quadratic non-residue modulo p. It

is then clear that n2(p) ≤ G(p) because otherwise we can write n2(p), which is a non-residue, as a product

of smaller numbers (mod p) which are all quadratic resiudes, which is not possible since the product of

quadratic residues modulo p gives a residue.

Let us now see how small can n2(p) be. In 1918, Vinogradov [9] established a bound for a more general

function of p, nχ(p), which is the smallest n with χ(n) 6= 1, where χ is a non-principal character (mod p).

He showed that nχ(p)� p1/(2√e)(log p)2 (c.f. [5]). We know from elementary number theory that the Jacobi

symbol, which basically represents a number being a quadratic residue or not, is in fact a (real) Dirichlet

character. If we assume the χ to be the Jacobi symbol mod p, then nχ = n2(p) and we have an upper bound

on n2(p).

Later, Burgess [2–4] proved better bounds in a series of papers, the strongest of which is stated in

Proposition 2 (see also [5]). An accessible account of the proof of this proposition can be found in [6,

Corollary 9.19].

Proposition 2. Let χ be a non-principal character modulo p, and let nχ denote the smallest positive integer

n such that χ(n) 6= 1. Then, nχ � p1/(4√e)+o(1).

We will give a full proof to this proposition in Appendix A. We would like to note here that the exponent

1/(4√e) arises naturally in the proof and we will see it a lot in the upper bounds we find.

Before moving to the new problem, we will mention one more property for G(p): let g(p) be the smallest

primitive root modulo p (which we know exists), then G(p) ≤ g(p) because powers of g(p) generate the full

unit group (mod p). Hence, n2(p) ≤ G(p) ≤ g(p). Pollack [8] proves numerous interesting results on G(p)

and g(p).

1

Let’s move on to the next problem, which is more difficult and was proposed by Booker and Pomerance [1]

. This time, we require the product to satisfy some conditions.

Problem 3. For a prime p ≥ 3 and integer 1 ≤ m ≤ p, is it possible to write all elements of (Z/pZ)× as a

square-free, m-friable positive integer? How small can m be for this to be possible?

Let y(p) be the smallest m for which Problem 3 has a positive answer. Booker and Pomerance show in

their paper that y(p) indeed exists and is at most p for all primes p > 7. In fact, they show that for any such

prime and any a ∈ Z, there is a square-free, p-friable positive integer n ≤ pO(log p) such that n ≡ a (mod p).

Munsch and Shparlinski [7] improve this result and show that all residue classes can be represented by a

positive square-free integer s ≤ p2+o(1) which is p1/(4√e)+o(1)-friable.

When we first started working on this thesis, we tried to solve Problem 3 but we soon realized that it

is truly difficult. Fortunately, Munsch and Shparlinski’s paper [7], which is heavily computational and uses

partial character sums, was published during the time we were working on the problem and we found out

how difficult Problem 3 really is. That being said, we relaxed the conditions of the problem and made the

so-called “medium problem” (if we consider Problems 1 and 3 easy and difficult, respectively) in this context:

Problem 4. Let p ≥ 3 be a prime and m > 1 and integer. Given an integer b not divisible by p,

1. Is it possible for a sub-product of A = {1, 2, . . . ,m} to be congruent to b modulo p?

2. If yes, how many such sub-products exist?

Again, the answer to Problem 4 definitely depends on m. In fact, if we choose m too small, the answer

to the second question in Problem 4 will be zero. There are a few questions that we might ask here. First,

how small can m be? Second, if we choose m large enough, what is the answer to the second question?

For the former question, we expect the best lower bound on m to be p1/(4√e)+o(1). One must note that by

“the best lower bound,” we mean the best bound assuming our cutting-edge tecnology, which is the Burgess

inequality. In fact, the unusual power 1/(4√e) comes into our computations because of Burgess, and if we

can improve Burgess, all our results would be improved. However, improving Burgess is not easy and so the

actual best lower bound for this problem is possibly much smaller than p1/(4√e).

For m > p1/(4√e)+o(1) we prove that the main term in the answer to Problem 4 is 2m/(p − 1). This is

not unexpected because there are a total of 2m possible products with elements chosen from {1, 2, . . . ,m}and p− 1 non-zero residue classes mod p. This is our main theorem:

Theorem 5. Let p be a prime, m and b positive integers with b not divisible by p. Denote by Sm(b) the

number of sub-products of {1, 2, . . . ,m} which are congruent to b mod p. Then, if m > p1/(4√e)+o(1), for any

real ε > 0,

Sm(b) =2m

p− 1+O

(2m

p2

). (1.1)

2

1.2 Proof Strategies

Let’s consider Problem 4 in a more general form:

Problem 6. Let p ≥ 3 be a prime. Given an integer b not divisible by p and a set A = {a1, a2, . . . , ak}(taken mod p), define SA(b) as the number of subproducts (mod p) of A which are congruent to b modulo

p. Find an estimate for SA(b).

The basic main idea for solving Problem 6 is a counting argument using the well-known orthogonality

property of Dirichlet characters. As this is a famous fact, we will use it as a lemma here.

Lemma 7. Let q > 1 be an integer. For any integer n

1

φ(q)

∑χ (mod q)

χ(n) =

1, if n ≡ 1 (mod q),

0, otherwise.

Let us get back to Problem 6. We want b to be a product of elements of A mod p. That is, we

want to find ji ∈ {0, 1}, 1 ≤ i ≤ k, so that b ≡ aj11 aj22 · · · a

jkk (mod p). Applying Lemma 7 to n =

aj11 aj22 · · · a

jkk b−1 (mod p), we find that

1

p− 1

∑χ (mod p)

χ(aj11 aj22 · · · a

jkk b−1) =

1, if b ≡ aj11 aj22 · · · a

jkk (mod p),

0, otherwise.(1.2)

We can use equation (1.2) to count the number of sub-products of A which are congruent to b (mod p). In

fact,

SA(b) =1

p− 1

1∑j1=0

1∑j2=0

· · ·1∑

jk=0

∑χ (mod p)

χ(aj11 a

j22 · · · a

jkk b−1)

=1

p− 1

∑χ (mod p)

χ(b−1)

k∏i=1

(1 + χ(ai)) . (1.3)

In equation (1.3), the term corresponding to χ = χ0 (the principal character mod p) contributes 2k/(p− 1).

In the next chapters of this thesis, we will show that this term is the main term in the expression and the

contribution from other characters can be considered as the error term, as they are smaller.

If we now consider the more specific question in Problem 4 where A = {1, 2, . . . ,m}, the above discussion

tells us that we are trying to prove the following:∣∣∣∣∣∣∣∣1

p− 1

∑χ (mod p)χ 6=χ0

m∏j=1

|1 + χ(j)|

∣∣∣∣∣∣∣∣ = o

(2m

p− 1

). (1.4)

To prove this, we suppose on the contrary that for some character χ,∏mj=1 |1 + χ(j)| is large, say, � 2m/p.

Assuming this, we can prove that the number of integers j in {1, 2, . . . ,m} for which |1−χ(j)| > 1/ log log p

is� log p log log p. When seeing this on a unit circle as in Figure 1.1, one finds that there exists some integer

j and real θ > 0 for which χ(n) for all n ∈ {1, 2, . . . ,m} lies in the 2θ arc shown in the figure.

This in turn means that the values of χ are close to 1, and so the sum∑

1≤j≤m χ(j) would be large.

After taking care of the details of the proof, we arrive at the conclusion that for m � p1/(4√e)+o(1), this

latter sum is �ε y, and this contradicts the Burgess inequality [2].

3

O AC

χ(j)

χ(j)

θ

θ

|1 + χ(j)|

|1 + χ(j)|

|1−χ

(j)|

|1−χ

(j)|

Figure 1.1: Proof strategy: all χ values lie in the 2θ arc.

We will show the proof to equation (1.4) for the special case when m = p − 1. In other words, we will

show that

Theorem 8. For any b not divisible by p,

Sp−1(b) =2p−1

p− 1+O

(2p−1

p2

)(1.5)

Proof. Using equation (1.3) for A = {1, 2, . . . , p− 1}, we find that

Sp−1(b) =1

p− 1

∣∣∣∣∣∣∑

χ (mod p)

p−1∏j=1

(1 + χ(j))

∣∣∣∣∣∣ =2p−1

p− 1+

1

p− 1


p−1∏j=1

|1 + χ(j)| . (1.6)

To prove equation (1.5), it suffices to show that

1

p− 1


p−1∏j=1

|1 + χ(j)| � 2p−1

p2(1.7)

4

Suppose that χ 6= χ0 is a character of order d. That is, d is the smallest positive integer that χ(n)d = 1 for

all integers n with (n, p) = 1. Since we assumed χ is not principal, d > 1. Since the values of χ(n) are all

dth roots of unity, and d | p− 1, we can write

p−1∏j=1

|1 + χ(j)| =

∣∣∣∣∣d∏k=1

(1 + e

(kd

))∣∣∣∣∣(p−1)/d

, (1.8)

where e(x) = e2πix. On the other hand, for all real x,

xd − 1 =

d∏k=1

(x− e

(kd

)). (1.9)

Plugging x = −1 in equation (1.9) and using equation (1.8), we get

p−1∏j=1

|1 + χ(j)| =∣∣(−1)d − 1

∣∣(p−1)/d . (1.10)

Depending on the parity of d, this is

p−1∏j=1

|1 + χ(j)| =

0, if d ≡ 0 (mod 2),

2(p−1)/d, if d ≡ 1 (mod 2).

Hence, we can write the left-hand side of equation (1.7) as

1

p− 1


p−1∏j=1

|1 + χ(j)| = 1

p− 1

∑d|p−1d oddd>1

∑χ (mod p)ord(χ)=d

2(p−1)/d � 2p−1

p2, (1.11)

and the proof is complete.

5

Chapter 2

The General Case:

Subproducts Congruent to b (mod p)

Problem 9. Let p and b be defined as in Problem 6 and suppose that

A = {a1, a2, . . . , ak} = {1, 2, . . . ,m},

where m < p. We aim to find an estimate similar to that of equation (1.5) for this case as well. Let Sm(b)

be the number of subproducts of {1, 2, . . . ,m} which are congruent to b mod p.

We do not really expect Sb(m) to be large (or even positive) when m is too small. So, it is natural to

expect some lower bound on m for Problem 9 to have a positive answer. We will prove that the best lower

bound for m is p1/(4√e)+ε (which happens to be the bound for various other problems in number theory as

said before; cf. e.g., [6, Corollary 9.19]). Here, one should notice that we mean p1/(4√e)+ε is the best lower

bound we can expect to get given certain other known current obstacles – not the best lower bound that is

theoretically possible.

We first proved the theorem with a weaker bound on m. In fact, we removed the√e in the exponent

and found a proof when m > p1/4+ε. Later, we found an interesting proof for the original problem (when

m > p1/(4√e)+ε), but it was different from the previous proof. We are going to provide both proofs here.

2.1 The Case m > p1/4+ε

Theorem 10. For any prime p, real number ε > 0, and integer m > p1/4+ε,

Sm(b) =2m

p− 1+O

(2m

p2

), (2.1)

with Sm(b) as in Problem 9.

The following lemma and the corollary right after it help us in the proof of Theorem 10.

Lemma 11. Let p be a prime number, ε > 0 any real number, m an integer with m > p1/4+ε, and

0 < θ < π/2. Also, suppose that t > 1 be a real number such that t > 1/ cos θ + 1. Then, there are at least

m/t integers j ∈ {1, 2, . . . ,m} such that χ(j) lies on the arc MCN of the unit circle, as shown in Figure 2.1.

6

O AC

M

N

θ

θ

Figure 2.1: Angle θ and the arc MCN in the unit circle in Lemma 11.

Proof. Let R(m) and L(m) be the number of values of χ(j) which lie on the arcs MAN and MCN , respec-

tively. Suppose for the sake of contradiction that L(m) < m/t. This means that R(m) > (t− 1)m/t. Note

that <(χ(j)) ≥ cos θ when χ(j) is on the arc MAN and χ(j) ≥ −1 for all j. Therefore,

<

m∑j=1

χ(j)

≥ R(m) · cos θ − L(m) >(t− 1)m

t· cos θ − m

t= m

((t− 1) cos θ − 1

t

). (2.2)

By our choice of t in the statement of the lemma (t > 1/ cos θ + 1), the coefficient of m in equation (2.2) is

positive. However, by Burgess [6, Theorem 9.27], we know that

m∑j=1

χ(j) = o(m), when m > p1/4+ε, (2.3)

which is in contradiction with equation (2.2). You can see a proof of equation (2.3) in Appendix A (follow

equations (A.10) to (A.13)). The proof is complete.

Corollary 12. Choose θ such that cos θ =√

2− 1. Also, choose t = 4. Then, for any m > p1/4+ε, at least

m/4 integers j ∈ {1, 2, . . . ,m} have χ(j) on the arc MCN .

Proof of Theorem 10. The main term, 2m/(p− 1), comes from the principal character (in a similar manner

to equation (1.6)). In fact,

Sm(b) =1

p− 1

∣∣∣∣∣∣∑

χ (mod p)

m∏j=1

(1 + χ(j))

∣∣∣∣∣∣ =2m

p− 1+

1

p− 1


m∏j=1

|1 + χ(j)| . (2.4)

7

Choose θ and t in Lemma 11 as given in the corollary. We know that for each j for which χ(j) lies on the

arc MCN , the maximum value of |1 + χ(j)| is the length of the segment CM , which is equal to√(1 + cos θ)2 + sin2 θ =

√2 + 2 cos θ =

√2 + 2(

√2− 1) = 23/4.

For each j such that χ(j) lies on the arc MAN , we will use the trivial bound |1 + χ(j)| ≤ 2.

Then, since for at least m/4 integers j, χ(j) lies on the arc MCN , the product in the error term is

m∏j=1

|1 + χ(j)| ≤(

23/4)m/4

︸︷︷︸Contribution from MCN

· 23m/4︸︷︷︸Contribution from MAN

= 215m/16.

Since there are p− 2 non-principal characters modulo p, the second term on the rightmost side of equation

(2.4) is

1

p− 1

∑χ mod pχ 6=χ0

m∏j=1

|1 + χ(j)| ≤ 1

p− 1· (p− 2) · 215m/16 =

p− 2

p− 1· 215m/16 ≤ 2m

p2.

The last step holds since

p− 2

p− 1· 215m/16 ≤ 2m

p2⇐⇒ m ≥

16 log(p2(p−2)p−1

)log 2

,

which obviously holds because m > p1/4+ε � log p.

2.2 The Case m > p1/(4√e)+ε

As mentioned before, we would like to improve the bound on m in Theorem 10 from m > p1/4+ε to

m > p1/(4√e)+ε. In other words, we want to prove

Theorem 13. For any prime p, real number ε > 0, and integer m > p1/(4√e)+ε,

Sm(b) =2m

p− 1+O

(2m

p2

), (2.5)

with Sm(b) as defined in Problem 9.

We will prove two propositions which will help us in the proof of Theorem 13. We postpone the proofs

of these propositions to Appendix B.

Proposition 14. Let k and m be positive integers. If n is an m-friable number that is less than m(k+1)/2,

then n can be factored as n = b1b2 · · · bk so that each bj ≤ m.

Remark. The exponent (k + 1)/2 is best possible: if n is the product of k + 1 primes each slightly larger

than m1/2, then no such k-way factorization is possible.

Proposition 15. Let k and m be positive integers and let ε be a positive real number that is sufficiently

small in terms of k. If n is an m-friable number such that m(1+ε)k/2 < n < m(k+1)/2, then n can be factored

as n = b1b2 · · · bk′ with k′ ≤ k so that each mε < bj ≤ m.

8

Remark. The lower bound (1+ε)k/2 is best possible: if m is q times the product of k/2 primes each of which

is very slightly less than m, where q is a prime slightly smaller than mε, then no such k-way factorization is

possible. Note also that we just need (k + 1)/2 larger than e1/2 for our application, so k = 3 suffices.

We are ready to prove Theorem 13 right after we prove the following lemmas. Lemma 16 is implicitly

proved in Appendix A but we give a short proof here as well. You can find more details in the appendix.

Lemma 16. Let m and y be positive integers such that m ≤ y ≤ m2. Define as usual by ψ(y,m) the

number of m-friable positive integers up to y. Then,

ψ(y,m) = y

(1− log

log y

logm

)+O

(y

log y

). (2.6)

Consequently, the number of non-m-friable positive integers up to y is y log(log y/ logm) +O(y/ log y).

Proof. Since m ≤ y ≤ m2, any positive integer n ≤ y has at most one prime factor in (√y, y]. Therefore,

ψ(y,m) = y −∑

m<p≤y

∑n≤yp|n

1 = y −∑

m<p≤y

⌊y

p

⌋= y − y

∑m<p≤y

1

p+O(π(y)). (2.7)

Using prime number theorem and Mertens’ estimate for∑

1/p [6, Theorem 2.7] (also stated in equation

(A.5) in Appendix A), we can verify the equation (2.6). The number of non-m-friable positive integers up

to y is then y − ψ(y,m) = y log(log y/ logm) +O(y/ log y).

Lemma 17. Suppose that∏mj=1 |1 +χ(j)| � 2m/p. Let N be the number of integers j ∈ {1, 2, . . . ,m} such

that |χ(j)− 1| > 1/ log log p. Then, N � log p(log log p)2.

Proof. By Pythagorean theorem, we know for all integers j that:

|1 + χ(j)|2 + |χ(j)− 1|2 = 22.

For each integer j with |χ(j)− 1| > 1/ log log p, we thus have

|1 + χ(j)| ≤

√4− 1

(log log p)2= 2− 1

4(log log p)2+O

(1

(log log p)2

).

To avoid the error terms, we can write

|1 + χ(j)| ≤ 2− 1

5(log log p)2for all j with |χ(j)− 1| > 1/ log log p. (2.8)

Let us estimate∏mj=1 |1 + χ(j)| now. For integers j such that |χ(j)− 1| > 1/ log log p, we use the inequality

in equation (2.8). For other integers j among {1, 2, . . . ,m} (there are m − N such integers), we use the

trivial bound |1 + χ(j)| ≤ 2. So,

m∏j=1

|1 + χ(j)| ≤ 2m−N ·(

2− 1

5(log log p)2

)N= 2m

(1− 1

10(log log p)2

)N. (2.9)

Notice that

log

(1− 1

10(log log p)2

)N= N log

(1− 1

10(log log p)2

)<

−N10(log log p)2

. (2.10)

9

In the last step we have used the fact that log(1 + x) < x for all real x > −1. If N ≥ 11 log p(log log p)2,

then by equation (2.10),

log

(1− 1

10(log log p)2

)N� −11 log p(log log p)2

10(log log p)2=−11

10log p = log

(1

p11/10

).

This, together with equation (2.9), implies that

m∏j=1

|1 + χ(j)| ≤ 2m(

1− 1

10(log log p)2

)N� 2m

p11/10,

which, is in contradiction with the assumption of lemma that the product is � 2m/p. This means that

N � log p(log log p)2.

Now we have all the tools for proving Theorem 13.

Proof of Theorem 13. We will use the same strategy as in the proof of Theorem 10 but the techniques used

here are different. The main term is clearly 2m/(p− 1). The error term is also the same as that of Theorem

10 (which is the right-most term in equation 2.4). Assume the error term is � 2m/p2. This means that for

some non-principal Dirichlet character χ,

m∏j=1

|1 + χ(j)| � 2m

p. (2.11)

Let m < y ≤ m2 be an integer. Note that

<∑n≤y

χ(n) = <∑n≤y

n is m-friable

χ(n) + <∑n≤y

n is non-m-friable

χ(n)

≥ <∑n≤y

n is m-friable

χ(n)−∑n≤y

n is non-m-friable

1. (2.12)

We will break the first sum in equation (2.12) into three sums by categorizing m-friable positive integers n

up to y into three groups:

1. n ≤ m3(1+ε)/2. Let’s call the set of numbers in this category A1, so that

#A1 ≤ m3(1+ε)/2. (2.13)

2. n is divisible by some k > mε such that |χ(k) − 1| > 1/ log log p. We proved in Lemma 17 that the

number of such n is � log p(log log p)2 (note that the condition of the lemma is satisfied since we

assumed equation (2.11) in the beginning of the proof). If we denote by A2 the set of such integers,

then

#A2 ≤∑k>mε

|χ(k)−1|>1/ log log p

y

k≤ y

mε· log p(log log p)2. (2.14)

10

O

AC

M

χ(bj)

N

P

κjκj/2

1

loglogp

1loglogp

|χ(bj )−

1|

Figure 2.2: Angles κj in the proof of Theorem 13.

3. m3(1+ε)/2 < n < m2 and n is not divisible by any k > mε such that |χ(k) − 1| > 1/ log log p. We

know by Proposition 15 that n can be written as n = b1b2b3, where each bj (1 ≤ j ≤ 3) is either 1

or satisfies mε < bj ≤ m. By our assumption, for each 1 ≤ j ≤ 3, we have |χ(bj) − 1| ≤ 1/ log log p.

Therefore, according to Figure 2.2, χ(bj) lies on the arc MAN , and if we define the angle κj as shown

in the diagram, we would have |χ(bj)− 1| = 2 sin(κj/2). Combining this with |χ(bj)− 1| ≤ 1/ log log p

gives an upper bound on κj : κj � 1/ log log p. This is because when x is in a neighbourhood of zero,

arcsinx = x+O(x3). Note that for j = 1, 2, 3, we have χ(bj) = eiκj , thus

<χ(n) = < (χ(b1)χ(b2)χ(b3)) = <(ei(κ1+κ2+κ3)

)= cos(κ1 + κ2 + κ3)� cos

(3

log log p

)= 1 +O

(1

(log log p)2

). (2.15)

One must notice that in equation (2.15), the inequality comes from the fact that 0 < x < y < π/2

implies cosx > cos y (this is a particular property of the cosine function and does not necessarily hold

for a general function).

11

The number of integers in the third category, which we call A3, is

#A3 ≥ # {n ≤ y : n is m-friable} −#A1 −#A2

= y −# {n ≤ y : n is non-m-friable} −#A1 −#A2. (2.16)

Combining equations (2.15) and (2.16), we find

<∑n∈A3

χ(n) ≥(

1 +O

(1

(log log p)2

))(y −# {n ≤ y : n is non-m-friable} −#A1 −#A2) . (2.17)

From equation (2.12),

<∑n≤y

χ(n) ≥ <∑n≤y

n is m-friable

χ(n)−∑n≤y

n is not m-friable

1

= <∑n∈A1

χ(n) + <∑n∈A2

χ(n) + <∑n∈A3

χ(n)−∑n≤y

n is non-m-friable

1.

Using the trivial estimation for the first two sums, we find that above is

≥ <∑n∈A3

χ(n)−# {n ≤ y : n is non-m-friable} −#A1 −#A2. (2.18)

To make equations shorter, let us define

A4 = {n ≤ y : n is non-m-friable} .

From Lemma 16 we know that

#A4 = y loglog y

logm+O

(y

log y

). (2.19)

From equations (2.17) and (2.18),

<∑n≤y

χ(n) ≥ y − 2#A4 − 2#A1 − 2#A2 +O

(y

(log log p)2

). (2.20)

The numerator in the error term is in fact y −#A4 −#A1 −#A2 but since it is ≤ y, we simply wrote y in

the numerator. Finally, combining equations (2.13), (2.14), (2.19), and (2.20), we find

<∑n≤y

χ(n) ≥ y(

1− 2 loglog y

logm

)+O

(y

(log log p)2+m3(1+ε)/2 +

y

mεlog p(log log p)2 +

y

log y

). (2.21)

(2.13), Now, if we choose y = m√e−ε, the coefficient of y in the main term of equation (2.21) is a positive

constant. Moreover, the terms inside the big-O are all � y by our choice of m. However, we know from

Burgess inequatlity that the charachter sum in equation (2.21) is o(y), which is a contradiction. Hence, our

assumption (equation (2.11)) is false and the proof of Theorem 13 is complete.

12

Bibliography

[1] Andrew R. Booker and Carl Pomerance, Squarefree smooth numbers and Euclidean prime generators, Proc. Amer. Math.

Soc. 145 (2017), no. 12, 5035–5042. MR3717934

[2] D. A. Burgess, The distribution of quadratic residues and non-residues, Mathematika 4 (1957), 106–112. MR0093504

[3] , On character sums and L-series. II, Proc. London Math. Soc. (3) 13 (1963), 524–536. MR0148626

[4] , The character sum estimate with r = 3, Journal of the London Mathematical Society s2-33 (1986), no. 2, 219–226.

[5] Yuk-Kam Lau and Jie Wu, On the least quadratic non-residue, International Journal of Number Theory 4 (2008), no. 03,

423–435.

[6] Hugh L. Montgomery and Robert C. Vaughan, Multiplicative number theory. I. Classical theory, Cambridge Studies in

Advanced Mathematics, vol. 97, Cambridge University Press, Cambridge, 2007. MR2378655

[7] Marc Munsch and Igor E Shparlinski, On smooth square-free numbers in arithmetic progressions, arXiv preprint

arXiv:1710.04705 (2017).

[8] Paul Pollack, The average least quadratic nonresidue modulo m and other variations on a theme of Erdos, J. Number

Theory 132 (2012), no. 6, 1185–1202.

[9] I. M. Vinogradov, Sur la distribution des residus et des non-residus des puissances, J. Phys.-Math. Soc. Perm 1 (1918),

94–96.

13

Appendix A

Proof of Proposition 2

In this appendix we establish the proof Proposition 2, the statement of which can be found on page 1. Prior

to the proof, we prove a key lemma which contains the heart of the proof.

Lemma 18. Let x and y be real numbers with y ≤ x < y2 such that χ(n) = 1 for all 1 ≤ n ≤ y. Then,∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣ ≥ x(

1− 2 loglog x

log y

)+O

(x

log x

). (A.1)

Proof. Consider the sum∑n≤x χ(n) and write it as∑n≤x

χ(n) =∑n≤x

n is y-friable

χ(n) +∑n≤x

n is not y-friable

χ(n). (A.2)

In the first sum on the right-hand side of equation (A.2), each y-friable n ≤ x can be factored as n =∏ki=1 p

αii

with each pi ≤ y and αi ≥ 1. Since χ is multiplicative and χ(pi) = 1 for 1 ≤ i ≤ k, we have χ(n) = 1.

Thus, the first sum is ψ(x, y), the number of y-friable positive integers up to x. For the second sum in

equation (A.2), note that a positive integer n ≤ x which is not y-friable has a prime divisor p > y and

n/p < y. Hence, since χ(n) = χ(n/p)χ(p),∑n≤x

n is not y-friable

χ(n) =∑

y<q≤x

χ(q)⌊xq

⌋,

where q denotes a prime. Therefore,∑n≤x

χ(n) = ψ(x, y) +∑

y<q≤x

χ(q)⌊xq

⌋. (A.3)

If we use the trivial bound |χ(n)| ≤ 1 for the sum on the righ-hand side of equation (A.3), we observe∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣ ≥ ψ(x, y)−∑

y<q≤x

⌊xq

⌋. (A.4)

We know that [6, Theorem 2.7] ∑1≤q≤x

1

q= log log x+ b+O

(1

log x

), (A.5)

14

where b is a constant. This immediately implies∑y<q≤x

1

q= log

log x

log y+O

(1

log x

). (A.6)

Thus,

∑y<q≤x

⌊xq

⌋=

∑y<q≤x

x

q+O

∑y<q≤x

1

= x loglog x

log y+O

(x

log x

). (A.7)

We note that in the above equations, the error coming from the sum∑y<q≤x 1 will be in the same as the

error from estimating the sum x∑y<q≤x

1q which we evaluated in equation (A.6). Furthermore,

ψ(x, y) = bxc −∑

y<p≤x

∑n≤xp|n

1 = x+O(1)−∑

y<p≤x

⌊xq

⌋. (A.8)

Combining equations (A.4), (A.7), and (A.8), we get the desired result.

Proof of Proposition 2. Let x and y be real numbers satisfying the conditions of Lemma 18. Then,∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣ ≥ x(

1− 2 loglog x

log y

)+O

(x

log x

). (A.9)

We know from the Burgess inequality [6, Theorem 9.27] that for any integer r > 2,∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣� rx1−1r p

r+1

4r2 (log p)12r , (A.10)

Choose x = p1/4+ε, where ε < 1/7, and a positive constant c such that r = c/ε is an integer. Then,

equation (A.10) becomes∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣� c

ε

(p1/4+ε

)1− εc

pc/ε+1

4c2/ε2 (log p)ε2c =

c

ε· p

14+ε−ε

2( 1c−

14c2

)(log p)ε2c . (A.11)

For c > 1/4, the 1c −

14c2 is positive and strictly decreasing as a function of c. Equation (A.11) then implies∣∣∣∣∣∣

∑n≤x

χ(n)

∣∣∣∣∣∣� c

ε· p

14+ε−ε

2( 1c−

14c2

)(log p)ε2c = o(p

14+ε), (A.12)

which means ∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣ = o(x) for x = p1/4+ε. (A.13)

Suppose that y > x1/√e+ε. Taking the logarithm from both sides and dividing by log x, we find

log y/ log x > 1/√e+ ε. Taking the logarithm again,

loglog y

log x> log

(1√e

+ ε

)> −1

2+ ε,

15

where the last equation holds for our choice of ε by monotonousity. Since log 1/a = − log a, this shows that

1− 2 loglog x

log y> 1 + 2

(−1

2+ ε

)= 2ε,

and so, by equation (A.9),∣∣∣∣∣∣∑n≤x

χ(n)

∣∣∣∣∣∣ ≥ 2εx+O

(x

log x

)� x for y > x1/

√e+ε, (A.14)

which is in contradiction with equation (A.13). Hence, y ≤ x1/√e+ε and,

nχ ≤ y ≤ x1/√e+ε =

(p1/4+ε

)1/√e+ε= p

14√

e+ε

(14+

1√e+ε

)< p1/(4

√e)+ε.

The last inequality holds for ε < 1− 14−

1√e, which means it is also true for ε < 1/7. The proof is complete.

16

Appendix B

Proof of Propositions 14 and 15

We establish Proposition 14 stated on page 8.

Proof of Proposition 14. Let n = p1p2 · · · pr be the factorization of n into prime factors of descending order.

That is, p1 ≥ p2 ≥ · · · ≥ pr. Let b1, b2, . . . , bk be positive integers all initially equal to 1. Find the first

bj not exceeding m/pi and multiply it by pi. At the end, we claim the bj form the desired factorization.

Algorithm 1 summarizes this process.

Algorithm 1 k-way Factorization

1: for 1 ≤ i ≤ r do

2: for 1 ≤ j ≤ k do

3: if pibj ≤ m then

4: bj ← pibj ;

5: break;

6: end if

7: end for

8: end for

We only need to prove that in the algorithm above, for each pi, there exists some bj such that pibj ≤ m.

Assume, for the sake of contradiction, that when running Algorithm 1, there is some p` for which all the

numbers p`b1, p`b2, . . . , p`bk exceed m. Then,

b1b2 · · · bk · p` = b1p` · b2p` · · · bkp` · p` · p−k` >mk

pk−1`

. (B.1)

Therefore,

m(k+1)/2 ≥ n ≥ b1b2 · · · bkp` >mk

pk−1`

.

This implies m(k−1)/2 < pk−1` or simply√m < p`. Notice that since each bj is initially equal to 1 and each

pi ≤ m, this situation only happens when each bj has been selected in step 3 of the algorithm at least once.

This means that each for each j ∈ {1, 2, . . . , k}, we have bj ≥ p` because bj is the product of elements of a

17

subset of {p1, p2, . . . , p`−1}, and pi ≥ p` for i = 1, . . . , `− 1. But then, since m <√p`,

n ≥ b1b2 · · · bkp` ≥ pk+1` > m(k+1)/2,

which is a contradiction.

We now prove Proposition 15 appearing on page 8.

Proof of Proposition 15. The result is obvious for k = 1, so let us assume k ≥ 2. If ε < 1/(k + 2), we

have n < m(k+1)/2 <(m1−ε)(k+2)/2

, and by Proposition 14 (with k and m replaced by k + 1 and m1−ε,

respectively), we may write n as

n = b1b2 · · · bk+1 with bj ≤ m1−ε for 1 ≤ j ≤ k + 1.

If there doesn’t exist bi and bj (1 ≤ i, j ≤ k + 1) such that bibj ≤ m, then n > (mk+1)1/2 = m(k+1)/2. This

can be verified easily: if k is odd, k+ 1 is even and we can divide bjs into two groups such that the product

of any two elements chosen from different groups is > m; if k is even, we can leave out the greatest bj (which

is obviously >√m), and divide the k remaining bjs into two groups and then proceed like the previous case.

In both cases, n > (mk+1)1/2 = m(k+1)/2, which is a contradiction. So, such bi, bj exist. Let’s suppose,

without loss of generality, that bkbk+1 ≤ m. Define ck = bkbk+1 and cj = bj for 1 ≤ j ≤ k − 1 so that

n = c1c2 · · · ck with ck ≤ m and cj ≤ m1−ε for 1 ≤ j ≤ k − 1. (B.2)

Let C be the number of cj (1 ≤ j ≤ k − 1) in equation (B.2) which are ≤ mε. If C = 0, there is nothing

to prove and (B.2) gives us the factorization with desired conditions. So, let’s suppose that C is a positive

integer. If C ≥ k/2,

n ≤ m · (mε)C ·(m1−ε)k−1−C = m1+εC+(1−ε)(k−1−C)

= mk−ε(k−1)−C(1−2ε) ≤ mk−ε(k−1)− k2 (1−2ε)

= mk/2+ε ≤ mk/2+εk/2 = m(1+ε)k/2,

which is a contradiction (in the last step, we have used k ≥ 2). Therefore, C < k/2. Without loss of

generality, suppose that c1, c2, . . . , cC be all cj (1 ≤ j ≤ k − 1) such that cj ≤ mε. Define, for 1 ≤ j ≤ C,

dj = cjcC+j , so that by the assumptions in equation (B.2), we have

mε < dj ≤ mε ·m1−ε = m. (B.3)

Notice that for dj to be well-defined, we need 2C ≤ k, which is implied from C < k/2 since C is an integer.

But then, n = d1 · · · dC · c2C+1 · · · ck, which is the product of at most k numbers all of which are in the

desired range (by equations (B.2) and (B.3)).

18

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