Vinogradovs theorem and its generalization on primes in arithmetic progression

VINOGRADOV’S THEOREM AND ITS
GENERALIZATION ON PRIMES IN
ARITHMETIC PROGRESSION

WONG WEI PIN
(B.Sc.(Hons.) NUS)

A THESIS SUBMITTED FOR THE DEGREE OF
MASTER OF SCIENCE

DEPARTMENT OF MATHEMATICS

NATIONAL UNIVERSITY OF SINGAPORE

2009


i

Acknowledgements
First and foremost, it is my great honor to work under Assistant Professor Chin
Chee Whye again, for he has been more than just a supervisor to me but as well as a
supportive friend; never in my life I have met another person who is so knowledgeable
but yet is extremely humble at the same time. Apart from the inspiring ideas and endless
support that Prof. Chin has given me, I would like to express my sincere thanks and
heartfelt appreciation for his patient and selfless sharing of his knowledge on algebraic
number theory, which has tremendously enlighten me. Also, I would like to thank him
for entertaining all my impromptu visits to his office for consultation and entrusting me
to be the grader for his Galois Theory module.
I would like to express my profound gratitude to Prof. Régis de la Bretèche, who
was my supervisor of my scientific internship at the Mathematics Institute of Jussieu,

Paris, for having equipped me with a solid foundation on understanding Vinogradov’s
theorem.
Many thanks to all the professors in the Mathematics department who have taught
me before. Also, special thanks to Professor Chan Heng Huat and Dr. Toh Pee Choon for
patiently attending my seminar series on this research as well as giving me constructive
suggestions to improve my thesis.
I would also like to take this opportunity to thank the administrative staff of the
Department of Mathematics for all their kindness in offering administrative assistant to
me throughout my Master’s study in NUS. Special mention goes to Ms. Shanthi D/O
D Devadas, Mdm. Tay Lee Lang and Mdm. Lum Yi Lei for always entertaining my
request with a smile on their face.
Last but not least, to my family and my fellow peers, Siong Thye, Jia Le, Jian Xing
and Tao Xi, thanks for all the laughter and support you have given me throughout my
Master’s study. It will be a memorable chapter of my life.
Wong Wei Pin
Spring 2009


Contents
Acknowledgements

i

Summary

iii

Notation

vi

1 Analytic tools

1

1.1

Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2

Arithmetic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.3

Dirichlet Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4

Infinite Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.5

Prime Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 The Ternary Goldbach Problem

22

2.1

The Minor Arcs m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2

The Major Arcs M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.3

Vinogradov’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Generalized Ternary Goldbach Problems

36

3.1

Ternary Goldbach Problem in Number Fields . . . . . . . . . . . . . . . . 36

3.2

The Minor Arcs m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3

The Major Arcs M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.4

Proof of Theorem 3.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Bibliography

63

ii


iii

Summary
Christian Goldbach first made his famous conjecture in 1742 that every even number
larger than 2 is a sum of two prime numbers. Although Goldbach’s Conjecture still
remains unsolved today, great progress has been achieved by many mathematical giants
such as Hardy, Littlewood, Vinogradov, Estermann, Chen Jinrun and Heath-Brown, in
proving weaker variants of the conjecture. One such variant is the Weak Goldbach’s
Conjecture (also known as the ternary Goldbach problem), which says that every odd
number larger than 5 is a sum of three prime numbers. The central idea in proving these
variants is to find a good estimate of the number of representation of an integer as a sum
of primes, using revolutionary and accurate counting methods.
One such method is the Hardy-Littlewood circle method, first invented by Hardy
and Ramanujan and then further developed and applied by Hardy and Littlewood in
solving the Waring’s problem. In fact, the circle method has far reaching applications in
other additive number theory problems, such as Birch’s theorem, Roth’s theorem and the
ternary Goldbach problem. In 1923, Hardy and Littlewood made a remarkable progress
on the ternary Goldbach problem by showing that every sufficiently large odd number

is a sum of three prime numbers, with the assumption of a zero free domain for the
Dirichlet L-functions (which is true if one assume the Generalized Riemann Hypothesis).
The ultimate breakthrough for the ternary Goldbach problem was done by Vinogradov
in 1937. With his ingenious estimation of the exponential sum on prime numbers as well
as his aptly application of the Siegel-Walfisz Theorem, Vinogradov managed to remove
the assumption in Hardy and Littlewood’s proof and thus proved unconditionally the
Vinogradov’s theorem: every sufficiently large odd number is a sum of three prime numbers. Subsequently, Chinese mathematicians Chen and Wang showed that the condition
for being sufficiently large is to be larger than 1043000 , but this astronomical number
is still far to be reached by numerical verification with computer programs in order to
prove the Weak Goldbach’s Conjecture completely.

∗∗∗


iv
The aim of this thesis is to first study a simplified proof of the Vinogradov’s theorem
given by Vaughan and to generalize the Vinogradov’s theorem to the quadratic fields.
This generalization will in turn give the motivation to formulate the Vinogradov’s theorem for primes in arithmetic progression: Let x1 , x2 , x3 and y be integers such that 1 < y
and (xi , y) = 1 for i = 1, 2, 3. Then for all sufficiently large odd integer N ≡ x1 + x2 + x3
mod y, there exist primes pi ≡ xi mod y for i = 1, 2, 3, such that N = p1 + p2 + p3 . This
was first conditionally proven in 1926 by Rademacher with the assumption of the Generalized Riemann Hypothesis (see [9]). By borrowing ideas from the proof of Vinogradov’s
theorem, Ayoub improved Rademacher’s argument to give an unconditional proof of this
theorem in 1953 (see [10]). However, the proof of this theorem presented in this thesis is mainly original. Once this theorem is established, the Vinogradov’s theorem for
quadratic fields will follow immediately as a corollary.

∗∗∗

This thesis is organized in three chapters. Chapter 1 concentrates on developing the
crucial analytical tools that will serve us in the later chapters. These include an important inequality for exponential sums, some typical arithmetic functions, Ramanujan’s
sums, Dirichlet’s series, infinite products and Euler products. The chapter winds up

with a short exposition on two important results of modern prime number theory: the
prime number theorem and the Siegel-Walfisz Theorem.
The entire Chapter 2 is dedicated to study Vaughan’s proof of Vinogradov’s theorem.
A general outline of the Hardy-Littlewood circle method in Vinogradov’s theorem will
be presented first, followed by the definition of the major arcs and the minor arcs. Then
the chapter proceeds on to find the asymptotic estimate of integrations over these two
arcs. Compiling all these estimations, the last section of this chapter will prove the
Vinogradov’s theorem by analyzing the behavior of the singular series S(N ).
At the beginning of Chapter 3, some possible generalizations of Goldbach’s Conjecture to number fields will be discussed and eventually we will focus our interest on
the ternary Goldbach problem on quadratic fields. After formulating our conjecture
on quadratic fields, we will explain why the conjecture is a direct consequence of the


v
Vinogradov’s theorem for primes in arithmetic progression and move on to prove this
theorem. The underlying idea of the proof follows exactly the one presented in Chapter
2, i.e. we will apply the Hardy-Littlewood circle method with the similar treatment of
integrations over the major arcs and the minor arcs. However, great effort is put in to
handle the twisted Ramanujan sum ηx,y that occurs in the estimation over the major
arcs. Once this is overcome, we conclude by proving Vinogradov’s theorem for primes in
arithmetic progression as well as Vinogradov’s theorem for quadratic fields.


vi

Notation
1. The letters a, b, d, j, k, , m, n, q, r, s always stand for integers. The letter p is always
reserved for prime numbers.
2. We abbreviate e2πiα as e(α).
3. If f (x), g(x) ≥ 0, then f (x)

g(x) means there exists an absolute constant C > 0

such that |f (x)| ≤ Cg(x). f

g means g

f.

y

means the constant C depends

on y.
4. α is the largest integer that is less than or equal to α.
5. {α} := α − α .
6. f (n) = O(g(n)) means lim

f (n)

n→∞ g(n)

< ∞. It is by default that the limit is taken

when n tends to ∞ unless stated otherwise.
7. f (n) = o(g(n)) means lim

f (n)

n→∞ g(n)

= 0. It is by default that the limit is taken when

n tends to ∞ unless stated otherwise.
8. The notation (n, m) refers to gcd(n, m).
9. [a, b] := {x ∈ ❘ : a ≤ x ≤ b}.
10.

always refers to the product over all prime numbers, unless stated otherwise.
p

11. log is the natural logarithm function.
12. ❩n is the set of classes of residues modulo n and ❩∗n is the set of multiplicative
invertible elements in ❩n .
13. The notation pr ||n means that pr is the highest power of p dividing n.


Chapter 1

Analytic tools
In this chapter, we develop some crucial analytical tools that will serve us in proving
Vinogradov’s theorem in Chapter 2 and the generalized Vinogradov’s theorem in Chapter
3. The most important of these tools is an inequality for exponential sums. We shall
also introduce some typical arithmetic functions, Ramanujan sums, infinite products and
Euler products. In the last section of this chapter, we state some important results of
modern prime number theory, which will be needed as well in the later chapters.

1.1

Inequalities

Lemma 1.1.1. (Dirichlet) Let α be a real number. Then for any real number X ≥ 1,
there exist integers a and q, such that (a, q) = 1, 1 ≤ q ≤ X and

α−

1
a
1
≤
≤ 2.
q
qX
q

Proof. It suffices to prove the first inequality, and without the condition (a, q) = 1. We
first let m = X and βq = αq − αq ∈ [0, 1) for q = 1, 2, . . . , m. Next, we consider
r−1
r
the m + 1 disjoint intervals Br = [ m+1
, m+1
), for r = 1, 2, . . . , m + 1. If there exists
1
m
a q such that βq ∈ [0, m+1
) (resp. [ m+1
, 1)), then we have α −

(resp.

α−

αq +1
q

<

1
q(m+1) ).

αq
q

<

<

1
(u−v)X ,

<

1
qX

Otherwise, by the pigeonhole principle, there exists

u > v and some r ∈ {2, 3, . . . , m} such that βu , βv ∈ Br , which implies α −
1
(u−v)(m+1)

1
q(m+1)

αu − αv
u−v

<

with u − v < X. The lemma then follows with a = αu − αv

1


CHAPTER 1. ANALYTIC TOOLS

2

and q = u − v.
Definition 1.1.2. Given any real number α, we denote the distance from α to the nearest
integer as :
||α|| := min |α − m|.
m∈❩

Remark: By the definition of ||α||, we have || − α|| = ||α|| ∈ [0, 1/2] and α = n ± ||α||
for some integer n.
Proposition 1.1.3. For all real numbers α, β, the following triangle inequalities hold:

||α|| − ||β|| ≤ ||α + β|| ≤ ||α|| + ||β||.

Proof. The first inequality in the assertion is an immediate consequence of the second,
so it suffices to prove the second inequality. Let m, n be integers such that α = m ± ||α||
and β = n±||β||. Without loss of generality, let ||α|| ≥ ||β||, so that ||α||−||β|| ∈ [0, 1/2]
and hence ||α|| − ||β|| = ||α|| − ||β||. With this, we obtain easily that

||α + β|| = m + n ± ||α|| ± ||β||
= ||α|| ± ||β||
≤ ||α|| + ||β||.

Lemma 1.1.4. For every real number α, we have

| sin(πα)| = sin(π||α||) ≥ 2||α||.

Proof. We have ||α|| ∈ [0, 21 ] and α = n ± ||α|| for some integer n. Hence
| sin(πα)| = | sin(πn ± π||α||)| = sin(π||α||) ≥ 2||α||,

as the function sin(πx) is concave when x ∈ [0, 12 ].


CHAPTER 1. ANALYTIC TOOLS

3

Lemma 1.1.5. For every real number α and all integers N1 < N2 , we have
N2

e(αn) ≤ min(N2 − N1 , ||α||−1 ).
n=N1 +1

Proof. It is obvious that

N2

N2

e(αn) ≤

1 = N2 − N1 .
n=N1 +1

n=N1 +1

For α ∈
/ ❩, we have ||α|| > 0 and e(α) = 1. Since the sum is a geometric series, we have
N2

e(αn) =
n=N1 +1

≤
=
=
=
≤

e(α(N2 + 1)) − e(α(N1 + 1))
e(α) − 1
2
|e(α/2) − e(−α/2)|
2
|2i sin(πα)|

1
| sin(πα)|
1
sin(π||α||)
1
1
≤
.
2||α||
||α||

Lemma 1.1.6. Let α, X, Y be real numbers, X ≥ 1, Y ≥ 1 and let q and a be integers
such that |α − aq | ≤ q −2 with (a, q) = 1 and q ≥ 1. Then
min
n≤X

XY
+ X + q log(2Xq).
q

XY
, ||αn||−1
n

Proof. The case of q = 1 for the assertion is trivial because

min
n≤X

XY

, ||αn||−1
n

≤ XY
n≤X

1
n

XY log X ≤ (XY + X + 1) log(2X).

So we assume q > 1 for the rest of this proof. By rewriting the summing index as residues


CHAPTER 1. ANALYTIC TOOLS

4

modulo q, we obtain the following:

S :=

XY
, ||αn||−1
n

min
n≤X

q

≤

min
r=1
0≤j≤ X
q

XY
, ||α(qj + r)||−1 .
qj + r

We now focus on finding a bound for most of the terms ||α(qj+r)||−1 . Define yj := αjq 2
for j = 0, . . . ,

X
q

and θ := q 2 α − qa ∈ [−1, 1], which then give the identity
αjq 2 + {αjq 2 } αq 2 r ar arq
+ 2 +
− 2
q
q
q
q
2
yj + ar {αjq } rθ
=
+

+ 2,
q
q
q

α(qj + r) =

We show that for a fixed j ∈ {0, . . . ,

X
q

j = 0, . . . ,

X
.
q

}, the inequality

||α(qj + r)|| ≥

1 (yj + ar)
2
q

(1)

fails to hold only for at most 7 exceptional values of r in {1, . . . , q}. By writing xr :=
yj +ar

q

and

:=

{αjq 2 }
q

+

rθ
q2

< 2q , it suffices to show that the inequality
1
||xr + || ≥ ||xr ||
2

fails to hold for at most 7 exceptional values of r if q ≥ 8 (so that | | <

1
4

and thus

|| || = | |). For those r’s where ||xr || ≥ 2| |; which is equivalent to ||xr || − | | ≥ 21 ||xr ||,
we have the desired inequality:
1
||xr + || ≥ ||xr || − || || = ||xr || − | | ≥ ||xr ||.

2
Thus, the r’s for which (1) might fail correspond to ||xr || < 2| | < 4q , which implies these
r’s can only correspond to the case yj + ar ≡ 0, ±1, ±2, ±3 mod q. So we conclude that
for q > 1, j ∈ {0, . . . ,

X
q

}, the inequality (1) holds for all r ∈ {1, . . . , q} with at most 7

exceptions. Notice that the 7 exceptions include also the unique r ∈ {1, . . . , q} for which
yj + ar ≡ 0 mod q.


CHAPTER 1. ANALYTIC TOOLS

5

In fact, in the case of j = 0, we need furthermore that the inequality

||αr|| = ||α(qj + r)|| ≥

holds for all r ∈ {1, . . . ,
|| ≤
|| rθ
q2

1
2q .

q
2

}. Indeed, for 1 ≤ r ≤

For q > 1, we have || ar
q || ≥

for r ∈ {1, . . . ,

q
2

1 (yj + ar)
1 ar
=
>0
2
q
2 q

1
q

q
2,

1
2q

| ≤
we get | rθ
q2

and thus

≥ 2|| rθ
||, because (a, q) = 1 and thus
q2

ar
q

∈
/❩

}. So we obtain
ar rθ
+ 2
q
q
rθ
ar
− 2
≥
q
q
1 ar
≥
> 0.

2 q

||α(qj + r)|| =

by Proposition 1.1.3

Now, we return to find an upper bound for S. For j = 0 or j = 0 and r ≥ 2q , we have
q(j + 1) ≤ 2(qj + r). This implies that for those r’s for which (1) does not hold, we can
bound the summing terms by
q

S≤

min
r=1
0≤j≤ X
q

2XY
q(j+1) .

XY
, ||α(qj + r)||−1
qj + r

q

≤

2

0≤j≤ X
q

≤2

≤4

≤

r=1
yj +ar≡0 mod q

X
+1
q
X
+1
q

q−1

s=1
q/2

s=1

Thus

s
q

(yj + ar)
q

−1

+ 14

XY
q

−1

+7
0≤j≤ X
q
X
j=0

1
j+1

q
XY
+ 14
2 log(2X)
s
q

X

XY
+ 1 q log(2q) +
log(2X)
q
q
XY
+ q + X log(2Xq).
q

2XY
q(j + 1)

by (1)


CHAPTER 1. ANALYTIC TOOLS

1.2

6

Arithmetic Functions

Definition 1.2.1.
I. An arithmetic function is a complex-valued function whose domain is the set of
all positive integers.
II. An arithmetic function f (n) is multiplicative if

f (mn) = f (m)f (n),

whenever (m, n) = 1.

In this case, it is easy to see that if f is not identically zero, then f (1) = 1.
III. An arithmetic function f (n) is completely multiplicative if

f (mn) = f (m)f (n),

for all positive integers m, n.

Theorem 1.2.2. Let f be a multiplicative function. If

lim f (pk ) = 0

pk →∞

as pk runs through the sequence of all prime powers (i.e. both p and k vary such that pk
tends to ∞), then
lim f (n) = 0.

n→∞

Proof. Since lim f (pk ) = 0, there are only finitely many prime powers pk such that
pk →∞

|f (pk )|

≥ 1. We denote
|f (pk )|.

A :=

|f (pk )|≥1

Now, given any > 0, there exists only finitely many prime powers pk for |f (pk )| >

A+

,

and thus for n ∈ ◆ large enough, the prime factorization of n must contain at least a
prime power pk with |f (pk )| ≤

A+

. Hence, n can be written as

r

r+s

pki i

n=
i=1

r+s+t

pki i
i=r+1

pki i ,

i=r+s+1


CHAPTER 1. ANALYTIC TOOLS

7

where pi are pairwise distinct prime numbers such that
1 ≤|f (pki i )|
A+

for i = 1, . . . , r

≤|f (pki i )| ≤ 1
|f (pki i )| ≤

for i = r + 1, . . . , r + s
for i = r + s + 1, . . . , r + s + t, t ≥ 1.

A+

Using the fact that f is multiplicative, we have for all n ∈ ◆ large enough,
r

r+s

r+s+t

|f (pki i )|

|f (n)| =
i=1

|f (pki i )|
i=r+1

|f (pki i )| ≤ A · 1 ·
i=r+s+1

A+

< .

We present here some arithmetic functions that appears frequently in the study of
analytic number theory.
Definition 1.2.3. The Möbius function is defined by:



1



µ(n) :=
0




 (−1)r

if n = 1,
if n is not square-free, i.e. divisible by the square of a prime ,
if n is the product of r distinct primes.

It is easy to check that the arithmetic function µ(n) is multiplicative and µ3 = µ.
Proposition 1.2.4. For any natural number n ∈ ◆>0 , we have the following identity:

µ(d) = δ(n) =



 1 if n = 1,

 0 otherwise.

d|n

Proof. The assertion is trivial for n = 1. For n > 1, we write n in its unique prime
decomposition :
r

pki i ,

n=
i=1

where pi are pairwise distinct primes and r ≥ 1. Since µ vanishes at non square-free
integers, we have

µ(d) =
d|n

µ(d) =
d|p1 ···pr

µ(d)+
d|p2 ···pr

µ(p1 d) =
d|p2 ···pr

µ(d)−
d|p2 ···pr

d|p2 ···pr

µ(d) = 0.


CHAPTER 1. ANALYTIC TOOLS

8

Definition 1.2.5. The von Mangoldt’s function is defined as:

Λ(n) :=




 log p

if n = pk , k ≥ 1,


 0

otherwise .

The arithmetic function Λ(n) is not multiplicative.
Proposition 1.2.6. For any natural number n ∈ ◆>0 , we have the following identity:

Λ(d) = log n.
d|n

Proof. The assertion is trivial for n = 1. For n > 1, we write n in its unique prime
decomposition :
r

pki i ,

n=
i=1

where pi are pairwise distinct primes and r ≥ 1. Since Λ is non zero only at prime
powers, we have
r

ki

r

Λ(pji ) =

Λ(d) =
d|n

i=1 j=1

r

log pki i = log n.

ki log pi =
i=1

i=1

Definition 1.2.7. The divisor function d(n) counts the number of positive divisors of
positive integer n. If we write n in its unique prime factorization:
n = pk11 · · · pkr r ,

where p1 , . . . , pr are distinct primes and k1 , . . . , kr are nonnegative integers, then it is
straight forward to deduce that

d(n) = (k1 + 1) · · · (kr + 1).

With this formula, we see that d(n) is multiplicative. In general, for any positive integer
m and n, d(mn) ≤ d(m)d(n). This inequality follows from the inequality (α + β + 1) ≤
(α + 1)(β + 1).

CHAPTER 1. ANALYTIC TOOLS

9

Theorem 1.2.8. For real number Z ≥ 2, we have

d(k)2

Z log3 Z.

k≤Z

Proof. For X ≥ 1, we have

m≤X

1≤

1=

d(m) =

d≤X m≤X
d|m

m≤X d|m

d≤X

X
d

X log(2X).

Now, if Z ≥ 2, we have

d(n)2 =
n≤Z

n≤Z

d(k)
k≤Z

d(mk) ≤

1=

d(n)
k|n

k≤Z

Z
log
k

2Z

k

m≤ Z
k

d(k)
k≤Z

Z log Z
k≤Z

d(m)
m≤ Z
k

d(k)
.
k

But

k≤Z

d(k)
=
k

k≤Z

1

k

1=
d|k

d≤Z k≤Z
d|k

1
=
k

d≤Z

1
d

m≤ Z
d

1
m

log2 Z.

This completes the proof.
Definition 1.2.9. The Euler φ-function is defined as:

φ(n) := Card{1 ≤ a ≤ n : (a, n) = 1},
which is also the order of the multiplicative group ❩∗n .

As a consequence of Chinese remainder theorem, φ is multiplicative and has the explicit
formula:
1−

φ(n) = n
p|n

1
p

Theorem 1.2.10. For any δ > 0, we always have
n1−δ
= 0.
n→∞ φ(n)
lim

.


CHAPTER 1. ANALYTIC TOOLS

10

Proof. Since φ is multiplicative, by Theorem 1.2.2, it is sufficient to prove
pk(1−δ)
= 0.
pk →∞ φ(pk )
lim

This is easily obtained, as the formula of φ gives

pk(1−δ)
2
pk(1−δ)
≤ kδ −→ 0.
=
p−1
k
φ(p )
p pk →∞
pk ( p )
Definition 1.2.11. Given any nonzero natural number n ∈ ◆>0 , there are φ(n) classes
of residues relatively prime to n. Any set of φ(n) residues, one from each class, is called
a complete set of residues relatively prime to n. If there is no ambiguity, we
denote any complete set of residues relatively prime to n as ❩×
n , which is different from
the notation of the multiplicative group ❩∗n .
Definition 1.2.12. Let q and n be integers with q ≥ 1. The exponential sum

cq (n) :=

e
a∈❩×
q

an
q

is called the Ramanujan sum.
Remark: It is straight forward that the sum is independent of the chosen complete
set of residues relatively prime to q but some authors prefer to define the function by

summing over the set {a ∈ ◆ : 1 ≤ a ≤ q, (a, q) = 1}. This function will appear
frequently in the proofs of Vinogradov’s theorem and generalized Vinogradov’s theorem.
Hence, an explicit formula of Ramanujan sum will be very useful.
Theorem 1.2.13. For a fixed integer n, the Ramanujan sum cq (n) is a multiplicative
function of q, i.e., if (a, b) = 1, then

cab (n) = ca (n)cb (n).
×
×
Proof. If (a, b) = 1, then {as + br : r ∈ ❩×
a , s ∈ ❩b } = ❩ab . In fact, if there exist
×
r1 , r2 ∈ ❩×
a and s1 , s2 ∈ ❩b such that

as1 + br1 ≡ as2 + br2

mod ab,


CHAPTER 1. ANALYTIC TOOLS
then

11



 as1 + br1 ≡ as2 + br2

mod a,


 as1 + br1 ≡ as2 + br2

mod b,

which implies


 r1 ≡ b−1 br1 ≡ b−1 br2 ≡ r2

mod a,


 s1 ≡ a−1 as1 ≡ a−1 as2 ≡ s2

mod b,

where b−1 is the inverse of b modulo a and a−1 is the inverse of a modulo b. Also,
(as + br, ab) = (as + br, a) · (as + br, b) = (br, a) · (as, b) = 1. This proves the claim and
thus
rn sn
+
=
a
b

e

ca (n)cb (n) =

r∈❩

×
a

s∈❩

×
b

e
rb+sa∈❩

×
ab

(rb + sa)n
ab

= cab (n).

Lemma 1.2.14. The Ramanujan sum can be expressed in the form

cq (n) =

µ
d|(q,n)

q
d.

d

Proof. Since the geometric series
d

e
=1

n
d

=



 d

if d|n,


 0

if d n,

it follows that
q

cq (n) =

e

k=1
(k,q)=1
q

=

e
k=1

kn
q

kn
q

µ(d)
d|(k,q)

by Proposition 1.2.4

(2)


CHAPTER 1. ANALYTIC TOOLS
q
d

µ(d)

=

dn
q

e
=1

d|q

q
d

µ(d)

=

n

e

q
d

=1

d|q

q
µ
d

=
d|q

=

d

n
d

e
=1

q
d
d

µ
d|q
d|n

by (2)

q
d
d

µ

=

12

d|(q,n)

Theorem 1.2.15. The Ramanujan sum has an explicit formula:
µ
cq (n) =

q
(q,n)

φ

φ(q)
.

q
(q,n)

Proof. We define
q :=

q
.
(q, n)

Then the formula of φ gives
1−

q
φ(q)
=
φ(q )

p|q

q
p |q

1
p

1
1−
p

1−

= (q, n)
p|q
pq

1
p

1−

= (q, n)

p|(q,n)
pq

1
p

.


CHAPTER 1. ANALYTIC TOOLS

13

Then

cq (n) =

µ

q
d
d

µ

q
(q, n)
·
(q, n)
d

d|(q,n)

=
d|(q,n)

µ(q c)

=
c|(q,n)

by Lemma 1.2.14

(q, n)
c
µ(q )µ(c)

= (q, n)
c|(q,n)
(q ,c)=1

= µ(q )(q, n)
c|(q,n)
(q ,c)=1

1
c

because µ(q c) = 0 if (q , c) > 1

µ(c)
c

1−

= µ(q )(q, n)
p|(q,n)
pq

=

d

1
p

µ(q )φ(q)
.
φ(q )

We remark that |cq (n)| ≤ min{φ(q), n} and cq (n) = µ(q) if (q, n) = 1.

1.3

Dirichlet Series

In this section, we give a brief introduction of Dirichlet series and state some of their
important properties without providing proofs. Readers who want to learn more about
this topic are advised to refer to [4].
Definition 1.3.1. A Dirichlet series is a series of the form

∞
n=1

Cn
,
ns

with Cn , s ∈ ❈.
Proposition 1.3.2. (Uniqueness of Dirichlet series) The function f defined by:
∞

f (s) :=
n=1

Cn
ns

is identically zero on the domain of convergence of the Dirichlet series if and only if all


CHAPTER 1. ANALYTIC TOOLS

14

the coefficients Cn = 0.
Example 1.3.3. The Riemann zeta function ζ, its derivative ζ and the derivative of
log ζ are the standard Dirichlet series that have been frequently studied in analytic
number theory. They have the explicit formula
∞

ζ(s) :=
m=1

1
,
ms

∞

−ζ (s) =
m=1

log m
,
ms

ζ
− (s) =
ζ

∞
m=1

Λ(m)
ms

(s) > 1 and they all converges absolutely on this domain. The proofs of these

for

identities can be found in [5] G. Tenenbaum and [6] G. H. Hardy and E. M. Wright.

1.4

Infinite Products

Definition 1.4.1.
I. Let (αk )k∈◆>0 be a sequence of complex numbers. The nth partial product of this
sequence is the number
n

α1 · · · αn =

αk .
k=1

II. If the sequence of nth partial product converges to a limit α when n tends to infinity,
∞

αk converges and we denote

we say that the infinite product
k=1
∞

n

αk := lim

n→∞

k=1

αk = α.
k=1

III. If the sequence of partial products does not converge when n tends to infinity, then
we say that the infinite product diverges.
∞

Theorem 1.4.2. Let ak ≥ 0 for all k ∈ ◆>0 . The infinite product

k=1

∞

ak converges.

if and only if the infinite series
n

Proof. Let sn :=

k=1
n

(1+ak ). Since ak ≥ 0, we have the strict inequality

ak and pn :=
k=1

(1 + ak ) converges

k=1

sn < pn . Also, since the inequality
1 + x ≤ ex


CHAPTER 1. ANALYTIC TOOLS

15

is true for all x ∈ ❘, we have
n

0≤

n

k=1

n

n

eak = exp

(1 + ak ) ≤

ak <

k=1

k=1

ak

,

k=1

i.e.
0 ≤ sn < pn ≤ esn .
Since both sequences {sn }n∈◆>0 and {pn }n∈◆>0 are monotone increasing, they converges
if and only if they are bounded. Thus the inequality above implies that the sequence
{sn }n∈◆>0 converges if and only if the sequence {pn }n∈◆>0 converges. Notice that pn ≥ 1
for all n and hence its limit, if it exists, is non zero.
∞

(1 + ak ) is said to converge absolutely if

Definition 1.4.3. The infinite product

k=1

∞

(1 + |ak |) converges.

the infinite product

k=1
∞

(1+ak ) converges absolutely, then it converges

Theorem 1.4.4. If the infinite product

k=1

and the limit is independent of the order of which the product is taken, i.e. for any
permutation σ of ◆>0 , we have
∞

∞

(1 + aσ(k) ).

(1 + ak ) =

k=1

k=1

Furthermore, the limit is zero if and only if 1 + ak = 0 for some k.
∞

∞

Proof. By Theorem 1.4.2,

ak converges

(1 + ak ) converges absolutely if and only if
k=1

k=1

absolutely. Thus, let
∞

|ak | .

C := exp
k=1

For any 0 < < 21 , there exists N0 such that
|ak | < .
k>N0

Let σ be a permutation of ◆>0 . Then for any integer N ≥ N0 , there exists an integer
M ≥ N such that
{1, . . . , N } ⊂ {σ(1), . . . , σ(M )}.


CHAPTER 1. ANALYTIC TOOLS

16

Then we have
M

N

(1 + aσ(k) ) −
k=1

(1 + ak ) =
k=1





N

(1 + ak ) − 1 ,

(1 + ak ) 
k=1

k∈E(M,N )

where E(M, N ) := {σ(1), . . . , σ(M )} \ {1, . . . , N }. Since E(M, N ) ⊂ ◆>N0 , thus

(1 + |ak |) − 1

(1 + ak ) − 1 ≤

k∈E(M,N )

k∈E(M,N )

e|ak | − 1

≤
k∈E(M,N )





|ak | − 1

≤ exp 
k∈E(M,N )





≤ exp 

|ak | − 1

k>N0

≤e −1≤2 .

This gives
M

N

(1 + aσ(k) ) −
k=1

N

N

(1 + ak ) ≤
k=1

|ak |

(1 + ak ) · 2 ≤ exp

· 2 ≤ 2C .

k=1

k=1

Now, if σ is the identity function, the inequality above is satisfied for any M ≥ N > N0
N

(1 + ak )

and hence the sequence of partial product

k=1

converges. Furthermore, for any M ≥ N = N0 , we have
N0

M

(1 + ak ) −
k=1

k=1

k∈◆>0

N0

M

(1 + ak ) ≤

is Cauchy and thus

(1 + aσ(k) ) −

N0

(1 + ak ) ≤ 2

k=1

k=1

which implies
N0

M

(1 + ak ) ≥ (1 − 2 )
k=1

(1 + ak ) .
k=1

By taking the limit when M tends to infinity, we obtain
∞

N0

(1 + ak ) ≥ (1 − 2 )
k=1

(1 + ak ) .
k=1

(1 + ak )
k=1

CHAPTER 1. ANALYTIC TOOLS

17

Hence, the limit of the infinite product is not zero if none of the 1 + ak = 0 and the
converse is trivially true. To complete the proof, let L be the limit of the sequence of
N

partial product

(1 + ak )
k=1

that

and by increasing N0 if necessary, we can assume
k∈◆>0
N0

(1 + ak ) − L ≤ .
k=1

Then for any arbitrarily permutation σ, let N = N0 and M ≥ N as defined before, then
for any K ≥ M , we have
K

N0

K

(1 + aσ(k) ) − L ≤
k=1

(1 + aσ(k) ) −
k=1

N0

(1 + ak ) − L

(1 + ak ) +
k=1

k=1

≤ 2C + ,

which yields the assertion that the limit of the infinite product is independent of the
order of which the product is taken.
Definition 1.4.5. An Euler product is an infinite product over the prime numbers.
Theorem 1.4.6. Let f (n) be a multiplicative function that is not identically zero. If the
∞

f (n) converges absolutely, then

series
n=1

∞

∞

1 + f (p) + f (p2 ) + . . . =

f (n) =
n=1

p

f (pk ) .

1+
p

k=1

Furthermore, the limit of the infinite product is independent of the ordering of the prime
index.
∞

f (n) converges absolutely, then the series

Proof. If
n=1

an :=







∞

f (pk ) if n = p is prime,
k=1



 0

otherwise,


CHAPTER 1. ANALYTIC TOOLS

18

is well-defined. Also, we have
∞

∞

∞
p

n=1

∞

|f (pk )| ≤

f (pk ) ≤

|an | =

p k=1

k=1

|f (n)| < ∞.
n=1

∞

∞
p

n=1

f (pk )

1+

(1 + an ) =

Thus, the infinite product

converges absolutely and

k=1

by Theorem 1.4.4, it converges and the limit is independent of the ordering of the prime
index. We proceed to complete the proof by establishing the equality. Let

> 0, then

there exists an integer N0 such that

|f (n)| < .
n≥N0

For every positive integer n, let P (n) denote the greatest prime factor of n. Since the
∞

|f (pk )| converges for any finite set I of prime numbers, therefore by Fubini’s

series
p∈I k=0

theorem and the fact that f is multiplicative, we have, for any N > N0 ,
∞

∞
k

1+
p≤N

f (p )

f (pk )

=
p≤N

k=1

=

k=0

f (n)
P (n)≤N

and so
∞

∞

f (n) −
n=1

1+
p≤N

∞
k

f (p )

f (n) −

=
n=1

k=1

≤

f (n)
P (n)≤N

|f (n)| ≤
n>N

|f (n)| < .
n>N0

By taking the limit when N tends to infinity, we obtain the desired equality.