MINISTRY OF EDUCATION AND TRAINING
HANOI PEDAGOGICAL UNIVERSITY N2

DEPARTMENT OF MATHEMATICS

NGUYEN THI TRANG

GAUSS SUMS AND JACOBI SUMS

GRADUATION THESIS
Major: Algebra

HA NOI - 2019


MINISTRY OF EDUCATION AND TRAINING
HANOI PEDAGOGICAL UNIVERSITY N2

DEPARTMENT OF MATHEMATICS

NGUYEN THI TRANG

GAUSS SUMS AND JACOBI SUMS

GRADUATION THESIS
Major: Algebra

Supervisor:
Dr. NGUYEN DUY TAN

HA NOI – 2019

1

Thesis Acknowledgement
I would like to express my gratitudes to the teachers of the Department
of Mathematics, Hanoi Pedagogical University 2, the teachers in the algebra group as well as the teachers involved. The lecturers have imparted
valuable knowledge and facilitated for me to complete the course and the
thesis.
In particular, I would like to express my deep respect and gratitude to
Dr. Nguyen Duy Tan, who has direct guidance, help me complete this
thesis.
Due to time, capacity and conditions are limited, so the thesis can not
avoid errors. Then, I look forward to receiving valuable comments from
teachers and friends.
Ha Noi, May 5, 2019
Student

Nguyen Thi Trang


Graduation thesis

NGUYEN THI TRANG

Thesis Assurance
I assure that the data and the results of this thesis are true and not
identical to other topics. I also assure that all the help for this thesis has
been acknowledged and that the results presented in the thesis has been
identified clearly.

Ha Noi, May 5, 2019
Student

Nguyen Thi Trang

i


Contents

Preface

1

1 Preliminaries

3

1.1

Quadratic residues . . . . . . . . . . . . . . . . . . . . . .

3

1.2

The matrix and its characteristic function . . . . . . . . .

6


1.3

Multiplicative characters . . . . . . . . . . . . . . . . . . .

8

2 Quadratic Gauss sums

13

2.1

The absolute value of quadratic Gauss sums . . . . . . . .

13

2.2

The sign of the quadratic Gauss sums . . . . . . . . . . . .

17

2.3

Extensions to composite moduli . . . . . . . . . . . . . . .

30

2.4


Application to the quadratic Gauss sums . . . . . . . . . .

33

2.4.1

The law of quadratic reciprocity . . . . . . . . . . .

33

2.4.2

Some related trigonometric problems . . . . . . . .

34

3 Gauss sums and Jacobi sums

36

3.1

Gauss sums . . . . . . . . . . . . . . . . . . . . . . . . . .

36

3.2

Jacobi sums . . . . . . . . . . . . . . . . . . . . . . . . . .


38

3.3

The equation xn + y n = 1 in Fp . . . . . . . . . . . . . . .

46

3.4

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

Conclusion

58

References

58
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Graduation thesis

NGUYEN THI TRANG

Preface
Devised in the 19th century, Gauss and Jacobi Sums are classical formulas that form the basis for contemporary research in many of today’s

sciences. In algebraic number theory, a Gauss sum is a particular kind
of finite sum of roots of unity. Gauss utilized the properties of the sum
to solve certain problem in number theory; a particular case is one of the
proofs of the quadratic reciprocity law. A Jacobi sum is a type of character
sum formed with Dirichlet characters. Simple examples would be Jacobi
sums J(χ, ψ) for Dirichlet chracters χ, ψ modulo a prime number p, defined
by J(χ, ψ) =

χ(a)ψ(1 − a), where the summation runs over all residues

a = 2; 3; · · · ; p − 1 (mod p) (for which neither a nor 1 − a is 0). The notions of Gauss and Jacobi sums are defined and applied to investigate the
number of solutions of polynomial equations over finite field.
Base on the basic knowledge about algebraic and desiring comprehensive
improvement of Mathematics, we has selected a topic “Gauss sums and
Jacobi sums” for Graduation Thesis.
This thesis is the general result of books [2],[3],[4]. The main aim in
this thesis is to know how to be Gauss sums and Jacobi sums, understand
some proofs about The sign of the Quadratic Gauss Sum, prove the laws
of quadratic Reciprocity and find the number of solutions to congruence
equations of the form xn + y n = 1 (mod p).
Let us describe the content of this thesis.
In Chapter 1, we will review the definition and some basic properties
of quadratic residues, Legendre symbol, prove Wilson’s theorem and recall
some knowledge about the matrix and its characteristic function. Then we
introduce the notion of multiplicative characters needed for the results of
this thesis.
1


Graduation thesis


NGUYEN THI TRANG

In Chapter 2, we introduce quadratic Gauss sums. We will present
the notion of Gauss sums, hence quadratic Gauss sums. We will give two
methods to prove the formula to calculate the absolute value of quadratic
Gauss sums. In particular, we also present three methods to prove about
the sign of the quadratic Gauss sums. Moreover, we also prove for two
remaining cases n ≡ 2 (mod 4) and n ≡ 0 (mod 4). Next, application to
the quadratic Gauss sums is given such as the law of quadratic reciprocity,
some related trigonometric problems.
In Chapter 3, a more general notion of Gauss sum and Jacobi sums will
be introduced. Here we shall consider a problem of counting the number
of solutions of equations with coefficients in a finite field. Finally, by using
Jacobi sums, we give some exercises to practice.

2


Chapter 1
Preliminaries
In this Chapter, we will review some results in number theory which will be
used in the other chapters. We begin with the notion of quadratic residues,
matrices and their characteristic functions. Multiplicative characters are
also introduced.

1.1

Quadratic residues


Theorem 1.1.1 (Wilson’s Theorem). A positive integer n > 1 is a prime
if and only if (n − 1)! ≡ −1 (mod n).
Proof. Firstly, we prove that: If a positive integer n > 1 is a prime, then
(n − 1)! ≡ −1 (mod n). Consider the field Z/nZ - the set of integers
modulo n.
By Fermat’s little theorem for n is a prime, we have
xn−1 − 1 ≡ 0

(mod n)

for all 1 ≤ x ≤ n − 1. So in Z/nZ, there must be the n − 1 roots of f .
Hence, we can write
n−1

x

n−1

−1=

n−1

(x − k) = (−1)
k=1

n−1

·

(k − x).

k=1

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Graduation thesis

NGUYEN THI TRANG

For odd prime n, n − 1 is even, and (−1)n−1 = 1. For even prime n = 2,
we get (−1)n−1 = −1 ≡ 1 (mod 2). Thus, xn−1 − 1 =
Let x = 0 we get

n−1
k=1 k

n−1
k=1 (k

− x).

= −1 = (n − 1)! in Z/nZ. That means

(n − 1)! ≡ −1 (mod n).
Secondly, we must prove that: For a positive integer n > 1, if (n − 1)! ≡
−1 (mod n), then n is a prime.
Indeed, assume that n is a composite number, then it has at least one
divisor d less than n, that is 2 ≤ d ≤ n − 1. But since (n − 1)! is the
product of all positive integers from 1 to n − 1, the product must contain d
and thus (n − 1)! is divisible by d. So we have (n − 1)! ≡ 0 (mod d). Since

d | n, (n − 1)! ≡ 0 ≡ −1 (mod d), contradicting the hypothesis. Then n
can not be composite, hence prime.
Let p be a prime number. We consider the equation
x2 ≡ k

(mod p).

Definition 1.1.2. We say that an integer k which is not divisible by p is
a quadratic residue modulo p if there exists an integer x such that x2 ≡ k
(mod p); we say that k is a quadratic non-residue if no such x exists.
The case p = 2 is trivial, so we assume p > 2, means that p is an odd
prime number.
From Fermat’s little theorem, we have xp−1 ≡ 1 (mod p), with x ≡ 1
(mod p). Then
(xp−1 − 1)(xp−1 + 1) ≡ 0
Theorem 1.1.3. The congruent equation x

(mod p).
p−1
2

p−1
roots, are congruent classes 1 , 2 , · · · ,
2
2

2

4


− 1 ≡ 0 (mod p) has
2

p−1
2

(mod p). Other congru-


Graduation thesis

NGUYEN THI TRANG

ent classes (not class 0) are roots of the equation
x

p−1
2

+1≡0

(mod p).
p−1

Proof. Indeed, we know that the equation x 2 − 1 ≡ 0 (mod p) has at
p−1
most
roots in a residue system modulo p.
2
2

p−1
2 2
We see that the congruent classes 1 , 2 , · · · ,
(mod p) are dis2
tinguish and from Fermat’s little theorem, they are also roots of the equation x

p−1
2

roots of x

− 1 ≡ 0 (mod p). From that, other congruent classes are not
p−1
2 −1≡0

(mod p)

.

Moreover, (xp−1 − 1) = (x
are roots of the equation x

p−1
2

p−1
2

+ 1)(x


p−1
2

− 1). So, other congruent classes

+ 1 ≡ 0 (mod p).

Corollary 1.1.4. The number a is a quadratic residue modulo p if and
only if
a

p−1
2

≡1

(mod p).

Proof. Suppose that a is a quadratic residue modulo p, then a ≡ x2
(mod p) for some x ∈ Z. Thus, a

p−1
2

≡ xp−1 ≡ 1 (mod p).

Now, if a is not a quadratic residue modulo p, then from Theorem 1.1.3,
a is a root of the equation x

p−1

2

+1 ≡ 0 (mod p), this means that a

(mod p).
Definition 1.1.5 (Legendre symbol). We write



1
if a is a quadratic residue modulo p



a
= −1 if a is a quadratic non-residue modulo p

p



0
if a ≡ 0 (mod p).
a
p

Corollary 1.1.6 (Euler criterion).

5


≡a

p−1
2

(mod p).

p−1
2

≡ −1


Graduation thesis

NGUYEN THI TRANG

Corollary 1.1.7. One has
−1
p

= (−1)

p−1
2


1 if p ≡ 1 mod 4
=
−1 if p ≡ 3 mod 4.


From Euler’s criterion, we see that
−
p

: Z/(p)× → {±1}

is a homomorphism group. This means that for all a, b are coprime to p,
we have
a
p

b
p

=

ab
.
p

In particular, we have the following equality:
u2 a
p

1.2

=

a

.
p

The matrix and its characteristic function

Definition
 1.2.1.
α11 · · ·

M =
· · · · · ·
αn1 · · ·

A matrix
 with entry in C is a square array
α1n

2
···
 = αkl of n elements, αkl ∈ C.
αnn

Definition 1.2.2.
(a) The definition of multiplication:
αkl

βkl =

m αkm βml .


(b) M 2 = M · M .

1 for k = l;
We denote : ekl =
and In = (ekl ) the identity matrix of
0 for k = l.
degree n.
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Graduation thesis

NGUYEN THI TRANG

Definition 1.2.3. The characteristic function of a matrix M = αkl , is
the determinant Φ(ξ) = |ξ(ekl ) − (αkl )|.
We note that Φ(ξ) is a polynomial of exactly n-th degree with leading
n
r=1 (ξ

coefficient 1, and therefore Φ(ξ) =

− ξr ) for some ξr ∈ C, the ξr

are called the characteristic roots of the matrix. The coefficient of ξ n−1 is
−

k

αkk , so that


n
r=1 ξr

=

k

αkk , and

k

αkk is called the trace of the

matrix.
Theorem 1.2.4. The characteristic roots of M 2 are the ξr2 , and therefore
those of (M 2 )2 are the ξr4 .
Proof. We must prove that
n

ξ(ekl ) −

(ξ − ξr2 ).

αkm αml =
m

r=1

Indeed, let θ be a number for which θ2 = ξ. Then we have

(θekl + αkl )(θekl − αkl ) = (

(θekm + αkm )(θeml − αml ))
m

= (θ2

ekm eml − θ
m

αkm eml −

ekm αml + θ
m

m

= (θ2 ekl − θαkl + θαkl −

αkm αml )
m

αkm αml ) = (ξekl −
m

αkm αml ).
m

Therefore
|θ(ekl ) + (αkl )||θ(ekl ) − (αkl )| = |ξ(ekl ) −


αkm αml |.
m

Moreover we get
n

|θ(ekl ) − (αkl )| = Φ(θ) =

(θ − ξr ),
r=1

7


Graduation thesis

NGUYEN THI TRANG

and
n
n

n

|θ(ekl ) + (αkl )| = (−1) Φ(−θ) = (−1)

n

(−θ − ξr ) =

r=1

(θ + ξr ).
r=1

So,
n

ξ(ekl ) −

αkm αml =
m

n
2

(θ +
r=1

ξr2 )

(ξ − ξr2 ).

=
r=1

Therefore, the characteristic roots of M 2 are the ξr2 .

1.3


Multiplicative characters

×
A multiplicative character on Fp is a map χ : F×
p → C satisfying:

χ(ab) = χ(a)χ(b),

∀a, b ∈ F×
p.

a
is an example of such a character if it is rep
garded as a function of the coset of a modulo p.
The Legendre symbol

The trivial multiplicative character ε is defined by ε(a) = 1, ∀a ∈ Fp . If
χ = ε, we define χ(0) = 0.
Proposition 1.3.1. Let χ be a multiplicative character and a ∈ F×
p . Then:
(a) χ(1) = 1;
(b) χ(a) is a (p − 1)st root of unity;
(c) χ(a−1 ) = χ(a)−1 = χ(a).
Proof. (a) One has χ(1) = χ(1.1) = χ(1).χ(1). Since χ(1) = 0, χ(1) = 1.
p−1
(b) For ∀a ∈ F×
= 1. So
p , we have a

χ(a)p−1 = χ(ap−1 ) = χ(1) = 1.

Therefore χ(a) is a (p − 1)st root of unity.
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Graduation thesis

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(c) One has
χ(1) = χ(a.a−1 ) = χ(a).χ(a−1 ) = 1.
Thus, χ(a−1 ) = χ(a)−1 . Since χ(a)p−1 = 1, |χ(a)|p−1 = 1. Therefore,
|χ(a)| = 1, and χ(a)−1 = χ(a).
Proposition 1.3.2. Let χ be a multiplicative character. Then

0 if χ = ε
χ(t) =
p if χ = ε.
t∈Fp

Proof. If χ = ε, then

t χ(t)

=

t ε(t)

= p. (Since ε(a) = 1, ∀a ∈ Fp ).

If χ = ε, then there exists an a ∈ F×

p such that χ(a) = 1. Put T =
t χ(t).

We have
χ(a).T = χ(a).

χ(t) =
t

Since χ(a) = 1, T = 0 =

χ(a).χ(t) =
t

χ(at) = T.
t

t χ(t).

We shall drop the use of the word ”multiplicative” for the remainder of
this chapter.
Definition 1.3.3. (1) If χ and λ are characters, then χλ is the map that
takes a ∈ F×
p to χ(a).λ(a).
−1
(2) If χ is a character, χ−1 is the map that takes a ∈ F×
p to χ(a) .

We may verify that χλ and χ−1 are characters and that these definitions
make the set of characters into a group. The trivial character is ε.

Proposition 1.3.4. The group of characters is a cyclic group of order
p − 1. If a ∈ F×
p and a = 1, then there is a character χ such that χ(a) = 1.
×
and χ be a character on F×
Proof. Let F×
p = characters of Fp
p.

9


Graduation thesis

NGUYEN THI TRANG

×
×
We know that F×
p is a cyclic, Let g be a generator of Fp , so that Fp = g .
l
l
Then ∀a ∈ F×
p , one has a = g for some l ∈ Z. Hence χ(a) = χ(g ) =

χ(g)l . This implies that χ is completely determined by the value of χ(g).
2πi

Note that χ(g) is a (p − 1)st of unity, i.e., χ(g) ∈ 1; e p−1 ; · · · ; e


2πi·(p−2)
p−1

.

Therefore, F×
p ≤ p − 1.
2πik

×
k
p−1 . We first show that λ is
Define a function λ : F×
p → C by λ(g ) = e

well-defined. Suppose g k =g l , then k = m(p − 1) + l for some m ∈ Z. Thus,
2πi[m(p−1)+l]
p−1

2πik

λ(g k ) = e p−1 = e

2πil

2πil

= e2πim · e p−1 = e p−1 = λ(g l ).

So λ is well-defined. It is easy to see that λ is a character.

We see that λp−1 = ε. Suppose λn = ε, it implies that λn (g) = ε(g) = 1.
2πin

Moreover, λn (g) = λ(g n ) = e p−1 = 1. So p − 1 | n. Thus the order of λ in
×
F×
p is ord(λ) = p − 1. Therefore Fp = λ , is a cyclic group of order p − 1.
l
If a ∈ F×
p and a = 1, then a = g with (p − 1)

l. Therefore λ(a) =

2πil

λ(g l ) = e p−1 = 1.
Corollary 1.3.5. If a ∈ F×
p and a = 1, then

χ χ(a)

= 0, where the

summation is over all characters.
Proof. Let T =

χ χ(a).

Since a = 1, so there exists a character λ such


that λ(a) = 1. Then
λ(a) · T = λ(a) ·

χ(a) =

λ(a)χ(a) =

χ

Since λχ runs over all characters,

χ
χ λχ(a)

λχ(a).
χ

= T. So (λ(a) − 1) · T = 0,

then T = 0.
n
Proposition 1.3.6. If a ∈ F×
p , n | p − 1, and x = a is not solvable, then

there is a character χ such that
(a) χn = ε,
10


Graduation thesis


NGUYEN THI TRANG

(b) χ(a) = 1.
Proof. Let g ∈ F×
p be a generator and define a function λ by the equation
λ(g k ) = e2πik/p−1 , then λ is a character.
Put χ = λp−1/n , then χn = λp−1 = ε. So
χ(g) = λ(p−1)/n (g) = λ(g)(p−1)/n = e2πi/n .
l
n
l
×
Since a ∈ F×
p and Fp = g , a = g for some l ∈ Z. Because x = a = g is

not solvable, so n l. Therefore,
χ(a) = χ(g l ) = χ(g)l = e2πil/n = 1.
For a ∈ Fp , let N (xn = a) denote the number of solutions of the equation
xn = a.
Proposition 1.3.7. If n | p − 1, then N (xn = a) =

χn =ε χ(a)

where the

sum is over all characters of order dividing n.
Proof. As in the proof of Proposition 1.3.6, we found a character χ such
that χ(g) = e2πi/n . Then {ε, χ, χ2 , · · · , χn−1 } are exactly n characters of
order dividing n.

If a = 0 then the equation xn = 0 has only one solution x = 0. And
χn =ε χ(0)

= ε(0)+ χ(0)+ · · · + χn−1 (0) = 1. (Since ε(0) = 1 and χ(0) = 0

∀χ = ε).
If a = 0 and xn = a is solvable, then there exists t such that tn = a. So
χ(a) = χ(tn ) = χ(t)n = χn (t) = ε(t) = 1.
Then χ(a) = ε(t) = 1, and

χn =ε χ(a)
n

= n = N (xn = a).

Suppose that a = 0 and x = a is not solvable. Let T =

χn =ε χ(a).

We must show that T = 0. Indeed, from Proposition 1.3.6, there is a

11


Graduation thesis

NGUYEN THI TRANG

character γ such that γ(a) = 1 and γ n = ε. Then,


χn =ε

χn =ε

χn =

Since γ(a) = 1, one has T = 0.

12

γχ(a) = T.

γχ(a) =

γ(a)·χ(a) =

χ(a) =

γ(a)·T = γ(a)·

(γχ)n =


Chapter 2
Quadratic Gauss sums
In this chapter, quadratic Gauss sums are introduced and we calculate
these sums by some methods. We will use quadratic Gauss sums to prove
the law of quadratic reciprocity.

2.1


The absolute value of quadratic Gauss sums

A quadratic Gauss sum is a particular kind of finite sum of roots of unity,
typically
p−1

G(a, p) = ga =
k=0

2πiak 2
.
exp
p

In particular, in the case a = 1, we get a quadratic Gauss sum
p−1

G(1, p) = g1 =
k=0

2πik 2
exp
.
p

Proposition 2.1.1.
p−1

G(a, p) =

k=0
p−1

Proof. We have

exp
k=0

2πiak 2
=
exp
p

p−1

k=0

k
p

exp

2πiak
.
p

2πik
= 0. For p is an odd prime, one considers two
p
13



Graduation thesis

NGUYEN THI TRANG

sums
A :=
k≡x2

2πiak
p

exp

2πiak
p

(mod p)

B :=
k≡x2

exp

(mod p)

Thus, 1 + A + B = 0.
a
If

= 1, then an integer k is a quadratic residue if and only if ak is
p
a quadratic residue. Then
p−1

G(a, p) = 1 + 2A = A − B =
k=1

a
2πiak
exp
.
p
p

a
= −1, then an integer k is a quadratic residue if and only if ak
p
is not a quadratic residue. Then
If

p−1

G(a, p) = 1 + 2B = B − A =
k=1

k
p

exp


2πiak
.
p

Proposition 2.1.2.
G(a, p) =

a
p

· G(1, p).

Proof. Denote ξ = e2πi/p . If a ≡ 0 (mod p), then ξ ak = 1, ∀ 0 ≤ k ≤ p − 1.
Thus,
p−1

G(a, p) =
k=0

k
p

a
p

=0=

· G(1, p).


If a ≡ 0 (mod p), then
a
p

p−1

· G(a, p) =
k=0

ak
p

· ξ ak =
x

14

x
p

· ξ x = G(1, p).


Graduation thesis

NGUYEN THI TRANG

It follows that
a
p


2

a
p

· G(1, p) =

· G(a, p) = G(a, p).

Let ξ = e2πi/p and we consider
p−1

a a
ξ .
p

G(1, p) = g1 = g =
a=1

Proposition 2.1.3. One has
g2 =

p−1
−1
p = (−1) 2 p.
p

First proof. One has
p−1


a a
ξ
p

g·g =
a=1

·
b=1

b b
ξ
p

=
a

b

a
p

b (a+b)
ξ
p

ab (a+b)
ξ
.

p

=
a

p−1

b

Substituting a by ab (mod p) we get

2

g =
a

=
b

=
=
=

b

−1
p

ab2
p


a (a+1)b
ξ
p

ξ (ab+b) =
a

a (a+1)b
ξ
p

+
a=−1

−1
(p − 1) +
p
−1
(p − 1) +
p
−1
(p − 1) +
p

b

b

a=−1


a=−1

−1
p

15

a
p

ξ (a+1)b
b

a
(−1)
p
=

p−1
−1
p = (−1) 2 p.
p


Graduation thesis

NGUYEN THI TRANG

Second proof. We have

x (ax)
ξ .
p

ga .g−a =
x

x

y

y

y a(x−y)
ξ
.
p

x
p

=

y (−ay)
ξ
p

Summing both sides over a we get
x
p


y
p

x
p

y
δ(x, y)p,
p

if x ≡ y

(mod p)

if x ≡ y

(mod p).

ga .g−a =
x

a

y

=
x

where


y


1
δ(x, y) =
0

ξ a(x−y)
a

So,
ga · g−a = (p − 1)p.
a

One also has
ga · g−a =

a
g·
p

−a
g=
p

−a2
p

g2 =


Thus
ga g−a =
a

−1
(p − 1)g 2 .
p

Therefore
−1
(p − 1)g 2 = (p − 1)p.
p
So we get
g2 =

−1
p.
p

16

−1 2
g .
p


Graduation thesis

2.2


NGUYEN THI TRANG

The sign of the quadratic Gauss sums

We see that


p
if p ≡ 1 (mod 4)
2
g =
−p if p ≡ 3 (mod 4).

So


±√p if p ≡ 1 (mod 4)
g=
±i√p if p ≡ 3 (mod 4).

Example 2.2.1. We will consider Gauss sum for p = 7 and p = 11.
(a) Quadratic Gauss sum for p = 7.
Denote ξ = e2πi/7 . We know that the quadratic residues modulo 7 are
{1, 2, 4} and {3, 5, 6} are quadratic non-residues modulo 7. Therefore,
from the quadratic Gauss sum we get
√
ξ + ξ 2 − ξ 3 + ξ 4 − ξ 5 − ξ 6 = ±i 7.
It is easy for us to check that this is correct. Indeed,
(ξ + ξ 2 − ξ 3 + ξ 4 − ξ 5 − ξ 6 )2 = ξ 2 + 2ξ 3 − ξ 4 + ξ 6 − 6ξ 7 + ξ 8 − ξ 10 +

+ 2ξ 11 + ξ 12
= −6 + ξ + ξ 2 + ξ 3 + ξ 4 + ξ 5 + ξ 6
= −6 − 1 = −7.
(b) Quadratic Gauss sum for p = 11.
Denote ξ = e2πi/11 . The quadratic residues modulo 11 are {1, 3, 4, 5, 9}
and {2, 6, 7, 8, 10}. Therefore, by the quadratic Gauss sum formula
2

3

4

5

6

7

8

9

ξ−ξ +ξ +ξ +ξ −ξ −ξ −ξ +ξ −ξ
Indeed,
17

10

√


= ±i 11.


Graduation thesis

NGUYEN THI TRANG

(ξ − ξ 2 + ξ 3 + ξ 4 + ξ 5 − ξ 6 − ξ 7 − ξ 8 + ξ 9 − ξ 10 )2
= −10 + ξ + ξ 2 + ξ 3 + ξ 4 + ξ 5 + ξ 6 + ξ 7 + ξ 8 + ξ 9 + ξ 10 = −10 − 1 = −11.
(Since ξ 11 = 1 and 1 + ξ + ξ 2 + · · · + ξ 10 = 0.)
Theorem 2.2.2. One has

√p
g=
i√p

if p ≡ 1

(mod 4)

if p ≡ 3

(mod 4).

First proof. [Use the Gaussian binomials]
For each integer n

0, define

(q)n = (1 − q)(1 − q 2 )...(1 − q n ),

where if n = 0, the empty product is understood to equal 1. For 0
n
n
(q)n
the Gauss coefficient
is defined by
=
.
(q)m (q)n−m
m
m
One has

m

n,

(1 − q)(1 − q 2 )...(1 − q n−1 )(1 − q m )
=
(1 − q)(1 − q 2 )...(1 − q m−1 )(1 − q)(1 − q 2 )...(1 − q n−m )(1 − q m )
m−1
n−1

n−1
m

(1 − q)(1 − q 2 )...(1 − q n−1 )(1 − q n−m )
=
(1 − q)(1 − q 2 )...(1 − q m )(1 − q)(1 − q 2 )...(1 − q n−m−1 )(1 − q n−m )


So
n−1
m−1

for 1

n−1

+ qm

m < n. Then we see that

m

n
m

=

(2.1)

m

is a polynomial in the variable q.

For a nonnegative integer n, define fn (q) =
18

n


n
k
k=0 (−1)

n
k

. Setting


Graduation thesis

β=e

2πi
p

NGUYEN THI TRANG

.

Firstly, we prove that: fp−1 (β) =
p−1

jβ

p

(β) = q −j .


j

−j(j+1)
2

p−1

−

j

. Indeed,

(β),

j−1

where
p

(q)p
(β) = 0.
(q)j .(q)p−j

(β) =

j
So,

p−1

j

(β) = −q −j .

p−1
j−1

(β).

Thus,
p−1

(β) = (−1)j .β

j
And

p−1

(−1)j .

fp−1 (β) =

p−1
j

j=0

Putting α =


p−1
2 ,

(β) =

β

−j(j+1)
2

.

j

we have

fp−1 (β) =

β

(p−1).j.(j+1)
2

β α(j

=

j
2


−j(j+1)
2

2

+j)

j
−2α

(Since β −2jα = (β jα )

= β jα

(−p+1)

β α(j−α)

=
j

= β jα )

Then,
fp−1 (β) = β

−α3

β


αj 2

=β

−(p2 −1)
8

j

=β

−(p2 −1)
8

β αj
j

α
G(2).
p

19

2

2

−α3

.



Graduation thesis

NGUYEN THI TRANG

Since
α
p

p−1
2

=

p
−2
p

=

=

p−1
2

4·

=


p

= (−1)

(p−1)(p−3)
8

2p − 2
p

.

We prove the recursion formula fn (q) = (1−q n−1 )fn−2 (q) for n ≥ 2. Indeed,
we have
n+1
1 − q n+1
(q)n+1
=
=
.
(q)m+1 (q)n−m
m 1 − q m+1
m+1
n

So,
n
m

(1 − q n+1 ) = (1 − q m+1 )


n+1
m+1

.

One has
n−2

(1 − q

n−1

(1 − q n−1 )

)fn−2 (q) =
j=0

n+1
m+1

n−2

(1 − q j+1 )(−1)j .

=
j=0
n−2

j


(−1) ·

=
j=0

=

n−1
1

n−1
j+1

n−1
j+1
n−2

(−1)j+1 · q j+1 ·

+
j=0

− · · · + (−1)n−2 ·

+ (−1)n−1 · q n−1 ·

n−1
n−1


20

.

n−1
n−1

−q·

n−1
j+1
n−1
1

+ ···+