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City University of New York (CUNY) City University of New York (CUNY)

Dissertations, Theses, and Capstone Projects CUNY Graduate Center

2-2015

Explicit Solutions of Imaginary Quadratic Norm Equations Explicit Solutions of Imaginary Quadratic Norm Equations

Sandra Sze

Graduate Center, City University of New York

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### Explicit Solutions of Imaginary Quadratic Norm Equations

by Sandra Sze

A dissertation submitted to the Graduate Faculty in Mathematics in partial fulfillment of the requirements for the degree of Doctor of Philosophy, The City University of New York.

2015

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ii

2015c

Sandra Sze

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iii This manuscript has been read and accepted for the Graduate Faculty in Mathematics in satisfaction of the dissertation requirements for the degree of Doctor of Philosophy.

Victor Kolyvagin

Date Chair of Examining Committee

Linda Keen

Date Executive Officer

Victor Kolyvagin

Alexander Gamburd

Kenneth Kramer

Supervisory Committee

THE CITY UNIVERSITY OF NEW YORK

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iv Abstract

Explicit Solutions of Imaginary Quadratic Norm Equations by

Sandra Sze

Let K = Q(

−d) be an imaginary quadratic extension of Q. Let h be the class number and D be the discriminant of the field K. Assume p is a prime such that D

p



= 1. Then psplits in K. The elements of the ring of integersOKare of the form x +

−dy if d ≡ 1, 2 (mod 4) and x +1 +

−d

2 yif d ≡ 3 (mod 4), where x and y ∈ Z. The norm NK/Q(x +

−dy) = x2+ dy2and NK/Q



x+1 +

−d

2 y



=(2x + y)2 4 +dy2

4 . In this the- sis, we find the elements of norm ph explicitly. We also prove certain congruences for solutions of norm equations.

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## Acknowledgements

I would like to first thank my advisor Victor Kolyvagin for coming up with the problem for this thesis, for his patience with me as I worked on it and for the many ideas he had in helping me solve this problem. I would also like to thank the NSF for their support of me through the RTG grant, as well as the National Physical Science Consortium for funding me during my first year at CUNY Graduate Center.

I thank the committee members Kenneth Kramer and Alexander Gamburd for taking the time to read my paper and providing feedback.

I am grateful for other professors who believed in my ability and helped me to learn mathematics: Lucien Szpiro, Jozef Dodziuk, Burton Randol, Raymond Hoobler, Delaram Kahrobaei, Carlos Moreno, Robert Thompson, the late Gilbert Baumslag, Roni Gouraige, Ravi Kulkarni, Wallace Goldberg, Russell Miller.

There are many other mathematicians who have helped me throughout the years. They were willing to talk math with me and helped me to become a better student and mathe- matician. Please forgive me if I forgot to add you to the list: Robert Suzzi-Valli, Ha Lam, Yunchun Hu, Viveka Erlandsson, Maggie Habeeb, Rebecca Steiner, Shlomo Ben-Har,

v

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vi Joey Hirsh, Phillip Williams, Nikita Miasnikov, Andrew Stout, Brian Stout, Lloyd West, Liang Zhao, Jorge Florez, Joe Kramer-Miller, John Basias, Ben Hutz, Adam Towsley, Blanca Marmolejo, Iryna Pavlyuk. I will always remember and cherish the hours we spent in the library, in empty classrooms, at Queens College, at different cafes, in the 8th floor cafeteria, working on homework, studying for the qualifiers and reviewing material our professors taught us.

I would like to thank my family and friends at Promise International Fellowship and at Resurrection Williamsburg. There are so many people at both congregations I would like to thank, so I’ll only name pastors Timothy Harris and David Stancil as well as their wives Kim Harris and Mia Stancil. Thank you to all my friends who prayed with and for me and gave me encouragement during my years as a graduate student.

Other friends I would like to thank are Monica Chung, Kaisa Ercillo, Khadijah Rentas, Meiling Wang, Stuart Fontek, Joebrielle Basiao, Justin Bernstein, Michael Byrd, Sean Bryd, Ann Chow, Patrick Donnelly.

I would like to thank my family for their support of me as I pursued my career: my parents Jerry Sze and Huang Huang Li, my sisters Cindy and Tiffeny, my brother Allen, my thirteen cousins and my many aunts and uncles and my god sister Maria Mo. I would also like to thank my in-laws, especially Robert, Janet and Ted Williams, Rita Williams and Pam Jacobstein. Lastly, I would like to thank my husband Phillip Williams, mentioned above, who is one of my biggest supporter and an excellent mathematician, for his love,

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vii patience, guidance, and our shared love for yummy food and love for Jesus.

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## Contents

1 Introduction 1

2 Preliminaries and Special Cases 8

2.1 The Gauss Sums and the Stickelberger Theorem . . . . 8 2.2 Solutions of Norm Equations for d = 1, 2, 3 . . . . 13

3 The General Case 22

3.1 The Gross-Koblitz Formula . . . . 22 3.2 Estimates of Solutions of Norm Equations . . . . 28 3.3 The Class Number Formulas . . . . 33 3.4 Representing an Element in Q(

−d) with Norm phas a Product of Gauss Sums . . . . 35 3.5 Solutions of Norm Equations in the case Q(

−d), d > 0, d square-free, d6= 1, 2, 3,

D p



= 1 . . . . 41

viii

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CONTENTS ix 3.6 Solutions of Norm Equations as Normalized Trace of a Ratio of Products

of Gauss Sums . . . . 44 3.7 Examples . . . . 48

4 Finding the Height of the Stickelberger Ideal 51

Bibliography 56

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## Chapter 1 Introduction

Let K = Q(

−d) be an imaginary quadratic extension of Q. Let h be the class number and D be the discriminant of the field K. (See Proposition 3.10 in section 3.3 for explicit formula for h.) Assume p is a prime such that D

p



= 1. Then p splits in K. The elements of the ring of integersOKare of the form x +

−dy if d ≡ 1, 2 (mod 4) and x +1 +

−d

2 y

if d ≡ 3 (mod 4), where x and y ∈ Z. The norm NK/Q(x +

−dy) = x2+ dy2and NK/Q



x+1 +

−d

2 y



=(2x + y)2 4 +dy2

4 . In this the- sis, we find the elements of norm ph explicitly. That is, we find the general integer solu- tions (x, y), p 6 |y, to the equations

x2+ dy2= ph (2x + y)2

4 +dy2 4 = ph

for any imaginary quadratic field K and prime p such that p splits in K.

We proceed as follows. By a theorem of Kronecker-Weber, K is contained in a cyclo-

1

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CHAPTER 1. INTRODUCTION 2 tomic field L = Q(ζm). We can choose m to be |D| if d 6= 3, m = 6 if d = 3.

We can find an element Z ∈OK such that the ideal (Z) = δ1h, where (p) = δ1δ2in K.

Let Fq be the residue field of a prime ℘ lying above δ1 in L and ω a Teichm¨uller character of Fqfor a prime p lying above ℘ in Q(ζq−1). Let τp(χ) = −

### ∑

a∈Fq

χ (a)ζpTr (a)be the Gauss sum associated to a character χ : F×q → µq−1, χ(0) = 0.

Let χ = ωq−1m . Then Z is a ratio W /V , where W is a product of τpbi), bi∈ Z and V is a power of p (d 6= 1, 2, 3) or a single Gauss sum (d = 1, 2, 3). We use Stickelberger relations, which give the prime decomposition of Gauss Sums, to find Z.

Let x, y ∈ Z such that x +

−dy = Z if d ≡ 1, 2 (mod 4), x +1 +

−d

2 y= Z if d ≡ 3 (mod 4). We use the Gross-Koblitz formula, which allows us to write Gauss Sums in terms of p−adic Gamma functions, to express the conjugate Z of Z as a product of values of p−adic Gamma functions up to a sign. This implies an explicit congruence for x or 2x + y (mod ph). Taking into account an estimate of the absolute value of x or 2x + y (see Proposition 3.6), this allows us to find x or 2x + y explicitly. We note that for the cases d = 1, 2, 3, one can use the classical Stickelberger congruences instead of the Gross- Koblitz formula. These cases were motivation for our study in the general case.

Let us describe our results. Let N be a natural number, t a rational N−integral number, then htiN denotes the integer such that htiN≡ t (mod N) and −N

2 < htiN N

2. We note that h = 1 when d = 1, 3, 2.

Let us first consider the case d = 1. If p ≡ 1 (mod 4), then there exist, and are deter-

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CHAPTER 1. INTRODUCTION 3

mined up to a sign and transposition, x, y ∈ Z such that x2+ y2= p. Letn r



= n!

r!(n − r)!

be a binomial coefficient. Let h i = h ip. We prove

Theorem 1.1. For a prime p ≡ 1 (mod 4)

* 1 2

3p−3

4 p−1

4

+2

+

* 1 2

(3p−3p )!

(p−14 )!

+2

= p. (1.1)

This equality withh i = h incharacterizes primes equivalent to 1 modulo 4 among the naturals, which are not squares and are equivalent to 1 modulo 4.

Corollary 1.2. Let p be a prime such that p ≡ 1 (mod 4) and p = n2+ m2. Then

3p−3

4 p−1

4



(mod p) ∈ {±2n, ±2m} (mod p).

Let d = 3. If p ≡ 1 (mod 3) and n2+ nm + m2= p, then the solutions of x2+ xy + y2= pare

±(n, m), ±(−n, n + m), ±(n + m, −n), ±(n + m, −m), ±(−n, n + m), ±(m, n).

Therefore, the possible values for 2x + y are

±(2n + m), ±(n − m), ±(n + 2m).

We prove

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CHAPTER 1. INTRODUCTION 4

Theorem 1.3. Let p ≡ 1 (mod 3), p 6= 7, α =

3p−3

6 p−1

6

 . Then

hαi,

r4p − hαi2 3

!

is a solution in integers of the equation x2+ 3y2= 4p.

Corollary 1.4. Let p be a prime such that p ≡ 1 (mod 3) and p = n2+ nm + m2. Then

3p−3 6 p−1

6



(mod p) ∈ {±(2n + m), ±(n − m), ±(n + 2m)} (mod p)

We also prove

Theorem 1.5. Let p ≡ 1 (mod 8) and α = 1 2

p−1 2 p−1

8



or p≡ 3 (mod 8) and

α =1 2

p−1

2 p−3

8

 . Then

hαi,

rp− hαi2 2

!

is a solution in integers of the equation x2+ 2y2= p.

Corollary 1.6. Suppose p ≡ 1, 3 (mod 8) and p = n2+ 2m2. Then

p−1

2 p−1

8



≡ ±2n (mod p), if p ≡ 1 (mod 8)

p−1

2 p−3

8



≡ ±2n (mod p), if p ≡ 3 (mod 8)

In fact, we first proved the congruences in the corollaries, which when combined with Proposition 3.6, imply Theorems 1.1, 1.3 and 1.5.

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CHAPTER 1. INTRODUCTION 5 Theorems 1.1, 1.3 and 1.5 will be proved in section 2.2 (Theorems 2.9, 2.10, 2.11), where we demonstrate our results and techniques for examples d = 1, 3, 2 .

For the case d 6= 1, 2, 3 we prove the following. Let p be a prime such that p splits in K, let m = |D|,

β =

### ∏

1 ≤ a < m χd(a) = 1

Γp

a m

 ,

where Γp: Zp→ Z×p is the p−adic Gamma function, χd: (ZmZ)×→ {±1} is the quadratic character associated to K (see Proposition 3.10). For an N−integral t, let [t]Nbe the integer in the interval [0, N) congruent to t (mod N).

Let c = c(d, p) be a natural number defined in Proposition 3.7. Note that if k = k(h) = h

2+ 1 if h is even, k = h+ 1

2 if h is odd, then c = k = h if h ≤ 2, c = k or k + 1 or (if p = 2) k+ 2 if h > 2, c = k if d ≡ 1, 2 (mod 4); c = k if d ≡ 3 (mod 4), d is not a prime, p ≥ 5;

c= k if d ≡ 3 (mod 4) is a prime and p ≥ 17.

Let c ≤ i ≤ h, i 6= 2 if p = 2, h i = h ipi, [ ] = [ ]pi. Let T = 1 m

 ,

α =

### ∏

1 ≤ a < m χd(a) = 1

(−1)[aT ]

### ∏

1 ≤ j < [aT ] ( j, p) = 1

j

We prove in section 3.5,

Theorem 1.7. Let d 6= 1, 2, 3, p be a prime that splits in K, p 6= 2 if d = 15, p 6= 5 if d = 11.

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CHAPTER 1. INTRODUCTION 6 Then for d≡ 1, 2 (mod 4),

x= 1 2α

= 1 2β

,

y= ±

rph− x2 d

is a solution in Z, p 6 |y, of the equation x2+ dy2= ph. For d≡ 3 (mod 4), x and y defined by

2x + y = hαi = hβ i, y= ±

r4ph− (2x + y)2 d

is a solution in Z, p 6 |y, of the equation (2x + y)2+ dy2= 4ph. Corollary 1.8. If ph= n2+ ds2or ph=

n+ s 2

2

+ ds2

4, with n, s ∈ Z, p 6 |s, when d ≡ 1, 2 (mod 4) or d ≡ 3 (mod 4), respectively, then

α ≡ β ≡ ±2n, or ± (2n + s) (mod pi),

respectively.

Theorems 1.1, 1.3, 1.5 and 1.7 provide us with explicitly determined solutions of norm equations. They also imply the corresponding congruences modulo ph. Theorems 1.1, 1.3, 1.5, 1.7 and their corollaries are the main results of this thesis.

We note that one can also express solutions x, x +1

2y (and then find y) of the norm equations by directly applying the normalized trace to the element Z or its image in Fq

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CHAPTER 1. INTRODUCTION 7 (see section 3.6 for the corresponding formulas).

Let us outline this thesis. In section 2.1, we describe the Stickelberger ideal and give the classical result about Gauss sums. Special cases d = 1, 2, 3 are computed in section 2.2.

In section 3.1, we give the Gross-Koblitz formula and use it to give the representation of Gauss sums in section 3.4. Section 3.2 contains estimates of solutions of norm equations.

We introduce the element Z and consider the cases d = 1, 2, 3 in sections 3.4 and 3.5. In section 3.7, we demonstrate examples of computations using our formulas. In chapter 4, we find the height of the Stickelberger ideal.

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## Preliminaries and Special Cases

### 2.1 The Gauss Sums and the Stickelberger Theorem

Let ζm denote a primitive mth of unity and let M/Q be a finite abelian extension. By a theorem of Kronecker-Weber, M ⊂ Q(ζm) for some m. We will assume that m is minimal.

Let G = Gal (M/Q), a quotient of Gm= Gal (Q(ζm)/Q) = ZmZ

×

.

For (a, m) = 1, let σadenote the element ζm7→ ζmaof Gmas well as its restriction to M.

Definition 2.1. The Stickelberger element in the group-ring Q[G] is defined to be

θ = θ (M) =

### ∑

1 ≤ a < m (a, m) = 1

a mσa−1.

Then I= I(M) = Z[G] ∩ θ Z[G] is an ideal of Z[G] consisting of Z[G] multiples of θ with coefficients in Z. This ideal is called the Stickelberger ideal.

Lemma 2.2. If M = Q(ζm) then I = I0θ , where I0 is the ideal of Z[G] generated by all elements of the form c− σcfor(c, m) = 1.

8

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 9 Proof. See [7].

Definition 2.3. We define the action of Z[G] on ideals by: if x = ∑ xσσ ∈ Z[G], then x acts on ideals A of M by

Ax=

### ∏

(Aσ)xσ.

We now give some definitions and properties of Gauss and Jacobi Sums.

Let p be an odd prime and q = pf. Let ζpbe a fixed primitive p−th root of unity. The Galois group Gal (Fq/Fp) is cyclic, generated by the Frobenius automorphism σp: x 7→ xp. Let λ : Fq → C× be the additive character, defined by λ (x) = ζpTr (x), where Tr (x) = TrFq/Fp(x) = x + xp+ · · · + xpf−1 is the trace map. Let χ : F×q → C× be a multiplicative character. We extend χ to Fqby defining χ(0) = 0, including the trivial character χ0.

Definition 2.4. The Gauss sum corresponding to χ is

τp(χ) = −

a∈Fq

χ (a)λ (a) = −

### ∑

a∈Fq

χ (a)ζpTr(a).

TheJacobi sum corresponding to χ1and χ2is

J(χ1, χ2) = −

### ∑

a∈Fq

χ1(a)χ2(1 − a).

It can be shown that the Gauss sum and Jacobi sum satisfy the following properties:

1. τp0) = 1 and J(χ0, χ0) = 2 − q.

2. τp(χ) = χ(−1)τp(χ).

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 10

3. If χ 6= χ0, τp(χ)τp(χ) = χ(−1)q. If χ1χ2 6= χ0, then J(χ1, χ2) = τp1p2) τp1χ2) . Thus, if χ1, χ2 are characters of order dividing m, then J(χ1, χ2) is an algebraic integer in Q(ζm).

4. If χ 6= χ0, then τp(χ)τp(χ) = q. If χ1, χ2, χ1χ26= χ0, then J(χ1, χ2)J(χ1, χ2) = q.

5. Let χ = χ1· · · χn. If ψ = χ|Fp = ψ0, the trivial character on Fp, then

n

### ∏

i=1

τpi) = (−1)n

Tr (a1+···+an)=0

χ1(a1) · · · χn(an) + (−1)n−1

Tr (a1+···+an)=1

χ1(a1) · · · χn(an).

If ψ 6= ψ0, then

n

### ∏

i=1

τpi) = (−1)n−1τp(ψ)

### ∑

Tr (a1+···+an)=1

χ1(a1) · · · χn(an).

(In these sums, (a1, . . . , an) ranges over all n−tuples of elements in F×q.)

Proof. The proof of properties 1 through 4 can be found in [14]. We prove property 5. Let a = (a1, · · · , an), z = za= Tr (a1+ · · · + an) and ci= ai

z if z 6= 0. Then

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 11

n

i=1

τpi) = (−1)n

### ∑

a∈(F×q)n

χ1(a1) · · · χn(anpz

= (−1)n

### ∑

{a∈(F×q)n:z=0}

χ1(a1) · · · χn(an) +

### ∑

{a∈(F×q)n:z6=0}

χ1(a1) · · · χn(anpz

= (−1)n

### ∑

{a∈(F×q)n:z=0}

χ1(a1) · · · χn(an) +

### ∑

{a∈(F×q)n:z6=0}

χ1(zc1) · · · χn(zcnpz

=(−1)n

{a∈(F×q)n:z=0}

χ1(a1) · · · χn(an) +

{a∈(F×q)n:z6=0}

χ1(z) · · · χn(z)χ1(c1) · · · χn(cnpz

=(−1)n

z=0

χ1(a1) · · · χn(an) + (−1)n−1

−

z∈F×p

χ (z)ζpz

Tr (c1+···+cn)=1

χ1(c1) · · · χn(cn)

! .

The last equality follows because the set

{(zc1, . . . , zcn) : z ∈ F×p, (c1, . . . , cn) ∈ (F×q)n, Tr (c1+ · · · + cn) = 1}

is equal to the set {(a1, . . . , an) ∈ (Fnq)×: z 6= 0}. If ψ = ψ0, then τp(ψ) = −

### ∑

z∈F×p

χ (z)ζpz= 1, otherwise there exists a b ∈ F×p such that ψ(b) 6= 1 so

Tr(ba1+ · · · ban) = bTr (a1+ · · · + an)

and

### ∑

Tr (a1+···+an)=0

χ1(a1) · · · χn(an) =

### ∑

bTr (a1+···+an)=0

χ1(ba1) · · · χn(ban)

= ψ(b)

### ∑

Tr (a1+···+an)=0

χ1(a1) · · · χn(an),

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 12 which implies that

### ∑

z=0

χ1(a1) · · · χn(an) = 0. Note that property 3 can be proved the same way by using z = a1+ · · · + anand ci= ai

z if z 6= 0.

The Teichm¨uller character is defined as follows.

Definition 2.5. Let q = pf. Let p be one of the ϕ (q − 1)

f prime ideals of Q(ζq−1) lying above the prime number p. Identify Z[ζq−1] (mod p) with Fq. The equation Xq−1− 1 = 0 has distinct roots modulo p. Thus, there is a group isomorphism

ω : F×q → µq−1,

such that ω(a) = a (mod p). ω is a character of F×q, called the Teich ¨muller character corresponding to p.

The Teichm¨uller character generates the character group. Thus, any character χ of F×q can be written as an integral power of ω. Note that since ω has order q − 1, the powers of ω can be expressed modulo q − 1.

Now let P be a prime above p in Q(ζq−1, ζp). Let k be an integer and first assume that 0 ≤ k < q − 1. We may represent k in its p−adic expansion by k = k0+ k1p+ · · · + kf−1pf−1, where 0 ≤ ki< p. Define s(k) and γ(k) by s(k) = k0+ k1+ · · · kf−1and γ(k) = k0!k1! · · · kf−1!. For k ≥ q − 1, use s(k0) and γ(k0) where k0≡ k (mod q − 1) and 0 ≤ k0<

q− 1. We have

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 13 Theorem 2.6 (Stickelberger). (See [9]) For any integer k,

τp−k)

p− 1)s(k) 1

γ (k) (mod P),

where ω is the Teichm¨uller character, which generates the character group.

### 2.2 Solutions of Norm Equations for d = 1, 2, 3

We begin by looking at three examples to motivate and introduce our study.

Example 2.1. Let K1= Q(ζ4), where ζ4 is a primitive fourth root of unity. Let p be a rational prime with p≡ 1 (mod 4). p splits in K1. That is, p= d1d2, where NK1/Q(di) = p for i= 1, 2. There exists x + ζ4yOK1 = Z[ζ4] such that (x + ζ4y) = di because the class number of K1is equal to 1. Therefore, NK1/Q((x + ζ4y)) = (NK1/Q(x + ζ4y)) = x2+ y2= NK1/Q(di) = p. Hence, any p ≡ 1 (mod 4) is representable as x2+ y2, where x, y∈ Z.

We can actually represent these elements of norm p in terms of Gauss sums. We re- call that τp(χ) = −

### ∑

a∈ZpZ

χ (a)ζpa for the character χ. Let ω be the Teichm¨uller char-

acter corresponding to a fixed prime ideal in K1 above p. Define χ1 = ωp−14 , χ2 = ω

p−1

2 and χ1χ2= ω3p−34 . From above, we see that J1, χ2) is an algebraic integer with J1, χ2)J(χ1, χ2) = p. Therefore, the element x + ζ4y=τ (χ1)τ(χ2)

τ (χ1χ2) in Z[ζ4] has norm p.

We may further characterize the x’s and y’s by using Theorem 2.6. Since p− 1 4 , p− 2

4 and 3p − 3

4 are all less than p− 1, they remain the same under p−adic expansion. We have,

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 14 using the property τp(χ)τp(χ) = χ(−1)p and the Stickelberger Theorem (Theorem 2.6),

x+ ζ4y= τ (χ1)τ(χ2)

τ (χ1χ2) = τ (ω

p−1

4 )τ(ωp−12 ) τ (ω

3p−3 4 )

= p−14 (−1) τ (ω

p−1 4 )

· p−12 (−1) τ (ω

p−1 2 )

· τ (ω

3p−3 4 ) 3p−34 (−1)

= pτ(ω3p−34 ) τ (ω

p−1

4 )τ(ωp−12 )

pp−1)

3p−3 4

(3p−34 )!

p−1)p−14

(p−14 )! ·p−1)

p−1 2

(p−12 )!

(mod P)

= p(p−14 )!(p−12 )!

(3p−34 )! (mod P), x+ ζ4−1y= τ (χ1)τ(χ2)

τ (χ1χ2) = τ (χ1)

χ1(−1)· τ (χ2)

χ2(−1)·χ1χ2(−1)

τ (χ1χ2) = τ (χ1)τ(χ2) τ (χ1χ2)

= τ (ω

p−1

4 )τ(ωp−12 ) τ (ω

3p−3 4 )

(3p−34 )!

(p−14 )!(p−12 )! (mod P).

Hence,

2x ≡ (3p−34 )!

(p−14 )!(p−12 )! (mod p), or

x (3p−34 )!

2 · (p−14 )!(p−12 )! (mod p)

= 1 2

3p−3 4 p−1

4



(mod p),

wheren r



= n!

r!(n − r)! is a binomial coefficient.

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 15 We can also get the congruence for y by subtracting x+ ζ4−1y from x+ ζ4y:

4y (3p−34 )!

(p−14 )!(p−12 )! (mod P).

Wilson’s theorem states that(p − 1)! ≡ −1 (mod p). That is,

(p − 1)! = (p − 1)(p − 2) · · ·



pp− 1 2

  p − 1 2



!

= (−1)p−12  p − 1 2



!2 (mod p)

≡ −1 (mod p),

or

 p − 1 2



!2≡ (−1)3−p2 (mod p).

Since p≡ 1 (mod 4), we see that p − 1 2



!2≡ −1 (mod p). Thus, we can choose P such that ζ4 p − 1

2



! (mod P). Finally,

y (3p−34 )!

2(p−14 )!(p−12 )!2 ≡ −(3p−34 )!

2(p−14 )! (mod p)

Now, NK1/Q((x +ζ4y)) = x2+y2= p implies that x2< p and y2< p so that x < p< p

2 and y<

p< p

2. Note that p< p

2 is true for p≥ 5.

Definition 2.7. Let α =a

b ∈ Q, a, b ∈ Z, (n, b) = 1 where n is odd. We define the notation hαi to be the integer such that hαi ≡ α (mod n) and −n− 1

2 ≤ hαi ≤ n− 1 2 .

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 16 Remark 1. Note that h−αi = −hαi. To see this, we have h−αi ≡ −α (mod n) and

−hαi ≡ −α (mod n). Also, n− 1

2 ≥ −hαi ≥ −n− 1

2 . Thus,h−αi = −hαi.

Lemma 2.8. Let n > 0 be a natural number such that n ≡ 1 (mod 4). Then n is prime if and only if

 n − 1 2



!2≡ −1 (mod n). (2.1)

Furthermore, when n is composite, n6= 9,

 n − 1 2



! ≡ 0 (mod n). (2.2)

Proof. We showed above that if n is a prime such that n ≡ 1 (mod 4), then n satisfies (2.1). Suppose now that n is composite and n ≡ 1 (mod 4). Let p be a prime such that p|n. Then p < n− 1

2 and n

p < n− 1

2 . If n is not the square of a prime, then p and n p are distinct and therefore  n − 1

2



! ≡ 0 (mod n). If n = p2 for some prime p, as long as p6= 3, then 2p < n− 1

2 so that p2| n − 1 2



! and  n − 1 2



! ≡ 0 (mod n) in this case as well. If n = 9, then 9 − 1

2



! = 24 ≡ 0 (mod 3).

Theorem 2.9. Let n > 0 be a natural number, not a square, such that n ≡ 1 (mod 4). The pair

*1 2

3n−3 4 n−1

4

+ ,

*(3n−34 )!

2(n−14 )!

+!

(2.3)

is a solution to x2+ y2= n if and only if n is prime.

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 17 Proof. We saw above that if p ≡ 1 (mod 4) and p is prime, then (2.3) is a solution to x2+ y2= p.

Suppose now that n is a composite number, not a square and n ≡ 1 (mod 4). Suppose further that

* 1 2

3n−3 4 n−1

4

+2

+

*(3n−34 )!

2(n−14 )!

+2

= n. (2.4)

It follows from congruence (2.2) and direct computations for n = 9 that (3n−34 )!

2(n−14 )! =1 2

3n−3 4 n−1

4

  n − 1 2



! ≡ 0 (mod n).

Equation (2.4) implies

*1 2

3n−3 4 n−1

4

+2

= n,

contradicting the condition that n is not a square.

Example 2.2. Now consider K2= Q(ζ6) = Q(ζ3) = Q(

−3) and let p be a rational prime with p≡ 1 (mod 3). p splits in K2, so p= d1d2 with NK2/Q(di) = p for i = 1, 2. Again, there exists x+ ζ6yOK2 = Z 1 +

−3 2



= Z[ζ6] such that (x + ζ6y) = d1. Therefore, NK2/Q((x + ζ6y)) = (x + ζ6y)(x + ζ6−1y) = x2+ xy + y2=

x+y 2

2

+3y2 4 = p.

Similar to the case K1= Q(ζ4), we define χ1= ωp−16 , χ2= ωp−13 , hence χ1χ2= ω3p−36

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CHAPTER 2. PRELIMINARIES AND SPECIAL CASES 18 and we have

x+ ζ6y= τp1p2)

τp1χ2) =τpp−16 )τ(ωp−13 ) τp3p−36 )

= p3p−36 ) τpp−16 pp−13 )

p(p−16 )!(p−13 )!

(3p−36 )! (mod P), x+ ζ6−1y (3p−36 )!

(p−16 )!(p−13 )! (mod P).

Adding the two equations,

2x + y ≡ (3p−36 )!

(p−16 )!(p−13 )! (mod p)

=

3p−3

6 p−1

6



(mod p)

Subtracting the two equations,

−3y ≡

3p−3

6 p−1

6



(mod p).

One may note that square roots modulo p can be computed. We will proceed without this fact. From(2x + y)2= 4p − 3y2, we see that(2x + y)2< 4p so that |2x + y| < 2

p< p 2 if4

p< p, or 16 < p. That is, if p ≥ 17, then |2x + y| < p

2. Let α =

3p−3

6 p−1

6



andhαi ≡ α (mod p) be the integer such that −p− 1

2 ≤ hαi ≤ p− 1

2 . In this case,2x + y = hαi and y2= 4p − (2x + y)2

3 implies

y= ±

r4p − hαi2

3 ,

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