SOME PROBLEMS IN NUMBER THEORY

by

Prem Prakash Pandey

(Math10200604008)

The Institute of Mathematical Sciences

Chennai 600113

A thesis submitted to the

Board of Studies in Mathematical Sciences

In partial fulfillment of requirements

For the Degree of

DOCTOR OF PHILOSOPHY

of

HOMI BHABHA NATIONAL INSTITUTE

July, 2012

Homi Bhabha National InstituteRecommendations of the Viva Voce Board

As members of the Viva Voce Board, we certify that we have read thedissertation prepared by Prem Prakash Pandey entitled “Some Problems inNumber Theory” and recommend that it maybe accepted as fulfilling thedissertation requirement for the Degree of Doctor of Philosophy.

Date:Chair and Convener - Prof. R. Balasubramanian

Date:Member - Prof. K. Srinivas

Date:Member - Dr. Sanoli Gun

Date:External Examiner - Prof. B. Sury

Final approval and acceptance of this dissertation is contingent upon thecandidate’s submission of the final copies of the dissertation to HBNI.

I hereby certify that I have read this dissertation prepared under mydirection and recommend that it may be accepted as fulfilling the dissertationrequirement.

Date:

Place: Guide: Prof. R. Balasubramanian

3

STATEMENT BY AUTHOR

This dissertation has been submitted in partial fulfillment of requirements

for an advanced degree at Homi Bhabha National Institute (HBNI) and is

deposited in the Library to be made available to borrowers under rules of

HBNI.

Brief quotations from this dissertation are allowable without permission,

provided that accurate acknowledgment of source is made. Requests for

permission for extended quotation from or reproduction of this manuscript

in whole or in part may be granted by the competent authority of HBNI

when in his or her judgment proposed use of the material is in the interests

of scholarship. In all other instances, however, permission must be obtained

from the author.

Prem Prakash Pandey

Candidate

4

DECLARATION

I, hereby declare that the investigation presented in the thesis has been car-

ried out by me. The work is original and has not been submitted earlier

as a whole or in part for a degree/diploma at this or any other Institu-

tion/University.

Prem Prakash Pandey

Candidate

5

Acknowledgement

First and foremost I would like to thank my advisor Prof. R. Balasubra-manian, who has been constant inspiration throughout. He granted me allthe freedom to enjoy life but also made sure that my stay in the Institute isnot limited to enjoyment. I could make any silly mistake and he will correctit cheerfuly and sometimes same mistake he will correct repeatedly with thesame smile. Whenever I had a rough phase of life he made sure that I am notgrilled in between mental and academic pressure. The enthusiasm he takesin doing any activity was very inspiring throughout. The way he explainsideas it looks that doing Mathematics is so easy. It is wonderful to see himin action. Jahanpanah Tussi Great ho!Sanoli has always been there to help. If a problem solved, I never cared towrite but she has constantly presurrized me to write down the things I workand finish the thesis quickly. Also she has always stood by me as a goodfriend and made me feel good at times when everything was going otherway.Dr. Purusottam Rath has helped everytime I landed with a problem. Hemade sure I do not feel inferior when I cant solve a problem and he solvedit quickly. I am thankful to Prof. K. Paranjape, Prof. K. Srinivas and Dr.Anirban Mukhopadhyay for their guidance at various stages. I will neverforget Umesh’s contribution. He was one person who was always availablethroughout. If I want to discuss any mathematics he will listen, and withhim I will be free to utter any nonsense, which was the case with few morefriends, but he was one of very few who will not take it till I accept that thereis a mistake or we correct it together. He did listen to non academic troublestoo but will try to get off these as soon as possible, I guess that eventuallyhelps.I got opportunity to discuss with Prof. Preda Mihailescu, it was an enjoyingconversation and I am thankful to him. I will like to thank Prof. JosephOesterle, Prof. Ram Murty, Prof. Kumar Murty, Prof. K. Soundararajan,Prof. Olivier Ramare, Prof. C. Gasabari for some useful comments. I amvery thankful to my teacher Prof. T. K. Das who inspired me to pursuea career in Mathematics. Thanks is due to HRI for being amazing host atmany occassions and my friends there.I am thankful to those friends who have accompained me at various wonder-ful treks. My Flatmates Rajeev and Somdeb, they both are wonderful personand one can always bank on them. My classmates Sundar, Krishna, Pooja,George, Ajay all were very helpful. It was good to have people like Alok,Mohan, Rohan and Bhavin around. Thanks are also due to some friendswhom I might not have named.Last but not the least, I am very thankful to my family members to have

6

patience for so long. In particular my mother who will be the happiest personto see me being doctorate. This thesis is dedicated to her.

7

Contents

1 Higher Residue Symbols 131.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.2 Determination Of The Degree . . . . . . . . . . . . . . . . . . 141.3 lth power residue symbol . . . . . . . . . . . . . . . . . . . . . 151.4 Another way to find the degree . . . . . . . . . . . . . . . . . 20

2 Catalan’s Conjecture 222.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 Cassels criteria . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3 Wieferich Criterion . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Proof of the Theorem 2.1.6 . . . . . . . . . . . . . . . . . . . . 282.5 Mihailescu and Cyclotomic Fields in the context of Catalan

Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.6 Number Field Analog . . . . . . . . . . . . . . . . . . . . . . . 30

3 Catalan Problem over Z[i] with even exponents 323.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2 Primes p > 5 and q > 5 . . . . . . . . . . . . . . . . . . . . . . 323.3 Elliptic curve case . . . . . . . . . . . . . . . . . . . . . . . . . 363.4 Equations x5 − y2 = 1 and x2 − y5 = 1 . . . . . . . . . . . . . 38

4 Cassels Criterion for Catalan Problem over Z[i] 404.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.2 Some preliminary results . . . . . . . . . . . . . . . . . . . . . 414.3 Cassels Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5 Catalan Problem over Z[i] 525.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.2 The Obstruction Group . . . . . . . . . . . . . . . . . . . . . 52

8

6 Inverse problems in Additive Number Theory 566.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566.2 Statements of Theorems . . . . . . . . . . . . . . . . . . . . . 576.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 586.4 Proof of Theorem 6.2.2 . . . . . . . . . . . . . . . . . . . . . . 626.5 Proof of Theorem 6.2.1 . . . . . . . . . . . . . . . . . . . . . . 73

A Some Facts from Algebraic Number Theory 78

9

Introduction

In this thesis the author has worked on three different problems. Someprogress is reported on these three problems.

The first problem considered is about “Higher Residue Symbols”. Givena finite set S of integers, the question of finding primes p such that eachinteger a ∈ S is a quadratic residue (non-residue) modulo p is dealt byvarious authors [19]. Many authors including M. Fried and S. Wright [49]have established the infinitude of primes p modulo which each a ∈ S is aquadratic residue. The density of such primes was considered in [1]. Wehave generalized the problem and studied the analogous questions. We takea prime number l and consider the lth cyclotomic field Q(ζl). For a prime pof Z[ζl] and an integer α ∈ Z[ζl] the lth residue symbol(

α

p

)l

have been studied by various mathematicians e.g.[21]. Given a finite setS = {si; 1 ≤ i ≤ n} of rational integers and a corresponding set T = {ti; 1 ≤i ≤ n} (with multiplicities allowed ) of lth roots of unity we determine thedensity of primes p of Z[ζl] for which(

sip

)l

= ti, for all i.

We also describe another approach to look at lth residue symbol using moresophisticated tools, and this might be useful in some contexts. This exposi-tion forms chapter one of the thesis.

The second problem considered is the Catalan’s conjecture/ Mihailescu’sTheorem. It was conjectured by Eugene Charles Catalan in 1844 that, theonly perfect powers among integers which differ by 1 are 8 and 9. Thusthe conjecture says that the exponential Diophantine equation xm − yn = 1,where x, y are positive integers and m,n are integers bigger than 1, has onlyone solution, viz, 32− 23 = 1. One observes that it is enough to consider theequation xp − yq = 1, when p, q are primes. The note stating the problemappeared among erratas of papers which appeared in an earlier volume of theCrelle Journal. Some particular cases of the problem were dealt by variousmathematicians including Euler and Lebesgue [18, 28]. Notably Lebesguehad solved the problem for the exponent q = 2. The progress was slow

10

till the work of J W S Cassels and Ko Chao [7, 8, 10]. In early 1960s KoChao proved the problem for the exponent p = 2 and q 6= 3 [10]. Thecase x2 − y3 = 1, which gives the non-trivial solution 32 − 23 = 1, wasalready settled by Euler using The method of descent. This reduced theCatalan’s conjecture to a stage when one can assume that both the primesp and q are odd. By a Catalan problem we will refer the diophantine equationxp − yq = 1 when p and q are odd primes and any solution will be writtenas a quadruple (x, y, p, q). Now we will let x, y ∈ Z and in this case when(x, y, p, q) is a solution then (−y,−x, q, p) is also a solution. A solution willbe called non-trivial if xy 6= 0. In 1960s Cassels proved that for any solution(x, y, p, q) of the Catalan problem one always has q|x and p|y. This is referredas Cassels criteria. The criteria of Cassels proved to be very important, andall subsequent work depended on this. Some immediate progress was made,most notably due to Inkeri and Tijdeman. In 1976, Tijdeman proved thatthe Catalan problem can have only finitely many possible solutions, actuallyCassels had made this weaker conjecture. The conjecture was finally solvedby Preda Mihailescu. Mihailescu used deep results from Algebraic NumberTheory, notably ‘theory of annihilators of class groups’, together with his‘power series method’. An overview of the proof is presented in chapter two.

As part of this thesis the author studies the equation xp − yq = 1 overa number field K, i.e. when x, y ∈ OK . Mainly the case considered here iswhen K is a quadratic imaginary field with class number one. It was provedby Brindza et al. [5] that for a fixed field K this equation has only finitelymany solutions, but the bounds obtained are very large. In the subsequentchapters we report certain progress made on this problem.

In chapter three we list all the solution of xp − yq = 1 when one of theprime is even and x and y run through integers in Q(i). The method is quitedifferent from that of Lebesgue and Ko Chao. The case x2−y3 = 1 is handledseparately using theory of torsion points on Elliptic curves.

In chapter four of the thesis we formulate an appropriate Cassels crite-ria and prove it partially for imaginary quadratic number fields with classnumber one. Because of the symmetry in solution (whenever (x, y, p, q) isa solution then so is (−y,−x, q, p)) we can assume p > q, without loss ofgenerality. There are two different cases coming naturally. When q does notsplit in K and when q splits in K. In the first case we are able to prove thatthere is a prime q1, above q, which divides x and there is a prime p1, abovep, which divides y. We also succeed in showing that under some assumptionon class number from this Cassels criterion we already obtain q|x and p|y.In the second case we have demonstrated that q1|x.

In chapter five we report further progress made on Catalan problem con-sidered here. We introduce a proper obstruction group, made up of solutions

11

of Catalan problem, and then trap it in a short exact sequence of fairlywell studied objects (namely class groups and unit groups). This is prettyanalogous to the work in the case of Catalan’s conjecture over Z.

Now we come to the chapter six of the thesis. If G is an abelian groupand A ⊂ G is finite then one defines the doubling constant of A to beD(A) = |A+A|

|A| . The characterization of A with D(A) ≤ 2 is well understood,

thanks to the work of Kneser [24]. The case when D(A) ≤ 2.04 was recentlyhandled by Deshouillers and Freiman [16]. These authors first study theproblem when A ⊂ Z × Z/dZ with D(A) ≤ 2.40 and then use it to provethe result ofr G = Z/nZ. We shall give simpler proof of this last result withD(A) ≤ 2.50 and then deduce the result for G = Z/nZ with D(A) ≤ 2.11.

List of Publications :

1. (joint with R. Balasubramanian) Density of Primes in lth Power Residues,accepted in Proceeding of Indian Academy of Sciences.

2. (joint with R. Balasubramanian) Catalan’s Equation for even primesover quadratic fields (submitted) http://arxiv.org/abs/1112.2688.

3. (joint with R. Balasubramanian) Catalan’s conjecture Revisited, In Prepa-ration.

4. (joint with R. Balasubramanian) A Remark on a Theorem of Deshouillersand Freiman, In preparation.

12

Chapter 1

Higher Residue Symbols

1.1 Introduction

In a recent paper [1] the authors have computed the relative ‘density’ (fordefinition see section 3 of this chapter) of primes for which a given finitestring S = {a1, . . . , am} of integers are quadratic residues simultaneously. Itturns out, via Chebotarev density theorem, that this density is reciprocalof the degree of the multiquadratic extension given by square roots of thefinite string of given integers over Q. Given a field K which contains an nth

root of unity and given a finite set of integers S = {a1, . . . , am} one can

determine the degree of the extension K(a1n1 , . . . , a

1nm)/K using Galois theory,

for instance see [48]. In this chapter, we study the distribution of primesp of Z[ζl] modulo which each of ai assumes a preassigned lth power residue

symbol and then relate it to the degree of the extension Q(a1l1 , . . . , a

1lm)/Q. We

give two methods to compute the degree of the extension Q(a1l1 , . . . , a

1lm)/Q

and either of the two methods may prove to be useful at a given instance. Insection 2, we use a ramification argument in place of the classical use of the

Eisenstein criterion to compute the degree of the extension Q(a1l1 , . . . , a

1lm)/Q.

Section 3 deals with the lth power residue symbols and study of the distri-bution of primes p modulo which they take a fixed value for each ai. Insection 4, we define a matrix T and then proceed to relate the degree of theextension to the rank of T . The tools we use are basic in nature but for thesake of completeness we will give some of the proofs.

We will fix an odd prime l and ζl will stand for a fixed primitive lth rootof unity in C. The main theorem proved in this chapter is the following;

Theorem 1.1.1. For integers r1, . . . , rm, the density of prime ideals p of Z[ζl]

13

satisfying (aip

)l = ζril for all i is [Q(a1l1 , . . . , a

1lm) : Q]−1, whenever there is at

least one prime ideal satisfying (aip

)l = ζril for all i.

The proof of this theorem appears at the end of section 3.

1.2 Determination Of The Degree

To start with, we can assume that a′si are lth power free and none of them is 1.

One has the following;

Lemma 1.2.1. If a ∈ Z is not a lth power of an element of Z, then X l − ais irreducible over Z.

Lemma 1.2.2. Let b1, b2, . . . , bi ∈ Z and b be an integer which is not a lth

power and such that there is a prime q 6= l which divides b but does not divide

any of bj. Then [Q(b1l1 , . . . , b

1li , b

1l ) : Q(b

1l1 , . . . , b

1li )] = l.

Proof. Let us write L = Q(b1l1 , . . . , b

1li , b

1l ) and K = Q(b

1l1 , . . . , b

1li ).

Since q does not divide bj for any j, q is unramified in each of Q(b1lj ) and

hence it is unramified in K as well. On the other hand looking at the fac-torization of X l − b over C, we find that if X l − b = f1(X)f2(X) in K[X]then f1(0) = b

rl ζcl ∈ K for some integers r and c in {0, . . . , l − 1}. Then the

discriminant of the field Q(brl ζcl ) is divisible by q (Theorem A.1.3) and hence

q ramifies in Q(brl ζcl ). Since Q(b

rl ζcl ) ⊂ K, we obtain that q ramifies in K, a

contradiction. So we have that the polynomial X l− b is irreducible in K[X].This proves the lemma.

Algorithm to compute the degree [Q(a1l1 , . . . , a

1lm) : Q]

Claim: There exists an integer t ≤ m and integers b1, . . . , bt with the followingproperties;(1) For every 1 ≤ i ≤ t,there exists a prime qi which divides bi but does notdivide bj for j 6= i,(2) None of bi is a lth power,

(3) The field Q(a1l1 , . . . , a

1lm) is the same as the field Q(b

1l1 , . . . , b

1lt ).

We will generate the numbers b′is in successive steps. Here upper index willindicate the number of steps.Let q1 be a prime divisor of a1 we put b

(1)1 = a1. For i > 1 if q1 - ai then we

will put b(1)i = ai and in case q1|ai then we will define b

(1)i as follows:

Let r1 and ri be the exponent of q1 in a1 and ai respectively. Without loss

14

of generality we can assume that 1 ≤ r1, ri ≤ l− 1. As mi runs modulo l thenumbers mir1 + ri are distinct modulo l and hence for some choice of mi we

will have mir1 + ri = λil and then we define b(1)i =

ami1 ai

qλil1

. Clearly q1 does not

divide b(1)i . If any of b

(1)i happens to be a lth power then we will omit it and

consider only those b(1)i which are not lth powers, say, b

(1)1 , . . . , b

(1)s1 . Now one

has q1|b(1)1 and q1 - b(1)

i for i > 1, and Q(a1l1 , . . . , a

1lm) = Q((b

(1)1 )

1l , . . . , (b

(1)s1 )

1l ).

Next we set b(2)2 = b

(1)2 and start with a prime divisor q2 of b

(2)2 and repeat

the same process to obtain b(2)i for all i 6= 2. Suppose this process stops at

kth step then b(k)1 , . . . b

(k)su are the required numbers. We will put t = su and

this proves the claim.If l = qi for any i then we call b1 = b

(k)i and the rest of t− 1 numbers can be

taken in any order and if p 6= qi for some i then we can take any ordering.

Now Lemma 1.2.1 gives that [Q(b1l1 ) : Q] = l. Then we use Lemma 1.2.2

successively to obtain [Q(b1l1 , . . . , b

1lt ) : Q] = lt, which is the required degree.

1.3 lth power residue symbol

Let p be a prime different from l and f be the inertia degree of p in Z[ζl]. Forany prime ideal p of Q(ζl) dividing p and an integer α ∈ Q(ζl) not containedin p one has l|pf − 1 and αp

f−1 ≡ 1 (mod p). Hence there is an lth root of

unity ζ il , 0 < i ≤ l such that αpf−1l ≡ ζ il (mod p). Since lth roots of unity are

distinct modulo p there is unique such i.Definition: We define the lth residue symbol of α with respect to p by(αp

)l = ζ il , where i is as defined above.

We state some results about the higher residue symbols [21, 35].

Theorem 1.3.1. (Kummer’s Criterion) (αp

)l ≡ αpf−1l (mod p).

Theorem 1.3.2. The lth power residue symbols are completely multiplicative.

Theorem 1.3.3. α ∈ Q(ζl) is an lth power modulo p if and only if (αp

)l = 1.

Given any ideal a of Q(ζl), we will define (αa )l =∏

p|a(αp

)l with multiplicity

counted. For β ∈ Q(ζl) we will define (αβ)l = ( α

<β >)l where < β > stands

for the principal ideal generated by β.An integer α ∈ Q(ζl) is called primary if it is congruent to a rational integermodulo (1− ζl)2.

15

Theorem 1.3.4. (Eisenstein’s Reciprocity Law) If α is a primary integer anda is a rational integer coprime to α and coprime to l then one has ( a

α)l = (α

a)l.

We will now introduce some more terminologies and see an alternate wayto define the lth residue symbol.Let L/K be a Galois extension of number fields. For an unramified prime pof primeOL we will write kL for the residue field of p and k will denote theresidue field of p ∩ OK . There is an isomorphism [37],

Dp∼= Gal(kL/k),

where Dp is the decomposition group of p. Let σp ∈ Gal(kL/k) be theFrobenius at p then its inverse image (in the above exact sequence) in Dp is

called Artin symbol of ℘ for the extension L/K and is written as(

℘L/K

).

We note that if p and p′

are primes in OL above the same prime of OK then(p

L/K

)and

(p′

L/K

)are conjugate, by an element of Gal(L/K) which maps

p to p′. In particular if L/K is abelian then, for prime p = p ∩ OK in OK ,

we can define(

pL/K

)=(

pL/K

).

Definition: For any set S of prime ideals of ring of integers OK in a numberfield K, the density of S is defined by

limx−→∞|{p ∈ S : NK/Q(p) ≤ x}||{p : NK/Q(p) ≤ x}|

,

if the limit exists. Here for any finite set A by |A| we denote its cardinalityand NK/Q is the norm for the extension K/Q.For any σ ∈ Gal(L/K) let PL/K(σ) denote the set of prime ideals p in OK

such that there is a prime ideal p of OL above p such that(

pL/K

)= σ. We

recall the theorem of Cebotarev [37]

Theorem 1.3.5. (Cebotarev Density Theorem) Let σ ∈ Gal(L/K) and Cσstand for the conjugacy class of σ. Then density of PL/K(σ) is |Cσ |

[L:K].

In particular, if L/K is abelian then the density of primes p of OK such

that(

pL/K

)= σ is 1

[L:K].

The following lemma can be easily verified.

16

Lemma 1.3.6. Let L/K be a Galois extension of number fields and let Fbe an intermediate field such that F/K is Galois. Then for any unramified

prime p of OL one has(

pL/K

)|F

=(

p∩OFL/K

).

For a fixed prime l, we want to define the lth residue symbol (ap)l for each

prime p 6= l and integer a coprime to p. We will write fa(X) = X l − a andlet Ka denote splitting field of fa(X). Then Ka ⊃ Q(ζl). For any prime p ofZ[ζl] above p we let σp,a ∈ Gal(Ka/Q(ζl)) denote the Artin symbol for the

prime p for the extension Ka/Q(ζl). Then σp,a(a1/l)

a1/lis an lth root of unity.

Note that this is independent of the choice of a1l .

Definition: We define the residue symbol(ap

)l

by σp,a(a1/l)

a1/l.

Lemma 1.3.7. The definition of lth residue symbol given here and the onegiven earlier are equivalent.

Proof. Let f be the inertia degree of p in the extension Q(ζl)/Q. Then theArtin symbol σp,a satisfies

σp,a(a1l ) ≡ a

pf

l ( mod p).

This givesσp,a(a

1l )

a1l

≡ apf−1l ( mod p),

which proves the lemma.

For any positive real number x let π(x) denote the number of rationalprimes not bigger than x. We will consider the following sum∑

p;Norm(p)≤x

(n

p)l,

where sum runs over the prime ideals p of Z[ζl].Let p be a prime number. Since Q(ζl)/Q is abelian, all the primes p abovep have same stabilizer in the Galois group Gal(Q(ζl)/Q). Hence the sumis invariant under the action of the Galois group and is a rational number.The following theorem is well known but we supply a proof for the sake ofcompleteness.

17

Theorem 1.3.8. If n is not an lth power of an integer then the estimate∑p;Norm(p)≤x

(n

p)l = o(π(x))

holds as x→∞. Here Norm denotes for Norm map of the extension Q(ζl)/Q.

Proof. One has

∑p;Norm(p)≤x

(n

p)l =

l∑b=1

∑Norm(p)≤x,σp,n=τb

(n

p)l

,

where τb is automorphism of Kn which sends n1l 7−→ (ζl)

bn1l . Hence

∑p;Norm(p)≤x

(n

p)l =

l∑b=1

∑Norm(p)≤x,σp,n=τb

ζbl

.

Thus we obtain

∑p;Norm(p)≤x

(n

p)l =

l∑b=1

ζbl (1

lπ(x) + o(π(x))) =

1

lπ(x)

l∑b=1

ζbl + o(π(x)).

The first term is zero and this proves the result.

Given an integer m, a set of m integers a1, . . . , am and m elements ζril inµl, not necessarily distinct, we want to determine density of primes p whichsatisfy (ai

p)l = ζril . For this, we will consider the counting function

Sx =1

ulm

∑p;Norm(p)≤x,p/∈S′

m∏k=1

(l∏

j=1,j 6=rk

(ζjl − (akp

)l)

).

Here S ′ is the set of primes dividing la1 . . . am and u is a unit satisfying

ulm =m∏k=1

l∏j=1,j 6=rk

(ζjl − ζrkl ).

We note that Sx exactly counts number of primes p of Norm up to x whichsatisfy (ai

p)l = ζril for all i. Note that the choices of ri can not be arbi-

trary because of the multiplicativity of lth power residue symbol. That is tosay that the assignment ai −→ ζril shall be restriction of some morphism of

18

semi groups Z∗/Z∗l −→ µl, but the counting function already takes care ofthis. To show this we note that any multiplicative relation among a

′is can

be brought into the form∏m

k=1 acii = cl for some integers ci and c. Now the

corresponding relation expected in µl is∏m

i=1 ζricil = 1. If this does not hold

then it is easy to see from Theorem 1.3.3 that for each prime p there is ani such that (ai

p)l 6= ζril . Thus if ri’s satisfy the required condition then Sx

exactly counts number of primes of Norm up to x which satisfy (aip

)l = ζrilfor all i. In case there is inconsistency among choices of ri, then Sx = 0.

Now to estimate Sx, we can actually pass down to the correspondingcounting function for b

′js which also will be denoted by Sx. When we change

from the set S = {a1, . . . , am} to the set T = {b1, . . . , bt} obtained as in thealgorithm in section 2, then, the given m elements ζril uniquely determine aset of t elements ζ

sjl such that (ai

p)l = ζril , for all 1 ≤ i ≤ m if and only if

(bjp

)l = ζsjl for all 1 ≤ j ≤ t. If the conditions (ai

p)l = ζril , for all 1 ≤ i ≤ m

lead to a condition of the form (bjp

)l = ζsjl for all 1 ≤ j ≤ t with bj an lth

power and sj 6= 0, then we can immediately conclude that there is no primep satisfying the condition. Studying the counting function with the bj makesit easier, since there will be only one main term with one root of unity in it(not a sum of roots of unity). Hence, it is enough to study the behavior of

primes p which satisfy (bjp

)l = ζsjl for all 1 ≤ j ≤ t. Now we consider the

counting function

Sx =1

vlt

∑p;Norm(p)≤x,p/∈S′

t∏k=1

(l∏

j=1,j 6=rk

(ζjl − (bkp

)l)

).

Here v is a unit satisfying

vlt =t∏

k=1

l∏j=1,j 6=sk

(ζjl − ζskl ).

We emphasize that Sx exactly counts number of primes of Norm up to xwhich satisfy (ai

p)l = ζril for all i. Because of multiplicativity of lth power

residue symbol one obtains

Sx =1

ult

∑p;Norm(p)≤x,p/∈S′

∑0≤di≤l−1,n=

∏bdii

ζtnl (n

p)l,

for some integer tn. Now we change the order of summation to obtain

Sx =1

ult

∑0≤di≤l−1,n=

∏bdii

ζtnl∑

p;Norm(p)≤x,p/∈S′(n

p)l.

19

By Theorem 1.3.8 if n is not an lth power then the contribution due to innersum is o(π(x)). From the construction of bj its clear that no n 6= 1 will be anlth power. Hence, the main term will give, in absolute value, 1

lt(π(x)− |S ′|).

Thus density of the primes p satisfying (bjp

)l = ζsjl for all 1 ≤ j ≤ t and

hence satisfying (aip

)l = ζril , for all 1 ≤ i ≤ m is 1lt

.

Remark 1.3.9. 1. Note that the density does not depend upon the choice ofri as long as there is the required consistency.2. One can also obtain the density of primes p of Z[ζl] which satisfy (αi

p)l =

ζril , where αi are integers of Q(ζl) and ri’s are as in the Introduction. Theabove proof may not work in this case, since the ring Z[ζl] need not be prin-cipal domain and hence the algorithm may not work. However we see fromthe second definition of lth residue symbol in section 3 that if the requirement(αip

)l= ζril is consistent (in the same sense as in section 3) then it uniquely

determines an element in Gal(Q(ζl, α1l1 , . . . , α

1lm)/Q(ζl)), and hence density of

such primes p is [Q(ζl, α1l1 , . . . , α

1lm) : Q(ζl)]

−1, see [26].

1.4 Another way to find the degree

Let p1, p2, . . . , pn be all the primes dividing a1 . . . am. Let us write λij forexact power of pj dividing ai. Then we will consider the m × n matrix Twhose (i, j)th entry is λij. Note that for our purpose we can assume that0 ≤ λij ≤ l − 1.

Lemma 1.4.1. The cardinality of the set A = {(λi)mi=1 : 0 ≤ λi ≤ l −1,∏

i aλii ∈ Zl} is a power of l.

Proof. Consider the Z/lZ vector space (Z/lZ)m with basis S = {a1, . . . , am}.Z/lZ acts on Q∗/(Q∗)l by α.x = xα. Consider the map T:(Z/lZ)m −→Q∗/(Q∗)l which sends ai → ai, where ai denotes the class represented by in-teger ai in Q∗/(Q∗)l and extend it linearly then

∑i λiai ∈ kerT iff (λi) ∈ A.

This proves that |A| is an lth power.

As mentioned in [48] the degree Q(a1l1 , . . . , a

1lm, ζl)/Q(ζl) is lm/lr where lr is

the cardinality of set A. Now we relate this degree to the rank of the matrixT .

Theorem 1.4.2. The rank of the matrix T is m− r.

Proof. If there are xi, 1 ≤ i ≤ m with 0 ≤ xi ≤ l−1 such that∏m

i=1 axii ∈ Zl,

then for all j we have x1λ1j + . . . + xmλmj = 0( mod l), i.e the row vectors

20

(λi1, . . . , λin), 1 ≤ i ≤ m in (Z/lZ)n are linearly dependent. Conversely anysuch linear dependence among row vectors of matrix T will give exactly lmany relations of the type in set A. Let r be the rank of the matrix T . Aftera rearrangement we can assume that the row vectors (λi1, . . . , λin), 1 ≤ i ≤ rare linearly independent. Then we see that for any choice of xi, 1 ≤ i ≤r with 1 ≤ xi ≤ l − 1 the condition

∏ri=1 a

xii ∈ Zl does not hold. On the

other hand for any selection of xj, j > r we have l many relation of the form∏mi=1 a

xii ∈ Zl (this can be seen by looking at the vectors (λi1, . . . , λin), 1 ≤

i ≤ r and xr+1(λr+11, . . . , λr+1n) + . . .+ xm(λm1, . . . , λmn) which are linearlydependent). This proves the theorem.

21

Chapter 2

Catalan’s Conjecture

This chapter gives a sketch of a proof of the Catalan’s conjecture over Z,mainly based on the exposition of Rene Schoof [41]. We shall also be referringto the articles by Yuri Bilu [3, 4]. In section one, we state the conjecture andmention some early developments made towards the solution. Section twois devoted to the Cassels’ criteria, the first significant progress towards thesolution of the Catalan’s conjecture. All the later results depend on Casselscriteria. Through section 3-5 we present, very briefly, Mihailescu’s threetheorems from which Catalan’s conjecture follows. In section 6 we mentionthe generalization of Catalan’s conjecture to the number fields. Most of theproofs will be suppressed due to the technicality.

2.1 Introduction

In 1844, Roman mathematician Eugene Charles Catalan, in a letter to theeditor of the Crelle’s journal, made the following;Conjecture: The only pair of consecutive integers both of which are perfectpowers is (8, 9).An integer is a perfect power if it can be written in the form ts for integerst 6= 0,±1 and s > 1. In terms of (exponential) Diophantine equation theCatalan’s conjecture amounts to the,Conjecture: The only solution (x, y,m, n) of the Diophantine equation xm−yn = 1 with x, y ∈ Z, xy 6= 0 and m ≥ 2, n ≥ 2 are (±3, 2, 2, 3).A Diophantine equation in which the exponents are also varying is calledexponential Diophantine equation. We see that if the equation xp − yq = 1with p, q primes and xy 6= 0 has only solution (±3, 2, 2, 3) then the equationxm − yn = 1 for general m,n ≥ 2 and xy 6= 0 has only the above solution.From now on by a Catalan equation over Z we will mean the equation xp −

22

yq = 1, where x, y ∈ Z and p, q are primes. A solution (x, y, p, q) of theCatalan’s equation is called trivial if xy = 0. Thus the Catalan’s conjecturereads as,Conjecture: The only non trivial solutions of Catalan’s equation over Z are(3, 2, 2, 3) and (−3, 2, 2, 3).The particular case “q = 2” was solved by V. A. Lebesgue in 1850.

Theorem 2.1.1. (Lebesgue) The Diophantine equation xp − y2 = 1 has nonon trivial solutions in integers.

For a proof we refer chapter 2 from [41] or [28]. Chapter 3 of this thesisalso contains a proof, in more general setup.Euler had shown, by the method of descent, that the only non trivial solu-tions of the equation x2 − y3 = 1 are (±3)2 − 23 = 1. Also the equationx2 = y3 + 1 represents an elliptic curve of rank 0. Using the method ofdescent one finds out that the group of rational points on this curve areprecisely (0,−1), (±1, 0), (±3, 2) and ‘the point at infinity’. For a completesolution we refer to [18] or [41].In 1965, Ko Chao proved that the Catalan’s equation has no non trivial so-lution when p = 2, q ≥ 5.

Theorem 2.1.2. (Ko Chao) The equation x2 − yq = 1 has no non trivialsolution in integers when q ≥ 5.

A proof can be found in chapter 3. Also [41, 10] contain a good accountof the proof.With these developments, we see that to solve the Catalan’s conjecture itremains to consider the case when both the exponents p and q are odd primes.Definition: Any non-trivial solution (x, y, p, q) of the Catalan’s equation withp and q odd primes will be referred as a Catalan tuple.Note that if (x, y, p, q) is a Catalan tuple then so is (−y,−x, q, p).The first major breakthrough for general Catalan’s equation was made by J.W. S. Cassels in 1960 [8]. Cassels gave the following criteria for a solution ofCatalan’s equation.

Theorem 2.1.3. (Cassels) Whenever (x, y, p, q) is a Catalan tuple then qdivides x and p divides y.

We remark that all the subsequent developments, including the Mihailescu’sproof, use Cassels criteria. In the next section we will sketch a proof of Cas-sels criteria. For a complete proof we refer the reader to [41] or [7, 8]. Alsochapter 4 contains a proof in some cases for Q(i), and it is easy to deduce

23

Theorem 2.1.3 from that.The work of Robert Tijdeman needs a special mention. In 1976, Tijdeman[46] proved, using the theory of linear forms in logarithms, that the Catalan’sequation has only finitely many solutions. The bound obtained in Tijdeman’swork are large and way beyond the reach of computer calculation.The conjecture was finally proven by Preda Mihailescu. Between the years2000-2003 Preda Mihailescu proved the following three theorems [32, 33, 34];

Theorem 2.1.4. (Mihailescu) For a Catalan tuple (x, y, p, q), we have

pq−1 ≡ 1 (mod q2) and qp−1 ≡ 1 (mod p2).

Theorem 2.1.5. (Mihailescu) Whenever (x, y, p, q) is a Catalan tuple, onehas

p ≡ 1 (mod q) or q ≡ 1 (mod p).

Theorem 2.1.6. (Mihailescu) If (x, y, p, q) is a Catalan tuple then

p < 4q2 and q < 4p2.

Now we show, how these three theorems give a complete proof of theCatalan’s conjecture.

Proof. (Catalan’s conjecture) It is sufficient to prove that there is no Catalantuple. Let (x, y, p, q) be a Catalan tuple. It is easy to establish p 6= q. Alsodue to symmetry of Catalan tuple, we can assume p > q. Now by Theorem2.1.5, we have p ≡ 1 (mod q), that is p = 1+kq for some integer k. Now usingTheorem 2.1.4 it is easy to see that q|k and thus we have p = 1 +k′q2. UsingTheorem 2.1.6 we see that k′ ∈ {0, 1, 2, 3}. But p is a prime so k′ 6= 0, 1, 3.On the other hand we see that 2q2 + 1 is divisible by 3 and hence p = 3, butp > q and both are odd primes. This is not possible.

We remark that theorem 2.1.4 and Theorem 2.1.5 already imply Catalan’sconjecture, when aided with the estimates due to linear form in logarithms [4].Mihailescu’s Theorem 2.1.6 completely removed the need of estimates andcomputer calculation. In order to make Theorem 2.1.5 and Theorem 2.1.6 towork, Mihailescu needed to rule out any solution of Catalan equation whenone of the primes p or q is smaller than or equal to 5. In this direction, heproved the following theorem.

Theorem 2.1.7. There is no Catalan tuple (x, y, p, q) with p, q ≤ 5.

The theorem proved by Mihailescu was much stronger, but this is suffi-cient to make Theorem 2.1.5 and Theorem 2.1.6 work. The proof of these4 theorems, of Mihailescu, uses two main ingredients: Runge method andtheory of cyclotomic fields [3, 4, 32, 33, 34, 41].

24

2.2 Cassels criteria

The proof of Cassels criteria is based on Runge’s method [39]. Here we willbriefly sketch the proof. A detailed proof of more general theorem appearsin chapter 4.For a Catalan tuple (x, y, p, q), it is easy to notice that p 6= q. So we assumep > q throughout this section.If q - x, then y + 1 and yq+1

y+1are co-prime and hence pth powers of integers,

say

y + 1 = bp andyq + 1

y + 1= vp.

The function f(X) = Xp − (bp − 1)q is strictly increasing function of X. Wesee at once that f(bq) > 1 and f(bq−1) < 1, and hence there is no integer Xsatisfying Xp − (bp − 1)q = 1. This contradiction shows that q|x. With thiswe obtain gcd(y

q+1y+1

, y + 1) 6= 1. Let

yq + 1

y + 1= qrup and y + 1 = qsbp.

One checks that r = 1 and s = kp − 1. Now using the fact yq+1y+1

≡ q

(mod y + 1) we obtain |x| ≥ q + qp−1. We record these as,

Proposition 2.2.1.(1)q divides x,(2)we have that |x| is at least as big as q + qp−1.

Now we recall the following,

Lemma 2.2.2. Let F (t) = ((t + 1)p − tp)1/q denote the function of realvariable t for |t| < 1. Put m = [p

q] + 1 and let Fm(t) denote the sum of terms

of degree at most m of Taylor expansion of F (t) around 0. Then we have

|F (t)− Fm(t)| ≤ |t|m+1

(1− |t|)2,

for |t| < 1.

In order to establish the Cassels criteria it remains to prove that p|y.Assume the contrary. Then we obtain x − 1 = aq for some integer a. Thus(aq + 1)p − 1 is a qth power, also apq is a qth power and hence |[(aq + 1)p −1]1/q − ap| is a non zero integer. Observe that y = apF ( 1

aq). Put z =

amq−py − amqFm( 1aq

), then for D = qm+ordq(m!) the number Dz is a non zerointeger. Using the lower bound on x in Proposition 2.2.1 and from Lemma

25

2.2.2 we conclude that |Dz| < 1, whenever q ≥ 3. This contradiction provesthat p|y.In next proposition we will record some important consequences of Casselscriteria;

Proposition 2.2.3. For any Catalan tuple (x, y, p, q) we have integers a, b, u, vsatisfyingyq+1y+1

= qup−1 and y + 1 = qp−1bp

xp−1x−1

= pvq and x− 1 = pq−1aq.Also |x| ≥ max(pq−1 − 1, qp−1 + q) and |y| ≥ max(qp−1 − 1, pq−1 + p).

2.3 Wieferich Criterion

In this section we will introduce the ’obstruction group’ and will sketch theproof of the Theorem 2.1.4 and the Theorem 2.1.7. We will need ideas from‘annihilator of class groups’. [47, 27]. Throughout this section, as earlier,(x, y, p, q) will denote a Catalan tuple. Let ζp denote a fixed primitive pth

root of unity. Let G denote the Galois group of the pth cyclotomic extensionQ(ζp)/Q. For integers a co-prime to p, we denote, by σa, the element of Gwhich sends ζp to ζap , e.g. σ−1 is complex conjugation. Consider the elements

θi =

p−1∑a=1

[ia

p]σ−1a ∈ Z[G].

Let J denote the ideal in Z[G] generated by θ′is. Then, as a consequence ofStickelberger’s theorem we have

Theorem 2.3.1. The ideal J annihilates the class group of Q(ζp).

We will put ei = (1−σ−1)(θi+1−θi) and denote the ideal in Z[G] generatedby e′is by I. The following lemma is a matter of routine verification,

Lemma 2.3.2. Let ei =∑

σ∈G niσσ−1, then niσ ∈ {±1}.

Now we define the obstruction group:

H = {α ∈ Q(ζp)∗ : ordτ (< α >) ≡ 0 (mod q) if τ - p}/Q(ζp) ∗q .

Here < α > denotes the principal ideal generated by α and ordτ (< α >)denotes the highest power of prime ideal τ dividing < α >. For α ∈ Q(ζp)

∗,the corresponding element of H will be denoted by α. Whenever α ∈ H, weimmediately have < α >= aq(1 − ζp)r, for some ideal a and integer r. Thefollowing lemma justifies the name obstruction group. The lemma will beproved, in a more general setting, in chapter 5.

26

Lemma 2.3.3. For any Catalan tuple (x, y, p, q) the class of (x − ζp) inQ(ζp) ∗ /Q(ζp)∗q is in H.

Let Ep = Z[ζp,1p] be the Z algebra generated by ζp and 1

p. We have

following,

Proposition 2.3.4. There is a natural exact sequence of Fq[G]-modules

0 7→ Ep/Eqp 7→ H 7→ Clq[p] 7→ 0,

where Clq[p] denotes the q−part of class group of Q(ζp) and the map H 7→Clq[p] is given by α 7→ a.

For any G module M , we will denote the G module M/σ−1M by M− andsimilarly M+ will denote Mσ−1M . Using Proposition 2.3.4 and Theorem2.3.1 we can deduce the following,

Proposition 2.3.5.(1) H− ∼= Cl−q [p],(2) The elements ei are in annihilator of the Z[G] module H.

With these ingredients in hand, now we are in a position to sketch theproof of the Theorem 2.1.4.

Theorem 2.3.6. If (x, y, p, q) is a Catalan tuple then q2 divides x and p2

divides y.

Proof. It is enough to prove q2|x and the other follows from symmetry . Letθ ∈ I, then by Proposition 2.3.5, we have (1− ζpx)θ = βq for some β ∈ Q(ζp)(note that x− ζp and 1− ζpx give same element in H). Since q|x we obtain1 ≡ βq (mod qZ[ζp]). From Lemma 4.2.9, to be proved in chapter 4, weimmediately obtain 1 ≡ βq (mod q2Z[ζp]). Now substituting (1 − ζpx)θ forβq and expanding it we obtain

1−∑σ∈G

nσσ(ζp)x ≡ 1 (mod q2Z[ζp]),

where θ =∑

σ∈G nσσ. If q2 - x then we see that q|θ but θ ∈ I was arbitrary.Taking θ = ei, we see that q - θ. So we obtain q2|x.

Proof. (Theorem 2.1.4) From Proposition 2.2.1 we have integer a such thatx − 1 = pq−1aq. Using Fermat’s little theorem and Cassels criteria we get−1 ≡ aq (mod q). Lemma 4.2.9 can be used to conclude −1 ≡ aq (mod q2).From this and Theorem 2.3.6 we derive pq−1 ≡ 1 (mod q2).

27

Proof. (Theorem 2.1.7) Due to symmetry in Catalan tuple we can assumep ≤ 5. For p ≤ 5 the group Cl−q [p] is trivial [47] and hence from (1) inProposition 2.3.5 we see that H− is trivial which implies that (x−ζp)1−σ−1 =αq for some α ∈ Q(ζp). But this last assertion is not true [41, 17]. Thiscontradiction establishes the theorem.

2.4 Proof of the Theorem 2.1.6

In this section we want to prove p < 4q2 and q < 4p2. We assume this isnot the case. Due to symmetry we can assume that q ≥ 4p2. We intend toarrive at a contradiction. In order to do this one needs to study the effectsof embeddings of Q(ζp) into C on the elements α which satisfy (x− ζp)θ = αq

for θ ∈ I. In this direction we will mention following result, proof of whichcan be found in [41].

Proposition 2.4.1. For any embedding φ : Q(ζp) 7→ C, there exists anelement θ ∈ I with ||θ|| ≤ 3q

p−1and with the property that (x− ζp)θ = αq for

some 1 6= α ∈ Q(ζp)∗ and

|φ(α)− 1| ≤ 2||θ||q(|x| − 1)

,

where for θ =∑nσσ we define ||θ|| =

∑|nσ|. Also ψ(α) is contained in the

unit circle, for any embedding ψ.

We now prove the Theorem 2.1.6.

Proof. (Theorem 2.1.6) We fix an embedding φ : Q(ζp) 7→ C and let θ ∈ Iand α ∈ Q(ζp)

∗ be as in the Proposition 2.4.1, then we have

|φ(α)− 1| ≤ 2||θ||q(|x| − 1)

.

The same inequality is true for the complex conjugate of φ(α). For any otherembedding ψ we have |ψ(α)− 1| ≤ 2. Thus we obtain

N(α− 1) ≤ 2q−1

q2

(||θ|||x| − 1

)2

.

Let < α >= aa′

for co-prime ideals a and a′. Note that < α − 1 > also hasdenominator a′. We have

(x− ζp)θ =∏σ∈G

(x− σ(ζp))nσ = aq/a′

q.

28

Since θ ∈ I the norm of (x− ζp)θ is 1. Hence N(a) = N(a′). Also

N(a′)2q = N(aa′)q =∏σ∈G

N((x− σ(ζp))|nσ |).

Now the latter one is at most (|x| + 1)(p−1)||θ|| and hence N(a′) ≤ (|x| +1)(p−1)||θ||. With this we conclude

(|x|+ 1)−(p−1)2q||θ|| ≤ N(α− 1) ≤ 2p−1

q2

(||θ|||x| − 1

)2

.

Since |x| > qp−1 > 80 we have (|x| + 1) ≤ 43(|x| − 1), also ||θ|| ≤ 3q

p−1so we

obtain

(|x|+ 1)1/2 ≤ 16

(p− 1)22p−1.

This gives q(p−1)/2 < 2p−1, which is impossible as q ≥ 5. With this contra-diction Theorem 2.1.6 is established.

2.5 Mihailescu and Cyclotomic Fields in the

context of Catalan Conjecture

In this section we aim to sketch the proof of the Theorem 2.1.5. For anys ∈ Q we let (1 + T )s denote the power series in Q[[T ]] defined by

(1 + T )s =∑k≥0

(s

k

)T k.

For an element θ =∑

σ∈G nσσ ∈ Z[G] we define

F (T ) = (1− ζpT )θ/q =∏σ

(1− σ(ζp)T )nσ/q ∈ Q(ζp)[[T ]].

For each embedding φ : Q(ζp) 7→ C we let F φ(T ) denote the power seriesobtained by applying φ to the coefficients of F (T ). If, for a complex numbert the power series F (T ) converges at T = t, then its limit will be denotedby F (t). Among many results of Mihailescu concerning this power series thefollowing is of utmost importance;

Proposition 2.5.1. Suppose that for some θ in the ideal of Z[G] generatedby (1 + σ−1) and for some t ∈ Q satisfying |t| < 1 we have (1 − ζpt)θ = βq

for some β ∈ Q(ζp)+, the maximal real sub field of Q(ζp). Then we have that

F φ(t) = φ(β), for every embedding φ : Q(ζp)+ 7→ R.

29

The proof is quite technical. Let G+ denote the Galois group of theextension Q(ζp)

+/Q. Using Proposition 2.5.1 Mihailescu proves the following;

Theorem 2.5.2. For any Catalan tuple (x, y, p, q) with p, q ≥ 7, the Fq[G]submodule of H generated by (x − ζp)1+σ−1 is a free module over the groupring Fq[G+].

The proof is based on the Runge’s method on the line of the proof ofCassels criteria. Next we have the following;Definition: An α ∈ Q(ζp)

∗ is called a q − adic qth power if it is qth power inthe completion of Q(ζp) with respect to every prime ideal dividing q.Next we consider the Selmer group

S = {α ∈ H : α is a q − adic qth power }.Then, as was the case with the negative component, we see that (x −ζp)

1+σ−1 ∈ S+.Using theory of semi simple group rings Mihailescu obtains an injection ofFq[G+]−modules:

S+ ↪→ C(q)Eq/Eq × E/CEq × Cl+q [p],

where E is the subgroup of p−units of Q(ζp)∗, C is the subgroup of p−cyclotomic

units which is the multiplicative Z[G] module generated by 1 − ζp. Inabove C(q), E(q) stands for the elements in C and E respectively which areq − adic qth powers. Now, using the theorem of Thaine [45], Mihailescuestablishes the following;

Theorem 2.5.3. If p > q are odd primes and p 6= 1 (mod q) then asan Fq[G+]−modules the annihilators of C(q)Eq/Eq ×E/CEq also annihilateCl+q [p].

Next he shows that C(q)Eq/Eq × E/CEq has non-trivial annihilatorswhenever p > q. Thus given any Catalan tuple (x, y, p, q), with p > q, and ifp 6= 1 (mod q) then by the Theorem 2.5.3 we see that as an Fq[G+]−modulethe submodule of S+ generated by (x− ζp)1+σ−1 has a non trivial annihilatorand this contradicts the Theorem 2.5.2. This contradiction establishes p ≡ 1(mod q) and we are done with the proof of the Theorem 2.1.5.

2.6 Number Field Analog

In this section we mention the analog of the Catalan conjecture over numberfields. Some, albeit small, progress is made by the author. These results

30

are main theme of chapter 3-5. Here we give the summary of those threechapters.We will use letter K to denote a number field. There are accounts of studiesof the Catalan’s equation over number fields [5, 38]. Given any number fieldK one can ask the followingCatalan Problem: Describe all the solution of xp − yq = 1 with p, q primesand x, y ∈ OK , xy 6= 0.There is no precise conjecture made for the number field analog of the prob-lem. It is known [5] that for any fixed number field K the Catalan problemhas only finitely many solutions. But the bounds are too big to solve theproblem completely. The problem is extremely difficult and no progress hasbeen made till the date, apart from the finiteness established by Brindza et.al.[5]We focus on the field K = Q(i). Here i is the complex number satisfy-ing i2 = −1. It seems that the only solutions of Catalan problem are(±3, 2, 2, 3), (−2,±3i, 3, 2) but we are unable to say something conclusive.Some of the results are true in more general setting (quadratic imaginarynumber field with class number one) and we will take liberty to state theresults for these class of fields, whenever possible. But our main focus willbe on the field Q(i).Even primes exhibited different behavior in rational case [10, 28] and so isthe case for Z[i]. In our attempt to study the Catalan problem for Z[i] wefirst need to dispose off the case when one of the primes is even. This isthe content of chapter 3. In Chapter 3 we present a solution of the Catalanproblem when one of the exponents is even.In Chapter 4 we give an analog of Cassels criterion. Chapter 5 introduces theobstruction group, and some more results are reported in this investigation.

31

Chapter 3

Catalan Problem over Z[i] witheven exponents

3.1 Introduction

Consider a tuple (x, y, p, q), with x, y ∈ Z[i] satisfying xp − yq = 1. In thischapter we are interested in the case when one of the primes is even. Recallthat we call a solution to be trivial if xy = 0. When both p = q = 2 thenx−y and x+y are units and this gives a trivial solution. Thus we can assumethat one of p and q is even and the other is odd. The equation xp − y2 = 1translates to y2 − xp by the change of coordinates x −→ −x and y −→ iy.Thus, its enough to study any one of the equations xp−y2 = 1 and x2−yq = 1in order to solve the Catalan equation with even exponent.In the first section we assume that the prime p > 5. The method is on theline of Liouville’s idea of approximations of algebraic numbers by rationals.In section 2 we give solution for the equations x2 − y3 = 1 and x3 − y2 = 1,which exhibit some non-trivial solutions. This computation is based on someideas on elliptic curves. In section 3 we handle the left cases x2 − y5 = 1and x5 − y2 = 1 which were not covered earlier and require delicate analysisthan in section 1. For the primes p > 5 the approach is to reduce thesolvability of the equation xp − y2 = 1 to solvability of an equation of theform (x

′)p − 4(y

′)p = 4, and then solve the latter one.

3.2 Primes p > 5 and q > 5

In this section, we prove

Theorem 3.2.1. The equation xp−y2 = 1 for p > 5 has only trivial solutionsx, y ∈ Z[i].

32

Proof. Suppose that xp − y2 = 1 has a solution. Then we have;xp = y2 + 1 = (y + i)(y − i).Note that y can not be real. If y is real, then y is a rational integer andxp = y2 + 1 is a positive integer. If y2 + 1 is a pth power of a rational integerthen we obtain a solution to the Catalan equation xp − y2 = 1 in rationalintegers, but such a solution does not exist. If y2 + 1 is not pth power of arational integer then the polynomial Xp − (y2 + 1) is irreducible over Z, asmentioned in Lemma 1.2.1, and hence can not have a solution in Z[i].We consider following two cases;Case (1): y + i and y − i are coprime.This will give y + i and y− i are pth powers up to a unit. Since all the unitsin Z[i] are pth powers so y + i and y − i are pth powers themselves.

One has y + i = xp1 and y − i = xp2, which leads to;

xp1 − xp2 = 2i. (3.1)

Since y is not real, |y + i| 6= |y − i|. So one has |x1| 6= |x2|. Without loss ofgenerality we can assume that |x1| > |x2| =

√n for some positive integer n.

So one has |x1| ≥√n+ 1. Now using equation (3.1) we obtain;

2 = |2i| = |xp1 − xp2|

≥ |xp1| − |xp2|

≥ (n+ 1)p/2 − np/2≥ 5/2.

This is a contradiction.case (2): y + i and y − i are not coprime.Claim: gcd(y + i, y − i) = 2iAny common divisor of y+i and y−i will divide 2i. If gcd(y+i, y−i) = 1+ithen at least one of y + i and y− i is divisible by (1 + i)2 = 2i, as the powerof 1 + i in (y + i)(y − i) is at least p > 3. Also y + i and y − i differ by 2i sothe other one too is divisible by 2i. This proves the claim.Hence one has,

y + i = (1 + i)r1xp1, y − i = (1 + i)r2xp2, x = (1 + i)kx1x2,

where r1, r2 are positive integers satisfying r1+r2 ≡ 0 (mod p), min{r1, r2} =2 and k is a positive integer.Let us assume that min{r1, r2} = r2, so one gets (1+ i)r1xp1− (1+ i)2xp2 = 2i,which in turn, by putting x3 = −(1 + i)kx1 with r1 + r2 = kp, leads to14xp3 − x

p2 = 1 for some integers x3, x2 in Z[i]. Also x3 = 0 or x2 = 0 will lead

to y = ±i, which corresponds to a trivial solution. The case min{r1, r2} = r1

is similar. Thus the theorem is proved once we show that the equation

33

14xp3− x

p2 = 1 has no non-trivial solutions in Z[i]. This is done in next propo-

sition.

Proposition 3.2.2. Let p ≥ 7 be a rational prime. The equation 14xp3−x

p2 = 1

has no non-trivial solution in Z[i].

Lemma 3.2.3. For any solution (x, y, p, q) of the Catalan’s equation with

y + i and y − i not coprime one has |x| > 212 (2

p−42 − 1).

Proof. As earlier we will obtain −2i = (1 + i)2xp2 − (1 + i)kp−2xp1, whichgives (−x2)p ≡ 1 (mod (1 + i)p−4), since k ≥ 1. As p does not divide2p−4 − 2p−5 = #(Z[i]/(1 + i)p−4)∗ we get −x2 ≡ 1 (mod (1 + i)p−4) and

hence |x2| ≥ 2(p−4)/2 − 1. Now |x| ≥ 212 |x2| ≥ 21/2(2(p−4)/2 − 1), proving the

lemma.

Lemma 3.2.4. Let x3, x2 be as in Proposition 3.2.2, then one has

|x3 − 41/px2| ≤ 41/p1

p

1

|x2|p−1|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)|.

Proof. Since x3, x2 satisfy xp3 − 4xp2 = 4, we have x3x2

= 41/pζnp (1 + 1xp2

)1/p, for

some integer n, here ζp = e2πip . Using binomial expansion we see that,

|x3 − 41/pζnp x2| = 41/p1

p

1

|x2|p−1|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)|.

We note that the following claim will establish the lemma.Claim: n = 0 (mod p).Clearly |x3| ≤ |x2|2, this gives us, after multiplying by x3,

||x3|2 − 41/pζnp x3x2| ≤ 41/p1

p

1

|x2|p−3|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)|.

Let x3x2 = a+ ib, then Im(|x3|2− 41/pζnp x3x2) = −41p

[b(ζnp+ζpn

)+a(ζnp−ζpn

)]

2and

hence one obtains

|41p [b(ζnp + ζp

n) + a(ζnp − ζp

n)]| ≤ 41/p2

p

1

|x2|p−3|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)|.

Let λ be a solution of λ2+1λ2−1

+ ab

= 0, then λ2 = a−bb+a

is a real number. We have

|ζnp + ζp

n

ζnp − ζpn −

λ2 + 1

λ2 − 1| ≤ 2

p

1

|x2|p−3|(1 + (1/p−1)

2!1

x2p−12

+ . . .)

ζnp − ζpn |,

34

i.e.

|2(λ2 − ζ2n

p )

(ζ2np − 1)(λ2 − 1)

| ≤ 2

p

1

|x2|p−3|(1 + (1/p−1)

2!1

x2p−12

+ . . .)

ζnp − ζpn |.

i.e.

|(λ2 − ζ2n

p )

(λ2 − 1)| ≤ 1

p

1

|x2|p−3|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)|.

Since p ≥ 7 and |x2| >√

3, the right side quantity in above inequality is atmost 2

p1√

3p−3 .

Now if |λ2 − 1| ≥ 3, then |λ2 − ζ2np | ≥ |λ2 − 1| − 2. This will force that left

hand side is at least 1/3, which is in contradiction to the upper bound.In case |λ2 − 1| ≤ 3, then we obtain |λ2 − ζ2n

p | ≤ 6p

1√3p−3 . Since λ2 is real

we obtain |Im(ζ2np )| ≤ 6

p1√

3p−3 . i.e |sin(2πn

p)| ≤ 6

p1√

3p−3 . If n 6= 0 (mod p)

then |sin(2πnp

)| ≥ |sin(2πp

)|. But since p ≥ 5 we have |sin(2πp

)| ≥ 4p

(when

x ≤ 14

then |sin(2πx)| ≥ 4x). This gives 4p< 6

p1√

3p−3 One checks that the last

inequality does not hold, thereby establishing the claim.

Now we give the proof of the Proposition 3.2.2.

Proof. (Proposition 3.2.2) One observes that if there is a non-trivial solution,then x3 is even and x2 is odd (i.e 1+ i|x3 and 1+ i - x2). As p > 5, x2 can notbe a unit. One checks that both|x3|, |x2| are bigger than

√3. Let us write

x3 = a3 + ib3, x2 = a2 + ib2.Since a3 − 41/pa2 = Re(x3 − 41/px2) so, using Lemma 3.2.4 one obtains

|a3 − 41/pa2| ≤ 41/p1

p

1

xp−12

(1 +(1/p− 1)

2!

1

x2p−12

+ . . .). (3.2)

One knows that 1√2|x2| ≥ min{|a2|, |b2|}. We make the following

Claim: min{|a2|, |b2|} 6= 0.If a2 = 0, then from inequality (3.2) we find that |a3| = 0. This will give(ib3)p − 4(ib2)p = 4, a contradiction as left hand side is not real.If b2 = 0 then considering the imaginary part we will obtain b3 = 0. Thus,in this case, x2 and x3 are real. Define y by y − i = (1 + i)2xp2, theny + i = −(1 + i)−2xp3 and y2 + 1 = (−x2x3)p. Thus y satisfies xp − y2 = 1with x = −x2x3 ∈ Z. Since y is purely imaginary, by putting x′ = −x andy′ = y/i we obtain an integral solution y′2−x′p = 1 of the Catalan’s equation,a contradiction. This contradiction establishes the claim.Let us assume that min{|a2|, |b2|} = |a2|. Now consider the function f(x) =

35

xp − 4, Then one has ,

1

|a2|p≤ |f(

a3

a2

)− f(41/p)| = |a3

a2

− 41/p||pξp−1|,

for some point ξ between a3a2

and 41/p. Now using above estimate we get

1

|a2|p≤ 1

|a2|41/p1

p

1

|x2|p−1|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)||pξp−1|.

Also we have |a2| ≤ 1√2|x2| and hence one obtains,

1

|a2|p≤ 1

|a2|41/p1

p

1

2p−12 |a2|p−1

|(1 +(1/p− 1)

2!

1

x2p−12

+ . . .)||pξp−1|.

i.e.

2p−12 ≤ 41/p|(1 +

(1/p− 1)

2!

1

x2p−12

+ . . .)||ξp−1|.

Now we note that 1 + (1/p−1)2!

1

x2p−12

+ . . . is dominated by the geometric series

1 + 1|x2|p + 1

|x2|2p + . . .. Now, using the lower bound on x2 obtained in Lemma

3.2.3 (|x2| ≥ 23/2 − 1), we see that 2p−12 < 41/p|ξp−1|(1 + 1

10). If |a3

a2| < 41/p

then we obtain 2p−12 < 4(1 + 1

10) but this is not possible for p ≥ 7.

Now we handle the case |a3a2| > 41/p. In this case we have ξ = 41/p + ε, where

|ε| ≤ |a3 − 41/pa2

|a2||.

So |ξ|p−1 ≤ (41/p + ε)p−1 ≤ 4(p−1)/p + 1. To see the last inequality we justnotice that |

(p−1k

)4(p−1−k)/pεk| ≤ 1/p and using this we obtain that p satisfies

2p−12 ≤ 4 + 41/p and this does not hold for p ≥ 7.

3.3 Elliptic curve case

In this section we intend to settle the equations x3− y2 = 1 and x2− y3 = 1.They both represent elliptic curves defined over Q.We will consider the equation x2− y3 = 1, which after change of co-ordinatetakes the form y2 = x3 + 1. The first one is dealt similarly. We will let Edenote the set of Q-rational points on the curve y2 = x3 + 1 and E(i) willdenote the Q(i)-rational point on the same. Both E(i) and E have a groupstructure under ‘elliptic curve addition +’. Given any point P = (a, b) in

36

E(i), the point P = (a, b) is also in E(i). Here z 7−→ z is complex conjuga-tion. The point P + P of E(i) is stable under complex conjugation and henceis in E. Thus we have the trace map T : E(i) −→ E sending P 7−→ P + P .To know the points in E(i) it is enough to find T−1(Q) for Q ∈ E. Thusif P = (a, b) with a, b ∈ Z[i] satisfies b2 = a3 + 1, then T (P ) is in E andhas integer co-ordinates. Hence to find all the solution of y2 = x3 + 1 withx, y ∈ Z[i] we shall find inverse images of the points in E with integer co-ordinates under the trace map T . Using Cremona’s table [12] we see that thatE is of rank 0 and the torsion group is of order 6. The six torsion points areR = (2, 3), 2R = (0, 1), 3R = (−1, 0), 4R = (0,−1), 5R = (2,−3), 6R =(∞,∞).

First consider 4R = (0,−1) ∈ E, assume that there is a point P =(a, b) ∈ E(i) such that T (P ) = 4R, i.e. P, P and (0, 1) are collinear. A linepassing through P, P and (0, 1) is given by y = mx + 1, with m = b−b

a−a . To

get the points P and P we solve the equations y = mx+ 1 and y2 = x3 + 1.This gives a cubic equation in x, namely, (mx + 1)2 = x3 + 1 . The threesolution of this equations are 0, a and a. Thus a, a satisfy x2−m2x−2m = 0.

Since we are interested in the case when a ∈ Z[i], we have a + a = m2

is an even integer. Further we want the point P in E(i) and not in E sothe equation x2 −m2x− 2m = 0 shall have two non real roots which are inZ[i]. Hence the discriminant m4 + 8m shall be −ve of square of an integer.One observes that this is impossible. Thus there are no points in E(i) withT (P ) = 4R. Since T is a homomorphism so there are no points on E(i)whose image under T is R, 2R (if P 7−→ R then 4P 7−→ 4R) and hence alsothere is no point on E(i) whose image is 5R = −R.Now consider the case 3R = (−1, 0). We consider the line through this point,which is given by y = m(x+1), we substitute this in the equation defining thecurve to obtain the points of the intersection. We have (m(x+ 1))2 = x3 + 1.Canceling the factorx + 1 we obtain x2 − (m2 + 1)x + (1 −m2) = 0. Againwe obtain m2 + 1 is an even integer and so m is an odd integer. Also thediscriminant (m2 + 1)2 − 4(1 − m2) is −ve of square of an integer. Thisis impossible for any integer m. Hence there are no points P on the curvemapping to 3R under T .Now we consider the last case, of point at infinity, the identity of the grouplaw. Here we are looking for points P on the curve E(i) such that P = −P .If we write P = (a+ ib, k + il) then at once we have b = 0, k = 0. But thenfrom the equation of the elliptic curve we obtain (il)2 = a3 + 1, i.e. (−a, l)is a solution to x3 − y2 = 1 in rational integers, this forces l = 0 and hence

37

P ∈ E. So there are no solution to the equation x2 − y3 = 1 in E(i) whichare not in E.For the equation x3− y2 = 1 we see that the point at infinity corresponds toone solution (−2,±3i) in E(i). There are no more solution.

3.4 Equations x5 − y2 = 1 and x2 − y5 = 1

We will work with one of the equations,viz. , x5 − y2 = 1. The other oneis analogous. Again as remarked in section 1 it boils down to proving ananalogue of Proposition 3.2.2 for p = 5. Here we prove the following;

Theorem 3.4.1. Equation x53 − 4x5

2 = 4 has no non-trivial solution exceptx3 = −(1 + i) and x2 = i and x3 = −1 + i and x2 = −i.

Proof. We note that if one of x3 and x2 is a unit then x3 = −(1+i) and x2 = ior x3 = −1 + i and x2 = −i. Now assume that none of them is a unit.So we can assume that both |x3|, |x2| are bigger than 2. Let us put x3 =a3 + ib3, x2 = a2 + ib2. Using the Lemma 3.2.4 we get |(x3 − 41/5x2)| < 1

20.

We begin with x53 − 4x5

2 = 4,i.e.

(x3 − 41/5x2)(x43 + 41/5x3

3x2 + . . .+ 44/5x42) = 4.

Let us put 41/5x2x3

= η, then |1− η| < 140

. One has

|(x3 − 41/5x2)| = 4

5|x3|4(

1− η1− η5

).

Similarly one also obtains

|b53 − 4b5

2| = |b3 − 41/5b2|5|b3|4|(1 + τ + . . .+ τ 4)|

for τ = 41/5b2b3

. Since |b3 − 41/5b2| ≤ |(x3 − 41/5x2)|, one gets, |1− τ | < 120

.

Using |b3 − 41/5b2| ≤ |(x3 − 41/5x2)|, we have

|b53 − 4b5

2| ≤4

5|x3|4

(1− η1− η5

)5|b3|4|(1 + τ + . . .+ τ 4)|. (3.3)

Note that (1 + τ + . . . + τ 4) = 1−(1−ε)5ε

, for ε = 1− τ . Now using the upperbound on ε we obtain |(1+ τ + . . .+ τ 4)| < 5+ 10

(20)+ 10

(20)2+ 5

(20)3+ 1

(20)4< 5.6.

38

Similarly we have |(

1−η1−η5

)| > 4.7.

Now if |b3| < 0.8|x3|, then from equation (3.3) we obtain

|b53 − 4b5

2| ≤ 4(0.80)4 5.6

4.7< 2.

We show that this is not possible. Since b53−4b5

2 is a non zero rational integerso we need to show that |b5

3 − 4b52| 6= 1.

Let us assume that |b53 − 4b5

2| = 1. Since x53 − 4x5

2 = 4, we obtain

(a3 + ib3)5 − 4(a2 + ib2)5 = 4.

Comparing the imaginary parts and taking |b53 − 4b5

2| = 1 into account wehave5a = 1, for some rational integer a, which is not possible. In case |b3| ≥0.8|x3|, then |a3| < 0.6|x3|. Now, as done with the imaginary part, using thereal part of x3 − 41/5x2 we obtain |a5

3 − 4a52| < 1, which is a contradiction.

Remark 3.4.2. We note that proof of this section also works for the section1 (in the sense that it works for any exponent p > 3) but we have includedthat proof, as it involves a different method.

Now using Lemma 3.2.3, Theorem 3.2.1, Theorem 3.4.1 we see that theequation xp − y2 = 1 and x2 − yq = 1 have no non-trivial solution in Z[i]for p ≥ 5 and q ≥ 5. This with section 2 shows that the non-trivial solutionto the equation xp − y2 = 1 in Z[i] are (−2,±3i) and that to the equationx2 − yq = 1 in Z[i] are (±3, 2).

39

Chapter 4

Cassels Criterion for CatalanProblem over Z[i]

4.1 Introduction

In this chapter we continue our study of Catalan problem over the imaginaryquadratic number field Q(i). Most of the results are true for any imaginaryquadratic number field K with class number one and we will state them inthat generality. We will follow the same notations as in section 2.3. Asremarked, in chapter 2, it is enough to consider the equation

xp − yq = 1, (4.1)

when p, q are rational primes. For K = Q(i), the cases p = 2 and q = 2have been dealt with in chapter 3. Thus it remains to find all the Catalantuples (x, y, p, q) with x, y ∈ Z[i]. It seems that there are no Catalan tupleswith x, y ∈ Z[i] and p, q odd primes. In this chapter we make some progresstowards this on the line of Cassels [7, 8].

It is worthwhile to remark that it was proved in 1986 by R. Tijdeman, K.Gyory and B. Brindza [5] that for a fixed number field K the equation (4.1)has only finitely many solutions.

In this chapter, we define Cassels’ criterion and prove it in a few cases.By the Cassels’ criterion for K we mean the following;

Criterion 4.1.1. For any Catalan tuple (x, y, p, q) for K, we have

gcd(q, x) 6= 1 and gcd(p, y) 6= 1.

As a consequence we have that q divides x and p divides y, for anyCatalan tuple (x, y, p, q). Here q is some prime above q and p is some prime

40

above p. When q does not divide x and p does not divide y, we will refer it asweak Cassels criterion. The case q divides x and p divides y will be referredas strong Cassels criterion.

4.2 Some preliminary results

Lemma 4.2.1. Let K be an imaginary quadratic number field and ε be a unitin OK. Let p > 3 be a rational prime and p be a prime ideal in OK above p.If ε ≡ 1(mod p), then ε = 1.

Proof. We note that the only units in K are roots of unity. Since K isquadratic, the only possible roots of unity in K are ±1, ±i, ±ω,±ω2, whereω is a cube root of unity. Now Norm(ε−1) = 2−2Re ε. From this it followsthat Norm(ε− 1) ≤ 4, hence if ε− 1 6= 0 then only possible primes dividingNorm(ε− 1) are 2 and 3. Hence ε ≡ 1(mod p) can not hold for p > 3 unlessε = 1.

Lemma 4.2.2. Let K be an imaginary quadratic number field with classnumber one. Then the equation xp − yq = 1 with p and q odd primes and xand y in OK has no non trivial solution for p = q.

Proof. On the contrary, let us assume p = q, then we have xp−yp = 1. Since(x− y) divides the left side, we see that ε = x− y is a unit in OK .In case p = 3, the only exception to Lemma 4.2.1 is K = Q(ω) and ε =ω or ε = ω2. One easily rules out this possibilty, so we can assume p ≥ 5.Case-1. p is not inert in K. In this case the residue field OK/p has orderp, where p is a prime ideal in OK lying above p. Hence (x − y)p ≡ x − y(mod p).

Since for each 1 ≤ j ≤ p− 1 the binomial coefficient(pj

)is divisible by p

we have (x− y)p ≡ xp − yp (mod p). Thus we get ε ≡ 1 (mod p). Then byLemma 4.2.1 we have x = 1 + y. Substituting this in the Catalan equation,we get (1 + y)p− yp = 1. This leads to (n+ 1)p/2−np/2 ≤ 1 for some positiveinteger n, as ||1 + y|2 − |y|2| ≥ 1. But this is not possible.Case- 2. p is inert in K. In this case one has σ(x − y) ≡ 1 (mod p) whereσ is an element of Gal(K/Q), which reduces to the Frobenius at p Thisgives x − y ≡ 1(mod p), as σ fixes p and is of order 2. From this we getε ≡ 1(mod p). A contradiction is obtained as above.

Lemma 4.2.3. Let K be an imaginary quadratic number field with classnumber one then 3 does not split in K as product of two distinct primesexcept for K = Q(

√−11).

41

Proof. If 3 factors in two distinct primes then 12 = (a2 − db2) with d ∈{−3,−4,−7,−8,−11,−19,−43,−67,−163} and a, b ∈ Z. This implies d =−3 or d = −11. In case d = −3, 3 is ramified. In case d = −11, 3 splits asproduct of two distinct primes.

Lemma 4.2.4. Let K be a number field and p a rational prime then for anyx ∈ OK one has gcd(< xp ± 1/x± 1 >,< x± 1 >) divides p.

Proof. We will give the proof for the negative sign, and the proof for thepositive sign is similar.If the gcd(< xp−1/x− 1 >,< x−1 >) = OK then we are done, otherwise wehave x ≡ 1 (mod ℘r) where r is the exact power of ℘ in gcd(< xp−1/x− 1 >,< x−1 >) . Now we have xp−1

x−1= xp−1 +xp−2 + ...+1. Reading this equation

modulo ℘r we get ℘r|p, as needed.

Remark 4.2.5. using Chinese remainder theorem one can see that in general℘ 6= p in above lemma.

Lemma 4.2.6. Let a ∈ R+ and z ∈ C with |z| � 1 and m ∈ N. Consider(1 + z)a =

∑m−1r=0

(ar

)zr + E1(a,m). and (1 − |z|)−a =

∑m−1r=0

(−ar

)(−|z|)r +

E2(a,m). Then one has|E1(a,m)| ≤ |E2(a,m)|.

Proof. First note that every term in the expansion of (1− |z|)−a is positive.rth term in (1 + z)a is

(ar

)zr and rth term in the expansion of (1 − |z|)−a is(−a

r

)(−|z|)r and one has |

(ar

)zr| ≤

(−ar

)(−|z|)r hence the result follows.

Lemma 4.2.7. For x ∈ R+ and a ∈ R+ with |x| ≤ 1 one has |E2(a,m)| ≤a(a+1)...(a+(m−1))

m!|x|m(1− |x|)−a−m.

Proof. We will use induction on m. For m = 1 we want to prove ((1−x)−a−1) ≤ ax(1− x)−a−1

Here the left side is −∫ 1

1−xddt

(t−a)dt = a∫ 1

1−x t−a−1dt ≤ a(1 − x)−a−1x. as

wanted. Let us write E(x) ≡ E(a,m)(x) ≡ (1 − x)−a −∑m−1

r=0

(−ar

)(−x)r

then one has

42

∂E(x)

∂x= a(1− x)−a−1 −

m−1∑r=1

r

(−ar

)(−1)rxr−1

= a(1− x)−a−1 −m−2∑r=0

(r + 1)

(−ar + 1

)(−1)r+1xr

= a(1− x)−a−1 −m−2∑r=0

−a(−a− 1

r

)(−1)r+1xr

= aE(a+ 1,m− 1)(x).

Now

E(a,m)(x) =

∫ x

0

∂E(a,m)(t)

∂tdt

= a

∫ x

0

E(a+ 1,m− 1)(t)dt.

By induction hypothesis

E(a,m)(x) ≤ a(a+ 1)...(a+ (m− 1))

(m− 1)!

∫ x

0

tm−1(1− t)−a−mdt

≤ a(a+ 1)...(a+ (m− 1))

(m− 1)!(1− x)−a−m

∫ x

0

tm−1dt

=a(a+ 1)...(a+ (m− 1))

(m− 1)!(1− x)−a−m

xm

m

=a(a+ 1)...(a+ (m− 1))

m!(1− x)−a−mxm

as needed.

Remark 4.2.8. By Lemma 4.2.6 one has |E1(a,m)| ≤ a(a+1)...(a+(m−1))m!

|x|m(1−|x|)−a−m.

Lemma 4.2.9. Let R be a commutative ring with unity and q be a rationalprime such that R/qR has no nil-potent elements. For x, y ∈ R satisfyingxq ≡ yq (mod qR) we have xq ≡ yq (mod q2R).

43

Proof. We are given xq ≡ yq (mod qR). Since q|(qk

)for 0 < k < q, so we

obtain (x − y)q = 0 in R/qR. But R/qR has no nil-potent elements, hencex − y = 0 in R/qR. Thus x = y + qα, for some α ∈ R. Then one easilychecks that xq ≡ yq (mod q2R).

Lemma 4.2.10. (Kronecker) If α is an algebraic integer all whose conjugateshave absolute value less than or equal to 1 then α is a root of unity.

Proof. Let α be of degree n over Q. If mα(X) is the minimal polynomial ofα, then

mα(X) =n∏j=1

(X − α(j)),

where α(j) denote the conjugates of α. Since |α(j)| ≤ 1 for all j, it followsthat all the coefficients in mα(X) are bounded by 2n. Similarly coefficientsof the minimal polynomial of any of αm are bounded by n. Thus there arefinitely many polynomials whose roots are αm,m ∈ N. This forces thatαm1 = αm2 for m1 6= m2, from which the lemma follows.

Theorem 4.2.11. Let b ∈ C and |b| ≥√

2 then for primes p > q ≥ 3 one

has 0 < |((bp − 1)q + 1)1p − bq| < 1.

Proof.

|((bp − 1)q + 1)1p − bq| = |bq(((1− b−p)q + b−pq)

1p − 1)|

= |bq((1− b−p)qp (1 + A)

1p − 1)|, (4.2)

where A = b−pq(1− b−p)−q.Now we will use Lemma 4.2.6 , Lemma 4.2.7 and Remark 4.2.8 successivelywith various values of m.From Remark 4.2.8 one gets

|(1− b−p)qp − (1− q

pb−p)| ≤

( qp)( qp

+ 1)

2|b|−2p(1− |b|−p)−

qp−2.

Hence one has

(1− b−p)qp = (1− q

pb−p)|+ λ1

( qp)( qp

+ 1)

2|b|−2p(1− |b|−p)−

qp−2,

where λ1 ∈ C and |λ1| < 1.Similarly one gets

(1 + A)1p = 1 + λ2

1

p|A|(1− A)

−1p−1,

44

where λ2 ∈ C and |λ2| � 1.Now since q < p and |b| ≥

√2 from above one has

(1− b−p)qp = 1− q

pb−p + 2λ

′

1|b|−2p,

for some λ′1 ∈ C with |λ′1| < 1 (As( qp

)( qp

+1)

2≤ 1 and (1− |b|−p)

−qp−2 < 2).

Similarly one obtains

(1 + A)1/p = 1 + 2λ′

2

|b|−(p+1)q

p

(|(1− A)−1p−1| < 2 and |A|

|b|−(p+1)q < 1)

Feeding this into equation (4.2) one obtains

|((bp − 1)q + 1)1p − bq| = |bq(−q

p+ 2λ

′

1|b|−2p + 4λ′′

2 |b|−(p+1)q)|

where λ′′2 ∈ C and |λ′′2 | < 1 Hence one has

|((bp − 1)q + 1)1p − bq|

≤ |b|q(| qp|b|−p + 2|b|−2p + 4|b|−(p+1)q)

= | qp|b|−p+q + 2|b|−2p+q + 4|b|−pq

≤ |b|−p+q + 2|b|−2p+q + 4|b|−pq< 1

(As |b| ≥√

2, p > q ≥ 3)

Of course the quantity is nonzero hence the theorem.

4.3 Cassels Criteria

Lemma 4.3.1. Let K be an imaginary quadratic field with class number one.For any Catalan tuple (x, y, p, q) in K with p > q we have |y| > 2.

Proof. On the contrary, let us assume that |y| ≤ 2. Since p > q, so byequation (4.1) we obtain |x| ≤ 2. But, for any fixed number field K inconsideration, there do not exist two co-prime integers x, y ∈ OK with 1 <|x|, |y| ≤ 2. If |x| > 1 then we immediately check that |y| = 1 and y is a unit.Thus we obtain either x or y is a unit, say x. Again from equation (4.1) wehave |y|q ≤ 2. This gives |y| = 1, as q ≥ 3, that is y is a unit. But it is easyto see that there are no Catalan tuple in K with both x and y being unit.The other case is similar.

45

Theorem 4.3.2. Let K be an imaginary quadratic number field with classnumber one. Then for any solution of Catalan’s equation

xp − yq = 1

with primes p > q ≥ 3 and x, y in OK one has(1) there is a prime ideal q in OK above q such that q|x ,

(2) either |x| ≤ 4(3q)pq or p|y for some prime ideal p in OK above p.

Proof. From Lemma 4.2.4 it follows that we need to prove

gcd(<yq + 1

y + 1>,< y + 1 >) 6= 1

wehre < t > denotes the ideal in OK generated by t.On the contrary assume that

gcd(<yq + 1

y + 1>,< y + 1 >) = 1

But one has (yq+1y+1

)(y + 1) = xp

Therefore both y + 1 and yq+1y+1

are pth powers up to a unit, since K has classnumber one.Also p ≥ 5 so every unit is a pth power and hence one obtainsy + 1 = bp yq+1

y+1= up for some b, u ∈ OK

Claim: One can assume that |b| ≥√

2.Note that |b| = 1 will give |y| ≤ 2, which contradicts Lemma 4.3.1. So nowone hasxp = (bp − 1)q + 1 with b ∈ OK and |b| ≥

√2. Then, x = ζjp((b

p − 1)q + 1)1p ,

for some j. Note that ζ−jp x also satisfies the Catalan equation, hence one canreplace x by ζ−jp x.Now, as in Theorem 4.2.11, we get

0 < |ζ−jp x− bq| < 4

p|b|−p.

As done in Lemma 3.2.4, we establish −j ≡ 0 (mod p). This gives |x− bq| <1, which is not possible.This proves (1) of the theorem.Thus one has q|x, let gcd(< x >,< q >) = q then one obtains

< y + 1 >=< bp > qp−1 <yq + 1

y + 1>=< up > q < x >=< ub > q

(4.3)

46

Now we will prove (2). In case we have |x| ≤ 4(3q)pq then nothing to prove.

So let us assume that |x| > 4(3q)pq .

From Lemma 4.2.4 we either have ℘|y or

gcd(<xp − 1

x− 1>,< x− 1 >) = 1.

Assume the latter. Then one obtains

x− 1 = aqxp − 1

x− 1= vq,

with a, v ∈ OK . Note that |x| > 4(3q)p/q and |a| ≥ 2. Now consider thepower series

F (t) = (1 + t)pq =

∞∑k=0

(pq

k

)tk,

series converges absolutely for |t| < 1

Let m be chosen so that p = (m− 1)q− µ for some 0 ≤ µ < q. Since |( pq

k

)| is

a decreasing function of k we obtain∑k≥m

|(pq

k

)tk| ≤ |

( pq

m

)|∑k≥m

|t|k ≤ 1

m|t|m(1− |t|)−1.

Thus

F (t) =m−1∑k=0

(pq

k

)tk +

λ

m|t|m(1− |t|)−1,

for some complex number λ of absolute value at most 1. Putting t = a−q

and multiplying by ap+µ, we get

aµ(aq + 1)p/q = ap+µm−1∑k=0

(pq

k

)a−qk +

λ

m|a|−q(1− |a|−q)−1.

Thus

aµxp/q = ap+µm−1∑k=0

(pq

k

)a−qk +

λ

m|a|−q(1− |a|−q)−1. (4.4)

Now

aµ(xp − 1)1/q = aµxp/q + 2xp/qλaµ

qxp.

47

The λ appearing here is some complex number (not necessarily the same asone in above expression) of absolute value smaller than 1. Using equation(4.4) we obtain

aµ(xp − 1)1/q = ap+µm−1∑k=0

(pq

k

)a−qk +

3λ

ma−q.

Multiplying by D = q(m−1)+ordq((m−1)!) we obtain

Daµ(xp − 1)1/q = Dap+µm−1∑k=0

(pq

k

)a−qk +

3λD

ma−q,

say S1 = S2 + S3. Clearly S1, S2 ∈ OK hence S3 ∈ OK . But from the boundon aq = x − 1 we see that |S3| < 1, hence S3 = 0. But S1 − S2 6= 0 as it isnot divisible by q. This contradiction proves the theorem.

Remark 4.3.3. We remark that 4(3q)pq ≤ 1

2qp−1 holds except when q = 3

and p = 5 or 7. In these two exceptions also we see that |x| ≥ 12qp−1 is

enough to achieve a contradiction in above theorem. Hence we have Cassels’criterion if we establish |x| ≥ 1

2qp−1.

Now we intend to obtain lower bounds on x, whenever there is a Catalantuple (x, y, p, q). If one can prove that |x| > 4(3q)

pq then by Theorem 4.3.1

we obtain the Cassels criteria as stated in the introduction. In this regardwe have following results.

Proposition 4.3.4. For a solution of the equation equation x5 − y3 = 1 infield K = Q(

√−11) the Cassels criterion is true.

Proof. In case |x| ≤ 30 then from x5 − y3 = 1 it follows that |y + 1| ≤((30)5 + 1)

13 + 1 but we also have that < y + 1 >=< bp > qp−1. Hence

|b|2 < 4, giving us |b|2 ∈ {1, 2, 3} but |b|2 = 2 does not hold for any b ∈ OK .Also |b|2 = 3 will give that gcd(b, 3) 6= 1, which is the Cassels criterion.Hence we are left with |b|2 = 1. Thus b is a unit which has been ruled outin proof of Theorem 4.3.2. This establishes the lower bound on x, for theTheorem 4.3.2 to be applicable.

Proposition 4.3.5. Consider a Catalan tuple (x, y, p, q) in an imaginaryquadratic number field K with class number one. If q does not split in Kthen we have |x| > 1

2qp−1.

48

Proof. By expanding one can see that

yq + 1

y + 1≡ q (mod (y + 1)). (4.5)

We consider following cases:Case (a): q is inert in K.Here we have q =< q >. From equation (4.3) and (4.5) one has upq ≡q (mod qp−1) Thus one obtains up ≡ 1 (mod qp−2). Since p > q, we seegcd(p, φ(qp−2)) = 1. This gives u ≡ 1 (mod qp−2) Now we will note thatu = 1 is not possible. If u = 1 thenyq+1y+1

= q i.e. |y|q − q|y| ≤ q − 1

But the left side is an increasing function of |y| for |y| ≥ 1 , but above in-equality does not hold if |y| > 2. Since |y| > 2, by Lemma 4.3.1 so we obtainthat u 6= 1.Hence |u| ≥ 1

2qp−2 giving us |x| ≥ 1

2qp−1 from which the bound follows.

Case (b): q ramifies in K. Here one has q = π2 and equation (4.3) and (4.5)lead toupπ = π2 (mod πp−1) giving us up = π (mod πp−2).So one has π|u hence πp|up giving us π2|π (note that p > 4). This contradic-tion shows that q divides x. So in this case we will have gcd(< yq+1

y+1>,<

y + 1 >) = q and then one can handle the bound on u as in case (a).

Proposition 4.3.6. Let (x, y, p, q) be a Catalan tuple for an imaginary quadraticnumber field K with class number one other than Q(ω), ω being cube root of unity.If gcd(< y >,< p >) = 1 then we have; q2|x for some prime ideal q in OK

above q.

Proof. Since gcd(< y >,< p >) = 1, hence by Lemma 4.2.4 we see that bothxq−1x−1

and x− 1 are qth powers. Thus there is some a ∈ OK with x− 1 = aq.We obtain −1 = aq (mod q). Now an application of Lemma 4.2.9 yields−1 = aq (mod q2). But aq = x− 1, from which the proposition follows.

In this section we aimed at proving a weaker Cassels criterion (Criterion4.1.1) for imaginary quadratic number field Q(i), viz, p1|y and q1|x. We suc-ceeded in establishing this only under the assumption on lower bound on x(Theorem 4.3.2). The lower bound was obtained only in few cases, namely,when the primes p and q do not split in Q(i).Next, we intend to show that under some hypothesis on the class number ofQ(ζp), the weak Cassels criterion implies the strong Cassels criterion, namelyq divides x and p divides y. We shall prove the first assertion (Theorem4.3.9) and the second one follows similarly.

49

Here we assume the weak Cassels criterion. As a consequence, for any Cata-lan tuple (x, y, p, q) in an imaginary quadratic number field K with classnumber one, we have

x− 1 = pq−11 aq

xp − 1

x− 1= p1u

p (4.6)

y + 1 = qp−11 bp

yq + 1

y + 1= q1v

p (4.7)

Where p1 is a prime in OK above p and q1 is a prime in OK above q.We will writeN(a) for the norm of ideal a for the relative extensionK(ζp)/Q(ζp).

Let hp− denote the relative class number of extension Q(ζp)/Q. We will write

R = Z[G], R− = (1 − σ−1)R, J− = (1 − σ−1)J . The following result can befound in [23].

Theorem 4.3.7. We have [R− : J−] = h−p . Further, if l is a rational primeand l - h−p then for α, β ∈ Q(ζ), αφ ≡ βφ (mod lr) for all φ ∈ J− givesαΘ ≡ βΘ (mod lr) for all Θ ∈ R−, where r is any positive integer.

Lemma 4.3.8. Suppose α = a0 +a1ζp+ ...+ap−1ζp−1p with ai ∈ Z and atleast

one ai = 0. If n ∈ Z divides α then n|aj for each j.

Proof. This follows because any subset of 1, ζp, ..., ζp−1p with p − 1 elements

forms a Z basis for Z[ζp].

Theorem 4.3.9. For a solution (x,y,p,q) of equation (4.1), with q - h−p onehas q|x.

Proof. We will put s = 1, if p splits in K and s = 2 otherwise. From equation(4.6) it follows that N(x−ζ)

(1−ζ)s ∈ Z[ζ] and < N(x−ζ)(1−ζ)s >= aq, for some ideal a in

z[ζ].For φ = (1− j)θ ∈ Z[G], where θ ∈ J , one has aφ =< β >.Thus N(x− ζ)φ = (1− ζ)sφηφβq for some unit η. But then (1− ζ)φ and ηφ

being roots of unity are qth powers. Hence we obtain,

N(x− ζ)φ = αq for some α ∈ Z[ζp]∗.

Also N(x− ζ) ≡ −ζ(t− ζ) (mod q) where t = x+ x.Since ζ is a qth power and t ∈ Z so tq ≡ t (mod q) , which results in −ζ(t−ζ) ≡ [−ζ

1q (t− ζ

1q )]q (mod q).

Thus we have

N(x− ζ)φ ≡ [−ζ(t− ζ)]φ (mod q),

50

and both N(x− ζ)φ and [−ζ(t− ζ)]φ are qth powers so Lemma 4.2.9 can beapplied to obtain

N(x− ζ)φ ≡ [−ζ(t− ζ)]φ (mod q2).

Now from Theorem 4.3.7 it follows thatN(x− ζ)1−j ≡ [−ζ(t− ζ)]1−j (mod q2)which gives q2|xx− xx(ζ + ζ). Now using Lemma 4.3.8 it follows that q2|xxfrom which we get q2

1|x or q|x.

As above, we have −ζ(t− ζ)φ ≡ [−ζ1q (t− ζ

1q )]qφ (mod q2)

Now using Theorem 4.3.7 we will obtain

−ζ(t− ζ)1−j ≡ [−ζ1q (t− ζ

1q )]q(1−j) (mod q2).

If we write F (z) = −z(t− z) then above expression simply says that

F (ζ)F (ζ1q )q = F (ζ)F (ζ

1q )q (mod q2).

Thus if we write above expression as a0 + a1ζ + ...+ ap−1ζp−1 = 0 (mod q2),

we have q2|a0, hence by Lemma 4.3.8 we will get q2|a1. Herea1 =

∑ms=0(−1)sp+1

(qsp

)tq−sp +

∑m−1s=0 (−1)sp+q0

(q

sp+q0

)tq−(q0+sp) +∑m

s=1(−1)sp−q0(

qsp−q0

)tq−(sp−q0) +

∑ms=1(−1)sp+1−2q0

(q

sp−2q0

)tq−(sp−2q0),

where m = [q/p] + 1 and − q ≡ q0 (mod q). Now all the term in a1 exceptfirst one, which is −tq, are divisible by q. Hence we get, q|t. This gives q|xand the theorem is proved.

51

Chapter 5

Catalan Problem over Z[i]

5.1 Introduction

Throughout this chapter we will have K = Q(i). All abelian groups willbe written multiplicatively. We will be assuming that neither p nor q splitsin K. In this chapter we intend to define the obstruction group and showthat it is part of a short exact sequence, the other objects of which are wellknown. Let (x, y, p, q) be a Catalan tuple for K. It was observed in chapter4 that p 6= q. Since q is inert in K, so by Proposition 4.3.5 and Theorem4.3.2 Cassels criteria holds and we have

x− 1 = pq−1aqxp − 1

x− 1= puq (5.1)

y + 1 = qp−1bpyq + 1

y + 1= qvp. (5.2)

5.2 The Obstruction Group

We will let G = Gal(Q(ζ4p)/Q) and σ−1 will stand for the complex conjuga-tion. For any positive integer n, µn will stand for group of nth roots of unityand ζn will denote a primitive nth root of unity. By our assumption the idealin Q(ζ4p) above p is principal and is generated by 1− ζp.

Using equation (5.1) we have∏ζp∈µp

< (x− ζp) >=< p >< a >q .

Note that 1− ζp|x− ζp. Also

<x− ζp1− ζp

> and <x− ζ ′p1− ζp

>

52

are coprime when ζp 6= ζ′p. So one obtains;

<x− ζp1− ζp

>= bq, for some integral ideal in Q(ζ4p). (5.3)

Now let us define E to be the multiplicative subgroup of (Q(ζ4p))∗ gener-

ated by (Z[ζ4p])∗ and 1− ζp i.e.

E =< (Z[ζ4p])∗, 1− ζp > .

We also consider the subgroup

Clq[4p] = {b : bq is a principal ideal }

of class group, where b is the ideal class of b in the ideal class group ofQ(ζ4p).The obstruction group is

H = {α ∈ Q(ζ4p)∗ : ordτ (< α >) ≡ o (mod q) if τ 6= ℘}/(Q(ζ4p)

∗)q,

where for prime ideal τ of Z[ζ4p] by ordτ (< α >) we mean the maximumpower of τ dividing < α >.Note that all three groups have Z[G] module structure. Before mention-ing any interconnection among them we shall recall following theorem fromappendix of [25];

Theorem 5.2.1. If Γ is a principal ideal domain and M is a free moduleof finite rank then for any submodule N of M there is a basis v1, . . . , vm ofM and elements d1, . . . , dn of Γ such that d1v1, . . . , dnvn is a basis of N anddi|di+1.

Now we have following;

Proposition 5.2.2. We have an exact sequence

0 −→ E/Eq −→ H −→ CLq[4p] −→ 0,

where the first map is induced from the inclusion of E ⊂ Q(ζ4p)∗ and the

second map is defined as follows;For any α ∈ H we have < α >= bq(1− ζp)r for some integer r and ideal bcoprime to 1− ζp, then we send α 7→ b.

Proof. (a) injectivity of the first map;Consider ε(1 − ζ)r ∈ Q(ζ4p)

∗q, for some ε ∈ Z[ζ4p]∗. We want to show that

ε(1− ζ)r ∈ Eq.

53

First let us assume that r = 0. In this case, we have ε = ξq, for someξ ∈ Q(ζ4p)

∗. But since ε ∈ Z[ζ4p]∗, we get ξ ∈ Z[ζ4p]

∗.Now we consider the general case. Again ε(1−ζp)r = ξq, for some ξ ∈ Q(ζ4p)

∗.First taking norm for the extension K(ζp)/Q(ζp) and then comparing the1− ζp-adic valuation gives us q|r. Thus, eventually, we are left to prove thatε ∈ Z[ζ4p]

∗q and this is already done.(b) surjectivity of the last map;For any b ∈ CL4p[q] we have bq =< α > for some α ∈ Q(ζ4p)

∗. Clearlyα ∈ H. We now write < α >= bq1(1 − ζp)qr for some integer r and ideal b1

coprime to 1 − ζp. Since b = b1 < 1 − ζp >, we get b = b1 and α 7→ b1.This proves the surjectivity of the last map.(c) Exactness at the centre;One is quick to see that Im ⊂ ker. On the other hand if α 7→ 1 then wehave α = bq(1 − ζp)r where b is a principal ideal. Let b =< β > for some

β ∈ Q(ζ4p)∗. Then α = ηβq(1−ζp)r, for some unit η. But then α = ˜η(1− ζp)r

and the right side is in the image. This proves that ker = Im.

Lemma 5.2.3. Let σ−1 ∈ G denote the element which maps ζ4p 7→ ζ−14p .

Then E/Eq is σ−1 invariant.

Proof. We want to prove that for any element ε(1 − ζp)r ∈ E the element(ε(1 − ζ)r)1−σ−1 ∈ Eq. Since ε ∈ Z[ζ4p]

∗ so (ε)1−σ−1 and all its conjugatehave absolute value 1 and hence by the Lemma 4.2.10 it is a root of unity inZ[ζ4p]. Note that all the roots of unity in Q(ζ4p) are qth powers. Also we seethat (1− ζ)1−σ−1 = −ζp is a qth power. This proves the lemma.

For any abelian group M on which Z[G] acts we will write M+ = M1+σ−1

and M− = M1−σ−1 , where M θ = {θ(m) : m ∈M} for any θ ∈ Z[G].

Lemma 5.2.4. For an abelian group M of odd order, one has M = M+M−,where the product on right is product of subgroups.

Proof. For any m ∈ M , we have mσ−1(m) ∈ M+ and m/σ−1(m) ∈ M−

so m2 ∈ M+M− hence M2 ⊂ M+M−. Since M is of odd order, the mapm 7→ m2 is an automorphism of M . One has M ⊂ M+M−. The other wayinclusion is obvious as M is G set.

Lemma 5.2.5. For an exact sequence

0 −→M′ −→M −→M

′′ −→ 0

of abelian groups of odd order on which Z[G] acts, and the maps involved inthe sequence commutes with the action of Z[G], the sequence

0 −→ (M′)− −→M− −→ (M

′′)− −→ 0

54

is exact.

Proof. Since the maps involved in the exact sequence commute with theaction of Z[G] so only thing to prove is that ker of the map M− −→ (M

′′)−

is contained in the image of the map (M′)− −→ M−. For this we note that

if (1 − σ−1)(m) ∈ ker then there is m′ ∈ M ′

whose image is (1 − σ−1)(m).So (1 − σ−1)(m

′) 7→ (1 − σ−1)2(m), but (1 − σ−1)2(m) = (1 − σ−1)(m2) (as

(1 − σ−1)2 = 2(1 − σ−1)). Now we use the trick used in Lemma 5.2.4, toconclude the proof.

Proposition 5.2.6. We have H− ∼= CLq[4p]−.

Proof. To prove this first we note that all the three abelian groups in theexact sequence;

0 −→ E/Eq −→ H −→ CLq[4p] −→ 0

are of odd order (the roots of unity in Z[ζ4p]∗ are qth powers and (1 − ζ)r

is not a qth power for any integer 0 < r < q. Hence using Dirichlet Unittheorem one gets |E/Eq| = qp+1/2. Also Clq[4p] is an abelian group withexponent q and hence its order is a power of q. Now we apply 1 − σ−1 toabove sequence to obtain following exact sequence;

0 −→ (E/Eq)− −→ H− −→ CLq[4p]− −→ 0.

Now Lemma 5.2.3 at once proves the proposition.

From equation (5.3) it is clear that (x−ζp) ∈ H hence (x−ζp)1−σ−1 ∈ H−.Thus for every Catalan tuple (x, y, p, q) of K we have an element in CLq[4p]

−.We are unable to prove the non triviality of (x−ζp)1−σ−1 . If one can establishthe non triviality of (x − ζp)

1−σ−1 , then the Inkeri type results [22] followimmediately.

55

Chapter 6

Inverse problems in AdditiveNumber Theory

6.1 Introduction

For finite set X, by |X| we will denote the number of elements in X. Fortwo subsets A and B of Z one immediately sees that |A+B| ≥ |A|+ |B|− 1,where A + B = {a + b : a ∈ A, b ∈ B}. In what follows, G is an abeliangroup, written additively. Let X be a subset of an abelian group G, the‘stabilizer’ of X is defined to be the subgroup {g ∈ G : g + x ∈ X, ∀x ∈ X}.Cauchy-Davenport theorem is the following assertion [36];For subsets A and B of Z/pZ, where p is a prime number, one has |A+B| ≥min{p, |A|+ |B| − 1}.

Chowla proved a ‘similar’ theorem for any cyclic group [11]. Kneser tooka big leap in this direction and proved following beautiful theorem [24].

Theorem 6.1.1. For finite subsets A and B of an abelian group G with|A + B| < |A|+ |B|, one has |A + B| = |A + H|+ |B + H| − |H|, where His the stabilizer of A+B.

Thus for any finite set A in an abelian group G one ‘almost’ always has|A+ A| ≥ 2|A| − 1.In the ‘Inverse problems in additive number theory’ one studies the conversequestion, i.e. if |A + A| is not too big in comparison with |A| then can we‘describe’ A? To be more precise, we define doubling constant of A to beconstant c satisfying |A + A| = c|A| [44]. Then one wants to ‘describe’ theset A when c is ‘small’. We mention few well known results along this line.The following two results are for the additive group of integers.

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Theorem 6.1.2. If A is a finite subset of Z with |A+A| ≤ 2|A| − 1 then Ais an arithmetic progression.

Theorem 6.1.3 (Freiman). Let A be a subset of integers such that |A| =k > 2. If |A+A| = 2k − 1 + b ≤ 3k − 4, then A is a subset of an arithmeticprogression of length k + b ≤ 2k − 3.

The theorem of Kneser is also a result in this direction for Z/nZ withc < 2 [36]. Freiman [13] improved the result by allowing c < 2.4, when Gis of prime order. Deshouillers and Freiman [16] extended Freiman’s resultfor any finite cyclic group, albeit with a smaller c. Their proof ‘essentially’proceeds along the following line;They prove a similar result with some doubling constant c′ when A is a subsetof Z×Z/dZ and then they use idea of rectification (what they call ‘a partiallift’) to obtain the result for Z/nZ with the doubling constant c. The proofis quite involved.Here we give an alternate proof of their result and on our way we make aslight improvement. We stress that the overall structure of our proof is quitesimilar to that of Deshouillers and Freiman. In the next section we stateour theorems and will mention some well known results needed for our proof.Also we will describe the line of proof quite elaborately, making the differenceof the two proofs more explicit. In section 3 and section 4 we will presentthe proof.

6.2 Statements of Theorems

The main result proved in this chapter is the following:

Theorem 6.2.1. Let n be a positive integer and A a non-empty subset ofZ/nZ with |A| < c′n (with c′ a ‘small’ absolute constant) which satisfies:

|A+A| ≤ c|A|.

When c = 2.11, there is a proper subgroup H of Z/nZ such that one of thefollowing three cases hold;(1) if the number of the cosets met by A, let us call it s, is different from 1and 3, then A is included in an arithmetic progression of l cosets modulo Hsuch that

(l − 1)|H| < |A+A| − |A|;

(2) if A exactly meets 3 cosets, i.e. s = 3, then above holds with l being

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replaced by min(l, 4);(3) if A is included in a single coset then we have

|A| > c′|H|.

Furthermore, when l ≥ 2 then there is a coset of H which contains more than23|H| elements from A.

Deshouillers and Freiman [16] proved this theorem for c = 2.04. LikeDeshouillers and Freiman we will deduce this theorem from following result,which is an analog of above result for Z× Z/dZ;

Theorem 6.2.2. For positive integers s ≥ 6 and d, consider integers a1 =0, a2, . . . , as with gcd of nonzero elements being 1, and let B1, . . . ,Bs be sub-sets of Z/dZ with 0 ∈ B1. We let Bi = ai × Bi and B = ∪si=1Bi. Under thecondition

|B + B| < 2.5|B|we have max ai < (1.5)s and there exists a subgroup H of Z/dZ and elementx ∈ Z/dZ such that Bi is included in the coset aix + H for each i and forsome j we have |Bj| ≥ 2

3|H|.

Moreover, we also have

(max ai)|H| < |B + B| − |B|.

In Deshouillers and Freiman [16], a corresponding theorem is proved fors ≥ 5. The cases s ≤ 4 are dealt separately. Our proof also works for smallervalues of s as well but under the condition |B + B| < c1|B|, where c1 is aconstant smaller than 2.5 and depends on s. In particular, Theorem 6.2.2 istrue with c1 = 2.4 for s = 5 and c1 = 2.25 for s = 4. We do not elaborateany more on the cases s ≤ 5 and refer the reader to [16]. In [16] existence ofsubgroup H is established under the hypothesis |B + B| < 5s−2

2s+1|B|. For the

assertion (max ai)|H| < |B + B| − |B|, it is assumed that |B + B| < 2.04|B|.Remark 6.2.3. In their proof, Deshouillers and Freiman use the orderinga1 < . . . < as critically, which one can always assume, to obtain a lowerbound on |B + B|. But we do not restrict our self to this particular orderingand use Hall’s marriage problem to get a better lower bound on |B + B|.

6.3 Preliminaries

In this section we mention some known results which will be needed for theproof and develop some preliminary results. Hall’s marriage problem statesthe following;

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Theorem 6.3.1. Given subsets G1, . . . , Gt contained in some set G, if forevery subset I of {1, . . . , t} we have | ∪i∈I Gi| ≥ |I| then there are elementsxi ∈ Gi such that xi 6= xj, whenever i 6= j.

In order to verify that the conditions of Hall’s marriage problem aresatisfied in our case, we need the following proposition which can be foundin [43]

Proposition 6.3.2. Let U and V be to non empty set of integers such that

V = {v1 < . . . < vt} ⊂ U = {0 = u1 < . . . < us}

and gcd(u2, . . . , us) = 1. Then we have|U + V| ≥ min(us + t, s+ 2t− 3);moreover, if U 6= V and us = s+ t− 2, then |U + V| ≥ us + t.

We will also need following result, which is a consequence of Kneser’sTheorem;

Proposition 6.3.3. For two subsets A and B of a finite abelian group Gwith |A| ≥ |B| and |A + B| < 3

2|A|, |B| > 3

4|A| there is a subgroup H of G

with |H| < 32|A| such that A+B lies in a single coset of H.

Proof. We have subgroup H of G, from Kneser’s theorem, satisfying;

|A+B| = |A+H|+ |B +H| − |H|. (6.1)

One has |A + H| ≥ |A| > 23|A + B| and |B + H| ≥ |B| > 3

4|A| > 1

2|A + B|.

Hence by equation (2) |H| > 16|A+B|.

Now H is stabilizer of A + B and so A + B is union of cosets of H, say µcosets. Then one has µ < 6.Let us assume that A intersects a many cosets of H, then a|H| ≥ |A| >23|A + B| ≥ 2µ

3|H|. This gives a > 2µ

3. Similarly if b is the number of cosets

of H which meets B, then one has b > µ2. Since |A + H| = a|H| and so on,

we also have µ|H| ≥ a|H| + b|H| − |H|. This yields a contradiction unlessµ = 1, which proves the proposition.

The following proposition is also a consequence of Kneser’s Theorem andthe proof run along the same line as of proposition 6.3.3.

Proposition 6.3.4. Consider two finite subsets A and B of an abelian groupG such that, |A + B| < 2|B| and |B| < 3/4|A|. Then A + B lies in a singlecoset modulo some subgroup H with |H| < 2|B|.

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Theorem 6.2.1 is deduced from Theorem 6.2.2 and this is achieved us-ing a ‘partial lift’. Deshouiller and Freiman are able to ‘rectify’ more than85 percentage of elements of A and then use Theorem 6.2.2 for this part.To complete the proof they use an argument to conclude that whatever isachieved is already enough. In our case we show that once more than 85percentage of elements of A are rectified and Theorem 6.2.2 can be appealedthen using the structure provided from Theorem 6.2.2 for this part we canconclude that one can rectify all the elements of A.We will also need the concept of Freiman isomorphism, in order to link The-orem 6.2.1 to Theorem 6.2.2. See [36],Definition: Let G1 and G2 be abelian groups written additively. For S ⊂G1, T ⊂ G2 a map f : S 7→ T is called 2 − isomorphism if f is a bijectionand s1+s2 = s3 +s4 holds in S if and only if and f(s1)+f(s2) = f(s3)+f(s4)holds in T .The following lemma will be of help in the sequel.

Lemma 6.3.5. Let G be a subset of [0, s− 1] with |G| ≥ 2s/3 + 1, then forany d ≤ s/3 there are elements g1, g2 ∈ G such that d = g1 − g2.

Proof. Suppose there are no solution of d = g1 − g2. Then observe that, fora ∈ G, a+ d is not in G. Thus every interval of length less than or equal to2d can have at most d elements in G. Hence by breaking the interval [1, s]into sub intervals of length 2d and one sub interval of length less than orequal to 2d, we see that |G| ≤ 2s

3; the proposition follows.

For any subset G of [0, s−1], we will define G(1) to be set of those elementsd of [0, s− 1] which satisfy a relation of the form d = b+ c− a for a, b, c ∈ G,not necessarily distinct. G(i+1) is defined from G(i) inductively. We will defineGgood = ∪i≥0G

(i), with G(0) = G. In case we have a subset A ⊂ [0, s − 1] athand and G ⊂ A, then Ggood shall be obtained by intersecting G(i) with A ateach step. The phrase ‘A misses an element a’ will be used in the sense thata is not in A. We have following useful proposition.

Proposition 6.3.6. For an integer s > 3, consider A ⊂ [0, s − 1] with|A| ≥ 2s/3+1 then we can choose a set G of two elements ai, aj from A suchthat Ggood = A and aj − ai = 1.

Proof. We use induction on s. Let t = [s/2], be the integral part of s/2.Let A′ = A ∩ [0, t − 1] or A′ = A ∩ [t, s − 1] depending upon if A has moreelements in [0, t−1] or in the other half. Let us consider the first case, Now byinduction hypothesis we can choose a G′ such that G′good∩A′ = A′ (in secondcase we translate by t). Now we see that the distance between maximum from

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A′ and minimum of A∩ [t, s−1] is not more than s/3, hence by Lemma 6.3.5,this difference is also achieved as difference of two elements from G′good. Thisshows that minimum of A ∩ [t, s − 1] is in G′good. Similar reasoning showsthat G′good ∩ A = A. Take G=G’, this proves the proposition.

The following theorem is an important result in itself and is essential forthe proof of Theorem 6.2.2. Here we will use some notations and boundsfrom the next section (Lemma 6.4.1).

Theorem 6.3.7. Let s ≥ 6 and A = {0 = a1, . . . , as} ⊂ [0, N − 1], withgvd(a2, . . . , as) = 1. Also consider integers x1, . . . , xs satisfying

aj − ai = ak − al = t =⇒ xj − xi = xk − xl,

whenever t ≤ s2. If |A+A| < 5

2s, then there exist x, y such that xi = aix+ y

for each i.

Our strategy to prove this theorem is to produce two elements ai, aj ∈ Awith aj − ai = 1 such that for G = {ai, aj} we have Ggood = A. Once wehave two elements ai, aj like this, then we can solve for x, y satisfying

xi = aix+ y and xj = ajx+ y.

By the definition of Ggood it is clear that for every xk ∈ Ggood we havexk = akx+ y.Let R be as defined in the beginning of section 6.4, then we can assumeN = s+R−2. By Lemma 6.4.1 we have R < s

2+3. Also if s ≥ 2

3(s+R−2)+1,

i.e. 2R ≤ s + 1, then by Proposition 6.3.6 we succeed. So we assumes+ 2 ≤ 2R ≤ s+ 5.Let a be the largest integer such that A misses a elements from the interval[0, 2a−1] (in case there are no a satisfying this then we take a = 0) and let b bethe largest integer such that A misses b elements from [s+R−2b−2, s+R−3].Finally let c be the number of elements A misses from [2a, s + R − 2b − 3].We have the following;Claim: |A+ A| ≥ 2s+R− 4 + c.To prove the claim, we make the following observation;for any n ≤ s + R − 3, if number of elements A misses from [0, n] is strictlyless than n+1

2, then n ∈ A+ A.

Using this we conclude that every element of the interval [2a, s+R− 3] is inA + A. Considering {s + R − 3− ai : ai ∈ A}, we see that every element ofthe interval [s + R − 3 + 2b, 2(s + R − 3)] appears in A + A. Also there area elements from [0, 2a − 1] and b elements from [s + R − 3, s + R − 3 + 2b]appearing in A+ A. But from the b elements of [s+R− 3, s+R− 3 + 2b],

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the element s+R− 3 is already considered. Hence |A+A| ≥ [(s+R− 3)−2a+ 1] + [2(s+ R− 3)− (s+ R− 3 + 2b) + 1] + a+ b = 2s+ R− 3 + c, asa+ b+ c = R− 2. The claim is established.Since |A+A| < 5

2s, we obtain c ≤ 1. We discuss the case c = 1, the case c = 0

being similar. A misses one element from the interval I = [2a, s+R−2b−3],hence for the set A′ = A∩I Proposition 6.3.6 is applicable provided 2s−12 ≥2R. Thus, for s ≥ 17 we can chose two elements ai, aj ∈ A′ with aj − ai = 1and for G′ = {ai, aj} we have G′good = A′. Even though A′ misses one elementfrom I but the set A′+A′−A′ coincides with the set I + I − I. The latter is[4a− s−R+ 2b+ 3, 2s+R− 2a− 4b− 6]. Now we consider two cases a ≤ band a > b.case 1- a ≤ b.Let k be an integer such that G′good = G′(k), then from above it is clear thatfor G = G′ ∈ A, we have G(k+1) = a ∩ [0, s + R − 2b − 3]. Also since anyelement of A∩ [s+R−2b−3, s+R−3] is at most at a distance of R−2−a−1from s + R − 2b − 3, hence using arguments as in Lemma 6.3.5 show thatG(k+2) = A.case 2- a > bHere the proof is similar, so we do not give the details.This finishes proof of the Theorem 6.3.7 for s ≥ 17. For smaller values of sit is possible to check case by case and we omit the proof.

6.4 Proof of Theorem 6.2.2

For A′ = {a1, . . . , as} ⊂ Z with 0 ∈ A′ and gcd of nonzero elements being 1,we shall define R = min{max ai − s + 3, s}. Clearly 2 ≤ R ≤ s. We havethe following;

Lemma 6.4.1. |A′ + A′| ≥ 2s+R− 3.

Proof. We will consider setsG1,1 = . . . = G1,s−1 = a1 + A′,G2,1 = G2,2 = a2 + A′, G3,1 = G3,2 = a3 + A′, . . . , GR,1 = GR,2 = aR + A′,GR+1 = aR+1 + A′, . . . Gs = as + A′.

It is easy to verify that the conditions of Hall’s marriage problem aresatisfied for the familyGij, which yields |A′+A′| ≥ (s−1)+2(R−1)+(s−R) =2s+R−3. The above proof also shows that there are distinct s−1 elementsin A′+A′ with a1 as a summand, 2 elements with ai as a summand, for each2 ≤ i ≤ R and one elements with ai, for i > R, as a summand.

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Now we will assume the setup as in Theorem 6.2.2. Just considering thefirst co-ordinate, for us the above lemma gives |π1(B + B)| ≥ 2s + R − 3,where π1(B + B) denotes the first co-ordinate of B + B. By considering thesecond coordinates, we give a lower bound on |B + B|.

Proposition 6.4.2. We have the following lower bound,

|B + B| ≥s−1∑j=1

|B1 + B1,j|+s+1∑j=s

|B2 + B2,j|+ . . .+s+2R−3∑j=s+2R−4

|BR + BR,j|

+|BR+1 + BR+1,j|+ . . .+ |Bs + Bs,j| (6.2)

where Bi,j ∈ {B1, . . . ,Bs} and for a fixed i the B′i,js are distinct.

Proof. In the proof of Lemma 6.4.1 we have produced 2s + R − 3 elementsin Π1(B+ B). There are s− 1 elements of the form a1 + aj, 2 elements of theform ai + aj for i = 2, . . . , R, and 1 elemet of the form ai + aj for i > R.

This gives us;

{(a1 + ai,B1 + Bi) for some s− 1 values of i, 1 ≤ i ≤ s}∪{(a2 + ai,B2 + Bi) for some 2 values of i, 1 ≤ i ≤ s}∪

B + B ⊃ . . . ∪ {(aR + ai,BR + Bi) for some 2 values of i, 1 ≤ i ≤ s}∪{(aR+1 + ai,BR+1 + Bi) for some 1 ≤ i ≤ s} ∪ . . .∪{(as + ai,Bs + Bi) for some 1 ≤ i ≤ s}

This way of listing elements of B + B plays a critical role in the proof. As aconsequence the proposition follows.

Mostly we will be assuming |B1| ≥ . . . ≥ |Bs| so that the lower boundobtained in the Proposition 6.4.2 is the best by this method, but at times wewill be assuming different ordering too.Since |Bi + Bj| ≥ max{|Bi|, |Bj|}, weobtain the following,

Corollary 6.4.3. |B + B| − |B| ≥ (s− 2)|B1|+ |B2|+ . . .+ |BR|.

Even though Proposition 6.4.2 holds for all values of R, for R = 2, 3we will need different consideration at times. For R = 3 we see that A′ =

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{0, 1, . . . , s} with one element i0 6= 0 omitted . Here we claim that;

{(a1 + ai,B1 + Bi) for some s values of i, 1 ≤ i ≤ s} ∪{(a2 + ai,B2 + Bi) for some 2 values of i, 1 ≤ i ≤ s} ∪

B + B ⊃ ∪{(a3 + ai,B3 + Bi) for some values of i, 1 ≤ i ≤ s} ∪{(a4 + ai,B4 + Bi) for some 1 ≤ i ≤ s} ∪ . . . ∪{(as + ai,Bs + Bi) for some 1 ≤ i ≤ s}.

This is achieved by considering the family of setsG1,1 = . . . = G1,s = a1 + A′,G2,1 = G2,2 = a2 + A′, G3 = a3 + A′, . . . , Gs = as + A′,and then applying the Hall’s marriage theorem. This immediately yields,

|B+B| ≥s∑j=1

|B1 +B1,j|+s+2∑j=s+1

|B2 +B2,j|+|B3 +B3,j|+. . .+|Bs+Bs,j|. (6.3)

For R = 2 a similar consideration yields,

|B + B| ≥s∑j=1

|B1 + B1,j|+ |B2 + B2,j|+ . . .+ |Bs + Bs,j|. (6.4)

Proposition 6.4.4. Under the assumption of Theorem 6.2.2 one has max ai <1.5s.

Proof. Using the trivial lower bound on |Bi+Bj| in equation (6.2) we obtain,

|B + B| ≥ (s− 1)|B1|+ 2|B2|+ . . .+ 2|BR|+ |BR+1|+ . . .+ |Bs|.

Since |A| =∑

i |Bi|, from above we obtain

|B + B| − |B| ≥ ((s− 2)|B1|+ |B2|+ . . .+ |BR|) .s+R− 3

s+R− 3.

Now(s− 2)|B1|+ |B2|+ . . .+ |BR|

s+R− 3,

being average of |B1|, s − 2 times, and |B2|, . . . , |BR|, is greater than theaverage of |B1|, . . . , |Bs|, namely ∑

i |Bi|s

(as |B1| ≥ |Bi| and s+R− 3 ≥ s). This gives

|B + B| − |B| ≥∑

i |Bi|s

(s+R− 3).

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Now because of the assumption we get s + R − 3 < 1.5s. Thus R 6= s andhence R = max ai − s+ 3, and this gives the proposition.

Now we intend to exhibit the subgroup H as sought in Theorem 6.2.2.This will be done through next few lemmas.

Lemma 6.4.5. There exists a subgroup H of Z/dZ such that B1 lies in asingle coset of H and |H| < 3/2|B1|.

Proof. If |B1 +Bi| < 3/2|B1| and |Bi| > 3/4|B1|, then from Proposition 6.3.3we get that B1 +Bi lies in a single coset of the stabilizer and consequently B1

also lies in a single coset. We will partition B′is in three different categories.U = {ai : |Bi| > 3

4|B1|},

V = {ai : 12|B1| < |Bi| ≤ 3

4|B1|} and

W = {ai : |Bi| ≤ 12|B1|}.

We let u, v and w denote the respective cardinality.If there is no subgroup as claimed in the lemma, then as seen in the Propo-sition 6.3.3 and Proposition 6.3.4, one has,|B1 + Bi| ≥ 3

2|B1|, if ai ∈ U ,

|B1 + Bi| ≥ 2|Bi|, if ai ∈ V ,|B1 + Bi| ≥ |B1| otherwise.

Then using equation (6.2) one obtains;

|B+B| ≥ 3(u− 1)

2|B1|+2

∑ai∈V

|Bi|+w|B1|+2(|B2|+. . .+|BR|)+|BR+1|+. . .+|Bs|.

This gives;

|B + B| − B ≥ 3

2(u− 1)|B1|+ 2

∑i

|Bi|+ w|B1| − |B1|+ (|B2|+ . . .+ |BR).

But we are given |B + B| − |B| < 1.5|B|. Comparing the two inequalities weget,

1.5|B| > 3

2(u− 1)|B1|+ 2

∑ai∈V

|Bi|+ w|B1| − |B1|+ (|B2|+ . . .+ |BR|).

Now we will assume that R ≥ 4. In case u + v = 1, then above inequalityalready yields a contradiction. So we assume that u+ v > 1.Let us put,|Bi|′ = |B1| if ai ∈ U , |Bi|′ = 3

4|B1| if ai ∈ V and |Bi|′ = 1

2|B1| if ai ∈ W .

Also we will write V = X ∪ Y , where X = V ∩ {a2, . . . , a4}. For ai ∈ V wehave |B1 + Bi| ≥ max{2|Bi|, |B1|} so we get |B1 + Bi| ≥ 0.5|Bi| + 0.75|B1|

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and |B1 + Bi| ≥ 32|Bi| + 1

4|B1| as well. We will use the first inequality when

ai ∈ X and the second one when ai ∈ Y . Then proceeding as above, fromequation (6.2) we obtain

1.5|B| ≥ 1.5(u− 1)|B1|+ 0.5∑ai∈X

|Bi|+ 0.75x|B1|+ 1.5∑ai∈Y

|Bi|

+0.25y|B1|+ w|B1| − |B1|+ (|B2|+ . . .+ |B4|),

with x = |X| and y = |Y |.Since coefficient of each |Bi|, i > 1 in above inequality (after shifting

everything to the left) is non negative so we can replace each |Bi| by |Bi|′.This argument leads to

0 > −2.5|B1|+ 0.25y|B1|+ 0.25w|B1|+ (|B2|′ + . . .+ |B4|′).

Note that u+x+y+w = s ≥ 5 and x ≤ 3. Using this in the above inequalitywe obtain a contradiction.

The same argument works for R = 3 by taking X = V ∩ {a2} and usingequation (6.3).We give the sketch of the proof for R = 2. Equation (6.4) leads to,

(1.5)u|B1|+ 2∑j∈V

|Bj|+ w|B1|+ |B2|+ . . .+ |Bs| ≤ |B + B|. (6.5)

Since |B + B| < 2.5|B|, the above inequality yields a contradiction wheneverw ≥ 4 or v ≥ 2. Also when v = 1 and w = 3, the above inequality leads toa contradiction. Thus we can assume v + w ≤ 3.Observe that for ai, aj ∈ U , we always have(|B1|+ |Bi|) + (|Bi|+ |Bj|) ≥ 3|Bi|. (For this we use, if Bi is in a single cosetthen |B1| + |Bi| ≥ 2|Bi| and otherwise |Bi| + |Bj| ≥ 3

2|Bi|). For any i let

Ai = {aj : j ≤ i}, then Ai + Ai ⊂ A′ + A′, and has at least 2i− 1 elements.Further, when there are 2i− 1 elements in Ai +Ai, there is some j ≤ i suchthat Ai+1 +Ai+1 = Ai +Ai ∪{ai+1 + ai+1, ai+1 + aj}. This gives us the lowerbound

(1.5)|B1|+ 3|B2|+ . . .+ 3|Bu|+ 2|Bu+1|+ . . .+ 2|Bu+3| < 2.5|B|. (6.6)

Since v +w ≤ 3, above yields a contradiction when u ≥ 7. So we assumeu < 7, v + w < 4. These cases can be worked out using the techniquesdeveloped so far, but for the sake of completeness we will sketch the proof.

Now we shall consider four cases;Case (1)- v + w = 3.If some Bi with a1 6= ai ∈ U lies in a single coset, then |B1 +Bi| ≥ 2|Bi|, and

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we obtain

3

2(u− 1)|B1|+ 2|Bi|+ 2

∑aj∈V

|Bj|+ w|B1|+ |B2|+ . . .+ |Bs| ≤ B + B.

This immediately yields a contradiction.In case all Bi with ai ∈ U lies in more than one coset (for any such subgroup),then we break A′ +A′ in U + U,U + V ∪W and V ∪W + V ∪W . We notethat in A′+A′, we can consider 3 elements from V ∪W +V ∪W , 3 elementsfrom U + V ∪W and rest from U + U . This consideration yields,

3|B1|+. . .+3|Bu−1|+3

2|Bu|+|Bu+Bu+1|+2|Bu+Bu+3|+2|Bu+2|+|Bu+3|) < 2.5|B|.

(6.7)Adding equation (6.6) and (6.7) we obtain

(1.5)u|B1| − 0.5|Bu|+ |Bu + Bu+1|+ |Bu + Bu+3|+ |Bu+2| − |Bu+1| <∑i

|Bi|,

i.e.1

2(u− 1)|B1|+ |Bu + Bu+1|+ |Bu + Bu+3| < 2|Bu+1|+ |Bu+3|,

this gives a contradiction.Case (2)- v + w = 2.If there are two or more Bi with a1 6= ai ∈ U which lie in a single coset, thenwe are through as in earlier case (these two Bi contributing 0.5|Bi| each and0.5|B1|, coming from V ∪W ).If there is only one such Bi which lies in a single coset. Let aj 6= a1, ai ∈ U ,then aj + U 6= a1 + U and hence we obtain

3

2(u− 1)|B1|+ 2|Bi|+

3

2|Bj|+ |B1|+ |B2|+ . . .+ |Bs| ≤ B + B.

(One |B1 + Bj| coming at the expense of one |B1 + Bt| with t ∈ V ∪W or|Bm + Bt| for m > 1).This yields a contradiction if 1.5|B1| + .5|Bi| > 1.5(|Bu+1| + |Bu+2|). If thisinequality does not hold then we have v = 2, w = 0 and from

3

2(u− 1)|B1|+ 2|Bi|+ 2

∑aj∈V

|Bj|+ w|B1|+ |B2|+ . . .+ |Bs| ≤ B + B

we get 0.5(|Bi| + |Bu+1| + |Bu+2|) < |B1|, this is not possible given that1.5|B1|+ 0.5|Bi| ≤ 1.5(|Bu+1|+ |Bu+2|).If there are no Bi with ai ∈ U which lies in a single coset, then by considering

67

the contribution of U + U,U + V ∪W and V ∪W + V ∪W separately indescending order we obtain

3|B1|+ . . .+ 3|Bu−1|+3

2|Bu|+

5

2(|Bu+1|+ |Bu+2|) < 2.5|B|,

which in turn gives 0.5(|B1|+ . . .+ |Bu−1|) < |Bu|, a contradiction as u ≥ 3.case (3)- v + w = 1.If two or more Bi with a1 6= ai ∈ U lie in a single coset then we are through.If there is only one Bi which lies in a single coset then we can obtain

3

2(u− 1)|B1|+ 2|Bi|+

3

2

∑aj 6=a1,au∈U

|Bj|+ |Bu|+ 2|Bu+1| ≤ B + B,

this gives a contradiction as u ≥ 4.case (4)- v + w = 0.If three or more Bi with a1 6= ai ∈ U lie in a single coset then we are through.If only two of Bi with ai 6= a1 lie in a single coset, say Bu−1,Bu. Among therest u− 2 let us consider the ordering |B1| ≥ . . . ≥ |Bu−2|. We can obtain

3(|B1|+ . . .+ |Bu−3|) +3

2|Bu−2|+ 2|Bu−1|+ |Bu−1|+ |Bu|+ |Bu| < 2.5|B|,

since u ≥ 5 we have a contradiction.The cases when one or none of Bi with a1 6= ai lie in a single coset are verysimilar to the calculation done earlier.

Next we shall show that each of the Bi lies in a single coset of H for somesubgroup H of Z/dZ with |H| < 3/2|B1|. This is content of Lemma 6.4.6.

Lemma 6.4.6. There is a subgroup H of Z/dZ with |H| < 3/2|B1| such thateach of the Bi is contained in a single coset of H.

Proof. For a clear exposition we shall give the proof when R ≥ 4. The casesR = 2, 3 can be handled with a bit more careful working. Here we shallconsider B in various different ordering. Let us write

B = {(ai, Ci) : 1 ≤ i ≤ r; |C1| ≥ |C2| ≥ . . .} ∪{(ar+j, Dj) : 1 ≤ j ≤ t; |D1| ≥ |D2| ≥ . . .},

where Ci lies in a singel coset modulo H for the subgroup H of lemma 3 andDj does not lie in a singel coset modulo H.By Lemma 6.4.5, we have C1 = B1 and one immediately has |C1 +Ci| ≥ |C1|

68

and |C1 +Dj| ≥ 2|C1|. Now using the description of B + B, we obtain;

|B + B| ≥ (r + 2t− 2)|B1|+ 2|B2|+ . . .+ 2|BR|+ |BR+1 + . . .+ |Bs|.

Next let us write (after a rearrangement),B = {(ai,Bi) : Bi = Di for 1 ≤ i ≤ t and Bt+i = Ci for 1 ≤ i ≤ r}.Proceeding in the same way for this listing of B, as we had done to obtainequation (6.2), we get

|B + B| ≥ t|D1|+ 2|C2|+ . . .+ 2|Cr|+ 2|D2|+ . . .+ 2|D4|+ |D5|+ . . .+ |Dt|+|C1|+ . . .+ |Cr|,

assuming R ≥ 4 and t ≥ 4. When t ≤ 3 then also the method works withappropriate changes. Now we assume c′ < 2.5 and add the two lower boundson |B + B| to obtain,

(r + 2t− 2)|B1|+ 2|B2|+ . . .+ 2|BR|+ |BR+1 + . . .+ |Bs|5|B| > +t|D1|+ 2|C2|+ . . .+ 2|Cr|+ 2|D2|+ . . .+ 2|D4|+ |D5|+ . . .+ |Dt|

+|C1|+ . . .+ |Cr|.

We note that

|B| =s∑i=1

|Bi| =∑i

|Ai| =r∑i=1

|Ci|+t∑i=1

|Di|.

Also |B1| = |C1|, this immediately yields,

3|B| > (r+2t−3)|B1|+|B2|+. . .+|BR|+(t−1)|D1|+2|C2|+. . .+2|Cr|+|D2|+. . .+|D4|.

Further we have s = r+ t and (s− 3)|B1|+ |B2|+ . . .+ |BR| ≥ |B|, so we get,

2|B| > t|B1|+ (t− 1)|D1|+ 2|C2|+ . . .+ 2|Cr|+ |D2|+ . . .+ |D4|.

Since (t − 2)|B1| + |D3| + |D4| ≥∑

i |Di| and (t − 1)|D1| + |D2| ≥∑

i |Di|,the above yields 2|B| > 2|B|, a contradiction.

Let xi ∈ Z/dZ be such that Bi ⊂ xi+H. Next we shall prove the existenceof x ∈ Z/dZ such that Bi ⊂ aix + H. Actually we shall exhibit x, y ∈ Z/dZsatisfying Bi ⊂ aix + y + H, but this is sufficient as translation by a fixedelement is a 2− isomorphism. We shall give the proof for R ≥ 4. The basicidea is to show that if such an x and y can not be obtained then it will resultin more terms on right side of equation (2), which will exceed the limit. Thisis made precise below.

69

First we note that, from Proposition 6.4.4, R < s/2 + 3. Also this gives usmax{ai} < 1.5s. For 1 ≤ k ≤ R− 2 we will consider,

Sk = {(ai, aj) ∈ A′ × A′ : aj − ai = k}.

Note that |Sk| ≥ s − R − k + 1. To see this we form pairs (ai, aj) with allchoices of 0 ≤ ai, aj ≤ max{ai} satisfying aj − ai = k, there are exactlys+R− 3− k many such pairs. Of these, at most 2(R− 2) of them can haveeither ai or aj not in A′. Now for (ai, aj) and (au, av) ∈ Sk we will define

(ai, aj) v (au, av) if xj − xi = xv − xu (mod H). We contend that underthis equivalence S1 has only one equivalence class. Let S1 = ttj=1S1j be thedecomposition of S1 in disjoint equivalence classes. We want to show thatt = 1. Let us assume t > 1. For j 6= j′ and (au, av) ∈ S1j, (aw, az) ∈ S1j′ wehave av+aw = au+az, where as Bu+Bz∩Bv+Bw = ∅. For the lower bound on|2B| in equation (6.2) we had considered at most one of Bj +Bv and Bi +Bu,corresponding to the first co-ordinate av + aw = au + az. This reasoningshows that, if S1 has more than one equivalence class then we can improveupon equation (6.2) to obtain a better lower bound.Let S1j0 have maximum cardinality among all equivalence classes, j0 neednot be unique and if there are more choices we fix any one. We arrangeelements of S1j0 with first co-ordinate in increasing order, and consider themas a row. Also we can arrange the elements of S1 which are not in S1j0 withfirst co-ordinate in increasing order, and consider them as a column. Forevery (au, av) in the row and every (aw, az) in the column we have an elementav+aw = au+az in A′+A′, and all such elements are distinct. Correspondingto the first co-ordinate av +aw = au+az, at most one of Bj +Bv and Bi+Buwas considered in equation (6.2). But Bj +Bv and Bi+Bu are disjoint. Thuswe get at least s−R− 1 many more summands in equation (6.2). Note thatthese summands are different because of the condition Bu+Bz∩Bv+Bw = ∅.We obtain

|B + B| ≥s−1∑j=1

|B1 + B1,j|+s+1∑j=s

|B2 + B2,j|+ . . .+s+2R−3∑j=s+2R−4

|BR + BR,j|

+|BR+1 + BR+1,j|+ . . .+ |Bs + Bs,j|+(|Bs|+ . . .+ |Bs−(s−R)|).

This lower bound leads to a contradiction, by noticing that each |Bi| can bereplaced by |B1| and there are at least 5s

2terms on the right side. This proves

that S1 shall have only one equivalence class. Similar analysis shows that Skhas only one equivalence class.

70

Thus for t ≤ R− 2, we have

aj − ai = ak − al = t =⇒ xj − xi = xk − xl.

Also it is easy to obtain |A′ + A′| < 5s2

. Now we can appeal Theorem 6.3.7to prove the existence of x and y as claimed. Next we intend to prove thelast assertion of the Theorem 6.2.2, namely;

max ai|H| < |B + B| − |B|.

Note that max ai = s+R− 3. We consider sets,U = {ai : |Ai| ≥ 2

3|H|}, V = {ai : 1

3|H| ≤ |Ai| < 2

3|H|},

W = {ai : |Ai| < 13|H|}

A first coordinate of B+ B will be referred as ‘good’ if it can be representedin the form u + v with u ∈ U, v ∈ U ∪ V . Every ‘good’ first coordinatecontributes |H| in equation (6.2). Showing that |W | ≤ 1 establishes theresult with the help of equation (6.2). We shall write u, v, w for the numberof elements in U, V,W respectively, also elements of U will be denoted byu1 < . . . < uu.

Lemma 6.4.7. Under the hypothesis of the Theorem 6.2.2 we have u ≥w + 2R− 3.

Proof. First we establish u ≥ R.On the contrary if R > u, then equation (6.2) gives us the inequality

(u+v−1)|H|+(w+u−2)|B1|+2/3(R−u)|H| < 1.5u|B1|+v|H|+ .75w|B1|.

This is proved by shifting B2, . . .BR on the right side of equation (6.2) andnoticing that the coefficient is positive so one can replace them by a biggerquantity, namely |B1|. Simplifying this yields

u|H|+ 0.25w|B1|+ 2/3(R− u)|H| < (0.5u+ 2)|B1|+ |H|,

Since R > u, u ≥ 1, w ≥ 2 we obtain a contradiction. Thus we have u ≥ R.Using this in equation (6.2) yields,

(u+ v − 1)|H|+ (w +R− 2)|B1| < 1.5u|B1|+ v|H|+ 0.5w|H|.

As 2/3|H| ≤ |B1| ≤ |H|, the above inequality shall be true for one of theextreme value of |B1|, as the inequality is linear in |B1|. Putting |B1| = 2/3|H|gives a contradiction and |B1| = |H| gives u ≥ w + 2R− 5. We need to gaina bit more. Since R ≥ 4, we have u ≥ R + 1. Let us partition W in twoparts,

71

W1 = {ai : |Bi| ≥ |H| − |BR|}, W2 = {ai : |Bi| < |H| − |BR|} with w1, w2

being their cardinality respectively. Again if w2 ≤ 1 then equation (6.2) willalready prove the assertion of the Theorem. As in this case in equation (6.2)there will be at most s first coordinates with summands from W2 which willcontribute at least |B| in equation (6.2) and rest will contribute (s+R−3)|H|.Equation (6.2) helps us in having,{

(u+ v + w1 − 1)|H|+(w2 +R− 3)|B1|+ |BR|

}<

{1.5(R− 1)|B1|+ 1.5(u−R + 1)|Br|+v|H|+ 0.5w1|H|+ 1.5w2(|H| − |BR|)

},

i.e.

[u+0.5w1−1.5w2−1]|H|+[w2−0.5R−1.5]|B1|+[1.5(w2−u+R)−0.5]|BR| < 0.

Since the last inequality does not hold for |B1| = 2/3|H| (which in turn willalso give |B1| = |BR|), so it shall be true when |B1| is replaced by |H|. Thisgives us,

(u+ 0.5w1 − 0.5w2 − 0.5R− 2.5)|H|+ (1.5w2 − 1.5u+ 1.5R− 0.5)|BR| < 0.

Again, as 2/3|H| ≤ |BR| ≤ |H|, as done earlier, we obtain u > w1 + 2w2 +2R− 6. As w2 ≥ 2 we get u ≥ w + 2R− 3, this proves the lemma.

We call an element of A′ ‘desirable’ if all the first coordinates (of B + B)it contributes to are ‘good’ and we say it is ‘almost desirable’ if all but onecoordinates it contributes to are ‘good’. Our aim is to show that there is atleast one ‘desirable’ and at least R−1 ‘almost desirable’ elements in A′. Thenrenaming desirable element as a1 and almost desirable elements as a2, . . . , aRin equation (6.2) yields the result. Towards this, we take T as complementof A′ in [0, s+R−3] and K = W ∪T . The cardinality of K is k = w+R−2.Let d1 be the number of elements of K which are smaller than uk+1 and d2

be the number of elements of K which are bigger than uk+1. We first assumethat none of di is zero and finish the proof.Claim: When none of di is zero then every element of U in the interval(ud1 , uu−d2) is ‘desirable’ and udi is ‘almost desirable’.Let uj ∈ U∩(ud1 , uu−d2) and w ∈ W then we wish to show that uj+w = u+vfor some u ∈ U, v ∈ U ∪ V .case (1)- uc+1 ≤ uj + w < uk+1.All elements uj + w − ur, 1 ≤ r ≤ c + 1 are smaller than uk+1 and are in[0, s+R− 3], hence can not be in K, proving that uj is ‘desirable’.case (2)- uk+1 ≤ uj + w ≤ s+R− 3.Here, the elements uj + w − ur, 1 ≤ r ≤ k + 1 can not all lie in K, makinguj a ‘desirable’ element.

72

case (3)- uj + w ≤ s+R− 3 + uu−k.In this case one of the elements from uj + w − ur, u − k ≤ r ≤ u makes uj‘desirable’.case (4)- uj + w > s+R− 3 + uu−k.Now one of the elements uj + w − ur, u − d2 ≤ r ≤ u assures that uj is‘desirable’.This produces R − 1 desirable elements. Similar analysis shows that theelements ud1 , ud2 are ‘almost desirable’. In case one of d1 and d2 is zero, sayd2 = 0. In this case clearly uj with uj + w ≥ uk+1 is ‘desirable’ and thereare R − 1 such uj, namely ur, for r ≥ k + 1. We claim that uk is almostdesirable. For this we notice that there can be at most one w such that{uk + w − uj : 1 ≤ j ≤ k} = K and thus uk contributes to all but one goodcoordinate, proving that uk is almost desirable. The case when d1 = 0 issimilar.

6.5 Proof of Theorem 6.2.1

We continue with the notations of the previous section. To prove the Theo-rem 6.2.1 we shall use a partial lift of A to a subset B ⊂ Z × Z/dZ, whichsatisfies conditions of the Theorem 6.2.2 and thus has the structures providedfrom the theorem. Using the structure of B we shall exhibit that on A. Alsowe shall give the proof for the case when A has 5 or more ‘layers’. For thedefinition of ‘layers’ and the proof of the Theorem 6.4.1 when A has less than5 ‘layers’ we refer the reader to [16]. Let us consider the following factoriza-tion n = md for some m ≥ 240. For Z/nZ we shall take {0, 1, . . . , n−1} andfor any t ∈ Z/nZ, let t = mη + ξ, with 0 ≤ ξ < m, be its representation inEuclidean algorithm for the pair (t,m).This gives a map Z/nZ 7→ [0,m]× Z/dZ.For any integer u coprime to n we shall compose above map with Z/nZ

u−→Z/nZ to obtain the map

φ = φu : Z/nZu−→ Z/nZ 7→ Z× Z/dZ.

We will write φ = (φ1, φ2) as the comma map, where φj is obtained aftercomposing with the jth projection. We make the following;Claim: There exists a u and a k such that for T = [k, k + m/2] we have|φ−1(T ) ∩ A| ≥ 0.844|A|.To establish the claim we shall use the theory of Fourier transform and aresult of Freiman [14].

73

For any integer u the Fourier transform of A is defined to be (see [44]),

A(u) =∑a∈A

exp(2πau/n).

The result of Freiman states the following;

Theorem 6.5.1. (Freiman) If u0 satisfies |A(u0)| ≥ l|A|. Then u0A modulon has at least 1+l

2|A| elements in some half circle modulo n, that is there

exists k, such that at least 1+l2|A| elements of u0A modulo n are contained

in [k, k + n/2].

There are improvements of this result due to Lev [29, 30]. These im-provements will be able to give the Theorem 4 for a higher value of c thanclaimed here. But for our purpose the result of Freiman suffices. The follow-ing lemma is already proved in [16], we reproduce the proof for the sake ofcompleteness.

Lemma 6.5.2. There exists u 6= 0 such that |A(u0)| ≥ 0.6868|A|.

Proof.

S1(u) =∑a∈A

exp(2πau/n), S2(u) =∑

b∈A+A

exp(2πbu/n),

for integers u coprime to n. Next let us put

S =n−1∑u=0

S21(u)S2(u).

One has,

S =n−1∑u=0

∑a1,a2∈A,b∈A+A

exp(2πiu(a1 + a2 − b)/n)

=∑a1∈A

∑a2∈A

n−1∑u=0,

b=a1+a2

exp(2πiu(a1 + a2 − b)/n)

+∑a1∈A

∑a2∈A

n−1∑u=0,

b 6=a1+a2

exp(2πiu(a1 + a2 − b)/n)

=∑a1∈A

∑a2∈A

n−1∑u=0

1

= n|A|2.

74

Let d(u) = gcd(u, n) and m(u) = nd(u)

. Then one has,

S =∑

u,m(u)<240

S21(u)S2(u) +

∑u,m(u)≥240

S21(u)S2(u) = T1 + T2.

The trivial bound on T1 yields

|T1| ≤ (240)2|A|2|A+A| ≤ (240)2(2.12)10−9n|A|2.

On the other hand,

|T2| ≤ maxu,m(u)≥240

|S1(u)||∑

u,m(u)≥240

S1(u)S2(u)|

≤ maxu,m(u)≥240

|S1(u)|∑

u,m(u)≥240

|S1(u)S2(u)|

≤ maxu,m(u)≥240

|S1(u)|( ∑u,m(u)≥240

|S1(u)|2)1/2( ∑

u,m(u)≥240

|S2(u)|2)1/2

≤ maxu,m(u)≥240

|S1(u)|(n|A)1/2(n|A+A|)1/2

≤ maxu,m(u)≥240

|S1(u)|√

2.11n|A|.

The third inequality follows from Holder’s and the fourth one is an applicationof Parseval’s identity. A comparison of the bounds for |T1| and |T2| withthe value of S suggests existence of some u with m(u) ≥ 240 such that|S1(u)| > 0.6868|A|. This establishes the claim with the help of the Freiman’stheorem. We remark that the condition m(u) ≥ 240 is of no importance tous.

Now we fix an u and some half circle T = [k, k+m/2) for which |φ−11 (T )∩

A| is maximum. Let us write

φ1(A) ∩ T = {β1 < . . . < βs}.

In case there are many choices of u then we will select the one correspondingto which the βs is minimal.

Lemma 6.5.3. gcd(β1, . . . , βs,m) > 1 or gcd(β1, . . . , βs) = 1.

Proof. Let gcd(β1, . . . , βs) = δ and gcd(δ,m) = 1. Choose ∆ coprime to dsatisfying ∆δ = 1 (mod m). Then for u′ = ∆u considering the map φu givesus

φu(A) ∩ T = {βi/δ : 1 ≤ i ≤ s},which is in contradiction to the minimality of βs.

75

Next we note that, without loss of generality, we can assume that A isnot contained in coset of a proper subgroup of Z/nZ. Because if it is, let K

be the smallest subgroup with the property that A is contained in a singlecoset. Then A is 2− isomorphic to a subset of Z/n′Z for n′ = |K|. Now theimage A′ of A is not contained in a coset of any proper subgroup of Z/n′Zand clearly we also have |A′+A′| < 2.11|A′|. It might happen that |A′| > cn′

in which case we have that (3) of Theorem 6.2.1 holds. In case |A′| < cn′,then what we shall prove will show that A′ has a structure given by (1) or(2) in Theorem 6.2.1 and consequently so is the case with A.We shall write φ1(A) = B ∪D where B = {β1 < . . . < βs} and D has emptyintersection with the half circle T . Let us assume that D is non empty. Ford ∈ D, where D consist of those elements of A whose first co-ordinate lies inD, we make the following;Claim:

(d+ B) ∩ (B + B) 6= ∅.Note that |A + A| ≥ |B + B|, where A = φ(A) and B ⊂ A such that firstco-ordinate lies in B. We observe that

|B + B| < |A+ A| < 2.11|A| ≤ 2.5|B|

and hence Theorem 6.2.2 can be appealed to give |B + B| − |B| ≥ βs|H|.Also if the claim is not true then we have |A+ A| ≥ |B|+ |B+ B|. This puttogether gives us

|A+ A| ≥ |B|+ βs|H|+ |B| > 3|B| > 2.53|A|.

This contradiction establishes the claim.Now we embark on the proof of the Theorem 6.2.1. Those elements of A ofthe form (j, jx + H) will be called good elements, where H is the subgroupof Z/dZ obtained by the Theorem 6.2.2 for B. Clearly elements of B and ofB + B are good elements. Since (d+ B)∩ (B + B) 6= ∅, any element d ∈ Dis of the form (j1 + j2− j3, (j1 + j2− j3)x+h), for some h ∈ H. We find thatthe elements of D are also good and hence every element of A is of the form(j, jx+ h), for some h ∈ H. We shall use this to establish the following;Claim: A is 2− isomorphic to A via φ.For this we see that,

(mη1 + ξ1) + (mη2 + ξ2) = (mη3 + ξ3) + (mη4 + ξ4) in A↓ φ ↓ φ ↓ φ ↓ φ

(ξ1, η1) (ξ2, η2) (ξ3, η3) (ξ4, η4) in A‖ ‖ ‖ ‖

(j1, j1x+ h1) (j2, j2x+ h2) (j3, j3x+ h3) (j4, j4x+ h4)

76

Note that ξi = ji and ηi = jix+hi. We want to prove that (mη1+ξ1)+(mη2+ξ2) = (mη3+ξ3)+(mη4+ξ4) if and only if (ξ1, η1)+(ξ2, η2) = (ξ3, η3)+(ξ4, η4).One way (the direct) follows easily, we shall prove the converse. We have(j1 + j2) +mx(j1 + j2) +m(h1 + h2) = (j3 + j4) +mx(j3 + j4) +m(h3 + h4)and we shall establish (ξ1, η1) + (ξ2, η2) = (ξ3, η3) + (ξ4, η4).If

(j1 + j2) < m and (j3 + j4) < m

or(j1 + j2) > m and (j3 + j4) > m,

then we immediately get (ξ1, η1) + (ξ2, η2) = (ξ3, η3) + (ξ4, η4). If we havej1 + j2 > m and j3 + j4 < m (or j1 + j2 < m and j3 + j4 > m), thenj1 + j2 = m + (j3 + j4). Now using (j1 + j2) + mx(j1 + j2) + m(h1 + h2) =(j3 + j4) + mx(j3 + j4) + m(h3 + h4) we obtain m(mx + 1) = mh for someh ∈ H. Since gcd(m, d) = 1, so m is invertible in Z/dZ and hence oneobtains 1 +mx ∈ H. But every element of A is of the form m(jx+ h) + j =j(1 +mx) +mh for some j ∈ [0,m] and h ∈ H and so A ⊂ H for H ⊂ Z/nZdefined by

H = {t ∈ Z/nZ : t (mod d) ∈ H}.

But we have assumed no such H exists. By now the claim is established.

77

Appendix A

Some Facts from AlgebraicNumber Theory

In this appendix, we shall recall some results from the Algebraic NumberTheory which are used in this thesis. Throughout this thesis, K will denotea number field of degree n, i.e. a subfield of C which is of dimension n overQ as a vector space. By OK we shall denote the ring of integers of K. Trwill always stand for the trace map from K to Q, i.e. for any α ∈ K theTr(α) is the trace of the map obtained by multiplying every element of K byα. For any n elements ω′1, . . . , ω

′n we will define their discriminant to be the

determinant of the matrix (Tr(ω′iω′j) and will be denoted by d(ω′1, . . . , ω

′n).

Also one notes that if gi denote all the Q−embeddings of K in Q, an algebraicclosure of Q, fixed throughout, then one has d(ω′1, . . . , ω

′n) = (det W )2 for the

matrix W = (gi(ω′j)).

Let ω1, . . . , ωn be an integral basis of K, i.e. a basis of OK as Z module. Thediscriminant of k is defined to be d(ω1, . . . , ωn) and is denoted by dK . If Ais the matrix satisfying (ω′i) = A(ωi) then one observes immediately that

d(ω′1, . . . , ω′n) = (det A)2dK .

Let M be a Z module inside K, then its complimentary module is defined by

M∗ = {α ∈ K : Tr(αM) ⊂ Z}.

The following fact is immediate to derive,

Fact A.0.4.1. M∗ is a Z module,2. If M1 ⊂M2, then M∗

2 ⊂M∗1 ,

3. OK ⊂ O∗K.

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Now let l be a prime integer and b be an integer which is lth power free.We consider the case K = Q(b1/l), the field obtained by attaching the real lth

root to Q. It is easy to establish that d(1, b1/l, . . . , b(l−1)/l) = llbl−1. We putδ = d(1, b1/l, . . . , b(l−1)/l), then we have the following lemma,

Lemma A.0.5. For every α ∈ OK, we have δα ∈ Z[b1/l].

Proof. Let α ∈ OK , then we have α =∑l−1

i=0 cibi/l, for rational numbers ci.

Now applying g′is to this, we obtain l equations and can solve for c′is, to seethat the denominator of ci is a divisor of δ. This proves the lemma.

Thus we have ωi = (δ)−1∑l−1

j=0 aijbj/l. From this we obtain

dK = (det A)2δ,

where A is the matrix (aijδ

). Since right side has odd power of δ so it followsthat δ|dK . We recall that a prime q ramifies in K if and only if it divides thediscriminant of K. All these we record as,

Theorem A.0.6. If q is a prime divisor of b, then q ramifies in Q(b1/l).

79

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