-3'$.""‘ ”MO“. . -..- . ' - —

SUMS OF DISTENCT DIVISORS OF
BATEONAL INTEGERS
AND

sums OF DESTINCT DIVISORS OF
QUADRATIC INTEGERS

Thesis for the Degree of PhD.
MICHIGAN STATE UNIVERSITY

Bernard Jacobson
1956

 

 
  

 

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Sums of Distinct Divisors of
Rational Integers and
Sums of Distinct Divisors of
Quadratic Integers

presented by

Bernard Jacobson

has been accepted towards fulfillment
of the requirements for

Doctor of Philosophy dqpmgnlrmthematics

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Date June 11, 1356

 

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MSU Is An Affirmative Action/Equal Opportunity Institution
emu“

 

 

 

 

 

 

Bernard Jacobson
candidate for the degree of

Doctor'of Philosophy

Final examination, June 1, 1956, 5:00 P.M., Physics-
Mathematics Building

Dissertation: Sums of Distinct Divisors of Rational
Integers and Sums of Distinct Divisors
of Quadratic Integers .

Outline of Studies

Undergraduate Studies, Western Reserve University
19u8-l951

Major-Subject: Mathematics

Minor Subject: Chemistry

Graduate Studies, Michigan State University
1951-1956
Major Subject: Mathematics (algebra)
Minor Subjects: Mathematics (analysis, tOpology,
applied mathematics)

Biographical Items
born, April 7, 1928, Cleveland, Ohio
Military Service, United States Army l9h6-l9h7
Experience: Graduate Assistant, 1951-1955
Instructor, 1955-1956
Michigan State University
Member of Phi Society, Pi Mu Epsilon, Sigma Xi,

Mathematical Association of America, American
Mathematical Society

SUMS OF DISTINCT DIVISOfiS OF RATIONAL INTEGERS AND SUMS

0F DISTINCT DIVISORS OF QUADRATIC INTEGERS

BY

BERNARD JACOBSON

A THESIS
Submitted to the School of Graduate Studies of Michigan
State University of Agriculture and Applied Science
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1956

ACKN Oh LEDG 1'.th T

The author wishes to express his sincere thanks
to Professor B.M. Stewart, who suggested this problem
and under whose supervision this investigation was

undertaken.

0‘

SUMS 0F DISTINCT DIVISORS OF RATIONAL INTEGERS AND SUMS

OF DISTINCT DIVISORS OF QUADRATIC INTEGERS

By

Bernard Jacobson

AN ABSTRACT

Submitted to the School of Graduate Studies of Michigan
State University of Agriculture and Applied Science
in partial fulfillment of the requirements
for the degree of

DOCTOR or PHILOSOPHY

Department of Mathematics

Year 1956

Approved A 744. yw

 

ABSTRACT

In a recent paper B.M. Stewart [6] discussed sums
of distinct positive divisors of rational integers. In
this dissertation these results are generalized. Let
0((M) be the number of positive integers n Which can be
written in the form.n =1Ed, where the d are distinct
positive or negative divisors of M. The author has
proved that “((M) = 0(M) if and only if n is of the form

b c k ‘11
n = 2 3 3:; p1 where b and c are not both zero,

... b c
5 < Pl < P2 < pk, p1 g 20(2 3 ) + 1 and pj+1 é

b c 3 ‘1
20(2 3 Trp1 )+ 1 for J = 1, 2, ..., k-l. The
i=1

function o((M)/0(M) is everywhere dense on the interval
0 to 1.

In the quadratic fields x + yvé and x + yJS every
integer in the field can be written as a finite sum of
distinct units, the algorithm produced depending upon
the representation of each integer of the field as a
lattice point in the plane. In any real quadratic field
there exist infinitely many integers n, having the
‘property that every integer in the field can be written
as a finite sum of distinct divisors of n. Explicitly

if a + me is the unit of smallest absolute value for

which a > 0 and b > 0, then any integer 2t+llm Where
at > a satisfies this condition. The proof again depends
upon the representation of each integer of the field as
a lattice point in the plane.
For the imaginary quadratic fields the set A(m)
is defined where an integer n belongs to A(m) if and only
if there exists a rational integer n' such that every
integer of the form.x + y m where -n' g x g n' and
-n' § y § n' and no other integer can be represented as
a sum of distinct divisors of n. It is shown that for
m = -2 numbers of the form RJ-Z belong to A(-2) Where

a
R 217131 1, p1 5 2, 5 or 7 mod 8 and 0((R) = 0(R)-

When m = -1, numbers of the form R = Trp1 1

belong to

A(-l), Whore r, E 5 mod n and uL(R) = 0(a). When m < -2
and m.£ 1 mod 4, integers of the form ZtJm belong to A(m).

If m < -2 and m E 1 mod 4, then the set A(m) is empty.

TABLE OF CONTENTS

page

INTRODUCTION 00.0.00.........OOIOOOOOOO.000...... 1

CHAPTER I
CHAPTER II

CHAPTER III

CHAPTER IV

CHAPTER V
CHAPTER VI

CONCLUSION

THE RATIONAIJ FIEIOD 00.000.000.00... 2

SPECIAL QUADRATIC FIELDS .......... 8

EIBESB

2 .0.........OOOOOOOOOOOOOOOOOO 9
......OOOOOOO00.00.000.000... 17
......OOOOOOOOOOOO0.0.0.0.... 22

7 COOOOOOOOOOOOOO00.000.000.00. 28

REAL QUADRATIC FIELDS ............. 5h
m>7andm~320r5modl+

RE
m.

m
m

ALQUADRATIC FIELDS MI

>

1 mod h

5 00.0.0000.........OOOOOOOOOOC 1+5

5 #9

THE GAUSSIAN FIELD ................ 61

IMAGINARY QUADRATIC FIELDS 61+
m.£ 1 mod h

m=-2 eoeeeeeeeeeeeeeeeeeeeeeeeeee 65
m<-2 eeeeeeeeeeeeeeeeeeeeeeeeeeee 67
eeeecoeeeeeoeeeeeeeeeeeeeeeeeeeeeee 69

BIBLImRAPM ......OOOOOOOO.......OOOOOOOOOOOOOO 70

INTRODUCTION

In a recent paper B.M. Stewart [6] discussed sums
of distinct positive divisors of rational integers. .In
this dissertation these results are generalized and
extended. In Chapter I these results are extended to
sums of distinct divisors of rational integers where the
divisors may be positive or negative. In Chapters II
throughlv sums of distinct divisors of integers in the
real quadratic fields are discussed. Sums of distinct
divisors of integers in the imaginary quadratic fields
are investigated in Chapters v and v1.

In each case a maximal set A or A(m) is defined
and the problem is to find which integers of a given
field belong to this set. For several fields a complete
characterization of the integers belonging to this set
is given. For the remaining fields we show that the
set A(m) is not empty by exhibiting infinitely many

integers which do belong to A(m).

CHAPTDR I
THE RATIONAL FISLD

For a given rational positive integer n let¢((n)
be the number of positive integers which can be written
as the sum of distinct divisors of n. The divisors may
be positive or negative. Let A‘be the set of all integers
for which<((n) = a(n).

LEMMA 1. If (n,2) = l, (n,5) = l and n f 1, then
n does not belong to A.

Lemma 1 is true because o(n)-5 cannot be written
as a sum of distinct divisors of n.

LEMMA 2. If n=2t, then n belongs to A.

Lemma 5. If n = 5t, then n belongs to A.

PROOF: The lemma is true for t = 1. Assume that
the lemma is true for t = k-l. Let t = X. Every divisor

of 5k-1 is a divisor of 5k. by the induction hypothesis

5 - can be written

and thus as divisors

every integer between 1 and 0(5

as a sum of distinct divisors of 5k-l

k , k k k k
of 5 . bvery integer from 5 -l + l = 5 +1 = 5 -(5 -l)

to 5k can be written as 5k minus a sum of distinct

k-l

divisors of 5 . Every integer from 5k to 0(5k) =

k+1

5 -l = 5-5k-l = 5k+ 5:-1 = 5k + 0(5k-1) can be written

4

as 5k plus a sum of distinct divisors of 5k-1. Thus by

induction n belongs to A.
LEMMA 4. If n belongs to A and p is an odd prime
with (n,p) = 1, then np belongs to A if and only if o(n)

g -le
PROOF: Suppose that n belongs to A and o(n) < 2-1.
The numbers 2-1 and 2+1 cannot be represented as sums of

distinct divisors of np since the largest number which

can be represented without using p is 0(n) < 2-1. The

smallest number which can be represented, using p, is

p-o(n) > p- 2-1 = 2+ . Thus the condition is necessary.

We now prove that the condition is sufficient. Suppose

that n belongs to A and that o(n)§ p-l. Let r be any

integer such that 0 g r g 0(n). Every integer between

rp and rp + 0(n)can be written as pZ d + Z. d'. Now
d/n d'/n

let r be any integer such that 0 g r+l g a(n). Every

integer from rp + 2-1 +1 = (r+l)p - (2-1) to (r+l)p

can be written as p d.-.2L.d'. Thus np belongs to A.
d n d'/n

LEMMA 5. If n belongs to A and (n,p) = 1, then

npt belongs to A if and only if c(n) g 2-1.

PROOF: The condition is necessary in order to

represent -1 and 2+ . Suppose that o(n) g 2-1. The

lemma was proved true for t = l in Lemma A. Let us assume

that the lemma is true for t = k-l and prove it true for

k-l

t = k by induction. Every divisor of np is a divisor

of npk. Thus by the induction hypothesis every integer

k-l) can be written as a sum of distinct

kel)

from 1 to o(np

divisors of npk. Every integer from pk - e(np to

pk can be written as pk - -ld° However we have that

d/np
k

k’l k-I k
o(np ) + l = l + o(n)o(p ) 2 l + (p-l)(p -12 = 2 +1

= pk - 2k-12 = pk - 2-1)(pk-l) 2 pk- 0(n)o(pk-1) =
p-
pk - 0(npk-l). Thus every integer up to pk - 0(npk-l)

can be represented. Let r be any integer such that
0 g r § o(n). Every integer from rpk to rpk + o(npk-1)

can be written as kad + Z d'. Now let r+l g o(n).
d/n d'/np

Every integer from (r+l)pk - 0(npk-1) to (r+l)pk can be

written as kad - _ld'. This proves the lemma
d/n d'/np

since rpk + o(npk-l) +1 g (r+l)pk - o(np

k-l)

t
LEMMA 6. Let n = -%-p11, where p1 < pJ for i<j.
i=1

If n does not belong to A, then nqs does not belong to

A for all primes q > pk and all s > O.

PROOF: Let n.j = ifiOpi for j _5_ k. Let p0 = r10 = l.

Since n does not belong to A there must be a smallest

integer 3 such that n does belong to A and n does not

J j+l
belong to A. By Lemmas 2, 5, and 5, 0(n3) < (pJ+l- l)/2.

Let R be the sum of all divisors of n greater than or
equal to pj+1. We will show that R - (pj+1- l)/2 cannot

be represented as a sum of distinct divisors of n. If all.
the divisors of n which are greater than or equal to p1+1
are used positively, the smallest integer which can be

represented is R - o(nJ) > R - (pj+l- l)/2. If any

divisor d g pJ+l is not used positively, the largest number
which can be written as a sum of distinct divisors 5? n is

R - d + 0(nj) < R - pJ+1 + (pj+l-l)/2 < R - (pj+l - l)/2.

As a direct result of Lemmas .1 through (5, we
can now state the following theorems: .

THEOREM 1. An even integer belongs to A if and
only if it has one of the two following factorizations as
a product of primes:

at, for all t g o.

t

t k i

ii) n = 2 p
i=1 1

i) n

with pr < pa for r < s; t g l,

 

THEOREM 2. An odd integer belongs to A if and only

if it has one of the two following factorizations as a

product of primes:
i) n = at for all t g o.

t k t1
ii)n—51Elpi with5<pr<psforr<s;t_2_1

t g l and p -l g a n for l g j s k.
1 ‘%*“ E E1 -
117

ieri

THEOREM 5. There exist arbitrarily large square

free integers in A.
k
PROOF: Let nk =i'Tl'p1 be the product of the first
i=1
k primes in their natural order. For every k,nk belongs

to A since p. g nj_1 + l g 0(nj-l) + l < 20(n ) + l

J 1-1

for all j.

LEMMA 7. If p is a prime such that p > 20(T) + 1,
then aL(pT) = 2-((T)[O((T) + l].

PROOF: Every positive sum of distinct divisors
of pT can be written as n = nlp + n2 where n1 is zero
or any positive number which can be written as a sum of
distinct divisors of T and n2 is zero or any positive or
negative integer which can be written as a sum of

distinct divisors of T. The maximum value for n2 is 0(T).

Since nlp + 0(T) < nlp + (2-1) = (n1+l)p - p + 2-1 =

(nl+1)p - (3+1; < (n1+l)p - (2-1) < (n1+1)p - 0(T),

there is no overlapping. We have B&(T)+l] choices for

nl and ZKIT) + 1 choices for n2 except when n1 is zero.

If n1 is zero, then n2 must be positive. Thus we have
0((p'r) = [OUTHIHZOUTHI] - [0((T)+l] = 2°<<T)[°<(T)+1].

DEFINITION: s(M) =°£(M).
oIMS

THEOREM A. s(M)_is everywhere dense on the
interval 0 to l. ‘-
PROOF: Given any x and any y such that O < x < y

l, we seek an integer R such that x < s(R) < y. Let

"A

R = pM where M belongs to A and p > 20(M) +1. By Lemma 7

s(R) = s(pM) = 20(M)[O(M)+l! = 2[o(M)+l]. We seek an
p+ o p+1

integer M and a prime p satisfying the above conditions

and such that l’< +1 <.l° Let u = 1, u(l+£) = l
ymfifinx 3? 2’

and v = 2u[o(M)+l]. .We know by Theorems 1 and 2 that we

 

can find an M belonging to A which is arbitrarily large
so that v = 2u[o(M)+l] can-be made arbitrarily large.
By the Cahen-Stieltjes theorem L1] we know that for
sufficiently large v there exists a prime p such that
v - l < p < (v-1)(l+£)== v(l+£) -l -£.< v(1+£) - 1. We
now have (v-l) < p < v(1+E) - l.

v < p + l <v(l+£),

2u[o(M)+l] < p + l < 2u[o(M)+l][1+£L

l< +1 <

y o M +

She

CHAPTER II
SPECIAL QUADRATIC FIELDS

In this chapter and in the chapter following this
we will focus our attention upon the integers in the
quadratic field R(m) = x + me , where m is square free,
greater than zero and not congruent to one modulo four.

The domain of integers D(m) of R(m) are the
numbers of the form x + me where x and y are rational
integers. The units are those numbers of D(m) where
x2 - my2 = :1. Each domain has an infinite number of
units and a basic unit such that if u1 = a1 + bifim is the
unit of smallest absolute value for which both x and y
are positive, then every unit in the domain can be written
:u? for n = O, :1, t2,"'.

Let uJ ‘ ui for j = O, l, 2,'°°. In this way we

may divide the units into one set S0 = {f1,-l} of two

units and infinitely many sets S1 = {u1,-u1,l ,;l; of
u u
i 1
four units. Every unit belongs to one and only one set.
DEFINITION: M111) is the set of integers{n} in
the field R(m) which have the following property: n
belongs to A(m) if and only if every integer in the domain
D(m) can be written as a finite sum of distinct divisors
of n.

DEFINITION: A'(m) is the set of integers {n} in

the field h(m) which have the following property: n
belongs to A'(m) if and only if every integer in the
domain D(m) can be written as a finite sum of distinct
associates of n.

We will show that for all values of m the set A(m)

is not empty and that for m > 2 the set A'(m)is empty.

m = 2

In this field u = 1 + wa, u2 = 5 + 2J2,-oo. Let

1

a1 be the rational part of u and b be the coefficient

1 i
. _ k
01 J2 in ui. Let Bk - fgibi.

Since the integers of D(2) are given as x + y¢2
where x and y are both rational integers we may consider
any integer P of D(m) as a lattice point P in the plane.
We will fill the plane with concentric squares whose

centers are the origin and whose diagonals are the

coordinate axes. The length of the diagonal of the k
square is Ask. It will be shown that every integer lying

in the k square can be represented as a sum of distinct
units belonging to the first k sets SO,Sl,°°°,Sk.

LEMMA 1. a =

k+l ak + 2bk and b = + b .

k+l “k k

PROOF: ak+l +bk+l¢2 = uk+l = uk(1+¢2) =
(ak+ka2)(l+J2) = (ak+2bk) + (ak+bk)¢2. Therefore ak-l =

' = +
ak + 20k and bk+l ak bk.

10

LLMhA 2. ak + bk g “Bk-l for all k g 5.

PROOF: The lemma is true for k = 5 and k = h.
We will assume that the lemma is true for 5 g k g n and

prove the lemma true by induction. Let k = n + l.Using

pan-l + 7bn-l

Lemma 1 we have: an+1 + bn+1 = 2an + 5bn

+an - “bn-l

+ Obn-l = an-l

“‘Dn-l + bn). Using the induction hypothesis we find that

the extreme right hand side of the equation is less than

or equal to “fin-2 + “(bn-l + bn) = th.

l

PROOF: ak ak_
PROOF: The lemma is true for k = 1. Assume that

the lemma is true for k = n. an+l + b = an +2bn + an + bn

By the induction hypothesis the right hand side of the

equation is greater than or equal to 2bn + 2Bn-l = 2Bn.

THEOREM 1. The integer 1 belongs to Mi).

PROOF: By inspection it can be seen that every
integer in the first three squares can be represented as
a sum of distinct units in the sate-30,81,82 and S5. Let

us assume that every integer in the first k-l squares can

11

be written as a sum of distinct units in the first k sets

of units, 30,8 --°,Sk_1 and prove the theorem by induction.

1,
Consider any point inside the k square. Since the square
and all sets of units are symmetric with respect to both
axes, there is no loss in generality by assuming that the
point lies in the first quadrant, on the positive x axis
or on the positive y axis. If the point lies inside the
k-l square a required representation is assured by the
induction hypothesis. Therefore consider a point with
rectangular coordinates (x,y) which lies in the first
quadrant, in the k square but not in the k-l square.

Thus we have: 25 < x + y g 23

k-l k’

y 2 O, x O.

IN

We divide the possibilities into three cases. In

Case 1 the point lies above the line y - x = 2(bk - Bk-l)'

In Case 2 the point lies below the line y-x=2(Bk_l-ak).

In Case 5 the point lies on or between these two lines.
Case 5 actually exists because Lemma h implies that

b - 5

k k-l é bk-I ' ak'

Case 1. Zbk-l < x + y g 23k:
y - x > 2(bk - Bk-l)’

ng, X 00

"V

If we subtract 2bk from y, we obtain a point P' with

12

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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x lrll A-l

 

15

rectangular coordinates x' = x and y' = y - 2bk. Thus

r. u. f = ' — h - ~ -r‘ =-- ‘
x' + y' g dbk dbk ZBk-l’ y x' g 4(bk bk-l) dbk 25k-1,
x' g 0. Thus the point P' lies inside the k-l square and

by the induction hypothesis can be represented as a sum Ok-l

of distinct units in the first k sets of units. P'=Ok_l
P' = P - ankle. Thus P = Ok-l +2ka2= ok-l + (ak+bk/2) +
(-ak+ka2) and the theorem is proved in Case 1.

Case d. aBk-l g x + y g 2Bk,

y-x

IIA

2(Bk-l-ak) 9

y g 0, x g 0.
If we subtract Zak from x,we obtain a point P' with

coordinates x' = x - 2ak and y' = y. Thus we have:

x' + y' g 2Bk - Zak g ZBk - 2bk = ZBk-l'
Y' ‘ x’ é akBk-l-ak) + 23k : in-l’
y' g 0. Thus the point P' lies inside the k-l

square and can be represented as a sum Ok-l of distinct

units in the first k sets S "°,S

0' k-l'P' = °k-1’

P' = P - 2a . Thus P =

+ =
k 2a o~

°k~1 k K-l + (ak+ka2) *

(ak-bk/Z). Thus the theorem is proved for Case 2.
Case 5. 23k_l g x + y g 23k.

2(bk-Bk_l) g y - x 3 2(Bk_l-ak).

 

 

‘04

 

 

 

 

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15

If we subtract ak from x and bk from y we obtain

a point P' whose coordinates are x' = x - ak and y' = y-b .

k

Using lemma 2 we obtain the following inequalities:

u t - - - “ = -
x + y 2 2bx-1 3k bk é ZBk-l 45k-i 2Bk-l’

x' + y' § 25k - ak - bk 5 abk - 2bk = 2Bk-l’

y, -x'

IN
N
m

k-l "ak ' bk + ak E aBk-l ' ak ' bk E “aBk-i'

y'ux'

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‘k-l
“Bk-1 ' 2Bk-1 = aBk-l'
Thus the point P' lies inside the k-l square and

can be represented as a sum 0 of distinct units in the

k-l

first k sets of units. P' = Ok-l' P' = P - (ak+ka2).
Thus P = Ok-l + (ak+ka2) and the theorem is proved.

COROLLARY 1. Every integer in DLZ) is in A12).

THEOREM 2. The only integers in A'(2) are units.

PhOOF: It was proved in theorem 1 that the units
belong to A'(2LLet X be any integer which is not a unit
and let Y be any integer which can be represented as a
sum of distinct associates of x. Using norms we obtain

from Y = 2X1 = 530.11 = xzui the following equation:
N(Y) = szu1) = N(X)N(Zui). If the norm of x

does not divide the norm of Y, the required representation

is not possible.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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17

m = 5

In this field u1 = 2 + .5, u2 = 7 + Warn. Let

b be the coefficient of

a 1’1

1 be the rational part of u

1 and Bk = ééibi' We will first prove that the

‘x/Einu
units do not belong to A(3).

LEMMA 1. 2bk > ak for all k > 1.

PROOF: dbk = dak_1 + “bk-l > 2ak_l + ak - 2ak_l = ak.

k-l
LEMMAZ. ak>22a1+ 3fork>2.
i=1

PROOF: The theorem is true for k = 5. Let us
assume that the theorem is true for 2 < k g n-1 and prove
the lemma by induction. Let k=n. Using Lemma 1 we obtain

n-2
n-1 + 5bn-l > dan-l + an-l > 2an-l + ZEEiai + 5 :

THEOhhM l. The numbers 2 and -2 cannot be written
as a sum of distinct units of D(5).
PROOF: We can see by inspection that neither 2 nor

-2 can be written as a sum of distinct units in S S

0’ 1'52

and S alone. Assume that 2 can be represented, 2 =Zv1

5

where the vi are units. If vi and -v;l both appear in the

summand their sum is a multiple of‘JB and contributes
nothing to the rational part. Therefore there must exist

a greatest positive integer 3 such that v1 is in the set 83’

18

v1 appears in the summand, and neither -v1 nor -v1'-1
appear in the summand. Lemma 2 tells us that j g 2 and
the theorem is proved.

We will now show that the prime p = (l + J5) belongs
to a(5). Let c2t + d2t¢5 = at + bivfi and °2t+1 + d2t+iJm

= pult. We now separate the divisors of p into the sets

Vk = [(ck+de5).(ck-dk/5).(-ck+de5),(-ck-ko3)1- Every

divisor of p belongs to exactly one of these sets. Each

set contains exactly one divisor c + dkyfi, with both

k

ck and dk non negative. Let Dk = fgldi'

We will make use of the following equations:

a
II

R 2°k-2 + 5dk-2 and dk = ck_2 + 2dk-2 for all k > 1;

and d = c + d

a
ll
0

+ 5d when k is odd;

k k-l k-l k k-l k-l
c1‘ = %1°k-l + Bdk-l] and dk = %[°k-l + dk-l] for k even.

mm 50 CR + dk g k-lo

PROOF: The lemma is true for k = 2 and k = 5.
Let us assume that the lemma is true for k g n and prove
that the lemma is true for k = n + 1. We have

°n+1 + dn+1 = 3°n-1 + 5dn-1 g °n-1 + dn-l + “an-1 + 2°n-1‘

Noting that 2cn_1 < hdn, and making use of the induction

hypothesis we see that the right hand side of the equation

hD

n-

is

"A

2 + “fin-l + hdn = hpn and the lemma is proved.

19

1‘ r.
LEMMA 40 Cl{ + Cik > 2yk’l.

PBCCF: The lBMLL is true for k = 2 and k = 5.
Let us asaune that the lemma is true for k g n and prove

the lemma by induction. Let k = n + l.

+ = + ’
°n+1 dn+1 cn-l dn-l + acn-l + “dn-l > 2Dn-2 + 2dn-1'

THEOREM 2. l +‘J5 belongs to A(3).

PROOF: We construct concentric squares with
center at the origin and with diagonal lengths equal to
auk. Because both the sets of divisors and the squares
are symmetric with respect to both coordinate axes there
is no loss in generality by considering only those points
in the first quadrant, on the positive x axis or on the
positive y axis. By inspection we can see that every
integer in the first square can be written as a sum of

distinct divisors in the sets V0 and V In order to

l.
' prove the theorem by induction we assume that every point
in the first k - 1 squares can be represented as a sum of

distinct divisors taken only from the first k sets V ---,V

0’ k-l'
Let P be any point with coordinates (x,y) which lies
inside or on the boundaries of the k square. If P lies in
the k - 1 square, the theorem is proved. If P does not

lie in the k - 1 square, we have the following inequalities:

2Dk_l < x + y é ZDk’

KEG, ngo

20

We divide the possibilities into three cases. In

Case 1 the point lies on or above the line y - x =

2dk - 2Dk-l' In. Case 2 the point lies on or below the

line y - x - -2ck + 2Dk-1 and.in Case 5 the point

lies between these lines. Lemma.h assures us that Case 3

actually exists.

Case 1. 2D -1 < x + y é 2D

k k’

y - x g 2dk - 2Dk-l’

XEOO
If we subtract 2dk from y we obtain a point P'

with rectangular coordinates y' = y - de and x' = x.
x' + y' g 2Dk - 2dk = 2Dk-l’
-I .’ - :.

y' x g 2dk de-l de 2Dk-l'

x' g 0.
The resulting point P' lies inside the k - 1

square and can be represented as a sum Ok-l of distinct

divisors in the first k sets V ;O-,V 1. P' = P - 2d and
O k-l k

+2d r.

P' = ok-l' Solving for P we obtain P = ok-l k

ok-l + (ck + ko5) + (-0k + kofi). This completes Case 1.
Case 2. de-l < y + x < 2D
y - x g -2c + 2D

y g 0.

21

If we subtract 2ck from x we obtain a point P'

with rectangular coordinates y' = y and x' = x - 20k.
1 I "' a. ' - =
y + x g 20k 20k < 2Dk 2dk 2Dk-l’
y' - x' g ~2ck + de-l + 20k = 2Dk_1,

y' g 0.
The resulting point P' lies inside the k-l square

and can be represented as a sum 0 of distinct divisors

k-l

in the first k sets of divisors vhf..’Vk-l' P' = P - 20k.

P' = Solving for P we obtain P = o + 2c =

°k-1° k-l k

ok-l +(ck + de5) + (ck - de5). This completes Case 2.
Case 5. 2Dk-l < y + x g 2Dk,

~20 + 2D x < 2d - 2D

k k-l < y ’ k k-l'

If we subtract dk from y and ok from x we obtain a

point P' with coordinates y' = y - d and x' = x - ck.

k
y' + x' g 2Dk - ck - dk < 2Dk - 2dk = ZDk-l' Applying
lemma 5 to each of the following inequalities we obtain:
y' + x' > 2Dk-l ' ck ‘dk E 2Dk-l ’ ”Du-1 = ’2Dk-1'

Y' - x' < 2d - 2D + c - d = c + d -'2D

22

The resulting point P' lies inside the k-l square

and can be represented as a sum On-l of distinct divisors

in the first k sets VO’VP'.°’Yk-l° P' = P - ck - deB.

I = . a - = ‘
P Solving Ior P we obtain P ok_l + ck + dkvj.

ok-l'
This completes Case 5 and the theorem is proved.

COROLLARY. If x + yJ} is an integer in D(5) with
x a y modulo 2, then x + yd} belongs to A(§).

Pacer: x + = (x - 3y) + (y - x)v3 . If we
I‘?*5%fi -2

have x E y modulo 2, then the right hand side is an

integer and x + yv3 is divisible by 1 +-¢5. It then

 

follows from Theorem 2 that x + y VB belongs to A(5).

m = o
In this field u1 = 5 + 2Vb, u2 = 49 + 2o¢6,--o.
Let bi be the coefficient of V% in ui and a1 be the

rational part of ui. We will first prove that the units
do not belong to A(6). .

LEMMA 1. 2 divides bk.

PROOF: The lemma is true for k = 1. Let us

assume that the lemma is true for k = n-1 and prove the

lemma by induction. bn = Sbn-l + 2an-l' by the induction

hypothesis 2 divides bn-l and thus divides theright hand

Side of the equation. Therefore 2 divides bn.

23

THEOREM 1. The units do not belong to A16).
PROOF: Lemma 1 tells us that in every sum of units
the coefficient of M0 is even. x + be can not be

represented as a sum of units when y is odd.

I

We will now show that the integer 2 belongs to AK5)-

Let us separate the divisors of 2 into sets V as follows:

1

v0 = [1, -l, 2, ~21
v1 = [2+Vo, 2~¢o, -2~Jo, ~2+Jol
-l -1
V2 = [“1' ul ' 'ul' ”“1 1
s -1 , , -1
V5 = [2u1, dul , -2ul, -2ul ]
v4_= [(2+vt)ul=22+9¢o, 22 ~9Vo, -22+9V6,-22-9¢6]

In every set Vk (k>0) there exists exactly one

divisor ok + dkvb with both ok and dk positive. The
divisors in the set Vk are ck + dkyh, ck - deb, ~ck - deb

k
and -c + d V%. Let D = 5:.d . We will make use of the
k k k i=1 1

following equations:

ck = 2ck_l and dk = 2dk_l when k : 0 mod 5, and k f O,

c = c if k E l or 2

+
R 3d

and dk = ck_1 + d
.2...

and 5dk > Ok for k > 1.

k-l k-l k-l

moaulo 5. In either case 2dk < 0k

LEMMA 2. ok + 2dk < 8Dk

-1.
PROOF: The lemma is true for k = 5. Let us

assume that the lemma is true for k = n-1 and prove the

lemma true by induction. Let k=n. If k = n E 0 mod 3:

0n + 2an = con-l + udn-l < cn-l + 7dn-l z cn-l + 2dn-l +
5on_l < 8D _2 + son-l < 8Dn_l. If k = n s 1 or 2 mod 5,
on + eon = 2cn_1 + 9on_l < cn_l + 8d _1 - cn_1 + 2d -1 +

Odn-l < 8Dn-2 + 0Qn_l < 8Dn_10
LEMMA 5. 0k + 2dk > th_1.
PROOF: The lemma is true for k = 2. Let us

assume that the lemma is true for k = n-1 and prove the

lemma true by induction. Let k = n. If k = n e 0 mod 3,

0n + 2dn = 20n-l +.udn-l > cn-l + 0dn-l = cn-l + 2dn-l +
”fin-l > th_2 + #dn-l = an-l' If k = n E l or 2 mod 5,
on + 2dn = 20n-l + 5dn-l > cn-l + 7dn-l = cn-l + 2dn-l +
5dn-i > “uh-2 + 5dn-i > “Dn-i‘

Thachsn 2. 2 belongs to A(6).
PROOF: We construct concentric diamonds with
centers at the origin, whose sides are given by the

equations 2y + x = :qu and 2y - x = thk. because the sets

of divisors and the diamonds are both symmetric to both
coordinate axes there is no loss in generality by
considering only those points lying in the first quadrant,
on the positive x axis or on the positive y axis. By
inspection we can see that every point in the first two

(iiamonds can be represented as a sum of distinct divisors

25

in the sets V0, V1 and V2. In order to prove the theorem

by induction we assume that every point in the first k-l
diamonds can be represented as a sum of distinct divisors

in the first k sets of divisors V0, Vl,°°',V Let P

k-l'
be any point with coordinates (x,y) which lies inside or
on the boundaries of the k diamond. If P lies in the k-l
diamond, the theorem is proved. If P does not lie in the

k-l diamond, we have the following inequalities:

uDk-1 < 23 + x g th,
KEG, yEOO
We divide the possibilities into three cases. In
Case 1 the point lies on or above the line 2y - x =

adk - uDk-l' In Case 2 the point lies on or below the
line 2y - x = -ck + th_l. In.Case 5 the point lies

between these lines. Lemma 5 assures us that Case 5

actually exists.
Case 1. “DR-l < 2y + x g th,
2y - X g ”dk " wk-lt
KEG.

If we subtract 2dk from ylwe obtain a point P'

with rectangular coordinates y' = y - 2dk and x' = x.
2y' + X' < ubk - 4dk " uDk-l. x' g 0.
2y' ‘ x' 2 ”Gk “bk-1 “Gk “bk-1°

26

The resulting point P' lies inside the k-l diamond
and thus can be represented as a sum Ok-l of distinct
divisors in the first k sets of divisors V0, Vl,°°°, Vk-l’
P' = P -2dk/o and P' = Ok-l’ Solving for P we obtain
P = Ok-l + 2deb = Ok-l + (0k + kob) + (-ck + deb).
This completes Case 1.

Case 2. th_l < 2y + x g nDk,

2y - x g ~2ck + uDk-l’

ngo
If we subtract 20k from x we obtain a point P'
with rectangular coordinates y' = y and x' = x - 2ck.

2y' + x' é ”Bk ' 20k < “Bk ’ “dk = LLDk-i'

2y' ‘ x' -ch + ”pk-1 * 20k.= “pk-1’

"A

y' g 0.
The resulting point P' lies inside the k-l diamond

and thus can be represented as a sum of distinct divisors

in the first k sets of divisors V0, V1, °--,Vk_l.
P"== P - 20k and P' = Ok-l’ Solving for P we obtain

P = c: +-2c = ok-l + (ok + dKVb) + (ok - dkyo). This

completes Case 2 .

Case 5. uDk-l < 2y + x_§ th,

520k + th_1 <‘2y - x < hdk - th_l.

27

If we subtract c from x and d

k k from y we obtain

a point P' with coordinates y' = y - (1k and x' = x - ck.

23'+ x' é “Bk ' de ' ck ‘ “Dk ' “fix 3 4Dk-i'
Applying Lemma 2 we also obtain the following inequalities;
2y' + x' > “DR-l - 2dk - ck > -th_1,

2y' ' x' < “dk ' “Bk-1 ' de + ck = Ck + 2dk ' “Dk-l < l“Die-1'

2y' - x' > -2ck + “Bk-l -2dk + 0k - ”DR-l - ck - 2dk > LDk_1.
The resulting point P' lies inside the k-l diamond

and thus can be represented as a sum ok-l of distinct

divisors in the first k sets of divisors V V ",V

O’ l’ k-l'

P' = P - c

k Solving for P we

" de6 and P' = ok‘l.

ootain P = Ok-l + 0k + deb. This completes Case 5
and the theorem is proved.

COROLLARY. If x 5 y E O modulo 2, then the integer

x + be belongs to A(b).

28

m = 7
In this field u1 = 8 + 3J7, u2 = 127 + g8/7 ....

Let a1 be the rational part of u1 and b1 be the

coefficient of J7 in ui. We will first prove that the

units do not belong to A(7)o

LEMMA 1. 5 divides bk for all k.

PROOF: The lemma is true for k = 1. Let us

assume that the lemma is true for k = n-1 and prove that

the lemma is true by induction. When k = n, we

have bn = 8bn_l + 5an_l. 5 divides bn-l and thus 5

divides the right hand side of the equation. Therefore

, 5 divides bn'

THEOREM 1. The units do not belong to A(7).

PROOF: In every sum of units the coefficient of

V7 is divisible by three and thus x + y¢7 cannot be

represented if y 5 l or 2 moaulo 5.
We will now show that the integer 2 belongs to A(7).

.Let us separate the divisors of 2 into the sets Vk as

.follows:

v = [1, -l, 2, -2]

= is + V7. 5 - M7. -5 - V7. -5 + V7]
V2 = [111, -ul, nil, -u;l]
V5 [2u1, ~2ul, 2uil, -2uI

V = nah/7W1 = 45 + 1N7. kWh/7. 46-11”, -t5+11\/7l

0
V1

11

N

29

In every set Vk (k>0) there exists exactly one

divisor ck + dRVW with both 0k and dk positive. The

divisors in the set Vk are ok + dk/7, ck - ko7, -ck - de7

and -ck + dk/7. Let Dk = fiidi. We will make use of the
following equations:

ck = ZCk-l and dk = 2dk_l when k f O modulo 5,
ok = (5ok_1 + 7dk_l)/2 and dk=(ck_l+ 5ok_l)/2 if ssh mod5.

Lemma 2. bdk < 20k ed

“A

k.
PROOF: The lemma is true for k = 1. Let us

assume that the lemma is true for k = n-1 and prove that

the lemma is true for k = n. then k = n E 0 mod 5 we have:

pdn = lOdn_l < 40 = 2cn = Re g 12d = bdn.

n-1 n-1

When k = n E l or 2 modulo 5 we have:

pdn = (jcn-l + 15dn_l)/2 < (bcn-J + l5dn_l)/2 < 2cn =
5°n-1 + 7dn-l < 5°n-1 + 9dn-l = bdn'

< 20D

k'l for k > 20

LEMMA 5. 2ck + 5dk
PROOF: The lemma is true for k = 5. Let us

assume that the lemma is true for k = n-1 and prove that

the lemma is true by induction. For the case k = n

and n E 0 mod 5 we apply Lemma 2 and the induction

hypothesis to obtain 2cn + an = hon-l + lOdn-l =

20n_1 + Sdn-l + 2cn_l + jdn-l < 20Dn_2 + lldn_1 < 20Dn_l.

50

When k = n i l or 2 modulo 5, we apply Lemma 2 and the

induction hypothesis to Obtain:

zen + son = (11on_l + 29on_l)/2 g (“ch-1 + 50dn_l)/2 =
2cn_1 + Sdn-l + 20dn_l < 20Dn_2 + 20dn_l = ZODn-l'

LEMMA a. 20k + jdk > lODk_l.

PROOF: The lemma is true for k = 2. Let us
assume that the lemma is true for k = n-1 and prove that
the lemma is true by induction. For the case k = n 5 O
mOdulo 5, we apply the induction hypothesis and Lemma 2

to Obtain 2cn + Bdn = “en-l + IOdn-l > lODn_2 + 1°dn-1 =

lODn-l' when k = n f O modulo 5, we apply Lemma 2 and
the induction hypothesis to obtain:

2g + jdn = (llc + 29dn_l)/2 = 20n_1 + Bdn-1 +

n n-l

('(cn_1 + l9dn_l)/2 > lODn_ +iodn = lODh

2 -1 -1°
THEOREM 2. The integer two belongs to 8(7).
PROOF: We construct concentric diamonds with

centers at the origin, whose sides are given by the

equations 5y + 2x = thDk and By - 2x = t 10Dk. Since the

sets of divisors and the diamonds are symmetric to both
coordinate axes there is no loss in generality in
considering only those points lying in the first quadrant,
on the positive x axis or on the positive y axis. By

inspection we can see that every point in the first two

51

diamonds can be represented as a sum of distinct divisors

in the sets V0, V1 and V2. In order to prove the theorem

by induction we assume that every point in the first k-l
diamonds can be represented as a sum of distinct divisors

in the first k sets of divisors V0, V '--,V Let P

l’ k-l'
be any point with coordinates (x,y) which lies inside or
on the boundaries of the k diamond. If P lies in the k-l
diamond, the theorem is proved. If P does not lie in the

k-l diamond, we have the following inequalities;

10D 1 < jy + 2x g 10D

k- k’
x g 0, y E O.
We divide the possibilities into three cases. In
Case 1 the point lies on or above the straight line

5y - 2x = 10dk - 10D . In Case 2 the point lies on

k-l

or below the line 5y - 2x = ~4ck + lOD In Case 5

k-l'
the point lies between the two lines. Lemma h assures
us that Case 5 actually exists.

Case 1. 10D < 5y + 2x g 10D

k-l k’
5y - 2x g 10dk - lODk-l’
XEO.
If we subtract 2dk from y we obtain a point P'
with rectangular coordinates y' = y - 2dk and x' = x.
by' + 2x' g 10Dk - lOdk = lODk-l' xi g 0,
.|_! - - :-
5y 2x g 10dk lODk_l 10dk lODk-l‘

52

The resulting point P' lies inside the k-l diamond

and thus can be represented as a sum 0 of distinct

k-l
divisors in the first k sets of divisors V V ---,V

o’ 1' k-l'

P' = P - 2dk/7 and P' = 0 Solving for P we obtain

kfl.
P = Ok-l + adkw = Ok-l + (ck + ek-m + (-ck + dkx/7).
This completes Case 1.

Case 2. 10D 1 < jy + 2x g 10D

k- k’
5y - 2x g -hck + lODk-l’
ng.
If we subtract 2ck from x, we obtain a point P'
with rectangular coordinates y' = y and x' = x - 2ck.

applying Lemma 2 we obtain:

Sy' + 2x' g lODk - hck < lODk - 10dk = lODk-l’
Sy' - 2x' g ~uck + lODk_1 + hck = lODk_1.
y. a O.

The resulting point P' lies inside the k-l diamond

and thus can be represented as a sum Ok-l of distinct

divisors in the first k sets of divisors V V

O,

P' = P -2c and P' =

k ok-l' Solving for P we obtain

2 r = - c- l
P ok-l + 2ck Ok-l + (0k + dkv7) + (CR de7)-

This completes Case 2.

55

Case 5. 10Dk_l < 5y + 2x g lODk,

-4ck + lCDk_l < 5y + 2x < 10ok - lODk_l.

If we subtract ck from x and dk from y we obtain

a point P' with rectangular coordinates y' = y - dk and
x' = x - ck. applying Lemma 2 we obtain:

"I "I '-'~“ —" -. :

9y + 2x g 10Dk )dk 20k < 10Dk 10dk lODk-l°

applying Lemma 5 to the following, we obtain:

'r' ' -r“-" -" .1.
5y + 2x > lODk_l )dk 20k > lODk-l CODk-l lODk-l'

5y! - 2x' < 10dk - lCDk-l - Sdk + 2ck = 2ok + Sdk 'lODk-l
< 20Dk_l - 109k-1 = lODk-l’

5y. _ 2x. > -uck + 10Dk_l + ask - Sdk = lODk_l - 20k - Sdk
> lODk_1 - 20Dk_1 = -10Dk-1-

The resulting point P' 1163 inside the k-l diamond

and thus can be represented as a sum ok-l of distinct

divisors in the first k sets of divisors V V '°°,V

O’ l’ k-l'

P' = P - c - de7 she P' =

k Solving for P we

ok-l'

obtain P = Ok-l + 0k + de7. This completes Case 5

and the theorem is proved.

COROLLARY. If x E y - O modulo 2, then the integer

x + yJ? belongs to A(7).

3h

CHAPTER III

REAL QUADRATIC FIELDS '
m > 7 and m E 2 or 5 mod 4

In this chapter we will focus our attention upon
the integers in the quadratic fields R(m) where m is
square free, greater than seven and not congruent to one
modulo four. We will show that if the norm of ul [N(ul)]
equals -1, then the number T = 2tvn belongs to A(m) and if

= +1, then the number T = 2t+l¢m belongs to A(m)
t+l

N(u1)
where t is chosen so that 2t § a2/al < 2
Let us separate the divisors of T into the sets

V0,Vl,°°' where VO = 80 and Vk = 2rsh+1 where h g 0,

0 g r g t and k = h(t+l) + r + 1. In every set Vk there

exists exactly one divisor ck + dkvm with both ok and dk

non-negative. A divisor of T belongs to at most one set.

‘ I I = +
We will make use of the fact that ak alak_ l blbk-l m

= +
and bk albk-l blak-l° We will first prove a series
of lemmas for the fields where N(ul) = ~l.

t

LEMhA l. 2 > a1 + l/2a1.

. t _ 2 2 _ 2
PROOF. 2 > a2/2al - (a1+b1m)/2a1 - (2a1+1)/2a1.
LENLIMX Z. bl/al > bk/ak for k > 10
2

,, 2 _ _ _
PROOP. Since a1 - blm l and bk-l > O, we have

2 2
0 > (8.1 "' blm)bk_lo

35

2 _ . 2
alblak-l + blbk-lm > a1018191 + albk-l°

bl(alak-l + blbk-lm) ’ al(blak-l + albk-1)'

bl‘k > albko
LEMMA 5. ak/bk < 5a1/b1 for all k.

2~b2m+2s‘:-) + 2a

PROOF: o < bk_1(a1 1

ibi‘k-1°

+ Dab

2
blak-l 1 k-lm < 5aibk-1 + Balblak-l'

a1

bl(alak-l + blbk-lm) < 581(aibk-1 + blak-l)’

blak < 5albk.

LEMMA 4 (2b a -b )a /a < 5(2t+l-l)b for k>1
' ° 1 1 1 k-l 1 k-l '

PROOF: By Lemma 5 we know that blak-l < aalbk-l'

Since 2al - l > O, we may multiply both sides by 2a -1,

l

. . 2 2
and obtain (2alol ol)ak-l< 5(2al-al)bk_l< 5(2a1-a1+l)bk_l.

Since al > O, we may divide both sides by a1 and obtain

, t+l
(2blal-bl)ak_l/al < 5(2alf% l)bk-l < 5(2 -l)bk_1.
l
Lshhn 5 b + b a < M2t+1 -1)s when k 1
° k 31 k = k-l ? °
1 .
PROOF: If k = 2 we have that bk + flak =
. a1
, 2 _ -t t+1_
dalbl + b1(2al + l) - Libl(al + l ) < h(2 )bl < h(2 1)b1.
3'; Hal

Let us assume that the lemma is true for k g n-1 and

prove the lemma by induction. Let k = n.

56

For k = n we have that bn + b a is equal to

..l n
81
_ 2
a'lbn-l + blan-l + b y‘al n- -l+ bl bn-lm) — bn-1(al+blm) +
81 81
dtklan-l = bn_l(2a1tal) + 2blan_l. By Lemma 1 the right
1
s . t+1. , _ t+1_
138u1d side is less than 2 bn-l + ablan-l - (2 l)bn-1 +
t51_1 +blan~l + an_l(2albl-bl)/al. Using Lemma h and the

ET'
1
idmduction hypothesis we find that the right hand side is

.< (2t+1-l)bn_l + h(2t+l-1)Bn_2 + 3(2t+1-1)bn_1. The

Zlatter is exactly h(2t+1- -l)i:sn.l which completes the proof.

to“ -r r t+1 ~ r
l
PROOF: If k = 1 and r = 1 the lemma is true. When
I? = l and k > 1, we use Lemmas 5 and 5 to see that

t+1 . _ t+1_

fabk + 2b1ak < h(2 -1)sk_l + bk + 5bk - h(2 1)Bk-l +
T1

l“bk. Thus the lemma is true for all values of k when r =1.

lLet us assume that the lemma is true for r g n-1 and prove

the lemma by induction. Using Lemma 5 and the induction

n - n-l n- --lb
hypothesis: 2 bk + 2 k - 2(2 ka + 213k) g
181 ElD1
+ - , -1 -1 _
tiat 1 - 1)sk_1 + 4(2n1-l)bk + 2n bk + 2n (5bk) -
t+l n .
#(2 - l)Bk_1 + h(2 - l)bk. This completes the proof

of the lemma.

37

k
Let Dk = §Egdi and s = bl. We can now summarize
a1
thee results of Lemmas 1 through 6 as follows:

LElUIVU't 70 d + SC

k g “Bk-l and s > dk/ck for k > 1.

k
We now turn our attention to the fields in which
N(L11) = +1 and prove the analogous lemmas. We will let
8 == Wm)"l and note that s > bk/ak for all k
LEMIJLA 80 2t > 8.1 "' 1/281.

t

. 1 r _ 2 2 , _ r 2 \
PROOP: 2 > a2/2al- (a1+blm)/2al - (dal'l)/2al.

LEMMA 9. sak <gbk for k > O.

. 2 2 _ 2
PROOF. ak - bkm - 1 < Ebkm.
2 2
ak <.2bkm.
a
ak < 5bk\/m .
2
LFHMA.10 2a2 - a < (2a2 - a - ’)
“ ' l l g 1 1 .3 °
5
PROOF: Since m g 10 we have that ai = l + mbi > 10

which implies that al > 5 and o < 2a? - al - 4- Thus

2 , 2 .
re 2 l" 2 I ‘
2a - a < (2a - a - 5).
1 l .3 1 1 3

q . t+l
LbdhA ll. bk_1/al + (2al-l)sak_1 < 5(2 -l)bk_1 for k>l.

PROOF: Making use of Lemmas 9 and lo, we have:

37

k
Let Dk = §Egdi and 8 2.3l. We can now summarize
81
the results of Lemmas 1 through 6 as follows:

LEMMA 7. d + sc

k g uDk-l and s > dk/ck for k > 1.

k
We now turn our attention to the fields in which

N(u = +1 and prove the analogous lemmas. We will let

1)
s = (Jm)-l and note that s > bk/ak for all k

LEMMA 8. 2t > a1 - l/2al.

., w t _ 2 2 , _ r 2 .
PROOP: 2 > a2/2a1- (al+blm)/2al - (2a1-1)/2a1.

LRMMA 9. sak <gbk for k > O.

PROOF: a

KN

2 _ 2
- bkm — l < Ebkm.

< ibim o
d

KN

ak < gbkvm.

._H. , 2 _ . 2 _ _ ~
Lehmn lO. 2&1 al <g(2al a1 3).

= 1 + mbi > 10

HNW

PROOF: Since m g 10 we have that a

which implies that al > 5 and O < 2a? - al - 4. Thus

2 , 2 .
hal - 2al < cal - 5al - a.
, 2 2 -
2a - a < (2a - a -5}.

1 1 g, l l 5

‘1‘... - t+l-
LadmA 11. bk_l/al + (2al l}sak_l < 5(2 l)bk_l for k>l.

PROOF: Making use of Lemmas 9 and 10, we have:

58

2 . x ,, 2 . 2 .
S‘Zal-al)ak-l < 9/4‘3a1'al7%)bk-l < 5(231'a17%)Dk-1' If

we divide both sides of the inequality by a1, we obtain

- l bk-l‘ Using Lemma 8

s(2a1-1)ak_l < 5(2al-lfé )bk_l a
l l
we have bk-l/al + s(2al - l)ak_l < 5(2al - 1 - l)bk_l <
a1
-ht+l
5‘4 l)bk_lo
LEMMA 12 b + sa < m2“l - 1)s for k > 1
o k k = _ k']. 0

PROOF: When k = 2 we have that b2 + sa2 =

2 2

, 2
2albl + s(al+b1m) 2albl + s(l+2blm) = 2 b

al 1 + ZbEJm + s
= 2a1bl(ltb£¢m) + s < haltl + s = hbl(al- l ) + 2b1 + 3.

al

a1 a1

Applying lemma 8 we see that the right hand side is

t+l

t . t .
<: abl(2 ) + 2bl/al + s < 401(2 ) +.5s < nb1(2 - 1).

This proves the lemma for k = 2. Let us assume that the
lemma is true for k g n-1 and prove the lemma by induction.
bn + 88n z a1°n-1 + blan-l + 8(alan-l+blbn-lm) z
bn_l(al+blvm) + (bl+sal)an_1 < 2albn_l + Zsalan_l.

By Lemma 8 the right hand side is less than

t+l _ t+l_
a a
l 1
as + s(2a -l)a = (2t+l-l)b + b + as +
n-l l n-1 n-1 n-1 n-1

bn-l/al + (2al-l)san_l. Making use of Lemma 11 and the

39

induction hypothesis we find that the right hand side is

t+l

nt+1 _ , _ t+l_
less than (2 l)bn_l + 4(2 1)Bn_2 + 3(2 l)bn

-1

= M2“

-l)bn_l which proves the lemma.

t+l

1.” r r a _ . _
Lanna 15. 2 bk + 2 sak g 4(2 l)bk~l + m2r 1)bk

for all r when k > 1 and r > 1 when k = 1.

PROOF: Lemma 9 tells us that sal < bl which
implies that hbl + hsa1 < n°5bl and the lemma is proved

for k =1 and r = 2. If k > 1 and r = 1, then we have‘

0 ,t+l_ - t+l_
20k + 288k < 4(2 l)Bk_l + bk + gbk < 4(2 1)Bk-l + hbk.

Let us assume that the lemma is true for r = n-l and prove
the lemma by induction. Using the induction hypothesis and

n-1

n ,n “ ,n-l
Lemma 9 we have 2 ok + 2 sak - 2(2 bk+2 sak) <

u<2t+1-1)sk_l + u(2“'1-1)bk + 2n‘1bk + g(2n'1)bk <

t+1

. . n-l ‘ n-l -
4(2 -l)3k_l + 4(2 -l)ok + u<2 )bk -

4(2t+1-l)bk_1 + 4(2n - l)bk. Thus the lemma is proved.

We can now summarize the results of Lemmas 8 to 15
as follows:

LEMMA 14. (1k + sck g th-l’ and s > dk/ck'

The first part of the lemma is true only for k g 5.
We will now proceed to show that the units do not
belong to AK“) and T does belong to Aim). It is obvious

that if bl # l, the units do not belong to A, for bl bk

1+0

implies that the coefficient of (m in any sum of units is
congruent to zero modulo bl' We make use of the following
lemma to show that the units do not belong to A(m) if b = l.

l
K-
LEMMA 15. ak > 2221a1 + 3 for k > 1.

t3!

PROOF: ai = :1 + m g m - l g 9 implies that algfi.

If k = 2, then a2 = a + blm > 2&1 + 3. Let us assume

that the lemma is true for k = n-1 and prove the lemma by

m-L
induction. a - alan-l + b m > Ban-l > 2an_l + 2§a1 +5

n n-l

'0'!
= 223 + 50
LN i

THEOREM 1. The units do not belong to A(m).

PROOF: We have shown that when b1 # l, the theorem
is true. By Lemma 15 no unit in Sj for j > 1 can be used
to represent the number two. It can be seen by inspection
that the number two cannot be represented using only units

in S and $1. Therefor the number two cannot be written

0
as a finite sum of distinct units.
THEOREM 2. The number T belongs to the set Ann).
PROOF: We may consider every integer P of D(m) as
a lattice point P in the plane with rectangular coordinates
x and y. We will fill the plane with concentric diamonds
with centers at the origin and whose diagonals are the
coordinate axes. The y intercepts of the sides of the

k diamond are 2Dk and --2Dk and the x intercepts are 2Dk/s

and -2Dk/s. For every point in the k diamond we have:

g y + sx g 2Dk,
-2D g y - sx § 2Dk.
Let V-l be the set 1,2,u,°°°, T ,Jm,2¢m,---,T .

m

It will be shown that every integer lying in the k diamond
can be represented as a sum of distinct divisors in the

first k+l sets V_1,V1,Va,~--,Vk. Since the divisors in

Vj and the diamonds are symmetric with respect to both
coordinate axes there is no loss of generality by considering
only points in the first quadrant on the positive x axis

and on the positive y axis. If N(u1) = -1, then we have

.t+l

2t > a1 and a - 1 > 2a - 1 > 2b1 - 1 > 2b - 1.

l 1

Thus every integer in the first diamond can be written as

a sum 01 of distinct divisors in the sets V_l and V1.

, a ,t+2
If h(ul) = +1, we have a - 1 > ual - 1 > ubl - 1 >

s
nt+2
ubl - l and 2al + 2 - l > 6al - l > 6bl - 1. also
8
as + 2t+2 1 6 , '
l - > bl - 1. Thus every integer in the

second diamond can be represented as a sum 02 of distinct

divisors in the sets V_1,Vl and V2.

Let us assume that every integer P' = x' + yivm
in the first k-l diamonds can be represented as a sum of

distinct divisors c of T without using any divisors in

k-l

1.42

the sets Vj for j > k-l and prove the theorem by induction.
Let P be any point (x,y) in the k diamond. If P lies in
the k-l diamond, then the theorem is proved. If P is not

in the k-l diamond we have 2Dk-l < y + sx g 2Dk’ x g 0

and y g 0. We divide the possibilities into three cases.
In Case 1 the point lies on or above the line y - sx =

2dk - 2D .In Case 2 the point lies on or below the

k-l

line y - ex = -2sck + 2Dk-1' In Case 3 the point

lies between these two lines.

Case 1. 2Dk-l < y + sx g 2Dk,
y - sx g 2dk - 2Dk-l’
x g 0, y g 0.
If we subtract 2dk from y we obtain a point P'
with rectangular coordinates x' = x and y' = y - 2dk.
y' + sx' g 2DK - 2dk = de-l'
y' - sx' g 2dk - 2DK_l - 2dk = 'ZDk-l'

Since x' = x g 0, the point P' lies in the K-1
diamond and P' = Ok-l' Since P' = P - 2kom, we have
P = ok-l + 2dkdm = Ok-l + (0k + dk/m) + (-ck + dk/m) and

the theorem is proved for Case 1.

Case 2. 2D < y +sx § 2Dk’

k-l
y - sx g 2Dk_l - deck,
y g 0, x g 0.

#3

If we subtract Zak from x we obtain a point P'

with rectangular coordinates x' = x - 2ck and y' = y.

l + l 9 - " O _ “‘ = “P
y sx é “Bk dsck < ‘Dk 20k (“k—1'

'- -’ ' :.
y sxf g 2Dk-l 28ck + 230k 2Dk-l'
Since y' = y g 0, the point P' lies in the k-l
diamond and P' = ok_1. From P' = P - 20k we obtain

P = ok-l + 20K — Ok-l + (0k + dkvm) + (0k - dkvm) and

the theorem is proved in Case 2.

Case 5. 2D

k-l < y + 8x g 2D

1,"

-2sc + 2D

k < y - sx < 2dk - 4D

k-l k-l'

If we subtract d from y and c from x we obtain

k k

a point P' with coordinates x' = x - CR and y' = y - dk’

making repeated use of Lemmas 7 and 14, we obtain:

y' + sx' é 2DK - d < 2Dk _ d - d = 2D

k ‘ sck k k k-l’

,-

y' + sx' > de-l - dk - sok g ‘Dk-l - “bk-l =n2Dk_1.

y' - sx' < 2d -2D

k l - d + sc g -2D + ”bk-l = 2D

k- k k k-l k-l’

- d + so -‘Dk-l'

, _ , _h ,
y sx p> csck + de-l k

k 2
Thus the point P' lies inside the'k-l diamond

and P' = Since P' = P - c - dKVm, we have that

°k-1' k

P = 0‘ + (c

K-l + kom) and the theorem is proved in Case

R

5. This completes the proof of the theorem.

CHAPTER IV

REAL QUADRATIC FIELDS
E l modulo h

In this chapter we will focus our attention upon
the integers in the quadratic fields R(m),where m is
square free, greater than zero and congruent to one modulo
four.

The domain of integers D(m) of R(m) is the set of
numbers of the form x + yvm, where x and y are both

rational integers or halves of odd rational integers.

The units are those numbers of D(m) where x2 - my2 = +1.

Each field has an infinite number of units and one basic

such that if u = a

unit 111 l

1 + bivm.is the unit of

smallest absolute value for which both x and y are
positive, then every unit in the field can be written

tu? for n = 0, fl, :2, --° .

Let uj = ui for j = O, 1, 2, °°° . In this way we

= [+1, -1] of two
-1
ui’uil]

may divide the units into one set SO

units and infinitely many sets S1 = [u1, -u1,-

of four units each. Every unit belongs to exactly one set.
DEFINITION: A(m) is the set of integers {n} in

the field R(m) which have the following property; n belongs

to a(m) if and only if every integer in the domain D(m)

can be written as a sum of distinct divisors of n.

We will show that for all m the set A(m) is not

#5

empty and that for m = 5 every integer belongs to A w‘

m = 5
In this field ul =(1 + V5)/2, u2 = (5 +-¢5)/2,
u} = 2 + V5, °°' . Let a1 be the rational part of 111 and

. 7.- _ k
b1 be the coefficient of J) in ui. Also let Bk - fElbi.

The integers of D(B) may be considered to be those
points in the plane for which x and y are rational
integers or halves of odd rational integers. We will
fill the plane with concentric squares with centers at
the origin, whose diagonals are the coordinate axes and
whose sides are given by the equations x + y = :2 B

k
and y - x = :ZBR. It will be shown that every integer

lying in the k square can be represented as a sum of

distinct units belonging to the first k + 1 sets S

000’s

0’ k°
We will make use of the following equations:
ak = (Bk-l + 5bk_1)/2 and bk = (ak-l + bk_l)/2.
1.13le% 10 8k + bk é “Bk-.10
PROOF: The lemma is true for k = 1. Let us
assume that the lemma is true for k = n-1 and prove the

lemma true by induction. Let k = n.

8n + bn = an-l + 3bn-l = an-l + bn-l + an-l <

th-2 + 2bn-l < uBn-l'

hb

- ‘ " P +
LblVlflA do ak bk > 2B

k-l'

PROOF: The lemma is true for k = 2. Let us
assume that the lemma is true for k = n-1 and prove that
the lemma is true by induction. Let k = n.

a + b = a
n

n + 5b = a

+ '
n-l bn + 2b

n-l -1 n-l

= 2B

> 2311- + 2b 11-10

2 n-l
THEOREM 1. The units belong to A(5).
PROOF: We can see by inspection that every integer
in the first square can be represented as a sum of

distinct units in S0 and 81' Let us assume that every

integer in the first k-l squares can be written as a sum
of distinct units taken from the first k sets of units,

SO’ 81’ °°',Sk_1 and prove the theorem by induction.

Consider any point inside the k square. Since the square and
the sets Sk are symmetric with respect to both axes

there is no loss in generality in considering only points

in the first quadrant, on the positive x axis and on the
positive y axis. If the point lies inside the k-l square,
the required representation is assured by the induction
hypothesis. Therefore we may consider only points in the
first quadrant, in the k square and not in the k-l square.

Thus we have; 25 < x + y g 23

k-l

y g 0, x

IN
0
o

1+?

We divide the possibilities into three cases. In

Case 1 the point lies on or above the line y - x = 2b ~28

k k-l'

In Case 2 the point lies on or below the line

y-x = -2ak + ZBk-l° In Case 5 the point lies between

these two lines. Lemma 2 tells us that Case 5 actually

exists.

Case 1. ZBk-l < x + y g 2Bk,

y - x g 2bk - 2B

k-l’
x g 0.
If we subtract 2bk from y we obtain a point P'
with rectangular coordinates x' = x and y' = y - 2bk’

x' t y' g 2B - 2bk = 23

k k-l’

y' - x' g 2bk - Zbk-l - 2bk = '25k-1’
x' g 0.
The resulting point P' lies in the k-l square and
thus by the induction hypothesis it can be represented as
a sum Ok-l of distinct units in the first k sets of units

SO, 31' °°', Sk-l' P' = Ok-l and P' = P - ZbEVS' Thus
P = Ok-l + 2bkv5 = ok-l + (ak + bk\/5) + (-8k + 1319/?)-

This completes Case 1.

Case 2. 25k~1 < x + y g 28k,

_ -2ak + 2Bk-l’

¢<
l
N
IA

1+8

If we subtract 2a from x we obtain a point P' with

k
rectangular coordinates x' = x,- 2ak and y' = y.
' ' r c n - =
x + y g 28k 2ak < 23k 2bk 2Bk-l’
y' - x' g --2ak + dBk-l + Zak = ZBk-l’

y'g00
The resulting point P' lies in the k-l square and
thus by the induction hypothesis it can be represented as

a sum Ok-l of distinct units in the first k sets of units

and P' = P - 2a . Therefore

3 S P' = k

000’s ok—l

(ak + kas) + (ak - bkyt). This

0’ l’ k-l'

+

P = Ok-l + Zak = Ok-l
completes Case 2.

Case 5. ZBk-l < x + y g 23k,

2B - 28 < y - x < 2b - EB

k-l k k k-l'

If we subtract ak from x and bk from y, we obtain

a point P' with coordinates x' = x - ak and y' = y - by.

I 0 - c. ' - :7.
x + y g 2Bk bk ak < 25k 2bk ZBk-l’

Applying Lemma 1 we obtain the following inequalities:
x' + y' > ZBk-l - ak - bk g -2Bk_1,

y. - x' < 2b - 2 25

k Bk-l k

”A

- b + ak = ak + bk - 28k_1

k-l’
y' - x' > aBk-l - Zak - bk + ak = ZBk-l - ak - bk g ~2Bk_l.

The resulting point P' lies in the k-l square and

h9

thus by the induction hypothesis can be represented as a

sum Ok-l of distinct units in the first k sets of units

" coo '= '= - u- I
S , S P ok-l and P P ak bky5.

o' 31' k-l'

Solving for P we obtain P = o + a + kaS. This

k-l k

completes Case 5 and the theorem is proved.

m > 5
In this section we will show that if N(ul) = -1,
then the number T = 2t-1(1+Jm) belongs to A(m) and if

N(ul) = +1, then the number T = 2t(1+¢h) belongs to A(m),

t t+l

where t is chosen so that 2 § a2/al < 2 . In either

case 2t divides T.
Let us separate the divisors of T into the sets

V = 30’ and V

.. P ..
0 - 2 sh+l where k - h(t + l) + r + 1,

k

h g 0, and 0 g r g t. In every set Vk there exists

+ dkvm with both c and d non-

exactly one divisor c k k

k
negative. A divisor of T belongs to at most one set.

alak-l + blbk-lm‘

We will make use of the fact that ak =
and bk = blah-l + albk_1. We will first prove a series

of lemmas for the fields in which N(ul) = -1.

LEMMA 5. 2t > a1 + l/2a1.

t _ 2 2 _ 2
PROOF: 2 > a2/2a1 - (a1 + blmJ/Za1 - (2al + l}/2a1.

LEMMA u. bl/al > bk/ak for k > 1.

SO

PROOF: Since ai - him = -l and bk-l > O, we have

0 > (a? - blm)bk-l°

b2b m > a b a

2
alblak-l + 1 k-l 1 1 albk-

k-l + 1'

+ a b

+ b b m) > al(blak-l l k-l)o

bl(alak-l 1 k-l

blak > albk.

Lawns a. ak/bk < 5al/b1 for all k > 1.

2 2 2
PROOF. O < bk_l(al - blm + 2&1) + 2a1blak_l.

2 ' 2
alblak-l + blbk-lm < aalbk-l + 5a1b

lak-l'
b1(“13k-1 + lek-lm) < 531(“1bk-1 + blak-1)°

blak-l < 5a1°k~1°
Lfimma 0 (2b a -b )a /a < 5(2t+1-l)b for k>1
' 1 l l k-l l k-l '

PROOF: From Lemma 5 we have blak-l < 5a1bk_1.
because (2al - l)/al > O, we may multiply both sides of the

above inequality by (2al - l)/al and obtain the following.

t+l _ 1)b

(2a1bl-bl)ak_1/al < 5(2al + % - l)bk-l < 5(2 k-l'

l
LEMMa 7 b + b ak/a s m2t+l - 1)s for k > 1
' k l l - k-l -
PROOF: When k = 2 we have that bk + blak/a1 =

. 2 _ , t t+1_
+ 01(2a1+1)/a1 - Lib1(al + 18‘ ) < h 2 b1 < h(2 1)b1.
1

2albl

Let us assume that the lemma is true for k g n - 1 and

51

prove the lemma true by induction. Let k = n.

bn + blan/al = albn-l + blah-l + b1(a1an-1 + blbn-lm)/al

_ 2 -
- bn_l(al + blm) + Zblan-l - bn-l(2a1 +‘% ) + ablan-l'

a1 1
By.Lemma 5 the right hand side is less than

2t+lb + 2b 2t+1

n-l lan-l = ( ’1)bn-1 + bn-l + blan-l/al +

an_l(2a1bl - bl)/al. applying Lemma b and the induction

hypothesis we find that the right hand side is less than

t+l t+l t+1
(2 - 1)bn_l + u(2 - 1)Bn-2 + 3(2 - l)bn_

l.

l - .r. r t+l r
LsMMA 8. a bk + 2 blak/al g u(2 -1)sk_1 + h(z -1)bk.
PROOF: The lemma is true when k =n2 and r = 1.

When r = l and k;> 2, we apply Lemmas 5 and 7 to obtain
, , t+1
2bk + 2blak/a1 < 4(2 - 1)sk_1 + bk + 5bk

t+1 .
11(2 dusk”1 + hbk.

Thus the lemma is true for all values of k > 1 when r = 1.

Let us assume that the lemma is true for r n - l and

"A

prove the lemma true by induction. Let r = n. Applying
Lemma 5 and the induction hypothesis we obtain

n _ . n-1 n-1
2nbk + 2 blak/al - 2(2 bk + 2 blak/al) g

11(2t+l - 1)sk_1 + h(an'l - 1)bk + en'lbk + 2n‘1(3bk) =

“(2t+l - l)bk_1 + 4(2n - l)bk. This completes the proof.
Let Dk = (11 and s = bl/al. We can now summarize
i=1

52

the results of Lemmas 5 through 8 as follows:

LEMMA 9. d sc g “pk-l and s > dk/ck for k > 1.

k + k

we now turn our attention to the fields where
N(u1) = + 1 and prove the analogous lemmas. We will let
s = (Vh)'1 and note that s > bk/ak for all k.

t

LEMMA 10. 2 > al - l/2al.

.A.,. t = 2 2 = 2 _
PROOF. 2 > a2/2al (a1 + blm)2al (2al 1)/2al.

LEMMA ll. sak < 5bk/2 for k g l.
. 2 _ 2
PROOF: bkm - 1 < 5bk m/u.

spm

a < 9bim/u-

KN

2
LEMMA 12 2a2 - a < (2a2 - a -‘3)
° 1 1 g 1 1 '
5
... 2 _ 2
PROOF: For m g 15, al — l + mb1 > h implies
a > 2 which in turn implies O < 2a? - al - h. Therefore
2 . 2
gal - 2al < oal - 5al - h.
2 2

t+l

LhMMA l5. bk-l/al + (2a1-l)sak_l < 5(2 -l)bk_l.

PROOF: Applying Lemmas 11 and 12, we obtain:

3(2ai - al)ak_1 < E(2a§ - a1 - %)bk-l < 3(2af - a1 - %)bk-l'

Dividing both sides of the inequality by a1, we obtain:

s(2al - l)ak~l < 5(2a1 -l -‘% )bk-l - bk_1/a1.
1

53

Applying Lemma 10 we obtain:

 

t+l
bk-l + s(2al -1)ak_l < 5(2al - % -1)bk-l < 5(2 -1)bk_l.
a1 1
LEMMA 14 b + as < 11(2t+l - 1)B for k > 1
' k k = k-l '

PROOF: When k = 2, we have that b2 + sa2 =

2a1bl

, 2 2 r 2 2
2albl + s(a1 + blm) = 2albl + s(l + 2b1m) + 2bIVh + s.

= 2albl(l + b1 ) + s < dalbl + s = Libl(al - l ) + 2b + s.
als
Applying Lemma 10 we see that the right hand side is <

t+l

sol(2t) + 2bl/al + s < sb1(2t) + 53 < ab1(2 - 1).

This proves the lemma for k = 2. Let us assume that the
lemma is true for k g n-1 and prove the lemma by induction.

bn + 83n = a1bn-1 + blan-l + 8(alan-l + blbn-lm) =

bn_l(al + bIVh) + (D1 + sal)an_1 < 2albn-l + 2salan_l.

Applying Lemma 10, the right hand side is less than

Ft+l . t+l
: - +
(2 + % )bn-l + 2salan_1 (2 l)bn_l +(l +‘% )bn-l
l 1
__ r t+1 - ‘
Ban-l + s(2al - 1)an__l — (2 l)bn_l + bn-l + Ban-l +

bn-l/al + (2al - 1)san_l. Applying Lemma 15 and the

induction hypothesis, the right hand side is less than

- + t+1 t+‘
(2t 1 - l)bn_l + 4(2 - l)bn-2 + 5(2 * - 1)bn =

~l

. . +
uk2t l - l)bn_l. This completes the proof of the lemma.
»..v _ .r r -t+l 2r
anmA 13. 2 bk + 2 sak g 4(2 -1)bk_l + h(2 -l)bk

for all r when k > 1 and r > 1 when k = l.

5h

PROOF: Lemma ll tells us that 321 < bl which
implies that hbl + heal < h(5bl) and the lemma is proved
for k = l and r = 2. If k > 1 and r = 1, then we have

t+1 t+1
2bk + 2ssk < u(2 - 1)Bk-l + bk egbk < h(z -1)Bk_1 + hbk.

Let us assume that the lemma is true for r = n-1 and prove
the lemma by induction. Let r = n. Applying Lemma 11
and the induction hypothesis we obtain:

2nb + 2nssk = 2(2n‘1b + 2n‘lsak) <

k
t+l

k
4(2

- 1)sk_1 + h(an‘l - 1)bk + 2n'1bk +£(2n‘1nk

< M2t+l - 1)Bk_1 + 11(2’1”1 - 1)bk + h(zn‘1)bk =

t+l n
h(2 - new1 + h(z - 1)bk.

We can now summarize Lemmas 10 to 15 as follows:

LEMMA 16. :1k + sok g th-l and s > dk/ck for k g 3.

We will now proceed to show that the units do not
belong to A(m) and that the integer T does belong to A(m).
It is obvious that if b1 # 1/2, then the units do not

belong to A(m). We make use of the following lemma to

show that the units do not belong to A(m) when b1 = 1/2
k-l
LEMMA 17. ak > 2 a1 + 5 when k > 1, m g 21
=1

and b1 = 1/2.

55

im : l g m/u - l > h implies al > 2.

a2 = a: + bim > 2al + h > 2al + 5. Therefore the lemma is

true for k = 2. Let us assume that the lemma is true for
k g n-1 and prove the lemma true by induction. Let k = n.

For m g 21 we have the following inequalities:

Li - m < ~11 < ~Li/mb;_l.

2 2 2
L‘Imbn-l - m bn‘1 < ”(+0
" 2 2 2 .2 2
dag-l - bn-lm = ”an-l - Limbn_l + Limbn_l - mzbfr1 < h-h = O.
&n_l < (bn_lm)/20

a = Elan-l + blbn-lm > 2an-1 + (b

n m)/2 > 2am

n—1 -1 + an-1°

Applying the induction hypothesis to the right hand side '
m-a. n-y
we obtain: an > 2an_l + 22ai 4- 5 = 22311 + 5.

L3: L5]
THLORBM 2. The units do not belong to A(m).
PROOF: If D1 = 1/2 and m g 21, then Lemma 17 tells
us that no unit in SJ for j > 1 can be used to represent
the number two as a sum of distinct units. It can be

seen by inspection that the number two cannot be written

and S .

as a sum of distinct units using only units in S0 1

«a
> 2§:a

when k = 15 we have that a
k ...1

+ b for k > 5. Thus

no unit in Sj for j > 5 can be used in the representation
of the number five. By inspection it can be seen that
the number five cannot be represented using only units in

30’ 31: 32 and S}. This completes the proof.

THEOREM 5. The number T belongs to A(m).

PROOF: ,We may consider every integer P of D(m) as
a point P in the plane with rectangular coordinates x and y
where x and y are either integers or halves of odd integers.
We will fill the plane with concentric diamonds whose
diagonals are the coordinate axes. The y intercepts of the
sides of the k diamond are : 2Dk and the x intercepts are
:ZDk/s. For every point in the k diamond we have:

k g y + sx g 2Dk’

I
n)
U
A

k=y-SX§2Dko

Let V be the set of all divisors of T which are not

contained in V for some 3 > 0. It will be shown that

3
every integer lying in the k diamond can be written as a

sum of distinct divisors of T contained in the first k + 1

sets v, v v --~,v

1’ 2’ k'

The divisors in V3 and the diamonds
are symmetric with respect to both coordinate axes. Thus
there is no loss in generality by considering only points
in the first quadrant, on the positive x axis and on the
positive y axis.

Let us first consider a representation for every
integer in the first diamond in the case where N(u1) = -l.

2t+1

Because - l > 2al - l > 2b1/s - 1 > 2bl - 1, every

rational integer between 0 and 2b1 and between 0 and 2bl/s

can be written as a sum of distinct divisors of 2t. Thus

57

every integer in the first diamond can be written in the
form: 25.2”(1 +\/m)/2 1%:22.

Let us now consider a representation for every

integer in the second diamond in the case where N(u1) = +1.
2t+2 - l > hal - l > hbl/s - 1 > hbl - l.
2al + 2t+2 - 1 > bal - 1 > 6bl/s - 1.
2bl + 2t+2 - 1 > 6bl - 1.

Applying the above inequalities we see that every

integer in the second diamond can be written either as
2r(1 +‘vsi/2 t 22 or as (2r + 2b1)(l +-¢h)/2 t 2z :261-
Therefore when N(ul) = - 1, every integer in the

first diamond can be written as a sum 01 of distinct

divisors in the sets V and V1. When N(ul) = + 1, every

integer in the second diamond can be written as a sum of

distinct divisors in the sets V, V and V .

1 2

Let us assume that every integer P' = x' + ytvm
in the first k-l diamonds can be represented as a sum °k-1
of distinct divisors of T without using any divisors in
the sets V3 for J > k-l, and prove the theorem by
induction. Let P be any point (x,y) in the k diamond
where x and y are either rational integers or both halves
of odd rational integers. If P lies in the k-l diamond,
then the theorem is proved. If P does not lie in the

k-l diamond, then we have the following inequalities:

58

< y + sx § 2Dk,

x g 0, y g 0.
We divide the possibilities into three cases. In
Case 1 the point lies on or above the straight line
y - sx = 2dk - 2Dk-l'
In Case 2. the point lies on or below the straight line
y - ex = -23ck + 2Dk-l'
In Case 5 the point lies between these two lines.

Case 1. ZDk-l < y + sx é 2Dk,
y " 8X .2- 2dk -2Dk‘l,
x g 0’ y g 00
If we subtract 2dk from y we obtain a point P'

with rectangular coordinates x' = x and y' = y - de.

y' + sx' é 2D

k - de = 2D

k-l’
y' - sx' g de - 2Dk-1 - 2dk = -‘Dk-l’

X'=X§Oo
The resulting point P' lies in the k-l diamond and

thus can be written as a sum ok-l of distinct divisors in
the first k sets of divisors V, V1, °--, Vk-l’ P' - Ok—l’
P' = P - ZdRJm. Solving for P we obtain the following:

P = Ok-l + adkvh = Ok-l + (°k + dkvh) + ('°k + dkvm)'

This completes the proof of the theorem for Case 1.

59

Case 2. 2Dk-l < y + sx 5 2D

y - sx ; aDk-l - 2sck,

y g 0, x g 0
If we subtract 2ck from x,we obtain a point P' whose
rectangular coordinates are x' = x - 2ck and y' = y.
I V - .. :g
y + 8x g 2Dk 2sck < 2Dk 2dk 2Dk-l'

y' - sx' g 2Dk-l - 2sck + 25ck = 2Dk-1’

y'=yg0.

The resulting point P' lies in the k-l diamond and

thus it can be represented as a sum Ok-l of distinct

divisors in the first k sets of divisors V, V -°°,V

1' k-l'

P' = P - 20 . Solving for P we obtain:

°k-1' k

P-= ok-l + 20k = ak_1 + (ck + dka) + (ck - kom).
Thus the point P can be represented as a sum of distinct

divisors taken from the sets V, V1, 00°,Vk. This completes

the proof of the theorem for Case 2.

Case 5. ZDk-l < y + BK § 2Dk'

-28ck + 2Dk-l < y - sx < 2dk - 2Dk-l'
If we subtract dk from y and ck from x, we obtain
a point P' whose rectangular coordinates are x' = x - ck

andy'=y-dk.

60

Applying Lemmas 9 and 15, we obtain:

' V r - - b =
y + sx g 2Dk dk sck < 2Dk 2dk 2Dk-l'
' - - - ‘ = -

y' - sx' < 2dk - 2Dk-l - dk + sck g th_1 - 2Dk-l = 2Dk-l'
y' - sx' > -2sck + 2Dk-l - dk + sck = 2Dk-l - d - sc
2 2Dk-i ‘ “Du-1 = 'ZDk-1°
The resulting point P' lies inside the k-l diamond
and thus it can be represented as a sum Ok-l of distinct
divisors in the first k sets of divisors V, V1, "',Vk_1.
I: I: .- — o
P Ok-l' P P ck dkvm. Solving for P we obtain.
P = Ok-l + ck + dkvm. Thus the point P can be represented

as a sum of distinct divisors taken from the sets V, -°- ,

Vk' This completes Case 5 and the theorem is proved.

61

CHAPTER V
THE GAUSSIAN FInLD

In this chapter we will focus our attention upon the
domain of integers G in the Gaussian field. The integers
in C are numbers of the form x + iy, where x and y are
rational integers and i =‘J-1. The units of G are the
numbers +1, -1, i and -i. because we have only four units
in G it is impossible to represent every integer in G as
a sum of distinct divisors of any one integer n of G.
However we can define a set A of integers in G whose
properties are analogous to the properties of the
rational integers belonging to the set A defined in
chapter one.

DEFINITION: A is the set of all integers {n} in
G which have the following properties: n belongs to A
if and only if there exists a rational integer n' such
that every integer x + iy in'G satisfying the inequalities

-n' g x'g +n' -n' g y g +n'
and no other integer in G, can be represented as a sum
of distinct divisors of n.

If we represent every Gaussian integer as a
lattice point in the plans we can see that geometrically
we have a situation analogous to the situation in
Chapter I . In Chapter I we represented every

integer on the real line between -a and +a as a sum of

62

distinct divisor of a rational integer. Here we are
going to represent as a sum of distinct divisors of a
Gaussian integer every lattice point inside a square whose
center is at the origin and whose sides are given by the
equations, y = :n' and x = tn'.

LEMMA 1. If n belongs to A and d divides n, then
d is either real or pure imaginary.

PROOF: Let n be a Gaussian integer belonging to

A. Let dj = a3. + ibj be the divisors of n. IfZZaJ = n',

every divisor lying in the first or fourth quadrants or
on the positive x axis must be included in the summand.
If E bj = n', every divisor lying in the first or second
quadrants or on the positive y axis must be included in
the summand. Thus the integer n' + in' cannot be
represented unless every divisor lying in the first
quadrant is used twice.

THEOREM 1. A Gaussian integer n belongs to A if and
only if n is an associate of a rational integer having one

of the following factorizations as a product of primes:

i) n = 5t for all t g 0.
t k t3
ii) n = 5 17-pj with 5 < pr < ps for r < s, t g l
1‘1

   

“A
0

ti g l, p. f 5 mod A and pi-l

J
T-

fori = l, 2,...,k0
J

PROOF: By Lemma l,n must be a rational integer

and all divisors of n must be rational integers or the

associates of rational integers. If we factor n as a
product of primes in the Gaussian field, each prime factor
must be a rational prime which is also a prime in G. Thus
if n belongs to A, it has the form 1) or ii). If n has
one of these forms, then.Theorem 2 of Chapter I tells us
that every rational integer between -o(n) and +c(n) can be
represented as a sum 2:dj of distinct rational divisors of
n. Every number of the form iy for -o(n) < y < +o(n) can

be represented as ide, where the C11: are distinct

rational divisors of n. If we let 0(n) = n', every number

in the square can be written as Zdj + iidk = 2dj +£idk.

THEOREM 2. There exist arbitrarily large square
free rational integers in A.

_ _ _ k
PROOF. Let nl — 5, r12 -21 and nk fg1p3 be the
prouuct of the first k rational primes which are congruent
+ 2 or n

to 5 modulo 4. Either n + h is congruent to

k k
5 mooulo u and therefore divisible by a prime q which is
congruent to 5 modulo b. However (h,pj) = 1 implies that

(PJ,Q) = 1. Thus for k > 1 we have;

"A

q nk + h = nk + 5 + l < 0(nk) < 20(nk) + l.

“A

pk+1
When k = l we have, 7 < 8 + 1 = 20(5) + 1. Thus nk

belongs to A for all k.

6h

CHAPTER VI
ItaGINARY QUADRATIC FIELDS
m.# l modulo 4

In this chapter we will focus our attention upon
the integers in the quadratic fields R(m), where m is
square free, less than -1 and congruent to two or three
modulo four.

The domain of integers D(m) in R(m) is the set of
numbers of the form x + yvm, where x and y are both
rational integers. The only units in D(m) are t 1.

DEFINITION: A(m) is the set of all integers {n}
in D(m) which have the following properties: n belongs
to A(m) if and only if there exists a rational integer
n' such that every integer x + me in D(m) satisfying the
inequalities -n' g x g n' and -n' g y g n'
and no other integers in D(m) can be represented as a
sum of distinct divisors of n.

We shall identify with every integer x + me a
point P = (x,y) in the plane. An integer n in D(m)
belongs to A(m) if and only if every point in a square,
whose center is the origin and whose sides are parallel
to the coordinate axes, and no point outside of this
square can be represented as a sum of distinct divisors

of n.

By using exactly the same argument as that used in

65

Lemma 1 of Chapter V"we can prove the following lemma.
LEMMA 1. If n belongs to A(m) and d divides n,
then either d = x or d = yVh, where x and y are rational

integers.

m = -2
Every integer in D(-2) can be factored uniquely as
a product of primes in D(-2). Lemma 1 tells us that if

k
n belongs to A(-2), then n = ZtTT'pivm, where each p1 is
i=1

a rational prime which is also a prime in D(-2). Therefore

J
p E 5 or 7 modulo 8. Let n = ZtIT'p for l 5 J 5 k-
1 J 1:1 1 -

The largest positive rational integer which can be written
as a sum of distinct divisors of n is 0(nk). In order to

represent every rational integer between -a(nk) and +a(nk)

as a sum of distinct divisors of n, nk must satisfy
Theorem I of Chapter I. We can now prove the following
theorem.

THEOREM 1. n belongs to A(-2) if and only if

n = nkvm, where nk satisfies Theorem 1 of Chapter I and

p1 = 5 or 7 modulo 8.
PROOF: If n belongs to A(-2), Lemma 1 tells us

that nk must have the form ZtTTpi, where p1 5 5 or 7

modulo 8. In order to represent every rational integer

between -o(nk) and +o(nk), nk must satisfy Theorem 1 of

66

Chapter I. In order to represent every integer of the

form yV-2 for -o(nk) g y

"A

+0(nk) we must have n = nkV¥2

Conversely if n = nkV-Z, every integer in the square

can be written as Z d +(Z: d' \/-2. Because the
d/nk d'/nk

maximum value for each sum is 0(nk), no integer outside

the square can be represented as a sum of distinct

divisors of n.

THEOREM 2. A(-2) contains integers n with

arbitrarily large square free nk.

k
PROOF: Let nk = 211'p1, where p1 = 5, p2 = 7 and
i=1
R
TT‘p1 is the product of the first k primes which are

i=1
congruent to 5 or 7 modulo 8. In order to prove the
theorem we must show that nk satisfies Theorem 1 of

Chapter I for all k. If k =1 or k = 2, n satisfies this

k
condition. For k > 2 we have that at least one of the

numbers nk/Z + 2, nk/2 + h, nk/2 + 6 or nk/Z + 8 must be

congruent to 5 modulo eight and therefore must be
divisible by a prime (q) congruent to 5 or 7 modulo 8.
Because (q,2) = (q,h) =(q,b) = (q,8} = l, we have (q,nk)
=l. Therefore pk+l g q g nk/Z + 8 < 20(nk) + l and nk

satisfies Theorem 1 of Chapter I for all k.

67

m < - 2

In this section we will consider the quadratic
fields R(m), where m is square free, less than -2 and not
congruent to l moaulo a. This implies that m is congruent
to 2, 5, 6 or 7 mOdulo 8. Let h = -m. h is congruent to
l, 2, 5 or b modulo 8.

.THLOREM 3. If a + me divides ZrMm, then either
a or b must be zero.

PROOF: Let us suppose that a + me divides vam
and that neither a nor b are zero. This implies that

ad - mb2 divides 22r 2 2

m or that a + hb divides 22

rh,
where a, b, h and r are rational integers and r and h are

positive.

2

a + hb2

= 2th',
(1)
0

MA

t §;2r , h"h.
We may rewrite equation (1) as
hue/w)2 +(h/h'>b2 = 2t, (2)
where a/h' and b are non-zero'rational integers. The
left hand side of equation (2) is greater than h' + h/h'
> h' + h/h' g 4. Thus t g 3.
If either a or b is an even rational integer, then

the other must also be even and equation (1) reduces to

(a2/4> + h(ba/u) = 2t'2h'. (5)

where ad/h and bZ/h are rational integers. If either

68

a2/h or bd/n are even rational integers, we may continue
our method of descent and obtain (a/L.)2 + h(b/M2 = 2t-hh'
where a2/lb and ba/lb are rational integers. Continuing

in this way we obtain
(ea/28> + h(bZ/23> = 2t'3hv (u)

where either t - s < 5 or both (a2/28) and (ha/23) are
odd rational integers. Because of the results following
equation (2), we cannot have t - s < 5. Therefore a
solution to equation (I) with a and b even rational
integers implies a solution to equation (1) with a and b
odd rational integers and t g 5.

When a and b are both odd rational integers we have

a2 + hb2 E h + l E 2, 5, 6 or 7 mod 8.

Therefore there exist no solutions in non-zero rational
integers to equation (I) and thus the theorem is proved.
THEOREM u. T = 2t¢m belongs to A(m) for all t.
PROOF: Theorem 5 tells us that the only divisors
of T are powers of 2 and Jm times powers of 2. Every
rational integer between -o(2t) and +o(2t) and no other
rational integer can be written as a sum of distinct
QiViSOPS or To Thus every integer x + me in D(m) where
-o(2t) g x g +o(2t) and ~o(2t) g y g +o(2t)

can be written as Z td + m2:

td'. No other integer
d/2- d'/2

can be so represented.

69

CONCLUSION

In this dissertation the problem of characterizing
the sets A and A(m) has been completely solved for the
rational field and the fields where m = 2, m = 5, m = -l
and m = -2. In every other quadratic field except where
m is negative and congruent to one modulo four, there
exist infinitely many integers which do belong to A(m)
and infinitely many integers which do not belong to A(m).
However in these fields we do not have a complete.
characterization of the integers which do belong to A(m).
One of the many difficulties which arises is that if
a + bvm divides a given integer, then a - me need not be
a divisor of this given integer. Thus it is impossible
to separate the divisors into sets which are symmetric
to both the x and y axes.

When m is negative and congruent to one modulo

four the set A(m) is empty. This arises from the fact
that integers of the form x + y¢m, where x and y are
halves of odd integers cannot be represented using only
divisors of the form di and divm where the di are
rational integers.

Another area for further investigation is in
the study of the behavior of sums of distinct divisors

of integers in fields of higher degree.

2.

70

BIhLIOGHAPHY

Cahen, M.E., Sur un theorme de M. Stiel es, Comptes
Rendus des Seances deLLTAcademie dis c ences,

116 (1895). #90.

Hardy, G.H. and wright, E.M., An Introduction to
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