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Continued Fractions in Approximation and Number Theory
CONTINUED FRACTIONS
IN
RATIONAL APPROXIMATIONS
AND
NUMBER THEORY
by
DAVID C. EDWARDS
Thesis for Kaster of Science
Department of Mathematics
McGill University
@
David C. üiwards
lm
January 1971
TABLE OF CONTENTS
INTRODUCTION.
i
PART I.
1
Some Basic Results
1.
Definitions, Notations, and the Basic Formulas
2
2.
Convergence of Continued Fractions with Unit
6
Numerators
3.
Simple Continued Fractions
10
4.
The Fibonacci Numbers
14
5.
Miscellaneous Results
16
PART II.
6.
7.
Applications to Rational Approximations
Best Approximations to Real Numbers
19
1-
Introduction and Motivation
19
2.
The Main Theorems
21
3.
A Restatement
30
Hurwitz's Theorem and Related Results
PART III.
18
Applications to Number Theory
32
37
8.
The Number of Steps in the Euc1idean A1gorithm
38
9.
Periodic Simple Continued Fractions
40
,fD
49
10.
The Expans ion of
11.
The PeU Equation
12.
The SolvabUity of
54
x
2
- Dy
2
= -1
60
CONCLUSION.
65
BIBLIOGRAPHY.
67
INTRODUCTION
This thesis surveys the elementary theory of continued
fractions and discusses in detail some important applications of
continued fractions to the theory of rational approximations to
real numbers and elementary number theory.
Chapters 1 to 5
introduce continued fractions and their basic properties, and provide
the results on which the following chapters are based.
Chapters 6
and 7 develop the elementary theory of rational approximations,
culminating with Hurwitz's Theorem, and using continued fractions as
the fundamental tool.
The approach is essentially that of Khintchine
[4], Sections l and II, but the theorems are stated and proved in
greater detail, an attempt is made to clearly motivate the definitions,
and some closely related results from [6], Chapter 7, are included.
Chapters 8 to 12 investigate closely the applications of continued
fractions to the Euclidean Algorithm and to the Pell Equation
2
2
x -Dy = N.
This involves a thorough examination of periodic continued fractions,
in particular the simple-continued-fraction expansion of
a positive nonsquare integer).
Perron [8],
Ol~s
.rD (D
being
The material is drawn primarily from
[7], and Niven and Zuckerman [6].
As far as l have
been able to discover, the bound on the period given in Theorems 9.4
and 10.1 is an original result, although admittedly a minor one.
Theorems 12.2 to 12.4 (from Perron [8]) are significant and fairly deep
results, rarely found in discussions of continued fractions or the Pell
Equation.
i
ii
Most recent texts on number theory treat the theory of continued
fractions fleetingly, or with a single application in mind ; in contrast,
an attempt has been made in this survey to demonstrate the richness and
wide applicability of the theory.
-1-
PART
l
SOME BASIC RESULTS
-2-
Chapter 1.
Let
DEFINITIONS, NOTATIONS, AND THE BASIC FORMULAS.
F
be a field.
(al' b , a , b 3 , a ,···)
2
2
3
F such that, for each
A continued fraction in
F is a sequence
of an odd or infinite number of elements of
n?2, each denominator in the following composite
fraction is not zero :
(1)
b _
n 1
+---~a
The fraction (1) is called the
and denoted by c
c
n
= a1
+ bn
an
nth
convergent of the continued fraction,
It is usually written in the more convenient form
n
+
n-l
b
-
a
b
2
2
+
a
b
3
3
+
+a
(2)
n
n
Additionally, the first convergent is
If the sequence is finite, say
cl = al .
(al' b 2 , a 2 ,··· ,bm' am)
where
m > 1 , then the continued fraction is called a finite continued fraction,
and its value is defined to be the 1ast convergent, c
m
.
If the sequence
is infinite, then we have an infinite sequence of convergents c ,c ,c .•. ,
1 2 3
and (assuming a metric on
convergent or divergent in
F) the continued fraction is said to be
F according as
~cn
exists or not in
In the former case, the limit is taken as the value of the continued
F.
-3-
fraction.
Normally
When
F
is the real field.
b2 , b3 ,··· are all
1 , the continued fraction is said
to have unit numerators, and we use the notation
(al' l, a 2 , l, a 3 , ••• )
=
(3)
[al' a 2 , a 3 ,···]
For a finite or convergent continued fraction, we often use the same
notation for the continued fraction and its value.
c
n
= [al'
a ,···,a ]
2
n
Thus we write
for the convergents of a continued fraction with
unit numerators.
A continued fraction
[al' a , ••• ] , with unit numerators and
2
infinitely many terms, is said to be periodic with period
exist
m>O
and t>l
n>m,
such that for al1
a +t
n
t
= an
if there
We write
(4)
The least period
any period
for any
t
0
is called the fundamental period.
t = k t +r, where
t , for i f not, let
0
0
divides
is the 1east period.
then
;
0
therefore
0
t
0
0 < r< t
n>m , we have a n = a n+t = a n+k t +r = a n+r
a period, contradicting that
t
is
r
The continued
fraction is said to be purely periodic if it is possible to take m = 0, i.e.
=
(5)
Given a continued fraction, it is desirab1e to have a systematic
method for expressing each convergent
such a method is to consider
c
n
c
n
as a simple fraction
as a function of
a
n
Now
p Iq
n
if we
simplify (1) starting at the lower right and working upwards, we find
n
-4-
That
C
a •
is a quotient of 1inear functions of
n
where
p
= k nn
a
n
and s a r e independent of
n
= kn
r
=
a
+ land
n
= r nn
a
n
+ s , and k , l ,r
n
n
n n
k, l , r , s
n
n
n
n
(i.e.
n
q
ln fact, suppose
n
depend on1y on
(an + b n+ 1 /a n + 1 + ln
n (an + b n+ 1 /a n+ 1 ) + sn
(k a + l ) a + + k b +
n 1
n n 1
n n
n
(r a + s ) a + + r b +
n 1
n n
n
n n 1
=
p a
+ k b
n n+1
n n+1
q a
+ r b
n n+1
n n+1
Thus we can take
and
c n+1
p
- p
n+2 -
Noting that
n+1
is a1so a quotient of 1inear functions of
the same argument to
where
and q
Pn+1 -- Pn a n+1 + k n b n+ 1
c n+ 1 ' we therefore have
c n+ 2
= q n a n+1 +r n b n+1
a + . App1ying
n 1
= Pn+2/ Qn+2'
a
n+1 n+2 + Pnbn+2 and
cl = (1.a
1
+ 0) / (0.a
1
+ 1)
and using induction, we
c1early have the fo11owing theorem :
Theorem 1.1
The convergents
(al' b 2 , a 2 , b 3 , a 3 ,···)
are
cl' c ' .•. of the continued fraction
2
c
n
= pn /q n ,
where p
n
and q
n
are
defined as fo11ows :
Pl = al ' q1
P2
= a la2
=1
+ b2 ,
q2
= a2
p
-ap
+bp
n - n n-l
n n-2
(n
?
3)
(6)
Q
=aq
+bq
n n-1
n n-2
(n
?
3)
(7)
n
'
-5-
Theorem 1. 2.
Fo11owing the notation of the previous theorem,
n
Pnq n-1 - pn- 1q n =
(_l)n
e -e
n n-1
=
e -e
n n-2
=
(-1) n
1i
i=2
b.1
(n
?
2)
(8)
(n
?
2)
(9)
(n
?
3)
(10)
n
bi
i=2
qnqn-1
n-1
a (-1) n-1 1i b
n
i=2 i
qnq n-2
1i
Proof:
true for
n = 2.
Assume (8) for
n-1.
Mu1tip1ying (6) by q
and
(9) fo11ows fram (8) by dividing by q q 1.
n n-
n-1
(8) is proved.
Using (9), e -e 2
n n-
n-1
(e n-e n- 1) + (e n- 1 - e n- 2) =
a (-1)
n
n-1 n-1
TC
b
. 2 i
1=
~
(-1) n-1
1i
i=2
qn-1
b. (
-------
sinee by (7),
1
b
-
-b q
+ q
n n+2
n
n + 1
qn
qn-2
= a nq n-1 .
)
=
1/
=
-6-
CONVERGENCE OF CONTINUED FRACTIONS WITH UNIT NUMERATORS.
Chapter 2.
From now on, a11 continued fractions will be assumed to have
unit numerators.
For the present chapter at 1east, this restriction
is not serious, because it is c1ear that any continued fraction
with
nonzero numerators can he converted to an equiva1ent (i.e. same
convergents) continued fraction with unit numerators.
A1so, for
definiteness, we sha11 work in the rea1 field.
Introducing
q
=
0
= p-1 and q-1 = 1 = Po ' it may he checked
that formulas
(6)
and that
is va1id for n = 0 and 1 , assuming
(8)
and
o
(7) of Chapter 1 are a1so va1id
for
hi = 1
n=l and 2,
for a11
i.
For the continued fraction
(1)
(where
a ,a ,a , ... are any real numbers) , the formulas of Chapter 1
1 2 3
therefore become :
cn
=
[a 1 ,a 2 ,··· ,an]
Pnq n-1- Pn-1q n = (-1)
c -c
n
c -c
n
Remark:
If
n-l
=
n-2
=
(_l)n
= Pn/qn
n
(n
?
1)
(2)
(n
?
1)
(3)
(n
?
1)
(4)
(n
?
0)
(5)
(n
?
2)
(6)
(n
?
3)
(7)
a , a , ••• are positive, then (1) is a valid continued
2
3
fraction (since all denominators are positive), although not necessarily
-7-
convergent, and induction on (3) shows that
Theorem 2.1
Let
ql' q2' .•• are positive.
a , a , •.• be positive real numbers.
2
3
Then the odd-
numbered convergents of (1) form a strictly increasing sequence, the
even-numbered convergents form a strictly decreasing sequence, and
every odd-numbered convergent is less than every even-numbered one ;
(1) is convergent if and only if
(r)
limqq
1=00
n n-
n -+ 00
Proof:
and (7).
The first three assertions follow immediately from formula (6)
It is then clear that (1) is convergent if and only if
c 2n - c 2n- l -+ 0 (equivalently,
equivalent to (8).
Corollary: If
c2n+l-c2n -+ 0 ), which, by (6), is
Il
x = [a ,a , •.. 1, where
l 2
a , a , ... > 0, then
2
3
x
lies
strictly between any two consecutive convergents (except the last two if
the continued fraction is finite).
The following important theorem provides a convenient necessary
and sufficient condition for convergence.
Theorem 2.2
Let a , a , ••• be positive real numbers.
3
2
if and only if the series
Then (1) converges
00
~
n=l
a
n
(9)
is divergent.
Proof: The proof uses the well known result that if 0 < t < 1 , then the
n
00
~
(l-t) is convergent (i.e. has a positive limit) if and only if
n=l
n
-8-
co
L t
is convergent.
n=l n
hence (a) qn- 1 < qn
N such that
Suppose (9) converges.
a
a n < 1 if n
n?N,
> N. For
By (3),
~O,
n
therefore there exists
(3) gives qn < a nq n + qn2- ,
i.e. q n < q n- 2/(1-a)
n , in case (a), and qn < a nq n- 1 + q n- 1 < qn- l/(l-a)
n
in case (b).
Therefore
q
n
<
( r = n-1 or n-2) •
Repeated application of this resu1t gives
r, ••• ,s,t such that n>r> ... >A?R
t = N-1 or N-2, and
q
n
qt
<
(10)
(l-an )(l-ar ) .•. (l-as )
co
~
(l-a ) = L > 0 ; the denominator of (10) exceeds L , therefore
i
i=N
1etting M be the larger of qN-l and qN-2' we have qn < M/L (n?N) ,
Now
hence qn+lqn < ~/L2 , so that (1) diverges, by Theorem 2.1.
Conversely, suppose (9) diverges.
(n even) and qn>qn-2 > •.• > ql (n odd).
gives q
> q
n -
n-
2 + ca
qn + qn-l
(0)2).
n-
Let
Now (3) gives qn > qn- 2> •.. >q2
c = max (ql,q2).
Then (3)
Therefore
? qn-2
+ qn-3 +c ( a n + a n- 1)
? qn-4
+ qn-S + c ( a n +an-l +an-2 +an-3 )
- ---
>
?t
n
L
+ q1 + c
i=3
n
q1 +q 0 +c
> c Sn
,
L
i=2
where
Sn
a.1
(n even)
ai
(n odd)
n
= L a.1
2=3
'
-9-
1
qu' qn-1 exceeds '2 c Sn' The other is at
1east c , hence qnqn-1 > C 2 S
By assumption, S -+ co ,
n
n
2
therefore (1) converges, by Theorem 2.1. Il
Therefore at 1east one of
Theorem 2.3.
Let a , a , •.• be positive rea1 numbers and let n>1 . The
3
2
continued fraction
is convergent if and on1y if (1) is convergent.
to values
x
If (1) and (11) converge
and m respective1y, then
n
(12)
Proof:
(By theorem 2.1, H > a > 0 for
n
n
The resu1t is trivial for
n
= 1.
For
n
n > 2,
hence (12) is a
= 2,
(13)
=
or
1
c -a
r
Now if
(11)
converges to
~
(14)
1
, then
~
+0
as noted, and taking 1imits
in (13) shows that (1) converges to a +1/m = [a ,m ] ; if (1) converges
2
1 2
1
to
x , then (x-al
+0
by Theorem 2.1) taking 1imits in (14) shows that
(11) converges, and again (12) ho1ds.
if [a , a , ••• ]
2
3
For n?3, (1) converges if and on1y
does, and, assuming the theorem for
n-1, the latter
converges if and on1y if (11) converges; a1so, using the theorem for
2 and n-1,
-10-
Chapter 3.
SIMPLE CONTlNUED FRACTIONS.
A simple continued fraction is one of the form [a ,a ,a , ... ],
l 2 3
is any integer and a ,a , ..• are positive integers. In the
where
2 3
following chapters we shall be concerned primarily with this special
type of continued fraction.
For a stmple continued fraction, the numbers
p ,q
n
are
n
integers, and furthermore it is clear fram (3) of Chapter 2 that
lim q q 1 = 00 , therefore
n- 00 n nby Theorem 2.1 any simple continued fraction is convergent; this also
In particular,
00
follows fram Theorem 2.2, since
a
Z
n=l
Theorem 3.1
Proof:
By
The integers
are relatively prime for any
and q
(5) of Chapter 2, any common divisor of
(_l)n , therefore
Theorem 3.2
Pn' qn
=00.
n
Let
(p , q )
n
x
n
=1
n.
divides
n
Il
.
be any real number.
Then
(finite or infinite) simple continued fraction
is the value of the
x
[al' a ,···], where
2
ai
are defined inductively as follows ( [x] denotes the integral part of x)
al
= [x]
1
The induction terminates with
a
n
if
a i +l
if
m
n
= [mi +l ]
(1)
.
is found to be an integer;
n+l, we have a >1
n
If no
m.
1
i8 an
integer, the continued fraction is infinite.
Proof:
It is clear that the process can be continued as long as
remaina nonintegral, and yields integers
al' a , .•• with
2
m.
1
a ,a , ..• positive.
2 3
-11-
Furthermore, if
Mi_l-a _ < 1.
i l
= ai
(i
?
m.(iFl)
ai = mi > 1, since
is an integer then
It remains to show that
x
=
[a ,a , ••. ] •
1 2
+ 1/mi + l , therefore it is c1ear by induction that x
0) •
If the process (1) terminates with
m
n
= an
Now mi
= [a1,a2, .•. ,ai,mi+l]
, this proves
If it does not terminate, formula (6) of Chapter 2
gives 1 x-c
Theorem 3.3
rational.
x
= [al'
The value of any finite simple continued fraction
Converse1y,
x = [al' a 2 ,··· ,an] , where al'··· ,an
r
= al
s
= a 2r l +
r
s + rI
r
2
s, r
rand
(0
< rI <
(0
< r 2 < rI)
s
s)
(2)
n-3
The first assertion is obvious.
equations of (2) by
is
(r,s integers, s > 0) , then the
if x ="rls
are the quotients in the Euclidean algorithm for
1
,···,
L
r
resu1ting 1eft hand side, by ml'
tœ same as in Theorem 3.2.
Theorem 3.4
where the
1
qiq i+l
Il
a 2 ,·.·]·
procedure in Theorem 3.2 gives
Proof:
1
=
Hence 1 x-co 1 < l/qi'" 0 ,
1.
prime refers to [al,···,a , m + ] •
i 1
i
therefore
i
1
n-
2' r
~,
n-
For the second, divide the
1 respectively, and denote the
•.. , mn
Clear1y mi
and
a.1.
Il
The representation of any irrational number as a simple
are
-12-
Any rational number is the value of
continued fraction is unique.
exactly two simple continued fractions, and these are of the form
Proof:
Let
x
be irrational and suppose
x = [al' a , .•. ]. This
2
continued fraction must be infinite, for otherwise
x
would be rational.
·To prove uniqueness, it is sufficient to prive that
integers as given by the procedure of Theorem 3.2.
because 1etting
and mi+1 > 1.
mi = [ai' a i + 1 ,···] we have
Now let
(finite or infinite).
x
Let
n
But this is clear,
ml = x , mi = a i +1/m i +1 '
be rational and suppose
x = [b , b , ... ]
2
1
Let x = [al' a 2 , ••• , anI (an> 1 if n FI),
e.g. the expansion given by Theorems 3.2 and 3.3.
induction on
are the same
that [b ,b 2 , ••• ] is [a ,a ,.··,a ]
l
l 2
n
We sha1l prove by
or
[a ,a 2 ,··· ,a _ ,l].
n 1
1
m (possib1y infinite) be the number of terms in [b ,b , •.. ]
1 2
For n = 1, x = al is an integer, and we consider 3 cases :
( i)
m= 1
b
l
= x = al
as required.
+ 1/b , therefore b = 1 , and
2
2
1
[b , b ] = [a - 1 , 1 ] as required
1
2
1
(iii) m > 2 : b < x < b + 1/b ' contradicting that x
1
1
2
(ii) m = 2
x = b
Therefore this case does not occur.
and we have
and
m?
s , b1 < x
~
b
l
For
n>1,
fo110ws by app1ying the resu1t for
Proof:
is not an integer
+ 1/ b 2 ~ b 1 + 1 , therefore b =[x] = al'
1
1/(x-b ) = [b , b , .•• ] = [a , ••• ,a ] ;
3
2
n
l
2
Corol1ary:
x
is an integer.
n-l.
now the required resu1t
Il
The value of any infinite simple continued fraction is irrationa1.
The value cannot be rational, since the on1y expansions of a
-13-
rational number are the two finite ones mentioned in the theorem.
Il
-14-
Chapter 4.
THE FIBONACCI NUMBERS.
In this chapter we consider some elementary properties of the
Fibonacci numbers and the closely related
[1,1,1, ••• ].
s~le
continued fraction
This continued fraction plays a special role in later
chapters.
Let
t
[1,1,1, ••• ] •
denote the value of
Clearly t=l+l/[l,l, ..• ]
= l+l/t , Le.
t
Solving
2
_ t - 1
=0
•
(1)
(1) for the positive root, we find
t
J"s
= 1+
2
s
1 = -l-J"s = - o. 618 ••.
= - -t
2 -.-
= 1.618 ...
(2)
the other root is
Formulas
(2) and
(3)
pn- p~1 + p~ 2 ' q n
ql = q2 = 1.
of Chapter 2, applied to
= q~ 1
+ q~ 2·
Furthermore
Now the Fibonacci numbers
F
F
1
n
= F
2
(3)
[1, 1, l, ... ] , give
Po = P1 = 1 and
FI' F 2 , F3' .. · are defined by
= 1
= F n-l
(4)
+ F
n-2
Therefore we have
(5)
and the convergents of
[1, 1, l, •.. ]
are ratios of consecutive Fibonacci
-15-
It is important to know how quick1y the Fibonacci numbers
grow with increasing
n •
and the same for
Therefore for any
s.
From
(1) we note that
t
n
= t
n-1
+ t
n-2
n
n
H = at + bs
n
b which resu1t in
a and b,
satisfies
H =H
+H
; solving for a and
n
n-1
n-2
= F , H = F ) , we get a = -b =
Ho = 0, Hl = 1 (Le. H0
01
1
l/.fs .
Therefore
H = F
and we have proved the very interesting formula
n
n
(ca11ed Binet's Formula)
F
n =
-
1
( 1 - J"s)n )
2
( (
.fs
(6)
It follows that
Hm
n
~oo
= -
1
(7)
.fs
so that
F , F , ••• approxtmates a geometric progression. In fact, it
l
2
is the nearest integer to tn,.fs.
is clear from (6) that F
n
The fo11owing theorem shows that the denominators of the
convergents of any simple continued fraction increase at 1east as fast
as the Fibonacci numbers.
Theorem 4.1.
Proof:
q
1
For any simple continued fraction,
q
>
n -
F
n
•
= 1 = F1
q n- 1 >
- F
n- I ' we have q n = a nqn- 1 + q n- 2 >
- q n- 1 + qn- 2 >
- Fn- 1 + Fn- 2 = F.
n
Il
-16-
Chapter 5.
MISCELlANEOUS RESULTS.
Here we mention some miscellaneous theorems which will be
used in later chapters, but which either are not of general interest
or are weIl known elementary results.
Theorem 5.1.
expansions of
x > 1 and consider the stmple continued fraction
Let
x
and
l/x.
is the reciprocal of the
Proof: If
For
(n-l) st convergent of
x = [al' a 2 ,···] , then
primed quantities refer to
Theorem 7.15.
Theorem 5.2.
x
l/x = [
0,
l/x
x.
al' a 2 ,···]·
Letting
l/x , and using (2) and (3) of Chapter 2,
,
one can verify by induction that
the result follows.
n ? , the nth convergent of
Pn = qn-l' qn
Details may be found in
=P
n- l ' from which
[6],
Il
Let
a , a , .•. be positive real numbers and assume that
2
3
= [al' a 2 ,···] F O. Denote [an' an+l""] by mn • Then for n? 1,
1
(1)
= [ a , a l " ' " al' - 1: ]
n
Proof:
n-
x
(Note: in the course of the proof, it must be shown that (1)
is indeed a valid continued fraction, i.e. has no zero denominators.)
For
n = 1 , [al' -
its value is
1
i]
is a valid continued fraction (since x ; 0) and
al - x , which, by Theorem 2.3, is
assume that the result for
mn
-11M
2
, For
n? 2 ,
n-l , i.e.
=
-1
(an_l,···,al,-l/x ]
(2)
-17-
Now
[an' .•• ,al' -l/x]
assomption
therefore
is a valid continued fraction since by
[a _ , ••• ,a , -l/x] is.
n 1
l
-l/Mn+l
= an
[an' ••• ,al' -l/x].
- mn
By Theorem 2.3, mn
= an +
l/[an _ l ,··· ,al,-l/x]
= [an ,mn+1]'
=
//
We sha1l have occasion to use some results from elementary
number theory.
If
p
is an odd prime and
cal1ed a quadratic residue of
x
2
1.
s:
a
p
~
0 (mod p) , then
if there exists
x
a
is
such that
(mod p).
-1 is a quadrat ic res idue of
2 is a quadratic residue of
2.
a
p
if and only if
p
if and only if
p
if and only i f
p
lE
1
(mod 4).
p le 1 or 7 (mod 8).
p == 1 or 3 (mod 8).
-2
is a quadratic residue of
If
N is a sum of two relatively prime squares, then
N is a
product of primes of the form 4k +1 or twice such a product.
A proof of 1. may be found in
Result 2. is proved in
[9], Chapter 5, Section 2.
[2], Chapter 20.
(The converse of 2. is also true).
-18-
PART II
APPLICATIONS TO RATIONAL APPROXIMATIONS
-19-
Chapter 6.
1.
BEST APPROXIMATIONS TO REAL NUMBERS.
Introduction and Motivation.
This first section is meant to motivate and c1arify the
definitions of best approximation of the first and second kind.
Given a rea1 number
x , it is natural to look for a rational
number which is a good approximation to
is "not too large".
x
More precisely,
and a positive integer
~
satisfying : (1) 1
1 ~ 8 ~ N , we have
b
~
x
(Problem A) given a real number
N, to find a11 pairs of integers a, b
N , and (2) for any integers
'x-a/bl < Ix-rlsl .
exists, and if one pair
easily be found, since
and yet whose denominator
a, b
cId
Definition. Rational number
r, s
Clear1y at least one such pair
is known, the others c, d
must equa1
a/b
(b > 0)
a/b
approximation of the first kind to
x
with
(if any) can
or 2x - a/b.
is 8aid to be a best
if
1
~
d
~
b,
cId
+ a/b
imply
1 x - cId 1 > J x - a/b 1 •
The relation of this definition to Problem A is indicated by
the fo1lowing e1ementary propositions.
Proposition.
Any best approximation
solution of Problem A for some
Proof:
For
rIs
Suppose
= cId
1 < s < d.
- -
cId
of the first kind i8 a
N ,namely N
= d.
Ix-c/dl < Ix-rlsl .
We must show that
this is tmmediate, while for
rIs
+ cId
it follows
from the definition of best approximation of the first kind.
Proposition:
Except for the case
N= 1
and
1
Il
x = n +2 (n an integer),
-20-
any solution
a, b
of Problem A with minimum b
of the first kind to
Proof:
Then
(by
Suppose
min~lity
Rence
t/b
a
x.
1 < d < b
= Ix-a/bl
Ix-c/dl
and
cid
(since
of b) ,and
c
x
+ a/b
,but
Ix-c/dl
5 Ix-a/bl.
=b
a, b satisfies Problem A) , d
is the average of
differ by l , for if
t
a/b
and
c/b
is strictly between
= cid.
C,
a and
wou1d be a better approximation to x •
Without loss of generality, assume
(r
is a best approximation
+ 0,
and then
for otherwise
x
=a
a/b
= S/l,
c
=a
hence
+ 1.
b
=1
a = qb+r, O<r<b
Write
by minima lit y of
+ 1/2, contrary to assumption.)
lt is easily
b,
checked
that
q
exceeds
a/b
but not
approximation as
(b-l) + r
b-1
(a+l)/b, therefore is at least as good an
a/b, contradicting the minima lit y of
Il
b.
lt may be noted that the situation is simplified if
x
is
irrational : then clear1y any solution of Prob1em A is a best
approx~tion
of the first kind.
Proposition:
Exclude the case
approx~tion
a/b
N
=1
,
x
of the first kind with
=n
+
1
2.
Then any best
1 < b < N and such that
b
is maxÛDWD, is a solution of Problem A •
Proof:
If
a, b
is not a solution of Problem A, let
solution with minimum d.
preceding proposition,
cId
We have
c, d
lx - c/df < lx - a/bl.
be a
Now by the
is a best approximation of the first kind,
-21-
therefore by maxima1ity of
of
a/b
b , we have
being a best approximation,
d < b.
But then by definition
Ix-c/dl>1 x - a/bl ,contradicting
Il
the first inequa1ity.
Thus (disregar.ding the trivial case
of Prob1em A for genera1
N
= 1,
x
= n +21 ),
solution
N is equiva1ent to knowing a11 the best
approximations of the first kind to
x.
For some purposes it is desirab1e to consider the difference
Ibx-al
rather than lx - a/bl.
This 1eads to a second king of best
approximation, as fo11ows :
Definition.
Rational number
a/b
approximation of the second kind
imp1y
(b>O)
to
x
is said to be a best
if 1 < d < b, cId F a/b
Idx-cl > Ibx-al.
Theorem 6.1.
Any best approximation of the second kind is a1so one of
the first kind.
Proof:
Mu1tip1ying
Idx-cl > Ibx-al by 1/d > lIb gives Ix-c/dl>lx-a/bl,
Il
as required.
Examp1es are readi1y found to show that the converse of Theorem
6.1 is fa1se.
e.g. let
x
= 5/12
- e , where 0 < e < 1/60 ; then
1/3
is a best approximation of the first kind but not of the second, since
it may be verified that
2.
12x - 11 < 13x - 11 •
The Main Theorems.
We shal1 see that the convergents of the simple continued fraction
expansion of
x
provide a very convenient means of finding al1 the best
approximations of the first and second kind to
x.
We continue to use
-22-
the notation of Chapter 2 (formulas (1) to (7»: x
= [a 1 ,a2 , ..• ]
is a finite or infinite simple continued fraction, with convergents
c
n
= [a 1 ,a 2 , .•• ,a n ] = pn Iq.
n
We begin by examining the difference
between the continued fraction and its convergents.
Theorem 6.2.
If
x
has at 1east
1 x - cn 1
Proof:
c
n
n + 2 terms, then
By the coro11ary to Theorem 2.1,
and
therefore
of Chapter 2).
there are on1y
x
lies strict1y between
=
(formula (6)
Il
(1)
n + 1 terms, for then
n + 1 terms, then
Theorem 6.3.
(1)
Ix-c n 1 < Ic n-c n+1 1
It shou1d be noted that
1east
1
<
Let
m =
n
is rep1aced by an inequa1ity if
x
= c n +1 •
Thus, if
x
has at
lx - c 1 < 1/q q +1 •
n
-
n n
Then
[a n , a n+ 1 ,· •.]
(_1)n+1
x - cn =
(2)
qn-1
qn2fmn+1 +
qn
Proof:
x =
=
Il
(al'· .. ,an ,mn+
Pnmn+1 + Pn-1
qnmn+1 + q n- 1
)
-23-
Using this expression for
x , we find that
Pn-1q n - PnQn-1
=
x -
+
2
Qn mn+1
and
(2)
fo11ows by use of formula
Theorem 6.4.
If
x
has at 1east
QnQn-1
(5)
of Chapter 2.
Il
n + 1 terms and its 1ast term
(if there is one) is not l , then
1x
Proof:
Comparing
1
- cl>
n
(2). and
(3)
(3), it is sufficient to prove that
this becomes
mn+1 < an+1 + l , which is c1ear1y true.
Theorem 6.5.
Let
(if any)
not 1.
x
have at 1east
n + 2
Il
terms, with the 1ast term
~hen:
(4)
Therefore by Theorem 6.4,
Ix-cnl > 1/qnqn+2
Iqn x - pl>
n
1
But, as noted after Theorem 6.2,
(5)
Ix-cn+11 < 1/qn+1qn+2'
1
fo11ows fram
i.e.
(6)
qn+2
(4)
Le.
(5)
and
(6).
Il
-24-
Corollary:
c +
n l
any convergent
c
n
of
Except for the convergents
is closer to
x
[ al' 1, 1 ] ,
than the preceding convergent
,Le.
(7)
1 x - c ni> 1 x - c n+1 1 •
Proof:
(i)
(ii)
formula
(7)
If there are only
n + 1
terms then
If there are only
n + 2
terms, then x = c + '
n 2
of Chapter 2,
=
1 x - cn
also,
1 x - c n +l 1
therefore
(7)
= 1,
a
2
= 1,
a
n
3
=1
= 1;
terms, this means that
(iii)
and
a
=
1
a n+2 qn+l > qn •
is equivalent to
(equality only if
n
and using
2
= 1),
hence
(7)
i8 false only for
sinee the present case is assuming only
x =
If there are
n+2
[al' 1, 1 ] , i.e. the exception noted.
n + 3 terms or more, we can absorb any
final unit into the second to last term, therefore we always have
by Theorem 6.5.
Assuming
Multiplying by
b
median value) of a/b
and
and
d
l/~?
l/qn+l
gives
(7).
(4),
Il
positive, the mediant (also called the
cId
is defined to be
is, as one can trivially ver if y, strictly between
(a+c) 1 (b+d) , and
a/b
and
cId
if
-25-
the latter are distinct.
Consider the fractions
Pn + ip~+l
( 0
qn + iqn+l
We observe that the first of these is
<
i
< a n+2
) •
cn' the last is
(8)
c n+2 ' and
each after the first is the mediant of the preceding one and
The fractions (8), other than the first and last, are
Pn+l/qn+l'
called intermediate fractions (or secondary convergents) of the
continued fraction.
The reason for introducing intermediate fractions
is the following result : every best approximation of the first kind
to
x
is either a convergent or an intermediate fraction of the simple
continued fraction expansion of
x.
Since the proof is somewhat long,
but not difficult, and the result will not be needed in what follows,
we shall omit the proof ; it may be found in [4],
§ 6, Theorem 15.
As will be shawn, every convergent is a best approximation of the first
kind ; however, this is not true of every intermediate fraction.
For
best approximationrof the second kind, a much sharper result (presented
in the next two theorems) is true, and this constitutes the main reason
for considering best approximations of the second kind.
Theorem 6.6.
If
there exists
n > 1
Proof:
Let
L<
a/b
~
is a best approximation of the second kind, then
such that
a/b
= cn •
be the number of terms in the given continued fraction.
For the purposes of this proof, " c, d gives a coni.radiction " will mean
that 1 ~ d ~ b,
that a/b
cId
+ a/b
,and
1 dx-c 1 ~ 1 bx-a l ,
is a best approximation of the second kind.
thus contradicting
We shall have
-26-
occasion to use the fact that for any integers
e/f
F
g/h implies
le/f - g/hl ? l/fh;
e, f, g, h (f,h positive),
this follows since leh-fgl
is not zero, hence at least 1.
If
al' 1
L
= 1,
then
= al = cl
x
gives a contradiction.
= cl'
a/b
= Ibx-al
for otherwise
L > 2.
Therefore assume
Ix-aIl < Ix-a/bl < b Ix-a/bl
we have
,and
If
a/b < cl'
" so tbat al' 1 gives
a contradiction.
a/b > c
If
2
' then
>
Ix-.!
b
Le.
b
al>
x
;
therefore
contradiction.
Thus, assuming
cl < a/b < c
and
= cL.
(i)
(ii)
For some
L
n? l ,a/b
is finite, and
We note that if
al' 1 aga in gives a
is not equal to any convergent,
; also, a/b F x , for otherwise x is rational
2
Therefore we are clearly in one of two cases
we have
a/b
a/b
1
>
ajb
ajb
is strictly between
is strictly between
is strictly between
c
r
and
c
n
c n+ .
2
and
c _
L 1
and cL=x .
c r +1 ' then
_1_<
bq
therefore
r
qr+l < b •
In case
(ii), therefore,
qL < b;
also, !qLx-PLf
=0
, hence
-27-
PL' qL gives a contradiction.
(i), we show that
In case
Now
is a1so strict1y between
a/b
Furthermore, a/b
g ives a contradiction.
Pn+1' qn+1
c
and
n
c n+ 1 ' hence qn+1 < b.
is c1ear1y at 1east as close to
c +
n 2
as it is to
x ,
so that
1x - ~
i.e.
But
Iqn+1 x - Pn+1 1
=
>
b
1
b qn+2
1
1 bx - al>
qn+1 J x - c n+ 1 1 < 1/qn+2
Iqn+1 x - Pn+11 ~ Ibx_al.
Therefore
(from Theorem 6.2).
Il
Assume that the continued fraction used for
Theorem 6.7
have
a
>
1 as a 1ast terme Let
n
= l,
there are at 1east 2 terms, and
2.
n
= l,
and the fraction is
Pn/qn
does not
n? l, and exc1ude the fo110wing cases :
1.
Then the convergent
x
a
2
=1
•
[al' 2].
is a best approxLmation of the second kind
(and hence a1so of the first kind).
Remarks:
then
c
(i)
Iq nx - Pn
1 ~ qn-1 ~ qn'
' each being
(H) In case
therefore
n>2 , and the 1ast term is
-
a n+1
=1
;
is not a best approx Lmat ion of the second kind ;
Pn/qn
c
n-1 1: n
Suppose
Cl
= a1/1
and
(since
x
= c n+1) , 1qn- l
x- pn- 11
=
1/qn+1·
l, or 2, above, we have
is not a best approxLmation of the first or second
-28-
kind
(al + 1) /.1 ) •
(by comparison with
Proof of Theorem 6.7:
continued fraction.
then
pn /qn - x
Let
m<
Let
~
be the number of terms in the
First of aIl, we can assume
= 1,
integer and
v
by
1 vx
is equiva1ent to
2, .• . ,q
1.
- u
vx-t
Letting
~
u
~
F attain a
min~
of aIl e1ements of
"Cu , v)
o
0
where
Define a function
.
n
(u, v)
t
u
vx + t ;
v
may be any
F with domain
X
F , F(u,v) < t
be any value of
is bounded, therefore on1y a
finite (and nonzero) number of e1ements
hence
n =m ,
and the resu1t is immediate.
X be the set of aIl pairs
F(u,v) =
n < m, for if
(u, v) satisfy
at one of these e1ements.
F(u, v) < t
M be the set
Let
X at which this minUmwo is attained, and let
be an e1ement of
with 1east
M
v
0
Then we have
1.
II.
For aIl
If
F(u,v) > F(u , v ) .
in X,
(u, v)
-
0
F(u,v) < F(u ,v), then (u, v)
-
0
0
is in M , and
0
v>v
-
0
We a1so c1aim the fo11owing
III.
If
is in M , then u
(u, v )
o
Once III is proved, it will fol1ow that
x, for if
then II gives
and III gives
d = v
o
To prove III, suppose
vx-ul=lvx-u
o
0
0
u IV
o 0
1 < d < v
of the second kind to
-
and
F(c,d) < F(u ,v ) ,
i8 in M, but
(u, v )
o
-
0
0
u ~ u .
o
Then
land
x =
Ivox-u0 1 ~
0
is a best approxLmation
= uo
c
u+u
a180,
= uo
0 , hence
1 v0x
2 v
- u
0
o
o
1=
(u + u ) 1 2 - u /1 1 > l
o
0
- 2
;
-29-
is rational, so
x
2v
m is finite and
are relatively prime, for suppose
0
then
Iqx - pl
=o<
2v
0
0
q < v
k>2 ;
-
= qm
; qm > 1
shows that
therefore
= l,
(hence also n
is in
and
o
= kq
,
and
X
u+u
0
- Pm '
I~l x - Pm-ll = qm-l l cm-cm_ll
Now
therefore II gives a contradiction
~l
vo >
,2v
Therefore,
< v o • Now 2vo
~l
if we can show that
> 2,
m-
m > 2.
l ,
= kp
u+u
(p, q)
1/2 < 1 v ox - u 0
-
Now u + u
0
Ivox-u 0 1 contradicts I.
= l/~ = 1/2vo <
assumed a
so that
0
= cm
x
= am ~l + qm-2
unless a
= m)
since we dismissed n
m
;
=2
and we
m=2
and
this case was excluded,
therefore III is proved.
Theorem 6.6 now shows that there exists
u /v
00
= c.
s
If
and II gives
Thus,
u
o
u
= kp
and
v
q > v , so that
k
0
-
= ps
0
,v
0
= qs
=1
n
which is false by definition of
q
s
recalling
+ q
<q
s+l -
n-l
=1
1qx-pl < 1v x - u 1
,then
and
such that
-
u ,v
o
0
0
0
are relatively prime.
• We complete the proof by showing that
Since we excluded the case
hence
= kq
0
s > 1
+ q
n
;
and
X.
a
= l,
2
Suppose
using 1,
n < s
leads to
s < n ; then
s
q
n
= n.
<q
s
s + 1 < n ,
Theorems 6.2 and 6.4, and
n <m
1
> 1 qnx-p n 1 > 1 q s x-p s 1
1
>----
>
1
qs~s+l
Therefore
q
n+l
< q n + q n- l ' which is false since
Il
-30-
3.
A Restatement.
The results proved in this section are closely re1ated
to Theorems 6.6
and
As demonstrated by Niven and Zuckerman
convenient to use.
their
6.7, but are in a form which is sometimes more
[6] in
Theorem 7.13, tbey may a1so be proved direct1y without using
our Theorems 6.6 and 6.7.
Theorem 6.8.
at 1east
Proof:
Let n > 1 and assume that the fraction used for
n + 1
terms.
We can assume
b > 0, Ib~-al < Iqnx-Pnl imp1y
Then
a
and
b
re1ative1y prime, for if
b = kd , (c, d) = 1 , we could app1y the theorem to
therefore
If
If
c, d
n = 1
and
[al' S] , we have
If
m=n+2
a
into
m
for
qn+1 + qn > qn+l·
n = l, a
Fina11y, if
2
a
b
m
=
and the theorem still gives
m = n + 1, we can assume
= 1 bas been treated) ;
an+l
a
? q n+1 = (a n +1+1) qn + qn-l =
fo11owing the statement of Theorem 6.7,
therefore absorbing
m-1
m and
, absorbing a n+2 into a n+ 1 ' we get (where
,
primes refer to the nev fraction)
,
qn+1=1.
A1so, we can assume that the continued fraction does not
-
the case
d~n+1'
lb -al >1 = Iq x - Pli , cont.radicting the
x
- 2
1
m>n + 3 we can absorb
qn+1 •
= kc ,
b? 2 = qn+1
have 1 as a 1ast term, for suppose the number of terms is
?
b~n+1.
to get
and a 2 = l, the resu1t is trivial, since then
n = 1 and the fraction is
assumption.
b
has
b? qn+l.
otherwise b = 1
If
a
x
ioto
an'
as indicated
n > 1 (since
in Remark (i)
Iqn_Ix-Pn_l' = 'qnx - Pn' ;
we have
q n = (a n+1) qn-l + q~2 = qn + qn-l = qn+l •
,
b? q n'
but
1.
-31-
Thus, Theorem 6.7 gives that
A1so, a/b F p Iq
n n
the second kind.
since
= 1)
(a, b)
which is fa1se.
, therefore
(otherwise
b < q
- n
b > qn'
This shows
is a best approximation of
Pn/qn
lbx -al
Iq nx-p n 1
lb -al > Iq nx-p n 1
x
implies
Suppose
=
b < qn+1'
,
Then a/b
is not a convergent, hence by Theorem 6.6 it is not a best approximation
of the second kind.
Thus there exist
1dx-c 1 ~ 1bx-al
.
c, d
(c, d)
= 1,
and
a Fc
and
1< 1a- c 1 ~ 1dx- cl + 1bx-al ~
but except in the case
~ ~/qn+1 ~
We c1aim that
= 1, a 2 = 1,
Ibx-al < Iq x-p 1
n
n-
and
d > q ,then
Now if
and
1
'2 '
n
n
cId
such that
21 bx-al
l<d<b, a/b F cId,
d
=b
, hence 1bx-al
>.!
- 2
d F b, for
1 ~ f < d,
is contradicted.
Therefore
n
unti1 we eventua11y obtain
rIs,
rIs F Pn/qn)
1east
n
Let
terms.
n
,(r,s) = 1,
, contradicting that
Therefore
Pn/qn
b? qn+1'
and 1 sx-r
Il
b >
0,
Exclude the case
n = 1, a
2
= 1.
Then
has at
b > qn .
n + 1
terms, since the resu1t is empty for
b < q ,mu1tip1ying the given inequality by this gives
- n
fbx-af < Iqnx-Pn' ,therefore
contradiction,
since
~+1
<
is a best
x
Proof: Assume at 1east
If
-
n > 1 and assume that the fraction used for
Ix-a/bl < Ix-p n Iq n 1 tmp1y
terms.
-
e/f ,
This process may be continued
1 < s < q
approximation of the second kind.
Coro11ary:
d < b .
is not a best approximation of the second kind
(e,f) = 1, Ifx-el ~Idx-cl .
Iqnx-Pnl (hence
;
which has been treated) Iq x - pnl
n
Idx-cl < Iq x-p l , so that by the same argument, we get
n
gives
b? qn+1 ' by the theorem.
> qn'
Il
This 1s a
n
-32-
Chapter 7.
HURWITZ' S THEOREM AND REIATED RESULTS.
2
q +1 :< q ,Theorem 6.2 shows that
n
- n
Since
and therefore for any irrationa1 number
many rationa1s
a/b
b
2
n
< l/q n ,
there exist infinite1y
2
Ix-a/bl < 1/b .
such that
the possibi1ity of rep1acing
x
1x - c 1
Let us now consider
by some other function of
More specifica11y, for how large a value of
k
can b
2
b.
be rep1aced by kb
2
Investigation of this prob1em 1eads to the striking resu1t, first proved
by Hurwitz in 1891 ( [3] , using Farey sequences rather than continued
fractions), that
k
J5,
can be as large as
however, we sha11 examine the easier case
Theorem 7.1.
Exc1ude the case where
n
Given any two consecutive convergents
c
k
=1
n
but no 1arger.
First,
= 2.
and the fraction is [a ,1,1] .
1
and
at 1east one of
the fo110wing two inequa1ities i8 true
1 x - cn 1 <
(1)
1
(2)
2
2q n+1
Proof:
x
- lx - c
n
1
Now
lies between c
1=
and
l/q q
- lx - c l .
n n+l
n
x - c n +l '
<
2
2
l/ab - 1/2a < 1/2b
Le. a j: b.
n
But
c
n+1'
therefore
Assuming
Ix-c
n+1
(1) false,
1
=
Ic
n+1
-c 1
n
we obtain
1
(3)
is equivalent to
2ab-b
qn+l ~r q n ' proving (2), un1ess
2
2
< a , i.e. (a-b)
n = land
2
> 0,
?
-33-
and clearly one of
case the fraction
1s true unless
(1), (2)
must be
[al' l, 1]
Since this case was
Given any irrational number
rational numbers
a/b
Proof:
If
x, there are inf1nitely many
such that
1 x - a/b 1 <
Theorem 7.2.
1
+"2 ' 1n wh1ch
Il
excluded, the theorem is proved.
Corollary:
x = al
(4)
We can assume
holds, then
b > o.
(4)
1
2
2b
1s a convergent.
a/b
Given
(4), we prove that
best approximation of the second k1nd to
x;
1s a
a/b
the des1red result then
follows by Theorem 6.6.
Suppose
1 < d <b
cId f a/b;
and
we must prove
(5)
1dx- cl> 1bx- al·
By
(4), it 1s enough to show
dx-c
....!
< I~ - .! 1 <
db - d
b
-
c
x - -d
a
+ 1 x - -b
<
c
x-d
+....!....
2
2b
<
c
x-d
+ibd
Therefore
Le.
1 x-~I
d
1
1dx - cl> l/2b.
Theorem 7.3.
If
1
> db - 2db
> l/2b .
1
1 ..
=-
2db
Il
k > ,,; , then there 1s an irrat10nal number
x
-34-
(for example
x
= (1 + ../5) /2) such that
a
1x - b
<_l_
2
kb
(6)
is true for only a finite number of rational numbers
Proof:
Let
numbers
a/b.
= (1 +../5) / 2 and let Fl,F2,F3' .•• be the Fibonacci
x
1,1,2, •••
As was shawn in Chapter 4, x
= [l, l, l, ••• ] and
By Theorem 6.3,
1 x - c
But
n
x + Fn- l/F n ~ x + 11x
for all
1
=
1
=
Therefore
../5
(7)
•
k >J5
implies that
n
sufficiently large, lx - cl>
l/kq2.
But k > &, hence
n
n
2
by Theorem 7.2, if lx - albl < l/kb
then alb is a convergent.
/1
Therefore the theorem is proved.
In preparation for proving the next theorem, we introduce
some notation
m
n
= [an' an+l'···]
qn-2
n = qn-l
w = u +m
n
n
n
u
We observe that
mn
1
un+ l
Therefore
1
un+ l
(n ? 1)
(8)
?
a)
(9)
(n >
a)
(10)
(n
= a n + l/mn+l' and
qn
=
+
qn-l
1
mn+l
=
an qn-l + qn-2
= a n +un
qn-l
= wn
(
n>_ )
(11)
-35-
and
a
1
=
n
- u
(n
n
?'
(12)
2)
The formula proved in Theorem 6.3 becomes
1
1 x - cn 1 =
Lemma:
Let
un+1 > (
n > S.
J5 -
From w ~.rs
n
(13)
w + ~.rs
n 1
and
1 ) 1 2 .
Proof: Using (11) ,
we have
Therefore
= 1
Mu1tip1ying by
u +
n 1
and comp1eting the square,
( u +
n 1
1
u +
n 1
2
J5
1
< 4
-Z )
~J51 <
"2
2
J5
?'"2 -
1
2
un+1 un+1
But
it follows that
1
=
.fs-1
2
is rational, therefore the 1emma is proved.
Theorem 7.4.
Let n > 1
n + 2 terms.
Then the inequa1ity
and assume that
1 x - c.1 1 <
= { al'
x
n + 3
a , .•• 1 has at 1east
2
1
is true for at 1east one of the three values
Proof: We can assume at 1east
Il
(14)
i
=n
, i
=n
terms, for otherwise
+ l, i
(14)
=n
+ 2
is true
-36-
for
i
=n
+ 2.
Then by
(13),
lemma,
u > (
i
Assume
J5
w ~
i
J5 -
(14)
= n+2,
(i
2
J"s-l
which is false.
Corollary:
i.
= n+l, n+2, n+3) , therefore by the
(i
1)/2
false for aIl three values of
-
n+3).
But then
J"s-l = 1
-2-
(12) gives
,
Il
Given any irrational number
many'rational numbers
a/b
such that
x, there are infinitely
-37-
PART III
APPLICATIONS TO NUMBER THEORY
-38-
Chapter 8.
THE NUMBER OF STEPS IN THE EUCLIDEAN ALGORITHM
In this chapter we prove an elementary yet significant result
concerning the Euclidean algorithme
The ease of the proof is a good
illustration of the power of the theory of continued fractions.
Let
r, s
algorithm for
s > O.
be any integers, with
rand
Consider the Euclidean
s
l.
r = a
2.
s = a r + r
2 l
2
3.
rI
1
s + rI
= a 3r 2
< rI < s )
( 0
( 0 < r 2 < rI )
<
( 0
+ r3
r
3
<
r
2
)
n-l.
n
Here,
n > 1
notation
is called the number of steps, and we shaii use the
E (r,s) = n •
As shawn in Theorem 3.3,
continued fraction expansion of
[al' a , ••• ,a ]
2
n
rIs,
and
(if
is the simple
>
n
1)
a
>2 •
n -
Therefore we can write
r
s
Using
to
Fi
=
an - 1 ' l ]
for the Fibonacci numbers and letting
(1), we have
(Theorem 4.1)
qn+l? Fn+1 -
s
therefore
n + 1 < m, i.e. E (r,s) < m - 2.
? Fn+l .
Now suppose
s < F
therefore
m
Pi
But
(1)
and
qi
refer
rIs = Pn+l/qn+l
s < F ,
Fn+ 1 <
m
This proves the foilowing
Then
-39-
theorem :
Theorem 8.1.
r
Let
FI
= 1,
F2
= l, •••
is any integer and 0 < s < F
m
Euclidean algorithm for
rand
a = F
(m--ll's) , hence
E(Fm, Fm--1)
Example 2.
Let
m--l
r
= 18,
s
, then the number of steps in the
s
and
Example 1. Taking
be the Fibonacci numbers. If
is at most
r
m--2.
= Fm , we have
= m-2.
= Il.
Il is between the Fibonacci numbers
= 8 and F7 = 13, hence the theorem predicts E( 18,11 ) < 7-2 = 5.
In fact, E (18,11) = 5 :
-
F6
18
= 1.11 +7
11
= 1.7+4
7
= 1.4 + 3
4
= 1.3 + 1
3
= 3.1
-40-
PERIODIC SIMPLE CONTINUED FRACTIONS
Chapter 9.
Periodic simple continued fractions have many interesting
and useful properties, due primarily to the fact that a continued
fraction is periodic if and only if it represents a quadratic irrational.
This striking result was first proved by Lagrange in 1770.
the simple-continued-fraction expansion of
.JD
where
In particular,
D is a positive
nonsquare integer) provides the key to the solution of Pellls equation
x
2
2
- Dy
=~
l, to be discussed in later chapters.
By a quadratic irrational we mean a number of the form
where
A and
B are rational numbers
nonsquare integer.
and
B
=F
= E).
JO
The conjugate of x
=x
y
hence also
x
-1
= A + IWD
y = E +
= x -1 )
~
0)
and
JO = E
+ F
B
A+
BJD,
JO,
=E
and consequently
ls def ined to be
FJO,
then A
/ (B-F) ,
=F
=x
then x + y
x
=A
-
IWD.
+ y and
; in other words, the operation of
taking the conjugate is an automorphism of the field
field of elements
D is a positive
JO = (E-A)
hence
is irrational
One easily verifies that, if
xy
A+B
We note that if
(for if B ~ F, we would have
contradicting that
A
(B
A+Ptfn ,
where
A and
Q(,fn) (i.e. the
B are rational).
Clearly a root of a quadratic equation
with integer coefficients
and positive nonsquare discrÜDinant is a quadratic irrational.
Furthermore, it is not difficult to see that any given quadratic
2
irrational is a root of precisely one quadratic equation ex +bx+c
where
a,b,c
are integers, a > 0, and (a,b,c)
statement , x = A + B ·JD is a root of
2
= 1.
2
=0
To prove the last
2
x -2Ax + (A -B D)
= 0,
which
-41-
can clearly be put into the required form ; to show uniqueness,
suppose
ax
2
+ bx + c = 0 = dx
(d,e,f) ) ; eltminating
(since
x
x
2
is irrationa1)
2
+ ex + f (a, d > 0, (a,b,c) = 1 =
,we get
bd = ae
(bd-ae) x = - (cd-af), hence
and
cd = af ; assuming first
that e F 0 and f F 0 , we have .!
d -- È
e -- ~f ; denoting each fraction
by
k
that
hence
and letting
k
rd+se+tf = 1, we get
is an integer;
a = d, b
= e,
c
but (a,b,c)
= f;
the case
k
= 1,
e
= ra+sb+tc,
showing
therefore in fact
=0
or
f
=0
k
= 1,
is treated
stmilarly.
One
half of Lagrange's result is relatively easy, and we
state it in the following theorem :
Theorem 9.1
The value of any periodic simple continued fraction is
a quadratic irrational.
Proof: Let
x =
[al' a 2 , ..• ] ,where, for aIl
(here, m > 0 and r? 1).
n > m, an+r= an
Let Y = [am+l' am+2, .•. I.
Then
x =
[al'··· ,am' yI = [al' ••• ,am+r' y1 , therefore
x =
PmY + Pm-l =
~y+~l
Renee
y
Therefore
Pm+r Y + Pm+r-l
qm+r Y + ~-l
satisfies a quadratic equation with integer coefficients
y, and consequently x,
is a quadratic irrational.
Il
-42-
As a first step towards proving the converse, we sha11
determine which rea1 numbers have pure1y periodic expansions. These
turn out to be the reduced quadratic irrationa1, defined as fo11ows
Definition:
x
A reduced guadratic irrationa1 is a quadratic irrationa1
such that
x > 1
-1 <
and
x < o.
Theorem 9.2 Any pure1y periodic simple continued fraction
(1)
is a reduced quadratic irrationa1, and
y
Proof:
x> al? l ,
=[
xy
= -l,
where
(2)
a r ,ar- l' ••. , al ] .
and by Theorem 9.1
x
is a quadratic irrational.
App1ying Theorem 5.2, we have
1
=[
x
1
(3)
a r ,ar- 1'· •. ,al' - -x ] .
(2) and (3) show that when the proof of Theorem 9.1 is app1ied to
and
1
x
(using m
equation.
But
y
= 0)
these numbers satisfy the same quadratic
and - 1/x
are distinct (they are of opposite sign) ,
therefore each is the conjugate of the other, proving that
Fina11y, x
LeIllD8 1.
= -l/Y
and
y > l, therefore -1 < x <
Any quadratic irrationa1
x
o.
xy = -1.
Il
may be expressed in the form
(4)
x =
where
y
P, Q, D are integers,
For any such representation,
D > 0 ,and
x
D is not a square.
is reduced if and on1y if
-43-
Given positive nonsquare
P <.JD
(5)
Q-P <.JD
(6)
P-fQ > .JD
(7)
D, there are precisely
quadratic irrational of the form (4),
where
h
h(h+l) reduced
= [ .JD
J(integral
part of .JD) •
~d Jk
Proof:
.! +
Assume
(4).
b
2
cao he written as (ad + /b 2c k) 1 (bd).
(i)
let
x
be reduced, i.e.
p+.JD >1
(8)
Q
P-.JD< 0
(9)
Q
P -.JD
(10)
> -1 •
Q
Q < 0 ; then (8) shows
Suppose
contradicting (9).
(8)-(10).
(ii)
Subtracting
Let
(5) and
Therefore
P < 0;
but then P - ~D < 0 ,
Q > 0, and (5)-(7) follow easily from
(5)-(7) be satisfied.
(7) gives
Geometrically, (5) - (7)
Q > 0, so that (8) - (10) follow.
mean that
(P,Q) is a lattice point
strictly inside the triangle who se vertices are
(.JD, 0), (0 "rn),
(../D, 2.JD).
Clearly tbe number of such lattice points is
h(h + 1).
Il
Lemma 2.
If
x
2(1+2+ ... +h)
is a reduced quadratic irrational, then so is -l/x
Proof: -1 <x < 0, tberefore -l/x > 1 ; x > l, hence -l/x, which i8
the conjugate of -l/x, lies between -1 and O.
1/
=
-44-
Lemma 3.
Let
x
be a quadratic irrationa1 satisfying
2
ax + bx + c
where
a (Fa) , b, c
are integers , and let
2
2
b -4ac = k D and
integers such that
=0
(11)
k
D be any
and
k divides
2a, b, and 2c.
Then given any sequence of integers
and
(i
?
(12)
1)
are of the form
(i
where
Pi
Proof:
x
=
and
Qi
~~2~)
1 2a 1s of the form (13).
prove that each
Xi
(ai
b~
where
'1= 0)
=s +
we fi n d
f
k
= a,
1/y ,
an~
2d, e, 2f
x
y = 1/(x-s) (s any integer) does.
substituting this into (11) and simplifying,
0
=,
where d
a1so, one can check that
contradicting that
Thus, it is sufficient to
2
dy2 + ey + f
divides
a,b,c, it is c1ear that
satisfies an equation ai Xi + b i Xi + Ci = 0
2
- 4a i c i = k D and k divides 2a i , bi' Ci .
For this, it is enough to prove that
Now x
(13)
1)
are integers.
Given (11) and the assumption about
(-b
?
= as 2 + bs + c,
2
e _ 4df
= b2
- 4ac
e
= k 2D
fina11y, d ; 0, for otherwise
is irrational.
= 2as + b ,and
e
2
; clear1y
= k 2D
,
Il
Theorem 9.3. The simple-continued-fraction expansion of any reduced
quadratic irrat10nal is purely periodic.
irrationa1 i9
a(Fa) , b, c
2
[al' a , ••• ] and satisfies ax + bx + c = 0 where
2
2
2
are integers vith b - 4ac = k D and k dividing
x
=
Furthermore, if the quadratic
-45-
2a, b, and 2e, then eaeh mn = [an , a n+l, •.• J (n
irrational of the form
+JO) / Q.
(P
n
(P ,Q
n
fundamental period does not exeeed
h(h+l) ,
Proof:
Then
m
In lemma 3, take
is of the form
n
an+l
?
(P
n
si = ai'
+JO) /
Q.
n
Xi
Assume
n
?
l)
is a redueed quadratie
integers), and the
where h= [JOJ •
= mi
and therefore eaeh
mn is redueed.
Now
mn+l >
1 ; also
.,
1
(14)
m - a
n
n
but
0
< - mn < l,
Le. a
now
that
x
x> l, sinee
= ml
a
n
< a n-mn < a n+1 , henee by (14) l/(an+1) <
> 1
n -
x
n = 1 this follows fram the faet
(for
is redueed), therefore
is redueed, therefore every m
n
m +l
n
i8 redueed.
is redueed.
But
By lemma l,
there are only h(h +l} redueed quadratie irrational of the form (4) ,
henee there exist r > 1
t < h (h+l) •
-
suppose
b,
n
and
t > 1
-
Choose the smallest possible r.
By leuma 2,
Then
and
r = 1, for
n
br = a r- 1 + lIbr- l'
bn
m
=m
r+t
r
so that, denoting -l/m by
r > 1.
we have
sueh that
is redueed, henee
Similarly br + t
= a r +t - l
bn > l , therefore [b r ]
+ l/b r +t _ 1 ,
= a r- l'
br - br+t ,this shows that a r- 1 = a r+t- l'
and eonsequently m _ l = m +t - , eontradicting the minimality of r.
r
r
1
Sinee
Thus
ml =
h(h+l).
ml~'
proving that
x
is purely periodic with period t <
1/
The following result completes the proof of Lagrange's theorem,
-46-
The stmp1e-continued-fraction expansion of any
Theorem 9.4.
quadratic irrationa1 is periodic.
irrationa1 is
a, b, c
= [al'
x
and 2c, then each
mn
2
b -4ac
= [an'
(Pn' Qn integers), and a11
= k 2D and
dividing
k
n
In view of
for which
+ J'ô)/Q
n
n
from some point onward are reduced.
m
h(h+1) , where h
As in the proof of Theorem 9.3, the fact that
fo11ows from 1emma 3.
2a, b,
[p
an+1'···] is of the form
The fundamenta1 period does not exceed
Proof:
ax 2+bx+ c = 0 where
and satisfies
a 2 ,··.]
are integers with
In fact, if the quadratic
m
mn
=
J'ô] .
(Pn+ J'ô)/Q n
Theorems 9.3 and 9.2, it remains
on1y to find an
n
does not exceed
h(h +1) is c1ear from 1emma 1.).
n
=[
is reduced.
(The fact that the period
Since
x
=
[a 1 , .•. ,an ,mn+1 ],
we have
x
=
mn+1 Pn + Pn-1
(n
?
(15)
0)
mn+1 qn + qn-1
Taking conjugates and s01ving for
mn+1 '
=-
=-
(16)
x - c
Now as
n
n
increases, each convergent is a1tanate1y sma11er and 1arger
than the previous one, therefore
( x - c n- 1) / (
x - cn )
is a1ternate1y
sma11er and 1arger than 1 ; since this fraction converges to (x-x)/(x-x)
there exists
n
such that
o < x - c n-1 < 1 •
x - cn
qn-1
~
qn ' hence (16) and (17) give
-1 < mn+l < 0 .
(17)
= 1,
-47-
A1so
mn+1 > a n+1
Theorem 9.5.
?
Let
let the integers
1, therefore
x =
[8
1
,
8
2
mn+1
, .•• ]
Il
is reduced.
be a quadratic irrationa1, and
a, b, c, k, D, Pn' Qn
be as in Theorem 9.4.
Then :
P
n+1
(n
?
1)
(18)
(n
?
1)
(19)
[Pn+1 + .rD] (n
?
0)
(20)
= a nQn
D - Pn+1
Qn + 1 =
a n+1
=
- Pn
2
Qn
Q
n+1
(n
Proof:
(20) fo11ows from the fact that
To prove
(18)
and
(19)
a n+1
=
(P +!D)-a Q
n
(21) and (22)
x
by
1)
(23)
[mn+1].
we note that
-Q (a
=
=
?
n n
n
-P
t)
n'tl
(a Q -P )
n n
2
n
n
+ ~.1D)
- D
are easi1y obtained by solving (15) for
(Pl + .rD) 1 Q, stmp1ifying, and using
mn+1' rep1acing
PnQn-1 - Pn-1 Qn
(23) and (24) may be found from (15) by rep1acing
n
= (_l)n.
by n-1 ,x by
-48-
(P
n
+
JD)
With regard to the value of
x
JiN)
D = N.
= (A
take
+
For example let
not of the form
=2
D, it should be noted that if
al
= l,
x
and
(P + Ji)/Q
= (1 + Ji) 1 2 (which, incidentally,
m = l/(x-a ) = 2 + 2 Ji which is
2
l
_ As indicated in the statement of the
D should be chosen by examining the quadratic equation
satisfied by x.
k
Il
1 B (A, B integers) , it is not alwaya possible to
is reduced) ; then
theorems,
1 Qn , and crossmultiplying.
lt is always possible to t ake
is useful, notably for
x
k
=
l, but sometimes
= JN .
Formulas (18) to (20) above provide a very convenient algorithm
for expanding a quadratic irrational into a simple continued fraction
find
D,
get Pl
and then use
some pair
and
QI
from
x
= (Pl
+ ~D) 1 QI ' obtain al = lx] ,
(18) - (20) to compute P2 ' Q2' a 2 , P3 ' Q3' a 3 , .•. until
P ,Q
n
n
repeats a previous pair.
Formula (22) is the basis
for the solution of the pell equation, as will be seen in later chapters.
-49-
Chapter 10.
THE EXPANSION OF
In the case where
x
JO.
is the square root of an integer, the
simple-continued-fraction expansion has a particularly interesting
form, which we now examine.
Theorem 10.1 Let
[al' a 2 ,···] , mn
the form
D be a positive nonsquare integer, and let
= [an'
a n+ l ,.·.] ,
h
= [JO] .
x
= JO =
The expansion is of
(1)
where
0 5 r
~
h(h +1) - l ,
1 5 b i 5 h , and
(b l , b2 ,.·· ,br) is
is of the form
Each m
symmetric, i.e. b i = br +l - i •
n
(n
?
(2)
1)
where
Pn' Qn
are integers satisfying, for
1 5 Qn
~
After
2h.
QI
n > 2 , 1
5
Pn
~
h ,
= l, Qr+2 is the first Qi to equal 1.
We have the following formulas
Pn+ 1
= a nQ
(n
?
1)
(3)
Qn+l
=
(n
?
1)
(4)
(n
?
1)
(5)
n - Pn
2
D - Pn+l
Qn
An+l
(-1)
(-1)
Pn-l
Dqn-l
=[Pn+l
+ hJ
Qn+l
n
n
Pn+l
= Dq n
Qn+l
= Pn2
= qn-l
= Pn-l
(n
q n-l - Pn Pn-l
- D qn2
Pn + qn-2 Qn
Pn + Pn-2 Qn
? 0)
(6)
(n
?
0)
(7)
(n
?
1)
(8)
(n
?
1)
(9)
-50-
Proof: x
satisfies
we have (2).
Pl
(18) - (24)
that if
x
=0
2
=0
- D
= 1,
QI
and
, therefore taking
so that
k
=2
in Theorem 9.4,
(3) - (9) fo11ow from
of Theorem 9.5 (with regard to (5), it is easi1y seen
v
is any rea1 number and
m, n are integers with
m> 0 ,
then
(n : v]
hence the
JO
=
(n: [v]] ;
of formula (20), Chapter 9, can be rep1aced by h ) .
By 1emma 1 (Chaper 9),
h +
therefore by Theorem 9.3,
JD
is reduced ; a1so [h +
h +
JD = [2h,
b , .•• ,b ]
1
r
JO]
= 2h ,
(r ~ 0).
[2h, b , ••. ,b , 2 ] , hence JO = [h,b , .•• ,b ,2 l ,
1
r h
r
h
1
r + 1 to be the fundamenta1 period, Theorem 9.4
This can be rewritten as
proving (1).
Taking
r ~ h(h +1) -1.
states that
To prove symmetry, we have -h + ~D
[O,b 1 ,··· ,br' 2h] , or , taking reciproca1s, -1 / (h-
= [br' •.. ,b 1 ,2hl
but by Theorem 9.2, -l/(h-JD)
unique, we conc1ude that
=
is symmetric.
~,
(b , ••• ,b )
r
1
Since (by Theorem 9.2)
of Chapter 9 gives
(for n
P
~
.Ji» = [b 1 ,··· ,b r ,2 h l
; since expansions are
(b , •.• ,b ) , Le. (b , ... ,b )
1
r
r
1
m , ... are reduced, 1emma 1
3
2)
JO
<
n
=
(10)
(11)
(12)
P < h ; a1so
Therefore
n-
0
'n
-
subtracting (10) and (12) gives
gives
and
P
Q
n
n
> 0 ;
=1
P < h, hence adding gives
n-
Qn-< 2h ;
Qn > 0 ; subtracting (11) and (12)
thus 1 < P < h and
n -
; then (10) and (12) give
1
<
Q < 2h.
n-
JO -
1 < P <
n
Suppose
JO,
n
>2
therefore
-51-
P
n
=h
; thus
=h
m
n
+
JO ;
but
hence, considering the expansion
for which mr+2 = m;+2 =
n
is at 1east
r + 2.
therefore by (5) ,
(l~i~).
m~
= h +
r + 1 is the fundamenta1 period,
h +
JO =
JO
(primes refer to h +
[2h, b , ••• ,b ]
r
1
Fina11y, this shows that
~
aj
(Pj+h) 1 Qj
~
(h+h) 1 2
= h,
(2~j~+1),
i.e. b
i
~
h
Il
1.
J5 = [ 2,4
2.
J8 = [
Examp1es
3.
=
D
13,
P
n
a
Ji13
4.
n
=
=3
1
2
3
4
5
6
o
3
1
2
1
3
1
4
3
3
4
1
3
1
1
1
1
6
= 31,
n
a
h
[3, l, l, l, l, 6]
D
P
]
2, l, 4 ]
n
n
n
h
r
=4
=5
1
2
3
4
5
6
7
8
9
o
5
1
4
5
5
4
1
5
1
6
5
3
2
3
5
6
1
5
1
1
3
5
3
1
1
10
Ji31 = [5, l, l, 3, 5, 3, l, 1, 10]
Theorem 10.2
fo11owing :
Qj? 2
JD) ,
r
=
7
Using the same notation as in Theorem 10.1, we have the
-52-
(ii)
If
(Ui)
If
(iv)
(v)
Qn = Qn-l
and
n<r + 2,
then
r = 2n - 4
P = P
and n<r + 2, then r = 2n - 5
n
n-l
If Qn = 2 and n < r + 2, then P +
P ,hence r = 2n - 3.
n l =
n
If bi=h' then i = (r+l) 1 2. (Le. only a central term of the
symmetric part can equal
Proof:
Let
YI ~ Y2
P +
mean YI
y;
JD _ R + JD
Q
h.)
= -1
one can check that
<=> P = R, QS = D _ p 2 .
(13)
S
In particular, using
(4) , we have that
(for i, j
? 2)
<=>
Now let
2 < i < r + 2;
(14)
using Theorem 9.2 and the symmetry of
~
[ bi_l'··· ,br' 2 h , b l ,···,b i _ 2 ]
(b l ,··· ,br>' mi =
=
= m +4-i' therefore (14) clearly shows that
r
(QI' Q2,···,Qr+2)
are symmetric.
shows that
therefore
m = m
.
n
r+4-n'
shows that
n-l = r+4-n,
Q = 2
n
since
and
ml
~ ~
n
~
r + 1
m
n-l
and
r + 2, and
but, as just shawn
m-n ~mn- 1 ' hence
impossible (it would make
Assume
m -m ;
n
n
m-m +4
n r -n
is the fundamental period, this
n = r + 4 - n , i.e. r = 2n-4.
Pn = P
n- l ' then
conclude
Suppose
(P 2 , ••. ,P +2 )
r
Stmilarly, if
= mr+4-n
equal to a la ter
;
now
~+2
n = 2
is
mi) , therefore we
i.e. r = 2n-5 .
n < r + 2.
Formula
(3)
gives
(15)
and
-53-
The symmetry of
gives
(P2 ,··· ,Pr +2) , (Ql"" ,Qr+2) , and (a2, ••• ,ar +l )
Pn+l = Pr +3- n , Qn = Qr+3-n ' and an = a r +3- n ·
-
= Pn+l +.ID
Qn
Therefore:
Pn +.ID
Qn
= Pn+l - Pn
Q
_n
But
< Qn = 2
0 < mi - ai < l, hence
•
(15) shows that
Pn+l - Pn is even, therefore Pn = Pn+l , and r = 2n-3
follows by (iii). Finally, Theorem 10.1 shows that Q + 1 (z<n<r+l) ,
n
- -
while Q > 2 implies (by (5) ) a < Ih/Q < h ; therefore if
n
nn
we must have
b
i
= h,
Il
Qi+l =2 , hence r = 2(i+l) - 3 , i.e. i = (r+l) / 2.
Formulas (3) - (5) provide a practial method for expanding
JO
P
Q and a are bounded,
n'
n
and no irrational numbers appear in the formulas, this method 1s
as a simple continued fraction.
Because
n'
especially convenient for rapid automat1c calculation.
(1) - (1ii)
Furthermore,
of Theorem 10.2 show that only about half of the period
needlto be calculated.
When r
is odd, (1) is said to have a central term (namely b.,
where i = (r+l) /2) •
terme
1
When
r
1s even, (1) 1s said to have no central
-54-
Chapter 11.
THE PELL EQUATION.
The Diophantine equation
given integers and
equation
x
and
x
2
Dy2 = N, where
D and
N are
are unknowns, is known as Pe11 1 s
y
(John Pe11, 1611-1685), although Pell was not the first
to consider it.
It appears in the famous cattle problem of Archimedes
(see [1] p.249),and the Hindus, as long ago as 800 A.D., apparently
cou1d solve various cases of the equation, but it remained for Lagrange
(1736-1813) to give a complete and elegant analysis of it, about one
hundred years after Fermat (1601-1665) proposed the problem to the
English mathematicians of his day.
The pell equation is important for several reasons.
By means of
various substitutions, the solution of the general quadratic Diophantine
equation
ax
2
+ bxy + cy
2
+ dx + ey + f
the solution of Peilis equation.
=0
can be made to depend upon
Knowledge of the structure of the set
of units in the field extension of the rationalS by...ID
positive nonsquare integer ; x
=a +
b ~D
(D being a
is said to be a unit if
x
and l/x satisfy.:a monic quadratic equation with integer coefficients)
depends upon a thorough knowledge
+ 4.
N= + 1
and
Other applications include the minÜDization of indefinite
quadratic forms
(see LeVeque
We sball take
primarily on the case
(When
[51
, Chapter 8).
D to be a positive nonaquare integer, and concentrate
N
fraction expansion of ~D
exist.
of Peilis equation for
=+
1.
It will be seen tbat the stmple-continued-
conveniently furnishes aIl solutions, if any
D is negative or a square the solutions are finite in
number and usually not difficult to find directly, especially when
N
-55-
is sma11.)
x
2
2
- Dy
It shou1d be noted that if
=N
= k 2E
D
, the solutions of
fo11ow immediate1y fram the solutions of
thus it is sufficient to consider on1y square-free
x
D;
2
2
-Ey
=N
;
however,
it will be just as easy to treat the genera1 case.
Theorem 11.1.
of
x
for some
2
- Dy2
=N
with
P ,q
n
=1
(x, y)
, satisfies
x
N > O.
y
and factoring ,
N
"2
y
<
N
1
<
y
2y2
Therefore by Theorem 7.2, there exists
= pn ,
y
= q.
n
-N
For the case
y2 - Ex
c1ever argument.
=
1NI <
JD
as auove gives that y/x
which is
= qn
have
N
=
2
Dividing by
JD , and we
x/y?
x -
M > 0, and
y
refer to the simp1e-continued-fraction
n
=
hence x
= Pn'
JD.
First assume
Therefore
N
(i.e. x > 0, y > 0) solution
Then any positive
n?1 , where
expansion of
Proof:
D be a positive nonsquare integer, and let
1 N 1< JO.
satisfy
x, y
Let
0) ;
since
convergent of ~D.
Theorem 11.2.
Let
2
= M,
n > 1
x/y
= pn /q n ,
N < 0 , we use the fo11owing
where E = l/D
gives
M<
Ji:
is a convergent of
JD > l,
such that
and
M = -N/D ; now
therefore the same argument
l/Jn
(and not the first,
Theorem 5.1 gives that
x/y
1s a
/1
D be a positive nonsquare integer, assume the
notation of Theorem 10.1 and consider the Pe11 equations:
-56-
x
x
2
2
-
(1)
2
.- Dy
(2)
=-1
(1) has infinitely many solutions; if
JD
of
r
is odd (i.e. the expansion
has a central term), aIl positive solutions of (1) are
(Pn' qn) , n
if
r
= k (r+l ) , k
= 1,2,3,...
;
is even (i.e. no central term), aIl positive solutioœof (1) are:
(Pn' qn) , n
(2) is solvable if and only if
positive solutions of
r
1 <
JO,
(r+l) , k
= 2,
is even ; if
r
4, 6, •..
is even, aIl
(2) are
(Pn' qn) ,
Proof:
=k
n
=k
(r+l) ,
k
= 1,3,5, •.•
and any solution of (1) or (2) is relatively prime,
hence by Theorem Il.1, any positive solution of (1) or (2) is
for some n > 1.
(Pn' qn)
Formula (7) of Theorem 10.1 gives
(n ~ 1)
Therefore
and
Qn+l
(P
n
' qn)
= 1.
is a solution of (1) if and only if
But by Theorem 10.1, Pn+l
(3)
n
is even
= 1 (where n ? 1) is
equivalent to n
= k (r+l), where k? 1. The assertions about (1)
clearly fo1low.
For (2), we first note that no solution can have
x
=0
or y
= O.
if and on1y if
(k
? 1).
Now
n
is a solution of (2)
(3) shows that (p n , q n )
is odd and
Qn+1
= l,
i.e.
Therefore there are no solutions if
n
r
odd and
n
= k(r+1)
is odd, while for
even, the positive solutions are as stated in the theorem.
r
Il
The remainder of this =hapter shows how aIl positive solutions of
-57-
(l)and (2) may be found once the least positive solution is known.
First, let us clarify the notion of "least" positive solution by
(xl' YI) and (x2 , Y2) are different positive
2
x2 - Dy = N (where D > 0), then either xl < x and
2
observing that if
solutions of
,
YI < Y2
or
therefore assume
hence
and
x 2 < xl
YI < Y2 ·
= x 2 ' then also YI = Y2 '
2
2
= x 2 - xl > 0, and
Y2 < YI :
if xl
2
2
DY2 - Dy l
xl < x 2 ; then
Secondly, given xl and YI, an equation
x n + Yn JD= (xl + YI JD)n
X
uniquely determines the integers
n
and
(4)
Y ,since JD is
n
irrational and we can equate terms after expanding the power; in fact,
denoting
Xl + YI JD
by u
X - Y JD =
n
n
n
(un - v ) / (2JD).
we have
Yn
=
Le1ŒD&:
denote
Let
JD
and its conjugate Xl - YI JD
n
un = un = v , therefore
D be a positive nonsquare integer, and for convenience
lei = 1fi = 1
Let
by".
m
< s + t
Then there exists a positive solution
a + b IK < x
by
1
l
(Xl + YI,,)m
m
2
et
< (Xl + YI ., )
~l
.
(5)
2
2
m
(a,b) of x -Dy = e f , such
III •
1/ (Xl + YI
We have
l < e
+ Y
2
x 1 - DYI = e,
,
m? 1, and suppose
(xl + YI • )
Proof:
= x~
For example, x 2
(Xl' YI' s, t positive),
that
by v,
G(
)
= e (Xl - YI
1(
)
,
so that dividing (5)
gives
(s + tGl) (Xl - YI" )
m
< Xl + YI oC·
(6)
-58-
a + b CI(. Then a + b 0( < Xl + Yl4iW' ,
2m 2
2
2
2
m
(a + b«) (a - b-c) = e
(s - Dt ) (Xl - Dy l ) = e f.
Let the middle member of
2
2
=
(6) be
and
a
Now
1/ (a + beC) = 1a-b4Ifl , therefore
But
2a = (a + bit) + (a - b-C)
- Db
Theorem Il.3.
Let
and
b > O.
//
(xl,y ) is the least
l
(1), then aIl positive solutions are (x n ' Yn)
positive solution of
by
a >0
D be a positive nonsquare integer, ~ = JO, and
consider the Pell equations
determined
- ~I <1 < a+b« •
2b-c= (a + bec') - (a - b __ ),
and
2a > 0, 2bot >0, fram which
hence
1a
(6) gives
(1)
(4), where
and
(2). If
n = 1, 2, 3, •••
positive solution of (2), then aIl positive solutions are
determined by
(4), where
n
(x , y )
n
n
= 1,3,5, •.• , and furthermore (x 2 ,y2) is
the least positive solution of (1).
Proof:
If x n ' Y
n
=
(X
- y IJ()
if
(xl' YI)
n
n
n = 1,2,3, .•.
Also, if
(Xl' YI)
n
is a positive solution of
(2),
= 1,3,5, ••• , and furthermore (xn'Yn )
= 2,4,6, •.• are positive solutions of (1).
n
(i)
let
n
(xn ' Yn) for n
then so are
for
are defined by (4), then x2n - Din = (xn + Yn ,,)
n
n
2
2 n
(Xl + YI lit) (Xl - Yl «) = (Xl - Dy l ) . Therefore
i8 a positive solution of (1), so are (x , Y ) for
Suppose
(Xl' YI) is the least positive solution of (1), and
(s, t) be any positive solution.
s + t De
? Xl
+ YI II(
,
Now Xl + YI " > 1 and
therefore there exists
(x ' Y ) , for otherwise (5) holds for some
n
n
provides a positive solution
(a, b)
n > 1 such that (s ,t) =
m > l, and the lemma
of (1) less than
(Xl' YI) .
-59-
.
(xl' Yl ) is the least positive solution of
(sl' t l ) is a positive solution of (1) less than
Suppose
(ii)
(l) , and that
(x2 ' Y2 )·
If
> xl + Yl « • define (s,t)
sl + tIC
= (sl,t ) ;
l
sl + t « < xl + Y -< , and there is an r such that
l
l
r-l
r
(sl + tlOC)
< xl + Yl -< < (sl + tlOC) ,in which case define
if not, we have
= (sl + t l -<) r • In either case, (s, t) is a positive
solution of (1) satisfying (5) for m = l, therefore the lemma gives
s + t «
a positive solution
(a,b)
of (2) less than
(x 2 ' Y2)
contradiction proves that
(xl' Y ). This
l
is the least positive solution
of (1).
(iii)
Suppose
(xl' Y ) is the least positive solution of (2) ,
l
and (s, t) is a positive solution of (2) not among (xl' Y ) ,
l
(x ' y ),... Then s + t" exceeds
xl + Y ~ and is not a power of
3
3
l
xl + Y10( , therefore (5)
positive solution
less than
holds for
(a, b) of (2) less than (xl' Yl ) , or else of (1)
(x ' Y ) (since xl + Y e( < x + Y 0(
2
2
l
2
2
even or odd.
),
according as
M is
The former is tmpossible by assumption, the latter by (ii).
It may be noted that is
and
The lelllDa gives a
m>l.
SOUle
(a, b) is a solution of
2
x2 - Dy = N
(xl' Y ) is a solution of
l
x
is also a solution of
+ y
n
x
2
n
JD =. (x1
- Dy
2
+ y
i
JD)
x 2 - Dy2 = 1
then (xn ' Yn ) given by
(a
+ b
JO) n
= N. However, there is no assurance that
all solutions can be obtained in this way from one known solution.
/1
-60-
Cbapter 12.
THE SOLVABILITY OF x
Since the Pell equation x
2
2
2
- Dy
= -1.
- Dy2 = -1 is solvable if and only
if the sfmple-continued-fraction expansion of
(Theorem Il.2),
those
JD
bas no central term
there naturally arises the problem of characterizing
D for which
JD
bas no central term.
As yet, there is no
complete solution of this problem, but some important partial results
are known, and this chapter is devoted to presenting these results.
The proofs given here are based on the material in Perron
[a] ,
Cbapter 3, Theorems 20-22.
Theorem 12.1.
Let the expansion of JiD
assume the notat im of Theorem 10. 1.
=
where n
have no central term, and
Then
2
2
D = Qn+l + Pn+l
(r + 2)/2, and (P n+ l , Pn+ l ) = 1.
(1)
Therefore (see Chapter 5)
D is a product (possibly zero factors, i.e. the product is 1) of
primes of the form 4k+l
Proof:
gives
Since
Q
n
r
= Q.n+l
To show that
or twice such a product.
is even, symmetryof
(QI' Q2"" ,Qr+2)
(Theorem 10.2)
,fram which (1) follows by formula (4) of Chapter 10.
Pn+l and
Qn+l are relatively prime, we note fram (7)
of Cbapter 10 that
2
Qn+l = pn - D qn2
2
(_l)n-l Q = 2
Pn-l - D qn-l
n
(-1)
Therefore if a prime
D
by (1) , and
p
p
n
divides
divides
Pn
Qn = Qn+l
and
Pn-l
and Pn+l' then
by
(2) and (3) ;
(2)
(3)
p
divides
but
-61-
= l,
(Pn ' Pn- l )
sinee
Pnqn-l - Pn- l qn
= (_l)n
formula (5) of
Il
Chapter 2).
The converse of the above theorem is false.
205
= 5.41
and
34
= 2.17
For example
are sums of two relatively prime squares,
but
Ji05 =
[l~, 3, 6, l, 4, l, 6, 3, 28 ]
~34 = [5, l, 4, l, 10 ]
and
have central terms.
Rowever, we have the following results, leading
up to the main theorem, Theorem 12.4.
Theorem 12.2.
D > 3 be nonsquare.
Let
x
x
x
2
2
2
- Dy
2
2
(4)
=-1
- Dy2 =
- Dy
Of the three equations
2
(5)
= -2
(6)
at most one has a solution.
Proof:
(mod 3) ;
Any square is convergent to
henee for
D = 3,
0 or 1 (mod 4) and to
(4) and (5)
0 or 1
are not solvable (eonsider
the terms module 4 and 3 respect ive ly), while (6) is solvable:
2
12 - 3.1 = -2.
Therefore assume
no solution has
x
=0
or
y
D? 5 ,henee
= 0,
2 <
.JD • Clearly
and also any solution has (x,y)
= 1.
Therefore, using the notation of Theorems 10.1 and 10.2, Theorem 11.1
gives that any solution
for seme
n > 1.
x,y
Renee, if
of Theorem 10.1 shows that
is of the form
(5) or (6)
~+l
= 2,
Ixl
= pn , 1yi
= qn
is solvable, then formula (7)
and Theorem 10.2 (iv)
shows that
-62-
n + 1
where
k > O.
Therefore
sinee
(-1) Qn+1
n
=2
=
r + 3 + k (r+1)
(7)
r
is odd, and (4) is not solvable.
A1so,
2
that (5) is not solvable if
solvable if
(r +3) / 2
is solvable.
n
for (5) and
(-1) Qn+1
(r +3) /2
is odd.
= -2
for (6), (7) shows
is even, whi1e (6) is not
Therefore at most one of (4) - (6)
Il
Theorem 12.3. Let nonsquare
D be a power (first or higher) of an
odd prime, or twiee sueh a power.
Then one and on1y one of (4) - (6)
is solvable.
Proof:
Let
Sinee the ease
• where
2p,
D = P• or
is an odd prime and
D>5
(4) - (6)
(so that
2 <
JO)
n
= a nQn
?1
.
and show at 1east
ia solvable. Suppose (4) i8 not solvable.
is odd, and by Theorem 10.2 and (3) of Theorem 10.1,
2P
~
D = 3 wa8 treated in the proof of Theorem 12.2, it
is suffieient to assume
one of
p
,where
n
= (r+3)/2.
Then r
Pn = Pn+1 and
Thus (8) and (9) of Theorem 10.1
give :
(8)
2pn- 1 = (qn- 1 an + 2qn- 2) Qn
(9)
Therefore
Qn / 2Pn_l' 2Dqn_1'
k / 2Pn_1 ' qn-1;
ia fa1se sinee
(i)
Suppose
k / Qn' qn-l ;
k >1i8 odd, then k 1 Pn-l' qn- l' whieh
if
= 1;
(Pn-1' qn-1)
1f k > 1 is even, then
(11)
2 1 Qn' ~-1 ' henee by (8) 4 12Pn_1 ' therefore
1s fa1se.
Therefore
(0, q
'Il
then
n-
(formula (7) of Theorem 10.1)
therefore
1)
21pn- l' qn- l' wh1eh
= 1 ,henee Qn 12pn- l' 2D. Now
(-1) n-1
~
2
2
= Pn-1 - D qn-1 '
-63-
2
(_l)n-l 4= Q" ( :n-l)
2
(10)
n
It follows that
(Q
,2D/Q) = 1, 2, or 4;
n
n
.,
(since 2 <p < r + 1), and 2D = 2p
that
(A)
Q
n
Qn
Qn
Q +1
n
therefore one finds
4p
= 2D, D, D/2, 4, or 2.
cannot be
2D or D
if
Q > D, hence P > D/2 .
n n (B)
.,
or
- -
also
cannot be
even, and
P
n
= 2D
or D, then
But (Theorem 10.1)
2P
n
P < h <
n -
= a nQn ->
JO < D/2.
Q = D/2, then Qn is odd, hence a
n
n
(a 12) Q > Q = D/2, leading to a contradiction
n
n - n
D/2 :
=
Q
n
if
as in (A).
(C)
Qn
if Q
n
cannot be 4 :
= 4,
then
2Pn_l ' hence 2/Pn_l' so that
(10)
2D/Q
qn-l
n
is odd, 4
= Qn
divides
is odd, and each term of
except the last is divisible by 4, which is impossible.
The on 1y rema i n i ng poss ib il i ty i s Q =.
2 Then (-1) n-l 2 = P2 2
n
n-l
Dqn-l (formula (7) of Theorem 10.1), so that (5) or (6) is solvable.
Theorem 12.4.
Let nonsquare
D he a power (first or higher) of a prime
of the form 4n+l , or twice a power (first or higher) of a prime of
the form
8n+s.
Then x
2
-
Dl
=-1
is solvable.
Proof: In view of Theorem 12.3, it is sufficient to prove that (5) and
(6) are not solvable •
.,
(i)
Let
D=P
Therefore
i.e.
(et:? l, p = 4n+l) .
x
2
Then D = 1 (mod 4).
2'
- Dy = 0-0,0-1, 1-0, or 1-1
0,1, or 3 (mod 4).
(6) are not solvable.
Il
But + 2
=2
(mod
(mod 4),
4), therefore (5) and
-64-
(ii)
x
2
Let
ii:!"
(~) and
= 8n +
Then x
2
(~)= 1 or (-~) =
(-!) given
in Chapter 5,
2
- Dy =
~2
tmplies
But,recalling the
1.
this is impossibe for
Il
5.
Corollary:
= 8n+5).
(G(? l , p
2 (mod p) ,therefore
values of
p
c
D = 2p
If P
is a prime of the form
4n+l, then x
2
solvable, therefore by Theorem 12.1 the representation of
sum of two squares can be found by expanding ~p
2
- py = -1 is
p as the
as a simple continued
fraction.
For example, let
have
r
= 4,
p = 13.
From Example 3 after Theorem 10.1, we
(r + 2)/2 = 3, therefore
_ Q2 + p2
P - 4
4
This construction for expressing a prime
=
p = 4n + 1 as the sum of
two squares is attributed to Legendre (1808) (see [7], Appendix l ).
The converse of Theorem 12.4 is false.
and
2
2
x -2.41 y
For example, x 2-5.l7y2 =-1
= -1 are solvable, since
Jas = [
and
J82 =
have no central terms.
9, 4, l, l, 4, 18 ]
[9, 18 ]
-65-
CONCLUSION
It is hoped that the preceding chapters have demonstrated the
utility and elegance of the theory of continued fractions as a tool
in approximation theory and the theory of numbers.
For although new
methods have recently been developed in the field of Diophantine
approximations, continued fractions remain the basic stepping stone,
while in elementary number the ory they provide one of the very few
direct methods.
There are, of course, many topics in this area which the
present survey has not discussed.
For example, continued fractions
can be used to assist in factoring numbers (see [1] p.266), and no
mention has been made of the beautiful subject of the geometry of
numbers, which is closely related to continued fractions.
are many directions for further study.
There
Hurwitz's Theorem (Chapter 7)
is the first of a whole series of related theorems and problems.
One could explore continued fractions themselves in greater detail by
referring to such books as Perron [8].
Alternatively, there is the
extension to analytic continued fractions.
Finally, there are two challenging problems introduced by the
material presented in this survey, problems which may provide subjects
for further research.
We have shawn that the length of period of the
sfmple-continued-fraction expansion of a quadratic irrational does not
exceed
h (h+l)
quite crude.
(see Theorem 9.4).
However, this bound appears to be
For example, it means that the period of
JD
is less
-66-
daa .....
D, wbi.le for
D = 91.9 - t D
(fiK'
= 991),
D ~ 1000
the 1argest period is 60
and .ost periods are much less than 60.
Yery Uttle see.a bI he ....... about this topie.
~ï..
zZ _
• ia Clapter 12, characterization of those
~Z = -1
-
of ....:.
deep
Seeondly, as
is solvable
(i.e. those
is 0IId) is far fraa cOliplete.
~,
D
D
for whieh
for wbieh the period
Tbeorem 12.4 is a fairly
bal: .-e iDl:lusboe resu1ts may be found.
-67-
BlBLlOGRAPHY
1-
Be 11er,
A. H.
Recreations in the Theory of Numbers, 2nd ed.
Dover, 1966, N.Y.
2.
Hardy, G.H. and Wright, E.M.
Chapter 22.
An Introduction to the Theory
of Numbers, 4th ed.
Oxford, 1960.
Chapters 10, H.
3.
"Ueber die angeniherte Darste Hung der
Hurwitz
Irrationalzahlen durch rationale Brûche,"
Mathematische Annalen,
4.
39(1891),279-284.
Continued Fractions
Khintchine, A. Ya.
3rd ed.
Noordhoff, 1963, Groningen. Sections l, II.
5.
LeVeque, W.J.
Topics in Humber Theory, Vol. 1.
Addison-Wesley, 1956, Reading, Mass. Chapter 8.
6.
Niven, 1. and Zuckerman, B.S.
An Introduction to the Theory
of Numbers, 2nd ed. Wiley, 1966, N.Y. Chapter 7.
7.
Olds, C. D.
8.
Perron,
o.
Continued Fractions.
RandOm House, 1963, N.Y.
Die Lehre von den Kettenbrüchen,
Chelsea,
1929, N. Y.
2nd ed.
Chapters 1-3.
9.
Vinogradov, I.M.
Elements of Number Theory.
10.
Vorob'ev, N. N.
Fibonacci Numbers.
Dover, 1954, N.Y.
Blaisdell.