Inverse problem for Pell equation and real quadratic fields of the least type

# Inverse problem for Pell equation and real quadratic fields of the least type

Park, Jeongho Department of Mathematics (Room 117), Pohang University of Science and Technology, San 31 Hyoja Dong, Nam-Gu, Pohang 790-784, KOREA. Tel. 82-10-3047-7793.

###### Key words and phrases:
Real quadratic field, Continued fraction, Pell equation, Symmetric sequence, Least type
###### 1991 Mathematics Subject Classification:
Primary:11J68, Secondary: 11R29, 11Y40
Supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MEST) (2010-0026473).
Supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MEST) (2010-0026473).

## 1. Introduction

Let be a non-square positive integer, and

 D={dif d≡1 mod 4,4dotherwiseωd={1+√d2if d≡1 mod 4√dotherwise.

By Dirichlet’s unit theorem, the set of positive solutions to the Pell’s equation forms a cyclic group generated by the fundamental unit . Consider the continued fraction expansion . It is well known that the expansion of is of the form where is a symmetric sequence and is the (minimal) period of the expansion . By proposition 2.6 is of the form or where . Here is readily determined for any given . Therefore the direct problem, i.e., to find , is essentially the same as the following: given , find corresponding symmetric sequence .

In the direct problem, the length of the symmetric sequence is of great interest but we know very little about it. Assume is square-free so that is the field discriminant of . Dirichlet’s class number formula for is

 (1.1) hd=√DL(1,χ)2logεd

where the -value is bounded in a relatively narrow range for any . It is a fact that the class number varies in a very wide range , which is necessarily equivalent to saying that varies as much as this. Using , for (proposition 2.6) and , it can be shown that . Therefore the size of is closely related to the length of the symmetric sequence. It is believed that the period of is as large as fairly often, but there is absolutely no result even close to this according to the author’s knowledge.

On the other hand, the inverse problem is as follows: given a symmetric sequence of positive integers, find all ’s such that for some . Let be the set of positive non-square integers mod such that for some . In ,, a polynomial of degree 2 is associated to so that either is an empty set or it consists of all positive integers of the form with , where depends on the symmetric sequence. Similarly, denotes the set of positive non-square integers such that , which either is empty or consists of all positive integers of the form with .

Like the length is of interest in the direct problem, the size of is of interest in the inverse problem. It seems that the smallest elements of and are very different from the remaining elements, as some studies show. For example, let be a prime mod and the fundamental unit of . Ankeny, Artin and Chowla conjectured that  mod , and Hashimoto showed that  unless is the smallest element of for some , is less than and so the conjecture is true for this . Another example is the notion of minimal type introduced in . Throughout section 3 of , it can be red off that whenever a non-square integer is of minimal type for or , it is the smallest element of or for some symmetric sequence . In that paper, it is shown that fundamental units of real quadratic fields that are not of minimal type are relatively small, and among such fields exactly 51 (with one more possible exception) have class number 1.

The purpose of this paper is to treat the inverse problem in a different way, and to show that almost all non-square integers are the least elements of or for some . In this paper we say ‘almost all’ to mean

 limN→∞#{exceptions between 1 and N}N=0.

Write so that the direct problem becomes to find corresponding rational number for a given . We forget about the symmetry of now, and formulate the inverse problem as follows: given any nonnegative rational number , find all ’s such that or . The basic idea in our approach is to examine the approximation quality of to the fractional part of . Roughly speaking, we will show that this approximation can be ‘sufficiently good’ only when mod and for some integer , where is a quadratic polynomial that depends on . Consequently we rediscover quadratic progressions related to the inverse problem.

The contents are as follows. In section 2 we list down several facts about square-free integers and continued fractions, together with prescribed results from ,. In section 3 we will explain what the meaning of ‘sufficiently good’ shall be, and define very narrow intervals assigned to each positive rational number . We determine exactly when such an interval contains an integer, and specifies that integer in terms of a quadratic polynomial. In section 4 we prove that the non-square integers that are not the smallest elements of or constitute a measure zero set among natural numbers, showing that almost all real quadratic number fields are of the ‘least type’.

## 2. Preliminary results

Let be the number of square-free integers between 1 and . It is well known(for example, see theorem 333 of ) that . In , under Riemann hypothesis it was proved that . This suggests that about of the integers between and shall be square-free for every sufficiently large . In addition to this, there is a useful

###### Theorem 2.1 ().

Let be the number of square-free integers between and and that are congruent to modulo . Assume and . Then

 S(x;c,k)∼6xπ2k∏p∣k(1−1p2)−1(x→∞).

In particular, and hence each third of square-free numbers is congruent to , and modulo .

Regarding continued fractions, we mostly use the conventions in chapter 10 of . We denote the simple continued fraction expansion of a positive real number by and its -th convergent by . By definition, we always assume for . We write and use the convention , . Given an expansion of , we call the -th partial quotient of , and the -th total quotient. It worths to mention that is determined only by because has nothing to do with the denominator .

In this manuscript we consider non-square integers and the expansion . For , let be its conjugate and . For the -th convergent of , put

 (2.1) ξn=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯pn−qnωd={pn−qn+qnωdif d≡1 mod 4pn+qnωdotherwise

and let . Recall that if and only if is odd (theorem 163 of ), so

 (2.2) νn=(−1)n+1N(ξn).

We say that a quadratic integer comes from a convergent to when for some . A quadratic integer with norm is called a quadratic unit. As usual denotes the greatest integer not exceeding .

On this setting, we have a basic

###### Proposition 2.2 (Theorem 150 in ).

.

The following appears on p.141 of :

 (2.3) [a0,a1,⋯,an,αn+1]=αn+1pn+pn−1αn+1qn+qn−1

and hence

 (2.4) [a0,⋯,an,αn+1]−[a0,⋯,an]=αn+1pn+pn−1αn+1qn+qn−1−pnqn=(−1)nqn(αn+1qn+qn−1).

We also include a

###### Lemma 2.3.

For

 αn+1=√Dνn−qn−1qn+δn<√Dνn

where .

###### Proof.

We first prove that for . There are four fundamental discriminants 5,8,12,13 that are less than 16, and we have , , , . One can easily check that for in all these cases, so we assume . Now suppose we have proved . For , since , the term is positive and . When , write

 α1=1ωd−⌊ωd⌋,ξ0=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯p0−q0ωd=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⌊ωd⌋−ωd, ν0={−(⌊ωd⌋−ωd)(⌊ωd⌋−1+ωd) if d≡1 mod 4−(⌊ωd⌋−ωd)(⌊ωd⌋+ωd) otherwise

and so

 α1=⎧⎪ ⎪⎨⎪ ⎪⎩⌊ωd⌋−1+ωdν0<1+√d2+−1+√d2ν0=√Dν0 if d≡1 mod 4⌊ωd⌋+ωdν0<√d+√dν0=√Dν0 otherwise.

Thus it suffices to prove .

The cases 2 and 3 (mod 4) are easier in computation, so here we assume (mod 4) so that . Recall that the continued fraction expansion of has a natural geometric interpretation on -plane. Let be the origin of the -plane, , , the intersection of and the line , and . Then and the area of is 1/2. Observe that the area of is . Let , .

We have

 ξn¯¯¯¯¯ξnq2n=(pnqn−1+ωd)(pnqn−1+1−ωd)=±νnq2n

or

 pnqn−ωd=±νnqn(pn−qn+ωdqn)

and therefore

 |□B′BDD′| =|(pn−qnωd)qn| =νnpn/qn−1+ωd =νn2ωd−1+(−1)n+1νnqn(pn−qn+ωdqn) =νn2ωd−1⎛⎜⎝11+(−1)n+1νn(2ωd−1)qn(pn−qn+ωdqn)⎞⎟⎠ =νn√d(1+ϵn)

where . Examining the ratios of the coordinates of and , it is easily deduced that the area of is , and hence

 |△OBC| =|△OBD|−|△BCD| =(1−1−qn−1/qn1+αn+1)νn2√d(1+ϵn) =(αn+1+qn−1/qn1+αn+1)νn2√d(1+ϵn)

But , whence . Thus where so . It follows that

 αn+1=√dνn−qn−1qn+ϵ′′n

where , which proves the lemma in case (mod 4).

When 2 or 3 (mod 4), exactly the same computation with continued fraction of completes the proof. ∎

The followings are well known facts which we quote in appropriate forms.

###### Proposition 2.4 (Theorem 162 of ).

A positive rational number which is not equal to 1 can be expressed as a finite simple continued fraction in exactly two ways, one with an even and the other with an odd number of convergents. In one form the last partial quotient is 1, in the other it is greater than 1.

For a rational number , we will usually write

 p/q=[a0,⋯,an,1]=[a0,⋯,an−1,1+an].
###### Proposition 2.5 ().

If (p,q) = 1 and

 ∣∣∣pq−x∣∣∣<12q2

then is a convergent to .

We also recall that a quadratic irrational is reduced if and , and that the continued fraction expansion of is purely periodic if and only if is reduced (for example, see theorem 7.20 of ). In particular and are reduced, so the continued fraction expansion of is of the form

 (2.5) √d=[a0,¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯a1,⋯,al−1,2a0],1+√d2=[a0,¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯a1,⋯,al−1,2a0−1].
###### Proposition 2.6.

Let be a non-square positive integer and assume

 √d=[a0,¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯a1,⋯,al−1,2a0]

where is the period of this expansion. Then for , and the fundamental unit of comes from the -th convergent to . Similarly, let mod be a non-square positive integer and let . Then for and the fundamental unit of comes from the -th convergent to .

This fact has been seen several times in the literature, but the author couldn’t find a compact proof so we include an elementary one here.

###### Proof.

For brevity, let be one of and . ( is allowed for mod too). Write and let be its period.

Referring to a table of continued fractions of and for small values of , the assertion can be easily verified for . So we assume , in which case (by lemma 2.3) it suffices to show that for .

By (2.3) one has . Recall (2.2), proposition 2.2 and write

 αn+1 =−ωqn−1+pn−1ωqn−pn=(ωqn−1−pn−1)(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯pn−ωqn)N(ξn) =−N(ω)qn−1qn−pn−1pn+pnqn−1ω+pn−1qn¯¯¯ω(−1)n+1νn =⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩d−14qn−1qn−pn−1pn+pn−1qn+(−1)n+1ω(−1)n+1νnif ω=1+√d2dqn−1qn−pn−1pn+(−1)n+1√d(−1)n+1νn% if ω=√d.

Therefore, if we have mod , i.e.,

 αn+1=[an+1,a1,a2,a3,⋯]

which implies that is a multiple of . Hence for . ∎

For a symmetric sequence of positive integers, let be the 2 by 2 identity matrix and

 Mn=(a1110)⋯(an110)for n≥1.

An induction argument easily proves that

 Mn=(qnqn−1rnrn−1),n≥0.

Here is symmetric, so is a symmetric matrix, i.e., . Therefore (mod 2) and (mod 2). Put

 f(a1,⋯,an;T)=q2nT2+AT+B,where
 A=4qn−1+2(−1)nqnqn−1rn−1,B=q2n−1r2n−1+4(−1)nr2n−1.

Observe that (mod ).

###### Theorem 2.7 (Corollary 1 and 1A in ).

Let be a symmetric sequence of positive integers and , , as above. The followings are equivalent:

1. Either (mod 2) or (mod 2).

In these cases, consists of all of the form

 d=14f(T)

where is any integer satisfying , mod .

Likewise, the followings are equivalent:

1. Either (mod 2) or (mod 2).

In these cases, consists of all , (mod 4) of the form

 d=f(T)

where is any integer satisfying , mod .

From now on we always assume whenever we write to denote a rational number. Recall that every irrational number has a unique continued fraction expansion. For convenience, write if is a convergent to . We use the convention . It is a simple observation that the set of positive real numbers is partitioned by the predecessors, i.e., for any positive integers the set

 {x∈R∣x>0,x=[a0,a1,a2,⋯,am,∗]}

is a closed interval. The following proposition quantifies these intervals in view of .

###### Proposition 2.8.

Let be positive integers and the number of non-square integers between 1 and such that

 √d=[a0,a1,a2,⋯,am,∗]

for some . Similarly let be the number of non-square integers between 1 and which are congruent to 1 modulo 4 such that

 ωd=[a0,a1,a2,⋯,am,∗]

for some . Then

 limN→∞f(N)N=limN→∞4f(1)(N)N=1qm(qm+qm−1).
###### Sketch of proof.

This is almost trivial. Consider the numbers in the interval where is large. For in the range , the curve may be approximated by a straight line of slope . The difference between and is , so there are approximately (non-square) integers between and that are counted by for . Similarly, for consider the curve which is close to a straight line of slope . Extracting integers congruent to 1 modulo 4, we get the result. ∎

## 3. The Attached Intervals

Let be any rational number. Note that

 ∣∣∣pq−ωd∣∣∣={ν(p+q(ωd−1))q if d≡1 mod 4ν(p+qωd)q otherwise

where . So we can interpret the norm of a quadratic integer as a measure of how successful the approximation of by is. Proposition 2.5 shows that when

 |N(p−qωd)|<⎧⎨⎩p/q+ωd−12 % if d≡1 mod 4p/q+ωd2 otherwise,

is a convergent to . Lemma 2.3 also shows that becomes a quadratic unit if and only if the -th convergent becomes as large as possible, namely . By (2.4) this means that is particularly close to ; in other words, if the approximation of by is ‘sufficiently good’.

###### Remark 3.1.

By proposition 2.6, such thing happens if and only if mod . Recall that where is symmetric. It follows that if a rational number satisfies for some , then for some symmetric sequence .

Based on the above context, for each rational number we assign tiny intervals , that consist of points where or is especially close to . More specifically, we want these intervals to satisfy following property: whenever a non-square integer falls into that interval, the quality of approximation of by is sufficiently good and hence or becomes a quadratic unit. Explicitly, we build the intervals for a fixed as follows.

Recall that there exists a unique sequence such that by proposition 2.4. From (2.4), for any positive number one has

 {[a0,⋯,am+1,λ]

Since is an integer, by lemma 2.3 a total quotient that appears in the continued fraction expansion of cannot assume any values between and . Observe that . Unless is too small, therefore, we can say that if and only if . Let

 A=A(p/q)=[a0,⋯,am+1,43a0−qm−1q], B=B(p/q)=[a0,⋯,am,1,43a