Inverse problem for Pell equation and real quadratic fields of the least type
Abstract.
The purpose of this article is to give the solutions of the inverse problem for Pellian equations. For any rational number , the fundamental discriminants satisfying are given in terms of a quadratic progression. There were studies about this problem based on symmetric sequences and periodic continued fractions , but in this article we solve the problem in a completely different way with simpler parameters. The result is obtained by measuring the quality of approximation of a rational number to or , and by defining a short interval attached to each rational number. On this formulation we also show that for almost all squarefree integer , is the least element of the prescribed quadratic progression for some .
Key words and phrases:
Real quadratic field, Continued fraction, Pell equation, Symmetric sequence, Least type1991 Mathematics Subject Classification:
Primary:11J68, Secondary: 11R29, 11Y401. Introduction
Let be a nonsquare positive integer, and
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 [4]. 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 squarefree so that is the field discriminant of . Dirichlet’s class number formula for is
(1.1) 
where the value is bounded in a relatively narrow range for any [9]. 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 [5], 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 nonsquare integers mod such that for some . In [3],[4], 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 nonsquare 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 [1] mod , and Hashimoto showed that [6] 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 [8]. Throughout section 3 of [8], it can be red off that whenever a nonsquare 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 nonsquare integers are the least elements of or for some . In this paper we say ‘almost all’ to mean
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 squarefree integers and continued fractions, together with prescribed results from [3],[4]. 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 nonsquare 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 squarefree integers between 1 and . It is well known(for example, see theorem 333 of [5]) that . In [2], under Riemann hypothesis it was proved that . This suggests that about of the integers between and shall be squarefree for every sufficiently large . In addition to this, there is a useful
Theorem 2.1 ([11]).
Let be the number of squarefree integers between and and that are congruent to modulo . Assume and . Then
In particular, and hence each third of squarefree numbers is congruent to , and modulo .
Regarding continued fractions, we mostly use the conventions in chapter 10 of [5]. 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 nonsquare integers and the expansion . For , let be its conjugate and . For the th convergent of , put
(2.1) 
and let . Recall that if and only if is odd (theorem 163 of [5]), so
(2.2) 
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 [5]).
.
(2.4) 
We also include a
Lemma 2.3.
For
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
and so
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
or
and therefore
where . Examining the ratios of the coordinates of and , it is easily deduced that the area of is , and hence
But , whence . Thus where so . It follows that
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 [5]).
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
Proposition 2.5 ([5]).
If (p,q) = 1 and
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 [10]). In particular and are reduced, so the continued fraction expansion of is of the form
(2.5) 
Proposition 2.6.
Let be a nonsquare positive integer and assume
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 nonsquare 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 .
For a symmetric sequence of positive integers, let be the 2 by 2 identity matrix and
An induction argument easily proves that
Here is symmetric, so is a symmetric matrix, i.e., . Therefore (mod 2) and (mod 2). Put
Observe that (mod ).
Theorem 2.7 (Corollary 1 and 1A in [4]).
Let be a symmetric sequence of positive integers and , , as above. The followings are equivalent:


Either (mod 2) or (mod 2).
In these cases, consists of all of the form
where is any integer satisfying , mod .
Likewise, the followings are equivalent:


Either (mod 2) or (mod 2).
In these cases, consists of all , (mod 4) of the form
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
is a closed interval. The following proposition quantifies these intervals in view of .
Proposition 2.8.
Let be positive integers and the number of nonsquare integers between 1 and such that
for some . Similarly let be the number of nonsquare integers between 1 and which are congruent to 1 modulo 4 such that
for some . Then
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 (nonsquare) 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
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
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 nonsquare 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
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