On the equivalence of certain quadratic irrationals

Abstract

This paper deals with quadratic irrationals of the form $m/q+\sqrt{v}$ for fixed positive integers v and q, v not a square, and varying integers m, $(m,q)=1$. Two numbers $m/q+\sqrt{v}$, $n/q+\sqrt{v}$ of this kind are equivalent (in a classical sense) if their continued fraction expansions can be written with the same period. We give a necessary and sufficient condition for the equivalence in terms of solutions of Pell’s equation. Moreover, we determine the number of equivalence classes to which these quadratic irrationals belong.

1 Introduction and results

Let v and q be positive integers, v not a square. In this paper we study the equivalence between numbers

$$\begin{aligned} x=m/q+\sqrt{v}, \end{aligned}$$

where m is an integer, $(m,q)=1$. Thus, v and q are fixed, whereas m may vary.

The equivalence of two numbers x, y of this kind means that the (regular) continued fractions of x and y can be written with the same period, say,

$$\begin{aligned} x=[a_0,\ldots ,a_{j-1},[b_1,\ldots ,b_k]], y=[c_0,\ldots ,c_{l-1},[b_1,\ldots ,b_k]], \end{aligned}$$

(1)

where $[b_1,\ldots ,b_k]$ is the common period. Here the pre-periods $a_0,\ldots ,a_{j-1}$ and $c_0,\ldots ,c_{l-1}$ need not occur. In general, it is more likely that you find equivalent numbers x and y than inequivalent ones, for example,

$$\begin{aligned} x= & {} \frac{1}{12}+\sqrt{7}=[2, [1, 2, 1, 2, 4, 5, 16, 47, 1, 1, 3, 1, 1, 4]],\\ y= & {} \frac{5}{12}+\sqrt{7}=[3, [16, 47, 1, 1, 3, 1, 1, 4, 1, 2, 1, 2, 4, 5]]\\= & {} [3,16,47,1,1,3,1,1,4,[1, 2, 1, 2, 4, 5, 16, 47, 1, 1, 3, 1, 1, 4]]. \end{aligned}$$

We write $x\sim y$ if x and y are equivalent. It is a classical result of Serret that $x\sim y$ is the same as

$$\begin{aligned} y=\frac{ax+b}{cx+d}, \text{ where } \left( \begin{array}{cc} a &{} b \\ c &{} d \\ \end{array} \right) \in \text{ GL }(2,\mathbb {Z}), \end{aligned}$$

(2)

i.e., $a,b,c,d\in \mathbb {Z}$ and $ad-bc=\pm 1$ (see [5, p. 54], [1, p. 38]).

Our first aim is the following theorem.

Theorem 1

Let $x=m/q+\sqrt{v}$, $y=n/q+\sqrt{v}$, $(m,q)=(n,q)=1$. Let $q_1=(m-n,q)$. Then $x\sim y$ if, and only if, the equation

$$\begin{aligned} r^2-c^2v=\pm 1 \end{aligned}$$

(3)

has a solution $(r,c)\in \mathbb {Z}^2$ such that $(c,q^2)=qq_1.$

Of course, (3) is known as Pell’s equation. Our next question concerns the number of equivalence classes to which our quadratic irrationals belong. Since $x+1\sim x$, we may restrict ourselves to numbers $x=m/q+\sqrt{v}$, $y=n/q+\sqrt{v}$ with $0\le m,n\le q-1$.

Theorem 2

Let $q_0$ be the smallest divisor of q such that there is a solution (r, c) of (3) with $(c,q^2)=qq_0$. Then the numbers $x=m/q+\sqrt{v}$, $(m,q)=1$, $0\le m\le q-1$, belong to exactly $\varphi (q_0)$ equivalence classes, each of which contains $\varphi (q)/\varphi (q_0)$ elements x.

Remarks

• 1. Note that every equivalence class contains many elements different from the numbers x in question. For instance, $1/3+\sqrt{2}\sim 9\sqrt{2}/2$, the latter not being of the appropriate form for $v=2$ and $q=3$. Equivalent numbers have the same discriminant (see [1, p. 41]). Since the discriminant of $x=m/q+\sqrt{v}$, $(m,q)=1$, equals $4q^4v$, x cannot be equivalent to a number $m'/q'+\sqrt{v}$, $(m',q')=1$, $q'>0$, $q'\ne q$.

• 2. The unit group $\mathbb {Z}[\sqrt{v}]^{\times }$ of the ring $\mathbb {Z}[\sqrt{v}]$ is isomorphic to $\mathbb {Z}\times \mathbb {Z}/2\mathbb {Z}$ and is generated by a fundamental unit $s+t\sqrt{v}$ together with $-1$. For $r+c\sqrt{v} \in \mathbb {Z}[\sqrt{v}]^{\times }$ we have $q\,|\,c$ if, and only if, $r+c\sqrt{v} \in \mathbb {Z}[q\sqrt{v}]^{\times }$, the unit group of the subring $\mathbb {Z}[q\sqrt{v}]$. This group has a finite index k in $\mathbb {Z}[\sqrt{v}]^{\times }$ (see [4, p. 296]). Accordingly, $(s+t\sqrt{v})^k$ is an element of $\mathbb {Z}[q\sqrt{v}]^{\times }$.

• 3. It may happen that $\mathbb {Z}[q\sqrt{v}]^{\times }$ coincides with $\mathbb {Z}[qq_1\sqrt{v}]^{\times }$ for some divisor $q_1$ of q, $q_1>2$. In this case c is divisible by $qq_1$ for each $r+c\sqrt{v} \in \mathbb {Z}[q\sqrt{v}]^{\times }$. In particular, $qq_1\,|\,(c,q^2)$ for all these units. Let $q_0$ be the smallest divisor of q such that there is a unit $r+c\sqrt{v}\in \mathbb {Z}[q\sqrt{v}]^{\times }$ with $(c,q^2)=qq_0$. Then $qq_1\,|\,(c,q^2)=qq_0$, and, accordingly, $q_1\,|\,q_0$. By Theorem 2, the numbers $m/q+\sqrt{v}$, $(m,q)=1$, $0\le m\le q-1$, belong to $\varphi (q_0)\ge \varphi (q_1)>1$ equivalence classes. In particular, not all of these numbers are equivalent.

Example

Let $v=979$ and $q=12$. The fundamental unit in $\mathbb {Z}[\sqrt{v}]$ is $s+t\sqrt{v}$ with $s=360449$ and $t=11520=q^2\cdot 80$. Hence $\mathbb {Z}[\sqrt{v}]^{\times }=\mathbb {Z}[q^2\sqrt{v}]^{\times }$ and for every $r+c\sqrt{v}\in \mathbb {Z}[\sqrt{v}]^{\times }$ we have $q^2\,|\,c$. Accordingly, the number $q_0$ of Theorem 2 equals $q=12$, and the four numbers $m/12+\sqrt{979}$, $m\in \{1,5,7,12\}$ belong to four different equivalence classes. Indeed,

$$\begin{aligned} \frac{1}{12}+\sqrt{979}= & {} [31, [2, 1, 2, 5, 2, 3, 6, 1, 4, 62]],\\ \frac{5}{12}+\sqrt{979}= & {} [31, [1, 2, 2, 1, 1, 13, 1, 4, 1, 6, 1, 61]]. \end{aligned}$$

So these numbers have periods of different lengths. The numbers $7/12+\sqrt{979}$ and $5/12+\sqrt{979}$ have inverse periods, and $11/12+\sqrt{979}$ and $1/12+\sqrt{979}$, too. In general, we say that x and y have inverse periods if they can be written as in (1), the period of y being $[b_k,b_{k-1},\ldots ,b_1]$, however.

Theorem 1 answers the question whether $x=m/q+\sqrt{v}$ and $y=n/q+\sqrt{v}$ have inverse periods. This happens if, and only if, $x\sim y'=n/q-\sqrt{v}$ (see [5, p. 77]). Since $y'\sim -y'=-n/q+\sqrt{v}$, we obtain the following corollary to Theorem 1.

Corollary 1

Let $x=m/q+\sqrt{v}$, $y=n/q+\sqrt{v}$ be as above. Let $q_1'=(m+n,q)$. Then x and y have inverse periods if, and only if, the equation (3) has a solution $(r,c)\in \mathbb {Z}^2$ such that $(c,q^2)=qq_1'.$

We say that x has a self-inverse  period if x can be written with a period $[b_1,\ldots ,b_k]$ but also with the period $[b_k,\ldots ,b_1]$ (see [5, p. 78], [2]). From Corollary 1 we obtain

Corollary 2

Let $x=m/q+\sqrt{v}$ be as above. Put $q_1'=2$ if q is even, and $q_1'=1$, otherwise. Then x has a self-inverse period if, and only if, the equation (3) has a solution $(r,c)\in \mathbb {Z}^2$ such that $(c,q^2)=qq_1'.$

Remarks

• 1. The reader may consult [3], where quadratic irrationals with self-inverse periods are classified by certain equivalences.

• 2. Many examples show the following tendency, for which we have no precise mathematical formulation. Namely, if the numbers $m/q+\sqrt{v}$, $(m,q)=1$, $0\le m\le q-1$, belong to many equivalence classes, then their periods are short. For instance, in the case $v=979$, $q=12$ of the above example we have the largest possible number of equivalence classes, which is 4. The corresponding period lengths of $m/q+\sqrt{v}$ are 10 or 12. If we choose $q=9$ instead, then all numbers $m/q+\sqrt{v}$ belong to the same equivalence class, and the common period of the 6 elements m/q has length 78.

2 Proofs

Proof of Theorem 1

Let $x=m/q+\sqrt{v}$, $y=n/q+\sqrt{v}$, $(m,q)=(n,q)=1$. First suppose $x\sim y$, i.e., there is a matrix $\left( \begin{array}{cc} a &{} b \\ c &{} d \\ \end{array} \right) \in \text{ GL }(2,\mathbb {Z})$ such that (2) holds. Then comparison of the coefficients with respect to the $\mathbb {Q}$-basis $(1,\sqrt{v})$ of $\mathbb {Q}[\sqrt{v}]$ shows that (2) is equivalent to the identities

$$\begin{aligned} a=\frac{c(m+n)}{q} +d \end{aligned}$$

(4)

and

$$\begin{aligned} b=\frac{-d(m-n)}{q}-\frac{cm^2}{q^2}+cv. \end{aligned}$$

(5)

Since $b\in \mathbb {Z}$, (5) implies

$$\begin{aligned} d(m-n)q +cm^2\equiv 0\mod q^2. \end{aligned}$$

(6)

However, $(m,q)=1$, so (6) requires $c\equiv 0\mod q$. Then $a\in \mathbb {Z}$, by (4). Let $m-n=q_1m_1$ with $q_1=(m-n,q)$ and $(m_1,q/q_1)=1$. Accordingly, (6) can be written

$$\begin{aligned} dm_1qq_1+cm^2\equiv 0\mod q^2. \end{aligned}$$

(7)

If $q_1=q$, this congruence yields $c\equiv 0\equiv qq_1\mod q^2$. If $q_1<q$, we have $qq_1\,|\,c$. In this case one easily checks that $(c,q^2)$ must be $qq_1$ (observe that $(d,q)=1$ since $q\,|\,c$ and the matrix in question has determinant $\pm 1$).

Moreover, the condition $ad-bc=\pm 1$ is the same as saying

$$\begin{aligned} d=-\frac{cm}{q}+\sqrt{c^2v\pm 1}, \end{aligned}$$

(8)

where we do not fix a sign for the square root. Hence there is an integer r such that (3) holds.

Conversely, suppose that (r, c) is a solution of (3) and $c=qq_1c_1$ for some integer $c_1$ with $(c_1, q/q_1)=1$ (observe $q^2/(qq_1)=q/q_1$). We define d in such a way that (8) is satisfied, i.e.,

$$\begin{aligned} d=-\frac{cm}{q}+r=-q_1c_1m+r. \end{aligned}$$

(9)

Then condition (6) reads

$$\begin{aligned} (-q_1c_1m+r)(m-n)q+qq_1c_1m^2\equiv 0\mod q^2. \end{aligned}$$

This congruence is equivalent to the congruence

$$\begin{aligned} c_1mn+rm_1\equiv 0\mod q/q_1. \end{aligned}$$

(10)

Here $(m,q)=(n,q)=(m_1,q/q_1)=1$. Observe that $(c_1,q/q_1)=1$. By (3), we have $(r,q)=1$, since $q\,|\,c$. Of course, it may happen that our pair (r, c) does not satisfy (10). In this case we consider $r'+c'\sqrt{v}=(r+c\sqrt{v})^k$ for some positive integer k prime to q. Since $q\,|\,c$, we obtain

$$\begin{aligned} r'+c'\sqrt{v}\equiv r^k+kr^{k-1}c\sqrt{v}\mod q^2, \end{aligned}$$

a congruence mod $\mathbb {Z}[\sqrt{v}]q^2$. It is easy to see that this congruence implies the congruences

$$\begin{aligned} r'\equiv r^k\mod q^2,\hspace{5.0pt}c'\equiv kr^{k-1}c \mod q^2, \end{aligned}$$

(11)

which are congruences mod $\mathbb {Z}q^2$. In particular, $(c',q^2)=(kc,q^2)=(c,q^2)$, since $(k,q)=1$. The second of the congruences (11) shows that we may write $c'=qq_1c_1'$ with $c_1'\equiv kr^{k-1}c_1\mod q/q_1$. The congruence (10), for $r'$ and $c'$ instead of r and c, reads

$$\begin{aligned} c_1'nm+r'm_1\equiv 0\mod q/q_1, \end{aligned}$$

(12)

or

$$\begin{aligned} kr^{k-1}c_1nm+r^km_1\equiv 0\mod q/q_1. \end{aligned}$$

Because $(r,q)=1$, this is equivalent to

$$\begin{aligned} kc_1nm+rm_1\equiv 0\mod q/q_1. \end{aligned}$$

(13)

Observe $(c_1,q/q_1)=(m_1,q/q_1)=(m,q)=(n,q)=1$ and $q/q_1\,|\,q$. Therefore, the number k (prime to q) can be chosen such that (13) holds. Then (12) holds, and, thus, the congruence (6).

We define d by (9) with $r'$, $c'$ instead of r, c. Finally, we define a, b by (4) and (5) with $c'$ instead of c. Then a, b are integers, $ad-bc'=\pm 1$, and (2) also holds. $\square$

Example

Let $v=7$, $q=12$, $m=1$, $n=5$. Hence $x=1/12+\sqrt{7}$, $y=5/12+\sqrt{7}$, $m-n=-4=q_1m_1$ with $q_1=4,m_1=-1$ (see the example at the beginning of this paper). The fundamental unit of $\mathbb {Z}[\sqrt{7}]$ is $s+t\sqrt{7}=8+3\sqrt{7}$. Since $(s+t\sqrt{7})^2=127+48\sqrt{7}$, we put $r=127$, $c=48$. In particular, $(c,q^2)=qq_1=48$. So x is equivalent to y (what we already know). We have $c_1=1$ and $q/q_1=3$. However, the congruence (10) does not hold, but (13) is true with $k=5$, $(k,q)=1$. This choice, however, leads to rather large numbers. But we may also choose $k=2$ (which is not prime to q). Indeed, define $r'$, $c'$ by

$$\begin{aligned} r'+c'\sqrt{7}=(r+c\sqrt{7})^{2}=32257+12192\sqrt{7}. \end{aligned}$$

Then $(c',q^2)=(12192,144)=48=qq_1$, as required. As in the proof of Theorem 1 we obtain $d=31241$, $a=37337$, and $b=95673$. In this way $ad-bc'=1$ and $(ax+b)/(c'x+d)=y$.

The proof of Theorem 2 requires the following lemmas.

Lemma 1

Let $(r_1,c_1)$ and $(r_2,c_2)$ be solutions of (3) such that $(c_1,q^2)=qq_1$ and $(c_2,q^2)$ $=qq_2$ for divisors $q_1$, $q_2$ of q. Let $q'=(q_1,q_2)$. Then there is a solution $(r',c')$ of (3) such that $(c',q^2)=q'$.

Proof

Let $j_1,j_2$ be positive integers. We have

$$\begin{aligned} (r_i+c_i\sqrt{v})^{j_i}\equiv r_i^{j_i}+j_ir_i^{j_i-1}c_i\sqrt{v} \mod q^2, \end{aligned}$$

$i=1,2$. We define $r'+c'\sqrt{v}$ by

$$\begin{aligned} r'+c'\sqrt{v}=(r_1+c_1\sqrt{v})^{j_1}(r_2+c_2\sqrt{v})^{j_2}. \end{aligned}$$

Since

$$\begin{aligned} (r_1+c_1\sqrt{v})^{j_1}(r_2+c_2\sqrt{v})^{j_2}\equiv r_1^{j_1}r_2^{j_2}+(j_1r_1^{j_1-1}r_2^{j_2}c_1+j_2r_1^{j_1}r_2^{j_2-1}c_2)\sqrt{v}\mod q^2, \end{aligned}$$

we obtain

$$\begin{aligned} c'\equiv r_1^{j_1-1}r_2^{j_2-1}(j_1r_2c_1+j_2r_1c_2)\mod q^2. \end{aligned}$$

(14)

Observe that $(r_1,q)=(r_2,q)=1$, since $q\,|\,c_1,c_2$. We consider the ideal $\mathbb {Z}r_2c_1+\mathbb {Z}q^2$ in $\mathbb {Z}$. We have $(r_2c_1,q^2)=qq_1$. This can be written as

$$\begin{aligned} \mathbb {Z}r_2c_1+\mathbb {Z}q^2=\mathbb {Z}qq_1. \end{aligned}$$

In the same way, we obtain

$$\begin{aligned} \mathbb {Z}r_1c_2+\mathbb {Z}q^2=\mathbb {Z}qq_2. \end{aligned}$$

However, $(qq_1,qq_2)=qq'$, and so $\mathbb {Z}qq_1+\mathbb {Z}qq_2=\mathbb {Z}qq'$. This yields

$$\begin{aligned} \mathbb {Z}r_2c_1+\mathbb {Z}r_1c_2 +\mathbb {Z}q^2=\mathbb {Z}qq'. \end{aligned}$$

Accordingly, there are integers $k_1$, $k_2$ such that

$$\begin{aligned} k_1r_2c_1+k_2r_1c_2 \equiv qq'\mod q^2. \end{aligned}$$

(15)

We put $j_1=k_1+lq^2$ and $j_2=k_2+lq^2$, where l is chosen such that both $j_1,j_2$ are positive. Then (14) and (15) show that $c'$ satisfies $(c',q^2)=qq'$. $\square$

Lemma 2

Let $q_1$ divide q. Let (r, c) be a solution of (3) such that $(c,q^2)=qq_1$. Let p be a prime dividing $q/q_1$. Define $(r',c')$ by $r'+c'\sqrt{v}=(r+c\sqrt{v})^p$. Then $(r',c')$ is a solution of (3) such that $(c',q^2)=qq_1p.$

Proof

Let l be a prime number. By $v_l(k)$ we denote the l-exponent of the integer k, i.e., $l^{v_l(k)}\,|\,k$, $l^{v_l(k)+1}\, \not \mid \,k$. If l divides q, we have $v_l(c)=v_l(qq_1)$, since $qq_1\,|\,q^2$ and $(c,q^2)=qq_1$.

First let $p=2$. Then $c'=2rc$ and $v_2(2rc)=v_2(c)+1$, since $v_2(r)=0$, by (3). If l is a prime divisor of q different from 2, we see $v_l(2rc)=v_l(c)$. Hence $(c',q^2)=qq_1p$.

If $p\ge 3$ we have

$$\begin{aligned} r'+c'\sqrt{v}\equiv r^p+pr^{p-1}c\sqrt{v} \mod c^2p, \end{aligned}$$

(16)

since $c^j\equiv 0\mod c^2p$ for $j\ge 3$ (recall $p\,|\,c$) and $\left( {\begin{array}{c}p\\ 2\end{array}}\right) c^2\equiv 0\mod c^2p$. Thus, $v_p(c')=v_p(pr^{p-1}c)=v_p(c)+1$, because $v_p(c^2p)>v_p(c)+1$. For a prime divisor l of q different from p we have $v_l(c^2p)=2v_l(c)>v_l(c) (\ge 1)$. From (16) we obtain $v_l(c')=v_l(pr^{p-1}c)=v_l(c)$. $\square$

Proof of Theorem 2

Let $q_0$ be the smallest divisor of q such that there is a solution (r, c) of (3) with $(c,q^2)=qq_0$.

If q is even, then $m-n$ is even for all m, n with $(m,q)=(n,q)=1$. Hence $(m-n,q)$ is even. Accordingly, $q_0$ cannot have the form $q_0=(m-n,q)$ if $q_0$ is odd. Suppose that this holds. Then we replace $q_0$ by $2q_0$. Since $\varphi (2q_0)=\varphi (q_0)$, this does not change the assertion of Theorem 2. Moreover, by Lemma 2, we have a solution $(r',c')$ of (3) such that $(c',q^2)=2qq_0$.

Accordingly, we may assume that $q_0$ is even if q is even and suppose that (r, c) is a solution of (3) with $(c,q^2)=qq_0$.

Let $y=n/q+\sqrt{v}, (n,q)=1$. We show that the sets

$$\begin{aligned} X_1=\{m/q+\sqrt{v}: (m,q)=1, m/q+\sqrt{v}\sim y\} \end{aligned}$$

and

$$\begin{aligned} X_2= \{m/q+\sqrt{v}: (m,q)=1, q_0\,|\,m-n\} \end{aligned}$$

coincide. Indeed, if $m/q+\sqrt{v}$ is in $X_1$ and $(m-n,q)=q_1$, then there is a solution $(r',c')$ of (3) such that $(c',q^2)=qq_1$. On the other hand, we have a solution (r, c) of (3) such that $(c,q^2)=qq_0$. By Lemma 1, there is a solution $(r'',c'')$ such that $(c'',q^2)=q(q_0,q_1)$. If $(q_0,q_1)\ne q_0$, then $(q_0,q_1)<q_0$, which contradicts the minimality of $q_0$. Accordingly, $(q_0,q_1)=q_0$ and $q_0\,|\,q_1=(m-n,q)$. In particular, $q_0$ divides $m-n$ and $m/q+\sqrt{v}\in X_2$.

Conversely, if $m/q+\sqrt{v}\in X_2$, then $(m-n,q)=q_0k$ for some positive integer k. By Lemma 2, there exists a solution $(r',c')$ of (3) such that $(c',q^2)=qq_0k$. This implies $m/q+\sqrt{v}\sim y$ and $m/q+\sqrt{v}\in X_1$.

We consider $X=\{m/q+\sqrt{v}: (m,q)=1, 0\le m\le q-1\}$ and $X_1'=X_1\cap X$. Now $m/q+\sqrt{v}\in X$ lies in $X_1'$ if, and only if, $q_0\,|\,m-n$, i.e., the canonical surjection

$$\begin{aligned} \pi :(\mathbb {Z}/q\mathbb {Z})^{\times }\rightarrow (\mathbb {Z}/q_0\mathbb {Z})^{\times }: \overline{k}\mapsto \overline{k} \end{aligned}$$

maps $\overline{m}$ onto $\overline{n}$. Thereby, $|X_1'|=|\pi ^{-1}(\overline{n})|=\varphi (q)/\varphi (q_0)$. Hence there are exactly $\varphi (q)/\varphi (q_0)$ elements of X that are equivalent to y. This, however, implies that there must be exactly $\varphi (q_0)$ equivalence classes whose intersections with X are not empty. $\square$