Disconnectedness of the subgraph $F^3$ for the group ${\Gamma }^3$

Abstract

In this paper we show that the subgraph $F^3$ is disconnected and that for all integers m, we find all integers a and b such that $(9m^2-4)a^2+4$ and $5b^2\pm 4$ are square. It turns out that the set of numbers b comprises the Fibonacci numbers.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let $\hat{\mathbb{Q}}=\mathbb{Q}\cup \{\mathrm{\infty }\}$ be the extended rationals, let Γ be the modular group acting on $\hat{\mathbb{Q}}$ as with the upper half-plane $\mathcal{H}=\{z\in \mathbb{C}:Imz>0\}$:

$$g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right):z=\frac{x}{y}\to \frac{az+b}{cz+d}=\frac{ax+by}{cx+dy},$$

where a, b, c, and d are rational integers, and let ${\Gamma }^3$ denote the group consisting of the cubes of the elements g of Γ, which is the group $\{g\in \Gamma :ab+cd\equiv 0(mod3)\}$; see [1].

Jones et al. [2] used the notion of the imprimitive action [3–5] for a Γ-invariant equivalence relation induced on $\hat{\mathbb{Q}}$ by the congruence subgroup ${\Gamma }_0(n)=\{g\in \Gamma :c\equiv 0(modn)\}$ to obtain some suborbital graphs and examined their connectedness and forest properties.

In [6], a ${\Gamma }^3$-invariant equivalence relation is introduced by using the subgroup ${\Gamma }_0^3(n)=\{g\in {\Gamma }^3:c\equiv 0(modn)\}$ to obtain suborbital graphs $F_{u,n}^3$. There, the connectivity properties of all subgraphs $F_{u,n}^3$ other than $F_{1,1}^3=F^3$ are examined.

In this paper we show that the subgraph $F^3$ is disconnected and give some results, which seem important from the point of view of number theory.

2 Preliminaries

Since ${\Gamma }^3=\{g^3:g\in \Gamma \}$, it is easily seen that the elements of ${\Gamma }^3$ are ones of the forms $\left(\begin{array}{cc}3a& b\\ c& 3d\end{array}\right)$, $\left(\begin{array}{cc}a& 3b\\ 3c& d\end{array}\right)$, and $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ where a, b, c, and $d\not\equiv 0(mod3)$ in the third matrix. Furthermore, ${\Gamma }_{\mathrm{\infty }}^3<{\Gamma }_0^3(n)\le {\Gamma }^3$ for each positive integer n, where ${\Gamma }_{\mathrm{\infty }}^3$ is the stabilizer of ∞ generated by the element $\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right)$, and second inclusion is strict if $n>1$.

Since the group ${\Gamma }^3$ is transitive on $\hat{\mathbb{Q}}$ in [6], any reduced fraction $\frac{r}{s}$ in $\hat{\mathbb{Q}}$ equals $g(\mathrm{\infty })$ for some $g\in {\Gamma }^3$. Similar to that of [2], we get the following ${\Gamma }^3$-invariant equivalence relation on $\hat{\mathbb{Q}}$ by ${\Gamma }_0^3(n)$ as $\frac{r}{s}\sim \frac{x}{y}$ if and only if $g^{-1}h\in {\Gamma }_0^3(n)$, where $g=\left(\begin{array}{cc}r& \ast \\ s& \ast \end{array}\right)$ and h is similar. Furthermore, the above equivalence relation is imprimitive, which means that it is different from the identity relation ($a\sim b$ if and only if $a=b$) and the universal relation ($a\sim b$ for all $a,b\in \hat{\mathbb{Q}}$).

From the above defined ${\Gamma }^3$-invariant equivalence relation, we can verify that $\frac{r}{s}\sim \frac{x}{y}$ if and only if $ry-sx\equiv 0(modn)$. The equivalence classes are called blocks, and a block containing the rational $\frac{x}{y}$ is denoted by $[\frac{x}{y}]$.

Although the equivalence relations are resulting almost the same as in [2], the subgraph $F_{1,1}$ in [2] is easily shown to be connected, but here we will see that the subgraph $F_{1,1}^3$ is disconnected. So, using different subgroups changes the characters of the subgraphs.

3 Subgraphs $F_{u,n}^3$

The group ${\Gamma }^3$ acts on $\hat{\mathbb{Q}}\times \hat{\mathbb{Q}}$ through $g:(\alpha ,\beta )\to (g(\alpha ),g(\beta ))$. The orbits are called suborbital. From the suborbital $O^3(\alpha ,\beta )$ containing $(\alpha ,\beta )$, we can form the suborbital graph $G^3(\alpha ,\beta )$ whose vertices are the elements of $\hat{\mathbb{Q}}$ and the edges are the pairs $(a,b)\in O^3(\alpha ,\beta )$, which we denote by $a\to b$ and represent as hyperbolic geodesics in ℋ.

Since ${\Gamma }^3$ acts transitively on $\hat{\mathbb{Q}}$, every suborbital contains a pair $(\mathrm{\infty },\frac{u}{n})$ for some $\frac{u}{n}$ in $\hat{\mathbb{Q}}$, $n\ge 0$, $(u,n)=1$. In this case, we denote the suborbital graph by $G_{u,n}^3$ for short. From now on, we assume that $n>0$.

As ${\Gamma }^3$ permutes the blocks transitively, all subgraphs corresponding to the blocks are isomorphic, as in [2]. Therefore, we will only consider the subgraph $F_{u,n}^3$ of $G_{u,n}^3$ whose vertices form the block $[\mathrm{\infty }]=\{\frac{x}{y}\in \hat{\mathbb{Q}}\mid y\equiv 0(modn)\}$. The following two results were proved in [6].

Theorem 1 $F_{u,n}^3=F_{u^{\prime },n^{\prime }}^3$ if and only if $n=n^{\prime }$ and $u\equiv u^{\prime }(mod3n)$.

Theorem 2 $\frac{r}{s}\to \frac{x}{y}$ is an edge in $F^3$ ($=F_{1,1}^3$) if and only if

• 1. if $r\equiv 0(mod3)$, then $y\equiv \pm s(mod3)$ and $ry-sx=\pm 1$, or

• 2. if $s\equiv 0(mod3)$, then $x\equiv \pm r(mod3)$, and $ry-sx=\pm 1$, or

• 3. if $r,s\not\equiv 0(mod3)$, then $x\not\equiv \pm r(mod3)$, $y\not\equiv \pm s(mod3)$ and $ry-sx=\pm 1$.

We can easily get the following lemmas.

Lemma 1 $\frac{r}{s}\to \frac{x}{y}$ is in $F^3$ if and only if $\frac{x}{y}\to \frac{r}{s}$ is in $F^3$.

Lemma 2 [2]

No edges of $F^3$ cross in ℋ.

4 Disconnectedness of $F^3$

Definition 1 For $m\in \mathbb{N}$ and $m\ge 2$, let $v_1,v_2,\dots ,v_m$ be a finite sequence of vertices of $F^3$. Then the configuration $v_1\to v_2\to \cdots \to v_m$ is called a finite path in $F^3$. A subgraph ∧ of $F^3$ is called connected if every two vertices x and y of ∧ are connected by a finite path in $F^3$. Otherwise, we call ∧ disconnected.

Now we give one of our main theorems.

Theorem 3 The graph $F^3$ is disconnected.

We prove Theorem 3 after giving some theorems, propositions and lemmas as follows.

By Theorem 2, it is easily seen that the graph $F^3$ is periodic with period 3. That is, if $a\to b$ is in $F^3$, then $a+3m\to b+3m$ is in $F^3$ for all $m\in \mathbb{Z}$, and therefore, for some m, $a+m$ or $b+m$ (not both) is ∞ or both $a+3m$ and $b+3m$ are in the interval $[1,4]$. Therefore we can only use the interval $[1,4]$ for our calculations as in Figure 1.

Figure 1

The subgraph ${\mathit{F}}^{\mathbf{3}}$ .

It is clear that $T=\left(\begin{array}{cc}1& -5\\ 1& -4\end{array}\right)$ is in ${\Gamma }^3$ and the corresponding transformation $T(z)=\frac{z-5}{z-4}$ is strictly increasing on $[1,4]\cap \mathbb{Q}$. Furthermore, it is easily seen that $T^m(\mathrm{\infty })\to T^m(\frac{1}{1})$ is an edge in $F^3$ for all non-negative integers m. From this, we get, as an example, a finite path in $F^3$ as $\mathrm{\infty }\to \frac{1}{1}\to \frac{4}{3}\to \frac{11}{8}$.

Lemma 3 Let T be as above, then the sequence $\{T^m(1)\}$ is strictly monotone increasing and $T(\frac{1}{0})\to T(\frac{1}{1})\to T^2(\frac{1}{1})\to \cdots \to T^m(\frac{1}{1})\to \cdots$ is an infinite path in $F^3$ in increasing order.

Proof The conclusion follows from Theorem 2 and from the fact that $T(z)=\frac{z-5}{z-4}$ is strictly increasing on $[1,4)\cap \mathbb{Q}$. □

Lemma 4 Let a and b be in ℕ and let $1\le \frac{a}{b}<\frac{5-\sqrt{5}}{2}$, then $\frac{a}{b}<T(\frac{a}{b})<\frac{5-\sqrt{5}}{2}$.

Proof From $\frac{a}{b}<\frac{5-\sqrt{5}}{2}$ we get $2a-5b<-\sqrt{5}b$. Then squaring gives the inequality $-a^2+4ab<-ab+5b^2$. That is, $\frac{a}{b}<\frac{-a+5b}{-a+4b}=T(\frac{a}{b})$. On the other hand, $a^2-5ab+5b^2>0$, then it is easily seen that $5{(a-4b)}^2<{(3a-10b)}^2$. As $\frac{a}{b}<2$, then taking square roots gives $\sqrt{5}(a-4b)>3a-10b$. This shows that $T(\frac{a}{b})<\frac{5-\sqrt{5}}{2}$. □

Proposition 1 Let T be as above and $1\le \frac{a}{b}<\frac{5-\sqrt{5}}{2}$. Then $\frac{a}{b}\to T(\frac{a}{b})$ is an edge in $F^3$ if and only if there exists a natural number u such that $u^2=5b^2+4$ and $a=\frac{5b-\sqrt{5b^2+4}}{2}$.

Proof Let $\frac{a}{b}\to T(\frac{a}{b})$ be an edge in $F^3$. Then, by using Theorem 2 and Lemma 4, we get $a^2-5ab+5b^2-1=0$. Since $\frac{a}{b}<\frac{5-\sqrt{5}}{2}$, we have $a=\frac{5b-\sqrt{5b^2+4}}{2}$. This concludes that $\sqrt{5b^2+4}$ is an integer u.

Conversely, it is clear to see $M=\left(\begin{array}{cc}\frac{-5b+\sqrt{5b^2+4}}{2}& 5b\\ -b& \frac{5b+\sqrt{5b^2+4}}{2}\end{array}\right)$ is in ${\Gamma }^3$ and that $M(\frac{1}{0})=\frac{a}{b}$ and $M(\frac{1}{1})=T(\frac{a}{b})$. Therefore, by the definition of edges of $F^3$, the configuration $\frac{a}{b}\to T(\frac{a}{b})$ is an edge in $F^3$. □

Theorem 4 The positive rational number $\frac{x}{y}$ is in $\{T^m(\frac{1}{0}):m\in \mathbb{N}\}=A$ if and only if there exists a natural number u such that $5y^2+4=u^2$ and $x=\frac{5y-\sqrt{5y^2+4}}{2}$.

Proof From Proposition 1, the ‘if’ part is clear.

Conversely, we show that under the hypothesis, $\frac{x}{y}$ is in the set A. Since $T^m(\frac{1}{0})=\frac{\frac{5y-\sqrt{5y^2+4}}{2}}{y}$ for any $m\in \mathbb{N}$, ${lim}_{m\to \mathrm{\infty }}{T}^m(\frac{1}{0})=\frac{5-\sqrt{5}}{2}$. As $1\le \frac{x}{y}<\frac{5-\sqrt{5}}{2}$, if $\frac{x}{y}$ is not in A, there exists $k\in \mathbb{N}$ such that $T^k(\frac{1}{0})<\frac{x}{y}<T^{k+1}(\frac{1}{0})$. From this, we get

$$T^k\left(\frac{1}{0}\right)=\frac{\frac{5a-\sqrt{5a^2+4}}{2}}{a}<\frac{\frac{5y-\sqrt{5y^2+4}}{2}}{y}<1+\frac{a}{3a-\frac{3a-\sqrt{5a^2+4}}{2}}=\frac{\frac{5a-v}{2}}{a}<\frac{\frac{5y-u}{2}}{y}<\frac{\frac{5a+v}{2}}{\frac{3a+v}{2}}.$$

We know that $\frac{\frac{5y-u}{2}}{y}\to T(\frac{\frac{5y-u}{2}}{y})=1+\frac{y}{3y-\frac{3y-u}{2}}$ is an edge in $F^3$. From Lemma 2 it must be that $T(\frac{\frac{5y-u}{2}}{y})=\frac{\frac{5y+u}{2}}{\frac{3y+u}{2}}<\frac{\frac{5a+v}{2}}{\frac{3a+v}{2}}$. This gives the inequality $vy<au$. But, from $\frac{\frac{5a-v}{2}}{a}<\frac{\frac{5y-u}{2}}{y}$, we arrive at the inequality $vy>au$, a contradiction. This concludes the proof of the theorem. □

Corollary 1 $\frac{1}{0}\to 0+\frac{1}{1}\to 1+\frac{1}{3}\to 1+\frac{3}{8}\stackrel{T}{\to }\cdots \stackrel{T}{\to }1+\frac{a_n}{b_n}\stackrel{T}{\to }1+\frac{b_n}{3b_n-a_n}\stackrel{T}{\to }\cdots$ is an infinite path in $F^3$, and all vertices of the path are smaller than $\frac{5-\sqrt{5}}{2}$, and the natural numbers x and y in the vertex $1+\frac{x}{y}$ are such that $5x^2+4$ and $5y^2+4$ are square. Furthermore, $a_n=\frac{3b_n-\sqrt{5b_n^2+4}}{2}$ and ${lim}_{n\to \mathrm{\infty }}(1+\frac{a_n}{b_n})=\frac{5-\sqrt{5}}{2}$, where T is as above.

Proof Lemma 3 and Theorem 4 conclude the proof. □

If we follow the way of the above, we arrive at the following two results without proofs.

Theorem 5 Let $T=\left(\begin{array}{cc}-1& 5\\ -1& 4\end{array}\right)$ and $\frac{5-\sqrt{5}}{2}<\frac{a}{b}\le 2$. Then $\frac{5-\sqrt{5}}{2}<T(\frac{a}{b})<\frac{a}{b}$ and $\frac{a}{b}\to T(\frac{a}{b})$ is an edge in $F^3$ if and only if $5b^2-4$ is a square and $a=\frac{5b-\sqrt{5b^2-4}}{2}$.

Corollary 2 $3\stackrel{T}{\to }2=1+\frac{1}{1}\stackrel{T}{\to }1+\frac{1}{2}\stackrel{T}{\to }1+\frac{2}{5}\stackrel{T}{\to }\cdots \stackrel{T}{\to }1+\frac{a_n}{b_n}\stackrel{T}{\to }1+\frac{b_n}{3b_n-a_n}\stackrel{T}{\to }\cdots$ is an infinite path in $F^3$ in decreasing order such that all vertices of the path are greater than $\frac{5-\sqrt{5}}{2}$, and the natural numbers x and y in the vertex $1+\frac{x}{y}$ are such that $5x^2-4$ and $5y^2-4$ are squares. Furthermore, $a_n=\frac{3b_n-\sqrt{5b_n^2-4}}{2}$ and ${lim}_{n\to \mathrm{\infty }}(1+\frac{a_n}{b_n})=\frac{5-\sqrt{5}}{2}$, where T is as above.

Theorem 6 Let k be a natural number and let the vertices $v_1,v_2,\dots ,v_k$ in $[1,3]$ of $F^3$ be such that at least one is smaller and one is greater than $\frac{5-\sqrt{5}}{2}$. Then the path $\mathrm{\infty }\to v_1\to v_2\to \cdots \to v_k$ does not occur in $F^3$.

Proof If the above situation occurs, since $\frac{5-\sqrt{5}}{2}$ is not a vertex in $F^3$, there exists $1\le m<k$ such that $v_m<\frac{5-\sqrt{5}}{2}<v_{m+1}$ and $v_m\to v_{m+1}$ is an edge in $F^3$.

Since the sequences of the vertices in Corollary 1 and Corollary 2 converge to $\frac{5-\sqrt{5}}{2}$, Lemma 2 gives that there exist naturals m and n such that

$$T^m\left(\frac{1}{0}\right)=v_m\to v_{m+1}=T^n\left(\frac{3}{1}\right).$$

Suppose first that $n>m$. Then, multiplying by $T^{-m}$ and using Theorem 5, we have $\frac{1}{0}\to T^{n-m}(\frac{3}{1})=\frac{\frac{5b-\sqrt{5b^2-4}}{2}}{b}$ for some $b\in \mathbb{N}$. Therefore, from Theorem 2, we get $b=1$. That is, $\frac{1}{0}\to \frac{3}{1}$ is an edge in $F^3$, a contradiction.

Suppose now that $m>n$. Then, again multiplying by $T^{-n}$, we have $T^{m-n}(\frac{1}{0})\to \frac{3}{1}$. In this case, $T^{m-n}(\frac{1}{0})\ge 2$. But, in any case, $T^{m-n}(\frac{1}{0})<\frac{5-\sqrt{5}}{2}$, a contradiction.

Finally, let $m=n$. Then $\frac{1}{0}\to \frac{3}{1}$ must be an edge in $F^3$, a contradiction. These complete the proof. □

Let $S=T^{-1}=\left(\begin{array}{cc}4& -5\\ 1& -1\end{array}\right)$. Then likewise we do before we give the following five results without a proof.

Lemma 5 Let $3\le \frac{a}{b}<\frac{5+\sqrt{5}}{2}$. Then $\frac{a}{b}<S(\frac{a}{b})<\frac{5+\sqrt{5}}{2}$. Furthermore, $\frac{a}{b}\to S(\frac{a}{b})$ is an edge in $F^3$ if and only if $\sqrt{5b^2-4}$ is a natural number and $a=\frac{5b+\sqrt{5b^2-4}}{2}$.

Corollary 3 $3=4-\frac{1}{1}\stackrel{S}{\to }4-\frac{1}{2}\stackrel{S}{\to }4-\frac{2}{5}\stackrel{S}{\to }\cdots \stackrel{S}{\to }4-\frac{\frac{3b-\sqrt{5b^2-4}}{2}}{b}\stackrel{S}{\to }\cdots$ is an infinite path in $F^3$ in increasing order, as seen Figure 2, and the limit of the sequence of vertices is $\frac{5+\sqrt{5}}{2}$.

Figure 2

A path in ${\mathit{F}}^{\mathbf{3}}$ .

Lemma 6 Let $\frac{5+\sqrt{5}}{2}<\frac{a}{b}\le 4$. Then $\frac{5+\sqrt{5}}{2}<S(\frac{a}{b})<\frac{a}{b}$. Furthermore, $\frac{a}{b}\to S(\frac{a}{b})$ is an edge in $F^3$ if and only if $5b^2+4$ is a square and $a=\frac{5b+\sqrt{5b^2+4}}{2}$.

Lemma 7 $4-\frac{0}{1}\stackrel{S}{\to }4-\frac{1}{3}\stackrel{S}{\to }4-\frac{3}{8}\stackrel{S}{\to }\cdots \stackrel{S}{\to }4-\frac{\frac{3b-\sqrt{5b^2+4}}{2}}{b}\stackrel{S}{\to }\cdots$ is an infinite path in $F^3$ in decreasing order, and the limit of the sequence of vertices is $\frac{5+\sqrt{5}}{2}$.

Theorem 7 Let k be a natural number and let the vertices $v_1,v_2,\dots ,v_k$, in $[3,4]$, of $F^3$ be such that at least one is smaller and one is greater than $\frac{5+\sqrt{5}}{2}$. Then $v_1\to v_2\to \cdots \to v_k\to \mathrm{\infty }$ does not occur in $F^3$.

Proof We conclude the proof as in Theorem 6. □

Proof of Theorem 3 Theorems 6 and 7 conclude that the vertices of $F^3$ in $(\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2})$ are not connected to the vertex ∞. That is, the graph $F^3$ is disconnected. □

We finally give one of our main results as follows.

Theorem 8 For all natural numbers m, the natural numbers b that make the number $(9m^2-4)b^2+4$ square are $0,1,3m,9m^2-1,3m(9m^2-1)-3m,\dots ,a,b,3mb-a,\dots$ .

Proof For the proof, we only use, as above, the interval $[1,4]$. It is clear that the matrix $M=\left(\begin{array}{cc}-1& 3m+2\\ -1& 3m+1\end{array}\right)$ is in ${\Gamma }^3$. Theorem 2 gives that $\frac{1}{1}\to M(\frac{1}{1})=\frac{3m+1}{3m}$ is an edge in $F^3$. Since the transformation $M(x)=\frac{-x+3m+2}{-x+3m+1}$ is an increasing function on $[1,4]$ and $\frac{1}{1}<M(\frac{1}{1})=\frac{3m+1}{3m}$, we can easily see that, for all $k\in \mathbb{N}$, $M^k(\frac{1}{1})<M^{k+1}(\frac{1}{1})$. That is, the sequence $\{M^k(1)\}$ is an increasing sequence. Furthermore, $M^k(1)=[1;\underset{k\mathrm{times}}{\underbrace{3m,3m,3m,\dots ,3m}}]$, or

$$M^k(1)=1+\frac{1}{3m-\frac{1}{3m-\frac{1}{3m-\cdots -\frac{1}{3m}}}}.$$

And more, ${lim}_{k\to \mathrm{\infty }}{M}^k(1)=\frac{3m+2-\sqrt{9m^2-4}}{2}$. So, for all $k\in \mathbb{N}$, $M^k(1)<\frac{3m+2-\sqrt{9m^2-4}}{2}$. If $\frac{a}{b}<\frac{3m+2-\sqrt{9m^2-4}}{2}$ we can easily see that $T(\frac{a}{b})<\frac{3m+2-\sqrt{9m^2-4}}{2}$, and furthermore, if $\frac{a}{b}\to T(\frac{a}{b})=\frac{-a+(3m+2)b}{-a+(3m+1)b}$ is an edge in $F^3$, then, by Theorem 2, $a^2-(3m+2)b+(3m+2)b^2-1=0$. Solving the equation, we have $a=\frac{(3m+2)b-\sqrt{(9m^2-4)b^2+4}}{2}$, where we get the sign ‘−’ since for all $k\in \mathbb{N}$, $M^k(1)<\frac{3m+2-\sqrt{9m^2-4}}{2}$. Because a is an integer, $\sqrt{(9m^2-4)b^2+4}$ must be an integer. According to Theorem 2, for all $k\in \mathbb{N}$, $M^k(\frac{1}{1})\to M^{k+1}(\frac{1}{1})$ is an edge in $F^3$. Therefore

$$\frac{1}{1}\stackrel{M}{\to }M\left(\frac{1}{1}\right)\stackrel{M}{\to }{M}^2\left(\frac{1}{1}\right)\stackrel{M}{\to }\cdots \stackrel{M}{\to }\left(\frac{a}{b}\right)\stackrel{M}{\to }M\left(\frac{a}{b}\right)\stackrel{M}{\to }\cdots$$

is an infinite path γ in $F^3$. All denominators of vertices $\frac{a}{b}$ of γ make $(9m^2-4)b^2+4$ square. We can rewrite γ as

$$\begin{array}{rcl}1+\frac{0}{1}& \to & 1+\frac{1}{3m}\to 1+\frac{3m}{9m^2-1}\to 1+\frac{9m^2-1}{3m(9m^2-1)-3m}\to \cdots \to 1+\frac{a^{\mathrm{\prime }}}{b^{\mathrm{\prime }}}\\ \to & 1+\frac{b^{\mathrm{\prime }}}{3m{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}}\to \cdots .\end{array}$$

And we conclude that the numbers $0,1,3m,9m^2-1,3m(9m^2-1)-3m,\dots ,a_0,b_0,3m{b}_0-a_0,\dots$ make $(9m^2-4)b^2+4$ square.

Let us now show the only non-negative integers b such that $(9m^2-4)b^2+4$ is square.

Conversely, suppose that there is a natural number t such that $(9m^2-4)t^2+4$ is square. Then $\frac{a_1}{b_1}=\frac{\frac{(3m+2)t-\sqrt{(9m^2-4)t^2+4}}{2}}{t}$ is smaller than $\frac{3m+2-\sqrt{9m^2-4}}{2}$ and, from Theorem 2, we get that $\frac{a_1}{b_1}\to T(\frac{a_1}{b_1})$ is an edge in $F^3$. Suppose, for some $k\in \mathbb{N}$, $M^k(\frac{1}{1})<\frac{a_1}{b_1}<M^{k+1}(\frac{1}{1})$. Lemma 2 says that $T^m(\frac{1}{1})<\frac{a_1}{b_1}<T^{m+1}(\frac{a_1}{b_1})<M^{k+1}(\frac{1}{1})$. Therefore, for some y,

$$\begin{array}{rl}{M}^k\left(\frac{1}{1}\right)& =1+\frac{\frac{3my-\sqrt{(9m^2-4)y^2+4}}{2}}{y}<\frac{a_1}{b_1}=1+\frac{\frac{3mt-\sqrt{(9m^2-4)t^2+4}}{2}}{t}\\ \to T\left(\frac{a_1}{b_1}\right)=1+\frac{t}{3mt-\frac{3mt-\sqrt{(9m^2-4)t^2+4}}{2}}<M^{k+1}\left(\frac{1}{1}\right)\\ =1+\frac{y}{3my-\frac{3my-\sqrt{(9m^2-4)y^2+4}}{2}}.\end{array}$$

From the first inequality, we conclude that $t\sqrt{(9m^2-4)y^2+4}>y\sqrt{(9m^2-4)y^2+4}$. From the second inequality, we just have $t\sqrt{(9m^2-4)y^2+4}<y\sqrt{(9m^2-4)y^2+4}$, a contradiction. Consequently, $\frac{a_1}{b_1}$ must be in the set $\{M^k(\frac{1}{1}):k\in \mathbb{N}\}$. This completes the proof of the theorem. □

From Corollaries 1 and 2, we get the following without a proof.

Corollary 4 The non-negative integers b such that $5b^2+4$ is square are $0,1,3,8,21,55,144,\dots ,a,b,3b-a,\dots$ .

Corollary 5 The non-negative integers b making $5b^2-4$ square are $1,2,5,13,34,89,\dots ,a,b,3b-a,\dots$ .

From Corollaries 4 and 5, we conclude the following important corollary.

Corollary 6 Let $\{a_n\}$ and $\{b_n\}$ be the sequences $(0,1,3,\dots ,a,b,3b-a,\dots )$ and $(1,2,5,\dots ,c,d,3d-c,\dots )$, respectively. Then the sequence $(a_1,b_1,a_2,b_2,\dots ,a_n,b_n,\dots )$ is the Fibonacci sequence.

Proof Let us see that $a_n+b_n=a_{n+1}$ and $b_n+a_{n+1}=b_{n+1}$ for all n in ℕ by induction. Suppose that the assertion is true up to the natural number k. Let us show that $a_{k+1}+b_{k+1}=a_{k+2}$. Since $a_{k+1}=3a_k-a_{k-1}$ and $b_{k+1}=3b_k-b_{k-1}$, $a_{k+1}+b_{k+1}=3(a_k+b_k)-(a_{k-1}+b_{k-1})=3a_{k+1}-a_k=a_{k+2}$. The other is similar. □