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

- Mehmet Akbaş
^{1}, - Tuncay Kör
^{1}Email author and - Yavuz Kesicioğlu
^{2}

**2013**:283

https://doi.org/10.1186/1029-242X-2013-283

© Akbaş et al.; licensee Springer 2013

**Received: **6 December 2012

**Accepted: **22 May 2013

**Published: **5 June 2013

## 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 $(9{m}^{2}-4){a}^{2}+4$ and $5{b}^{2}\pm 4$ are square. It turns out that the set of numbers *b* comprises the Fibonacci numbers.

## Keywords

## Dedication

Dedicated to Professor Hari M Srivastava

## 1 Introduction

*Γ*be the modular group acting on $\stackrel{\u02c6}{\mathbb{Q}}$ as with the upper half-plane $\mathcal{H}=\{z\in \mathbb{C}:Imz>0\}$:

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 $\stackrel{\u02c6}{\mathbb{Q}}$ 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 $\stackrel{\u02c6}{\mathbb{Q}}$ in [6], any reduced fraction $\frac{r}{s}$ in $\stackrel{\u02c6}{\mathbb{Q}}$ 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 $\stackrel{\u02c6}{\mathbb{Q}}$ 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 $a,b\in \stackrel{\u02c6}{\mathbb{Q}}$).

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 $\stackrel{\u02c6}{\mathbb{Q}}\times \stackrel{\u02c6}{\mathbb{Q}}$ 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 $\stackrel{\u02c6}{\mathbb{Q}}$ 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 $\stackrel{\u02c6}{\mathbb{Q}}$, every suborbital contains a pair $(\mathrm{\infty},\frac{u}{n})$ for some $\frac{u}{n}$ in $\stackrel{\u02c6}{\mathbb{Q}}$, $n\ge 0$, $(u,n)=1$. 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 $[\mathrm{\infty}]=\{\frac{x}{y}\in \stackrel{\u02c6}{\mathbb{Q}}\mid y\equiv 0(modn)\}$. 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.

*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.

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 $-{a}^{2}+4ab<-ab+5{b}^{2}$. That is, $\frac{a}{b}<\frac{-a+5b}{-a+4b}=T(\frac{a}{b})$. On the other hand, ${a}^{2}-5ab+5{b}^{2}>0$, 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* ${u}^{2}=5{b}^{2}+4$ *and* $a=\frac{5b-\sqrt{5{b}^{2}+4}}{2}$.

*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 ${a}^{2}-5ab+5{b}^{2}-1=0$. Since $\frac{a}{b}<\frac{5-\sqrt{5}}{2}$, we have $a=\frac{5b-\sqrt{5{b}^{2}+4}}{2}$. This concludes that $\sqrt{5{b}^{2}+4}$ is an integer *u*.

Conversely, it is clear to see $M=\left(\begin{array}{cc}\frac{-5b+\sqrt{5{b}^{2}+4}}{2}& 5b\\ -b& \frac{5b+\sqrt{5{b}^{2}+4}}{2}\end{array}\right)$ 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* $5{y}^{2}+4={u}^{2}$ *and* $x=\frac{5y-\sqrt{5{y}^{2}+4}}{2}$.

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

*A*. Since ${T}^{m}(\frac{1}{0})=\frac{\frac{5y-\sqrt{5{y}^{2}+4}}{2}}{y}$ 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

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}}{3{b}_{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* $5{x}^{2}+4$ *and* $5{y}^{2}+4$ *are square*. *Furthermore*, ${a}_{n}=\frac{3{b}_{n}-\sqrt{5{b}_{n}^{2}+4}}{2}$ *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* $5{b}^{2}-4$ *is a square and* $a=\frac{5b-\sqrt{5{b}^{2}-4}}{2}$.

**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}}{3{b}_{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* $5{x}^{2}-4$ *and* $5{y}^{2}-4$ *are squares*. *Furthermore*, ${a}_{n}=\frac{3{b}_{n}-\sqrt{5{b}_{n}^{2}-4}}{2}$ *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 ${F}^{3}$.

*m*and

*n*such that

Suppose first that $n>m$. Then, multiplying by ${T}^{-m}$ and using Theorem 5, we have $\frac{1}{0}\to {T}^{n-m}(\frac{3}{1})=\frac{\frac{5b-\sqrt{5{b}^{2}-4}}{2}}{b}$ 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* $\sqrt{5{b}^{2}-4}$ *is a natural number and* $a=\frac{5b+\sqrt{5{b}^{2}-4}}{2}$.

**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{5{b}^{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*$\frac{5+\sqrt{5}}{2}$.

**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* $5{b}^{2}+4$ *is a square and* $a=\frac{5b+\sqrt{5{b}^{2}+4}}{2}$.

**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{5{b}^{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* $(9{m}^{2}-4){b}^{2}+4$ *square are* $0,1,3m,9{m}^{2}-1,3m(9{m}^{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, ${M}^{k}(1)=[1;\underset{k\phantom{\rule{0.25em}{0ex}}\mathrm{times}}{\underset{\u23df}{3m,3m,3m,\dots ,3m}}]$, or

*a*is an integer, $\sqrt{(9{m}^{2}-4){b}^{2}+4}$ 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

*γ*in ${F}^{3}$. All denominators of vertices $\frac{a}{b}$ of

*γ*make $(9{m}^{2}-4){b}^{2}+4$ square. We can rewrite

*γ*as

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

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

*t*such that $(9{m}^{2}-4){t}^{2}+4$ is square. Then $\frac{{a}_{1}}{{b}_{1}}=\frac{\frac{(3m+2)t-\sqrt{(9{m}^{2}-4){t}^{2}+4}}{2}}{t}$ is smaller than $\frac{3m+2-\sqrt{9{m}^{2}-4}}{2}$ 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*,

From the first inequality, we conclude that $t\sqrt{(9{m}^{2}-4){y}^{2}+4}>y\sqrt{(9{m}^{2}-4){y}^{2}+4}$. From the second inequality, we just have $t\sqrt{(9{m}^{2}-4){y}^{2}+4}<y\sqrt{(9{m}^{2}-4){y}^{2}+4}$, 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* $5{b}^{2}+4$ *is square are* $0,1,3,8,21,55,144,\dots ,a,b,3b-a,\dots $ .

**Corollary 5** *The non*-*negative integers* *b* *making* $5{b}^{2}-4$ *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 ${a}_{k+1}=3{a}_{k}-{a}_{k-1}$ and ${b}_{k+1}=3{b}_{k}-{b}_{k-1}$, ${a}_{k+1}+{b}_{k+1}=3({a}_{k}+{b}_{k})-({a}_{k-1}+{b}_{k-1})=3{a}_{k+1}-{a}_{k}={a}_{k+2}$. The other is similar. □

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## Authors’ Affiliations

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