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Disconnectedness of the subgraph F 3 for the group Γ 3

Journal of Inequalities and Applications20132013:283

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

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 ± 4 are square. It turns out that the set of numbers b comprises the Fibonacci numbers.

Keywords

modular groupsuborbital graphdisconnectednessFibonacci numbers

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let Q ˆ = Q { } be the extended rationals, let Γ be the modular group acting on Q ˆ as with the upper half-plane H = { z C : Im z > 0 } :
g = ( a b c d ) : z = x y a z + b c z + d = a x + b y c x + d y ,

where a, b, c, and d are rational integers, and let Γ 3 denote the group consisting of the cubes of the elements g of Γ, which is the group { g Γ : a b + c d 0 ( mod 3 ) } ; see [1].

Jones et al. [2] used the notion of the imprimitive action [35] for a Γ-invariant equivalence relation induced on Q ˆ by the congruence subgroup Γ 0 ( n ) = { g Γ : c 0 ( mod n ) } to obtain some suborbital graphs and examined their connectedness and forest properties.

In [6], a Γ 3 -invariant equivalence relation is introduced by using the subgroup Γ 0 3 ( n ) = { g Γ 3 : c 0 ( mod n ) } 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 Γ 3 = { g 3 : g Γ } , it is easily seen that the elements of Γ 3 are ones of the forms ( 3 a b c 3 d ) , ( a 3 b 3 c d ) , and ( a b c d ) where a, b, c, and d 0 ( mod 3 ) in the third matrix. Furthermore, Γ 3 < Γ 0 3 ( n ) Γ 3 for each positive integer n, where Γ 3 is the stabilizer of ∞ generated by the element ( 1 3 0 1 ) , and second inclusion is strict if n > 1 .

Since the group Γ 3 is transitive on Q ˆ in [6], any reduced fraction r s in Q ˆ equals g ( ) for some g Γ 3 . Similar to that of [2], we get the following Γ 3 -invariant equivalence relation on Q ˆ by Γ 0 3 ( n ) as r s x y if and only if g 1 h Γ 0 3 ( n ) , where g = ( r s ) and h is similar. Furthermore, the above equivalence relation is imprimitive, which means that it is different from the identity relation ( a b if and only if a = b ) and the universal relation ( a b for all a , b Q ˆ ).

From the above defined Γ 3 -invariant equivalence relation, we can verify that r s x y if and only if r y s x 0 ( mod n ) . The equivalence classes are called blocks, and a block containing the rational x y is denoted by [ 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 Γ 3 acts on Q ˆ × Q ˆ through g : ( α , β ) ( g ( α ) , g ( β ) ) . The orbits are called suborbital. From the suborbital O 3 ( α , β ) containing ( α , β ) , we can form the suborbital graph G 3 ( α , β ) whose vertices are the elements of Q ˆ and the edges are the pairs ( a , b ) O 3 ( α , β ) , which we denote by a b and represent as hyperbolic geodesics in .

Since Γ 3 acts transitively on Q ˆ , every suborbital contains a pair ( , u n ) for some u n in Q ˆ , n 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 Γ 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 [ ] = { x y Q ˆ y 0 ( mod n ) } . The following two results were proved in [6].

Theorem 1 F u , n 3 = F u , n 3 if and only if n = n and u u ( mod 3 n ) .

Theorem 2 r s x y is an edge in F 3 ( = F 1 , 1 3 ) if and only if
  1. 1.

    if r 0 ( mod 3 ) , then y ± s ( mod 3 ) and r y s x = ± 1 , or

     
  2. 2.

    if s 0 ( mod 3 ) , then x ± r ( mod 3 ) , and r y s x = ± 1 , or

     
  3. 3.

    if r , s 0 ( mod 3 ) , then x ± r ( mod 3 ) , y ± s ( mod 3 ) and r y s x = ± 1 .

     

We can easily get the following lemmas.

Lemma 1 r s x y is in F 3 if and only if x y r s is in F 3 .

Lemma 2 [2]

No edges of F 3 cross in .

4 Disconnectedness of F 3

Definition 1 For m N and m 2 , let v 1 , v 2 , , v m be a finite sequence of vertices of F 3 . Then the configuration v 1 v 2 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 b is in F 3 , then a + 3 m b + 3 m is in F 3 for all m Z , and therefore, for some m, a + m or b + m (not both) is ∞ or both a + 3 m and b + 3 m are in the interval [ 1 , 4 ] . Therefore we can only use the interval [ 1 , 4 ] for our calculations as in Figure 1.
Figure 1
Figure 1

The subgraph F 3 .

It is clear that T = ( 1 5 1 4 ) is in Γ 3 and the corresponding transformation T ( z ) = z 5 z 4 is strictly increasing on [ 1 , 4 ] Q . Furthermore, it is easily seen that T m ( ) T m ( 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 1 1 4 3 11 8 .

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

Proof The conclusion follows from Theorem 2 and from the fact that T ( z ) = z 5 z 4 is strictly increasing on [ 1 , 4 ) Q . □

Lemma 4 Let a and b be in and let 1 a b < 5 5 2 , then a b < T ( a b ) < 5 5 2 .

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

Proposition 1 Let T be as above and 1 a b < 5 5 2 . Then a b T ( 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 = 5 b 5 b 2 + 4 2 .

Proof Let a b T ( a b ) be an edge in F 3 . Then, by using Theorem 2 and Lemma 4, we get a 2 5 a b + 5 b 2 1 = 0 . Since a b < 5 5 2 , we have a = 5 b 5 b 2 + 4 2 . This concludes that 5 b 2 + 4 is an integer u.

Conversely, it is clear to see M = ( 5 b + 5 b 2 + 4 2 5 b b 5 b + 5 b 2 + 4 2 ) is in Γ 3 and that M ( 1 0 ) = a b and M ( 1 1 ) = T ( a b ) . Therefore, by the definition of edges of F 3 , the configuration a b T ( a b ) is an edge in F 3 . □

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

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

Conversely, we show that under the hypothesis, x y is in the set A. Since T m ( 1 0 ) = 5 y 5 y 2 + 4 2 y for any m N , lim m T m ( 1 0 ) = 5 5 2 . As 1 x y < 5 5 2 , if x y is not in A, there exists k N such that T k ( 1 0 ) < x y < T k + 1 ( 1 0 ) . From this, we get
T k ( 1 0 ) = 5 a 5 a 2 + 4 2 a < 5 y 5 y 2 + 4 2 y < 1 + a 3 a 3 a 5 a 2 + 4 2 = 5 a v 2 a < 5 y u 2 y < 5 a + v 2 3 a + v 2 .

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

Corollary 1 1 0 0 + 1 1 1 + 1 3 1 + 3 8 T T 1 + a n b n T 1 + b n 3 b n a n T is an infinite path in F 3 , and all vertices of the path are smaller than 5 5 2 , and the natural numbers x and y in the vertex 1 + x y are such that 5 x 2 + 4 and 5 y 2 + 4 are square. Furthermore, a n = 3 b n 5 b n 2 + 4 2 and lim n ( 1 + a n b n ) = 5 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 = ( 1 5 1 4 ) and 5 5 2 < a b 2 . Then 5 5 2 < T ( a b ) < a b and a b T ( a b ) is an edge in F 3 if and only if 5 b 2 4 is a square and a = 5 b 5 b 2 4 2 .

Corollary 2 3 T 2 = 1 + 1 1 T 1 + 1 2 T 1 + 2 5 T T 1 + a n b n T 1 + b n 3 b n a n T is an infinite path in F 3 in decreasing order such that all vertices of the path are greater than 5 5 2 , and the natural numbers x and y in the vertex 1 + x y are such that 5 x 2 4 and 5 y 2 4 are squares. Furthermore, a n = 3 b n 5 b n 2 4 2 and lim n ( 1 + a n b n ) = 5 5 2 , where T is as above.

Theorem 6 Let k be a natural number and let the vertices v 1 , v 2 , , v k in [ 1 , 3 ] of F 3 be such that at least one is smaller and one is greater than 5 5 2 . Then the path v 1 v 2 v k does not occur in F 3 .

Proof If the above situation occurs, since 5 5 2 is not a vertex in F 3 , there exists 1 m < k such that v m < 5 5 2 < v m + 1 and v m v m + 1 is an edge in F 3 .

Since the sequences of the vertices in Corollary 1 and Corollary 2 converge to 5 5 2 , Lemma 2 gives that there exist naturals m and n such that
T m ( 1 0 ) = v m v m + 1 = T n ( 3 1 ) .

Suppose first that n > m . Then, multiplying by T m and using Theorem 5, we have 1 0 T n m ( 3 1 ) = 5 b 5 b 2 4 2 b for some b N . Therefore, from Theorem 2, we get b = 1 . That is, 1 0 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 ( 1 0 ) 3 1 . In this case, T m n ( 1 0 ) 2 . But, in any case, T m n ( 1 0 ) < 5 5 2 , a contradiction.

Finally, let m = n . Then 1 0 3 1 must be an edge in F 3 , a contradiction. These complete the proof. □

Let S = T 1 = ( 4 5 1 1 ) . Then likewise we do before we give the following five results without a proof.

Lemma 5 Let 3 a b < 5 + 5 2 . Then a b < S ( a b ) < 5 + 5 2 . Furthermore, a b S ( a b ) is an edge in F 3 if and only if 5 b 2 4 is a natural number and a = 5 b + 5 b 2 4 2 .

Corollary 3 3 = 4 1 1 S 4 1 2 S 4 2 5 S S 4 3 b 5 b 2 4 2 b S is an infinite path in F 3 in increasing order, as seen Figure 2, and the limit of the sequence of vertices is 5 + 5 2 .
Figure 2
Figure 2

A path in F 3 .

Lemma 6 Let 5 + 5 2 < a b 4 . Then 5 + 5 2 < S ( a b ) < a b . Furthermore, a b S ( a b ) is an edge in F 3 if and only if 5 b 2 + 4 is a square and a = 5 b + 5 b 2 + 4 2 .

Lemma 7 4 0 1 S 4 1 3 S 4 3 8 S S 4 3 b 5 b 2 + 4 2 b S is an infinite path in F 3 in decreasing order, and the limit of the sequence of vertices is 5 + 5 2 .

Theorem 7 Let k be a natural number and let the vertices v 1 , v 2 , , v k , in [ 3 , 4 ] , of F 3 be such that at least one is smaller and one is greater than 5 + 5 2 . Then v 1 v 2 v k 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 ( 5 5 2 , 5 + 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 , 3 m , 9 m 2 1 , 3 m ( 9 m 2 1 ) 3 m , , a , b , 3 m b a ,  .

Proof For the proof, we only use, as above, the interval [ 1 , 4 ] . It is clear that the matrix M = ( 1 3 m + 2 1 3 m + 1 ) is in Γ 3 . Theorem 2 gives that 1 1 M ( 1 1 ) = 3 m + 1 3 m is an edge in F 3 . Since the transformation M ( x ) = x + 3 m + 2 x + 3 m + 1 is an increasing function on [ 1 , 4 ] and 1 1 < M ( 1 1 ) = 3 m + 1 3 m , we can easily see that, for all k N , M k ( 1 1 ) < M k + 1 ( 1 1 ) . That is, the sequence { M k ( 1 ) } is an increasing sequence. Furthermore, M k ( 1 ) = [ 1 ; 3 m , 3 m , 3 m , , 3 m k times ] , or
M k ( 1 ) = 1 + 1 3 m 1 3 m 1 3 m 1 3 m .
And more, lim k M k ( 1 ) = 3 m + 2 9 m 2 4 2 . So, for all k N , M k ( 1 ) < 3 m + 2 9 m 2 4 2 . If a b < 3 m + 2 9 m 2 4 2 we can easily see that T ( a b ) < 3 m + 2 9 m 2 4 2 , and furthermore, if a b T ( a b ) = a + ( 3 m + 2 ) b a + ( 3 m + 1 ) b is an edge in F 3 , then, by Theorem 2, a 2 ( 3 m + 2 ) b + ( 3 m + 2 ) b 2 1 = 0 . Solving the equation, we have a = ( 3 m + 2 ) b ( 9 m 2 4 ) b 2 + 4 2 , where we get the sign ‘−’ since for all k N , M k ( 1 ) < 3 m + 2 9 m 2 4 2 . Because a is an integer, ( 9 m 2 4 ) b 2 + 4 must be an integer. According to Theorem 2, for all k N , M k ( 1 1 ) M k + 1 ( 1 1 ) is an edge in F 3 . Therefore
1 1 M M ( 1 1 ) M M 2 ( 1 1 ) M M ( a b ) M M ( a b ) M
is an infinite path γ in F 3 . All denominators of vertices a b of γ make ( 9 m 2 4 ) b 2 + 4 square. We can rewrite γ as
1 + 0 1 1 + 1 3 m 1 + 3 m 9 m 2 1 1 + 9 m 2 1 3 m ( 9 m 2 1 ) 3 m 1 + a b 1 + b 3 m b a .

And we conclude that the numbers 0 , 1 , 3 m , 9 m 2 1 , 3 m ( 9 m 2 1 ) 3 m , , a 0 , b 0 , 3 m b 0 a 0 , 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.

Conversely, suppose that there is a natural number t such that ( 9 m 2 4 ) t 2 + 4 is square. Then a 1 b 1 = ( 3 m + 2 ) t ( 9 m 2 4 ) t 2 + 4 2 t is smaller than 3 m + 2 9 m 2 4 2 and, from Theorem 2, we get that a 1 b 1 T ( a 1 b 1 ) is an edge in F 3 . Suppose, for some k N , M k ( 1 1 ) < a 1 b 1 < M k + 1 ( 1 1 ) . Lemma 2 says that T m ( 1 1 ) < a 1 b 1 < T m + 1 ( a 1 b 1 ) < M k + 1 ( 1 1 ) . Therefore, for some y,
M k ( 1 1 ) = 1 + 3 m y ( 9 m 2 4 ) y 2 + 4 2 y < a 1 b 1 = 1 + 3 m t ( 9 m 2 4 ) t 2 + 4 2 t T ( a 1 b 1 ) = 1 + t 3 m t 3 m t ( 9 m 2 4 ) t 2 + 4 2 < M k + 1 ( 1 1 ) = 1 + y 3 m y 3 m y ( 9 m 2 4 ) y 2 + 4 2 .

From the first inequality, we conclude that t ( 9 m 2 4 ) y 2 + 4 > y ( 9 m 2 4 ) y 2 + 4 . From the second inequality, we just have t ( 9 m 2 4 ) y 2 + 4 < y ( 9 m 2 4 ) y 2 + 4 , a contradiction. Consequently, a 1 b 1 must be in the set { M k ( 1 1 ) : k 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 , , a , b , 3 b a ,  .

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

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 , , a , b , 3 b a , ) and ( 1 , 2 , 5 , , c , d , 3 d c , ) , respectively. Then the sequence ( a 1 , b 1 , a 2 , b 2 , , a n , b n , ) 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 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. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Karadeniz Technical University, Trabzon, Turkey
(2)
Department of Mathematics, Recep Tayyip Erdoğan University, Rize, Turkey

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Copyright

© Akbaş et al.; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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