Markov Triples with Generalized Pell Numbers

Abstract

For an integer $k\ge 2$, let ${(P_n^{(k)})}_n$ be the k-generalized Pell sequence which starts with $0,\dots ,0,1$ (k terms), and each term afterwards is given by $P_n^{(k)}=2P_{n-1}^{(k)}+P_{n-2}^{(k)}+\cdots +P_{n-k}^{(k)}$. In this paper, we determine all solutions of the Markov equation $x^2+y^2+z^2=3xyz$, with x, y, and z being k-generalized Pell numbers. This paper continues and extends a previous work of Kafle, Srinivasan and Togbé, who found all Markov triples with Pell components.

1. Introduction

A triple of positive integers $(x,y,z)$ is known as a Markov triple if its entries satisfy the Markov Diophantine equation $x^2+y^2+z^2=3xyz$. The first triples with the above property are

$$(1,1,1),(1,1,2),(1,2,5),(1,5,13),(2,5,29),(1,13,34),(1,34,89),\dots .$$

Since any permutation of a Markov triple $(x,y,z)$ is a Markov triple, we can write them from now on in weakly increasing order, i.e., by assuming that $x\le y\le z$. The entries of a Markov triple are said to be a Markov number, and the sequence formed by these numbers coincides with the sequence A002559 in OEIS [1].

Markov triples are a mathematical object rich in properties. To begin, we state that there are infinitely many Markov triples of the form

$$(1,F_{2n-1},F_{2n+1})\mathrm{and}(2,P_{2n-1},P_{2n+1})\mathrm{for}n\ge 1,$$

where $F_n$ and $P_n$ represent the nth Fibonacci number and the nth Pell number, respectively (see ([2] Section 3.1)). Recall that these numbers are defined recurrently by

$$F_n=F_{n-1}+F_{n-2}\mathrm{and}{P}_n=2P_{n-1}+P_{n-2}\mathrm{for}\mathrm{all}n\ge 2,$$

together with the initial conditions $F_0=P_0=0$ and $F_1=P_1=1$.

In recent years, several authors have focused on studying relationships between Markov triples and terms of linear recurrence sequences. For example, Luca and Srinivasan [3] characterized all Markov triples whose entries are Fibonacci numbers, while Kafle et al. [4] solved the above problem with Pell numbers instead of Fibonacci numbers. Specifically, they proved that all Markov triples with Pell components are $(1,1,1)$, $(1,1,2)$, $(1,2,5)$ or triples of the form $(2,P_{2n-1},P_{2n+1})$ for all $n\ge 2$.

In 2020, Gómez et al. [5] extended the study of Markov triples to the k-generalized Fibonacci sequence $F_{}^{(k)}:={(F_n^{(k)})}_{n\ge 2-k}$. The sequence $F_{}^{(k)}$ is a generalization of the Fibonacci sequence defined for an integer $k\ge 2$ by the initial conditions $F_i^{(k)}=0$ for $i=2-k,\dots ,0$, $F_1^{(k)}=1$ and the kth order recursion

$$F_n^{(k)}=F_{n-1}^{(k)}+F_{n-2}^{(k)}+\cdots +F_{n-k}^{(k)}\mathrm{for}\mathrm{all}n\ge 2.$$

In a similar vein, the k-generalized Pell sequence or, for simplicity, the k-Pell sequence $P_{}^{(k)}:={\left(P_n^{(k)}\right)}_{n\ge 2-k}$ is given by the recurrence

$$P_n^{(k)}=2P_{n-1}^{(k)}+P_{n-2}^{(k)}+\cdots +P_{n-k}^{(k)}\mathrm{for}\mathrm{all}n\ge 2,$$

with $P_j^{(k)}=0$ for $j=2-k,\dots ,0$ and $P_1^{(k)}=1$. Note that, when $k=2$, this coincides with the classical Pell sequence.

In this paper, we determine all Markov triples whose entries are generalized Pell numbers, which continues and extends the work done by Kafle et al. [4]. More precisely, we solve the Diophantine equation

$${\left(P_s^{(k)}\right)}^2+{\left(P_m^{(k)}\right)}^2+{\left(P_n^{(k)}\right)}^2=3P_s^{(k)}{P}_m^{(k)}{P}_n^{(k)}$$

(1)

in positive integers $s,m,n$, and k with $n\ge m\ge s$ and $k\ge 2$.

Our main result is the following.

Theorem 1.

All the solutions $(s,m,n,k)$ of the Diophantine Equation (1) are:

(a)

$(s,m,n,k)=(1,1,1,k)$ for all $k\ge 2$.

(b)

$(s,m,n,k)=(2,2t-1,2t+1,2)$ for all $t\ge 2$.

(c)

$(s,m,n,k)=(1,t,t+1,k)$ for all $k\ge 2$ and all $t\in [1,k]$.

Before we get into the details of the paper, it is important to note that, from Theorem 1$(a)$ with $k=2$, we have that $(1,1,1)$ is a Markov triple with entries in the usual Pell sequence. It also follows from Theorem 1$(c)$ with $(k,t)=(2,1)$ and $(k,t)=(2,2)$ that $(1,1,2)$ and $(1,2,5)$ are Markov triples with Pell components, respectively. Finally, we have that $(2,P_{2t-1},P_{2t+1})$ are all Markov triples for all $t\ge 2$, which follows from Theorem 1$(b)$. In this sense, Theorem 1 generalizes that of [4].

2. Preliminary Results

The first interesting fact about the k-Pell sequence, shown by Kiliç in [6], is that the first $k+1$ non-zero terms in $P_{}^{(k)}$ are Fibonacci numbers with an odd index, namely

$$P_n^{(k)}=F_{2n-1}\mathrm{for}\mathrm{all}1\le n\le k+1.$$

(2)

Moreover, it was proved in [7] that

$$P_n^{(k)}<F_{2n-1}\mathrm{holds}\mathrm{for}\mathrm{all}n\ge k+2.$$

On the other hand, it is known that the characteristic polynomial of the sequence $P_{}^{(k)}$, namely

$${\Phi }_k(x)=x^k-2x^{k-1}-x^{k-2}-\cdots -x-1,$$

is irreducible over $\mathbb{Q}\left[x\right]$ and has just one real root outside the unit circle (for more details on ${\Phi }_k(x)$, see Section 2 of [7]). We shall denote this single root by $\gamma (k)$ and omit the dependence on k of $\gamma$ whenever no confusion may arise. Furthermore, it is known that $\gamma (k)$ is exponentially close to ${\phi }^2$, where $\phi$ denotes the golden section. In fact, it was proved in ([7] Lemma 3.2), that $\gamma (k)\in ({\phi }^2(1-{\phi }^{-k}),{\phi }^2)$.

We next introduce the fundamental properties of the generalized Pell sequence $P_{}^{(k)}$. These results were established in 2021 by Bravo, Herrera, and Luca [7].

Lemma 1

(Properties of $P_{}^{(k)}$). Let $k\ge 2$ be an integer. Then

(a)

${\gamma }^{n-2}\le P_n^{(k)}\le {\gamma }^{n-1}$ for all $n\ge 1$.

(b)

$P_{}^{(k)}$ satisfies the following “Binet-like” formula

$$P_n^{(k)}=\sum _{i=1}^k{g}_k({\gamma }_i){\gamma }_i^n,$$

where $\gamma ={\gamma }_1,\dots ,{\gamma }_k$ are the roots of ${\Phi }_k(x)$ and

$$g_k(z)=\frac{z-1}{(k+1)z^2-3kz+k-1}.$$

(c)

$0.276<g_k(\gamma )<0.5$ and $|g_k({\gamma }_i)|<1$ for $2\le i\le k$.

(d)

$|P_n^{(k)}-g_k(\gamma ){\gamma }^n|<1/2$ holds for all $n\ge 2-k$.

We finish this section of preliminaries with the following technical result, which will be used later.

Lemma 2.

Let $(1,x,y)$ be a Markov triple in weakly increasing order. Then

$$x=F_{2r-1}\mathit{and}y=F_{2r+1}\mathit{for}\mathit{some}\mathit{integer}r\ge 1.$$

Proof. 

Since $(1,x,y)$ is a Markov triple, it must hold that $1+x^2+y^2=3xy$, which can be rewritten as the Pell equation ${(2y-3x)}^2-5x^2=-4$. However, we already know (see, for example, ([8] Lemma 1)) that such an equation is solvable in positive integers if and only if $2y-3x=L_{2r-1}$ and $x=F_{2r-1}$ for some integer $r\ge 1$, where $L_t$ denotes the tth Lucas number (see ([1] A000032)). It remains only to prove that $y=F_{2r+1}$. To do that, we just recall the identity $L_t=F_{t-1}+F_{t+1}$, which holds for all $t\ge 1$, to obtain

$$\begin{array}{cc}\hfill y& =\frac{1}{2}\left(3F_{2r-1}+L_{2r-1}\right)=\frac{1}{2}\left(3F_{2r-1}+F_{2r-2}+F_{2r}\right)\hfill \\ \hfill & =\frac{1}{2}\left(2F_{2r-1}+(F_{2r-1}+F_{2r-2})+F_{2r}\right)=\frac{1}{2}\left(2F_{2r-1}+2F_{2r}\right)\hfill \\ \hfill & =F_{2r+1},\hfill \end{array}$$

as desired. □

3. Proof of Theorem 1

Assume throughout that $(s,m,n,k)$ is a solution to (1). Since the only Markov triples with repeated numbers are $(1,1,1)$ and $(1,1,2)$ (see ([2] Lemma 3.1)), whose entries are all generalized Pell numbers, we find that $(1,1,1,k)$ and $(1,1,2,k)$ are solutions to (1) for all $k\ge 2$. These solutions appear in Theorem 1 items $(a)$ and $(c)$ (with $t=1$), respectively. Thus, we can assume from now on that $n>m>s$. In addition, we can assume $k\ge 3$, since the case $k=2$ was completely solved in [4].

We now look for a relationship between our variables. Using Lemma 1$(a)$ and (1), we find that

$${\gamma }^{2(n-2)}\le {\left(P_n^{(k)}\right)}^2<3P_s^{(k)}{P}_m^{(k)}{P}_n^{(k)}\le 3{\gamma }^{s+m+n-3}<{\gamma }^{s+m+n}$$

and

$$3{\gamma }^{s+m+n-6}\le 3P_s^{(k)}{P}_m^{(k)}{P}_n^{(k)}={\left(P_s^{(k)}\right)}^2+{\left(P_m^{(k)}\right)}^2+{\left(P_n^{(k)}\right)}^2<3{\gamma }^{2(n-1)}.$$

The above chain of inequalities allows us to establish the following lemma.

Lemma 3.

If $(s,m,n,k)$ is a solution to the Diophantine Equation (1), then

$$|n-(m+s)|\le 3.$$

On the other hand, by Lemma 1$(d)$, we can write

$$P_n^{(k)}=g_k(\gamma ){\gamma }^n+e_n(k)\mathrm{where}|e_n(k)|<1/2,$$

(3)

and a simple computation gives

$$P_n^{(k)}=g_k(\gamma ){\gamma }^n\left(1+X_n^{(k)}\right)\mathrm{where}|X_n^{(k)}|<\frac{1.82}{{\gamma }^n}.$$

(4)

At this point, we use (3) and (4) to rewrite (1) as

$${\left(P_s^{(k)}\right)}^2+{\left(P_m^{(k)}\right)}^2+{\left(g_k(\gamma ){\gamma }^n+e_n(k)\right)}^2=3g_k{(\gamma )}^3{\gamma }^{n+m+s}(1+X),$$

where

$$X=X_s^{(k)}+X_m^{(k)}+X_n^{(k)}+X_s^{(k)}{X}_m^{(k)}+X_s^{(k)}{X}_n^{(k)}+X_m^{(k)}{X}_n^{(k)}+X_s^{(k)}{X}_m^{(k)}{X}_n^{(k)},$$

for which $|X|<21.43/{\gamma }^s$ holds. A straightforward calculation now leads to the following inequality:

$$|g_k{(\gamma )}^2{\gamma }^{2n}-3g_k{(\gamma )}^3{\gamma }^{n+m+s}|<2{(P_m^{(k)})}^2+g_k(\gamma ){\gamma }^n+3g_k{(\gamma )}^3{\gamma }^{n+m+s}|X|+\frac{1}{4},$$

and so

$$|{\gamma }^{2n}-3g_k(\gamma ){\gamma }^{n+m+s}|<\frac{2{\gamma }^{2(m-1)}}{g_k{(\gamma )}^2}+\frac{{\gamma }^n}{g_k(\gamma )}+3g_k(\gamma ){\gamma }^{n+m+s}|X|+\frac{1}{4g_k{(\gamma )}^2}.$$

Dividing both sides by $3g_k(\gamma ){\gamma }^{n+m+s}$ and using the fact that $g_k(\gamma )>0.276$ given by $(c)$ of Lemma 1 as well as the fact that $2.54682\dots =\gamma (3)\le \gamma$, we arrive at

$$|1-{(3g_k(\gamma ))}^{-1}{\gamma }^{n-(m+s)}|<\frac{35}{{\gamma }^s}.$$

(5)

For simplicity of notation, we put $t:=n-(m+s)$ so that $t\in \{0,\pm 1,\pm 2,\pm 3\}$ by Lemma 3. In order to obtain a lower bound of the left-hand side of (5), we use the inequalities $0.276<g_k(\gamma )<0.5$ once more to deduce that $2{\gamma }^t/3<{(3g_k(\gamma ))}^{-1}{\gamma }^t<1.208{\gamma }^t$, which holds for any t and $k\ge 3$. We next distinguish the following cases.

If $t>0$, then the following chain of inequalities is valid:

$$1.69<\frac{2}{3}\gamma (3)\le \frac{2}{3}{\gamma }^t<{(3g_k(\gamma ))}^{-1}{\gamma }^t<1.208{\gamma }^t<1.208{\phi }^6<22.$$

Thus,

$$|1-{(3g_k(\gamma ))}^{-1}{\gamma }^t|>0.69.$$

If, on the contrary, $t<0$, then

$$0.03<\frac{2}{3}{\phi }^{-6}<\frac{2}{3}{\gamma }^t<{(3g_k(\gamma ))}^{-1}{\gamma }^t<1.208{\gamma }^t<1.208\gamma {(3)}^{-1}<0.48,$$

which implies

$$|1-{(3g_k(\gamma ))}^{-1}{\gamma }^t|>0.52.$$

Finally, we handle the case when $t=0$. To do that, let us introduce the temporary function $h_k(x)$ given by

$$h_k(x)={(3g_k(x))}^{-1}-1=\frac{(k+1)x^2-3(k+1)x+k+2}{3(x-1)},$$

where $g_k(x)$ is given by Lemma 1. Note that $h_k(x)$ is a continuous and increasing function in $(1,\infty )$. Since ${\phi }^2(1-{\phi }^{-k})<\gamma <{\phi }^2$, we obtain that

$$h_k({\phi }^2(1-{\phi }^{-k}))<h_k(\gamma )<h_k({\phi }^2)=\frac{1}{3\phi }=0.206011\dots <0.207.$$

On the other hand, we computationally note that $h_k({\phi }^2(1-{\phi }^{-k}))<0$ for $3\le k\le 8$ and that this is not on our side, since we are interested in finding a nontrivial lower bound on $|h_k(\gamma )|$. However, for $k\ge 9$ we have

$$0.05<0.0508799\dots =h_9({\phi }^2(1-{\phi }^{-9}))\le h_k({\phi }^2(1-{\phi }^{-k}))<h_k(\gamma ).$$

Hence, $|h_k(\gamma )|>0.05$ for all $k\ge 9$. However, computationally, we find that

k	3	4	5	6	7	8
$h_k(\gamma (k))$	0.082…	0.149…	0.180…	0.194…	0.201…	0.203…

which shows that $|h_k(\gamma )|>0.05$ also holds for all $k\ge 3$. Consequently,

$$|1-{(3g_k(\gamma ))}^{-1}{\gamma }^t|>0.05\mathrm{for}\mathrm{all}k\ge 3\mathrm{and}t\in \{0,\pm 1,\pm 2,\pm 3\}.$$

(6)

Comparing inequalities (5) and (6), we obtain $2.{54}^s<700$, and so $s\le 7$. From this and Lemma 3, we conclude that

$$1\le s\le 7\mathrm{and}1\le n-m\le 10.$$

Turning back to the original Equation (1) and using (4) once more, we can rewrite it as

$${\left(P_s^{(k)}\right)}^2+g_k{(\gamma )}^2\left(({\gamma }^{2n}+{\gamma }^{2m})+({\gamma }^{2m}{Y}_1+{\gamma }^{2n}{Y}_2)\right)=3P_s^{(k)}{g}_k{(\gamma )}^2{\gamma }^{n+m}\left(1+Y_3\right),$$

(7)

where we put

$$Y_1:=2X_m^{(k)}+{(X_m^{(k)})}^2,Y_2:=2X_n^{(k)}+{(X_n^{(k)})}^2\mathrm{and}{Y}_3:=X_n^{(k)}+X_m^{(k)}+X_n^{(k)}{X}_m^{(k)}.$$

Using (4) once again, we can check that $|Y_i|<6.96/{\gamma }^m$ for $i=1,3$ and $|Y_2|<6.96/{\gamma }^n$. It then follows from (7) and some elementary algebra that

$$\begin{array}{c}|3P_s^{(k)}{g}_k{(\gamma )}^2{\gamma }^{n+m}-g_k{(\gamma )}^2({\gamma }^{2n}+{\gamma }^{2m})|\hfill \\ \hfill \le {\left(P_s^{(k)}\right)}^2+g_k{(\gamma )}^2({\gamma }^{2m}|Y_1|+{\gamma }^{2n}|Y_2|+3P_s^{(k)}{\gamma }^{n+m}|Y_3|).\end{array}$$

Dividing across by $3P_s^{(k)}{g}_k{(\gamma )}^2{\gamma }^{n+m}$, regrouping some terms, and using the inequalities $P_s^{(k)}\le P_7^{(k)}\le 233$ and $g_k(\gamma )>0.276$, which holds for all $k\ge 3$, we obtain

$$|{(3P_s^{(k)})}^{-1}({\gamma }^{\ell }+{\gamma }^{-\ell })-1|<\frac{1050}{{\gamma }^m},$$

(8)

where we have put $\ell :=n-m$. We now need to consider the following two cases depending on $(s,\ell )$.

3.1. Case $(s,\ell )=(1,1)$

In this case, Equation (1) is transformed into the simpler equation

$$1+{\left(P_m^{(k)}\right)}^2+{\left(P_{m+1}^{(k)}\right)}^2=3P_m^{(k)}{P}_{m+1}^{(k)},$$

which tells us that $(1,P_m^{(k)},P_{m+1}^{(k)})$ is a Markov triple. Hence, $P_m^{(k)}$ and $P_{m+1}^{(k)}$ are both Fibonacci numbers by Lemma 2. However, since the intersection between generalized Pell numbers and Fibonacci numbers is only trivial by the main theorem of [9], we can conclude that $m\le k$ and, therefore, $(1,P_m^{(k)},P_{m+1}^{(k)})=(1,F_{2m-1},F_{2m+1})$ by (2). This leads to the solutions listed in Theorem 1$(c)$ and completes the analysis of the case $(s,\ell )=(1,1)$.

3.2. Case $(s,\ell )\ne (1,1)$

To treat this case, we need to find the lower bound of the left-hand side of (8). In order to do so, let us consider the function

$$f_k(s,\ell ,x):={(3P_s^{(k)})}^{-1}(x^{\ell }+x^{-\ell })-1,$$

where $(s,\ell )\in \Omega :=([1,7]\times [1,10])\setminus \{(1,1)\}$ and $x\in [\gamma (3),{\phi }^2]$. For the cases $k\in \{3,4,5\}$, we can check computationally that ${min}_{(s,\ell )\in \Omega }|f_k(s,\ell ,\gamma )|>0.0793$.

Suppose next that $k\ge 6$. Then, $s\le k+1$ and so $P_s^{(k)}=F_{2s-1}$. Thus,

$$f(s,\ell ,x):=f_k(s,\ell ,x)={(3F_{2s-1})}^{-1}(x^{\ell }+x^{-\ell })-1.$$

Taking into account that $f(s,\ell ,x)$ is increasing in terms of x, one finds that

$$f(s,\ell ,\gamma (3))\le f(s,\ell ,\gamma )<f(s,\ell ,{\phi }^2).$$

(9)

Here, we also find computationally that $f(s,\ell ,\gamma (3)),f(s,\ell ,{\phi }^2)\ne 0$ for all $(s,\ell )\in \Omega$ and that

$$-0.996<\underset{(s,\ell )\in \Omega }{min}f(s,\ell ,\gamma (3))\mathrm{and}\underset{(s,\ell )\in \Omega }{max}f(s,\ell ,{\phi }^2)<5042.$$

From this and (9), we deduce that $|f(s,\ell ,\gamma )|>0.996$ for all $(s,\ell )\in \Omega$.

In summary, we have that $|f_k(s,\ell ,\gamma )|>0.0793$ for all $k\ge 3$ and all $(s,\ell )\in \Omega$. Comparing this with inequality (8), we get $2.{54}^m<13241$, giving $m\le 10$. We have thus proved that

$$m\le 10,s\le 7,\mathrm{and}n\le 20.$$

Assume, for the sake of contradiction, that $k\ge 19$. Then, $n\le k+1$ and so, by (2), Equation (1) takes the form

$$F_{2s-1}^2+F_{2m-1}^2+F_{2n-1}^2=3F_{2s-1}{F}_{2m-1}{F}_{2n-1},$$

which tells us that $(F_{2s-1},F_{2m-1},F_{2n-1})$ is a Markov triple. However, Markov triples with values in the Fibonacci sequence have already been studied, as mentioned before. Indeed, by the main theorem in [3], there must be a nonnegative integer $n_0$ such that $(2s-1,2m-1,2n-1)=(1,2n_0-1,2n_0+1)$, implying that $(s,\ell )=(1,1)$. However, this contradicts our assumption that $(s,\ell )\ne (1,1)$. Consequently, $k\in [3,18]$.

Finally, a brute-force search with Mathematica in the range

$$3\le k\le 18,1\le s\le 7,s+1\le m\le 10\mathrm{and}m+1\le n\le 20$$

gives no solutions to Equation (1). This completes the analysis of the case $(s,\ell )\ne (1,1)$ and therefore the proof of Theorem 1.