Carlitz’s Equations on Generalized Fibonacci Numbers

Abstract

Carlitz solved some Diophantine equations on Fibonacci or Lucas numbers. We extend his results to the sequence of generalized Fibonacci and Lucas numbers. In this paper, we solve the Diophantine equations of the form $A_{n_1}\cdots A_{n_k}=B_{m_1}\cdots B_{m_r}{C}_{t_1}\cdots C_{t_s}$, where ($A_n$), ($B_m$), and ($C_t$) are generalized Fibonacci or Lucas numbers. Especially, we also find all solutions of symmetric Diophantine equations $U_{a_1}{U}_{a_2}\cdots U_{a_m}=U_{b_1}{U}_{b_2}\cdots U_{b_n}$, where $1<a_1\le a_2\le \cdots \le a_m$, and $1<b_1\le b_2\le \cdots \le b_n$.

1. Introduction

Let P, Q be nonzero coprime integers with $D=P^2-4Q\ne 0$. The sequences of generalized Fibonacci numbers and Lucas numbers, $U_n$ and $V_n$ satisfy the following recurrence relation:

$$U_0=0,U_1=1,U_n=P{U}_{n-1}-Q{U}_{n-2}(n\ge 2)$$

(1)

$$V_0=2,V_1=P,V_n=P{V}_{n-1}-Q{V}_{n-2}(n\ge 2)$$

(2)

Their close forms are

$$U_n=\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta },V_n={\alpha }^n+{\beta }^n.$$

(3)

where

$$\alpha =\frac{P+\sqrt{P^2-4Q}}{2}\mathrm{and}\beta =\frac{P-\sqrt{P^2-4Q}}{2}$$

are roots of $x^2-Px+Q=0$.

For $m,n\in \mathbb{N}$. It is well known that these numbers have the following properties

• (a) $gcd(U_m,U_n)=U_d$;
• (b) If $m\mid n$, then $U_m\mid U_n$;
• (c) If $U_m\mid U_n$ and $m>2$, then $m\mid n$.

The generalized Fibonacci and Lucas numbers include many famous integer sequences such as Fibonacci numbers, Lucas numbers, Pell numbers, and Jacobsthal numbers. Their fascinated properties lead to abundant applications in totally surprising and unrelated fields (cf. [1,2,3,4,5,6]).

Consider equations:

$$U_n=U_m{V}_k,$$

(4)

$$U_n=U_m{U}_k,$$

(5)

$$U_n=V_m{V}_k,$$

(6)

$$V_n=U_m{V}_k,$$

(7)

$$V_n=V_m{V}_k,$$

(8)

$$V_n=U_m{U}_k,$$

(9)

where $n>m\ge k\ge 0$, $P>1$, and $Q<-1$.

In 1964, L. Carlitz [7] solved the above equations for $P=1$ and $Q=-1$, i.e., Fibonacci numbers and Lucas numbers. After half a century, M. Farrokhi D. G. [8] showed equation $F_n=k{F}_m$ has at most one solution $(n,m)$ for $k>1$. He gives the complete nontrivial solutions of the equation

$$F_m=F_{n_1}{F}_{n_2}{F}_{n_3}\cdots F_{n_k}.$$

Moreover, he also gives the complete nontrivial solutions of the symmetric Diophantine equation

$$F_{m_1}{F}_{m_2}\cdots F_{m_k}=F_{n_1}{F}_{n_2}{F}_{n_3}\cdots F_{n_s}.$$

In 2011, as a byproduct of Lucas square classes, R. Keskin and B. Demirturk [9] rediscovered that $L_n=L_m{L}_r$ is impossible if m, $r>1$. Two years later, R. Keskin and Z. Siar [10] proved that when $P>1$ and $Q=\pm 1$, there is no generalized Lucas number $V_n$ such that $V_n=V_m{V}_r$ for m, $r>1$ as a byproduct of Lucas square classes. Lastly, they show that there is no generalized Fibonacci number $U_n$ such that $U_n=U_m{U}_r$ for $Q=\pm 1$ and $1<r<m$. With the help of Carmichael’s primitive divisor theorem, P. Pongsriiam [11] solved equations:

$$F_m^a=F_{n_1}{F}_{n_2}{F}_{n_3}\cdots F_{n_k},$$

$$F_m^a=L_{n_1}{L}_{n_2}{L}_{n_3}\cdots L_{n_k},$$

$$L_m^a=F_{n_1}{F}_{n_2}{F}_{n_3}\cdots F_{n_k},$$

$$L_m^a=L_{n_1}{L}_{n_2}{L}_{n_3}\cdots L_{n_k},$$

where $a\ge 1,m\ge 0,k\ge 1$, and $0\le n_1\le n_2\le \cdots \le n_k$.

In 2017, P. Pongsriiam [12] considered the following Diophantine equations:

$$F_m=F_{n_1}{F}_{n_2}{F}_{n_3}\cdots F_{n_k}\pm 1,$$

$$F_m=L_{n_1}{L}_{n_2}{L}_{n_3}\cdots L_{n_k}\pm 1,$$

$$L_m=F_{n_1}{F}_{n_2}{F}_{n_3}\cdots F_{n_k}\pm 1,$$

$$L_m=L_{n_1}{L}_{n_2}{L}_{n_3}\cdots L_{n_k}\pm 1,$$

where $m\ge 0,k\ge 1$, and $0\le n_1\le n_2\le \cdots \le n_k$. There are various other types of equations involving generalized Fibonacci and Lucas numbers that many authors have also considered (cf. [13,14]).

Assume $P>1$ and $Q<-1$. In this paper, we find Equation (4) holds if and only if $n=2,m=k=1$ or $n=2m=2k$. Equation (6) holds if and only if $n=4,m=2,k=1$. Moreover, Equations (5) and (7)–(9) have no solution. Generally, we completely solved the Diophantine equations of the symmetric form

$$U_a=U_{b_1}{U}_{b_2}{U}_{b_3}\cdots U_{b_n},$$

(10)

$$U_a=V_{b_1}{V}_{b_2}{V}_{b_3}\cdots V_{b_n},$$

(11)

With the help of (10) and (11), we also find all solutions of symmetric Diophantine equations

$$U_{a_1}{U}_{a_2}\cdots U_{a_m}=U_{b_1}{U}_{b_2}\cdots U_{b_n},$$

(12)

where $a>1$, $1<a_1\le a_2\le \cdots \le a_m$, and $1<b_1\le b_2\le \cdots \le b_n$.

2. Preliminaries

In this section, we give some equalities and inequalities concerning generalized Fibonacci and Lucas numbers.

Lemma 1.

Let $n>2$, $P\ge 1$, and $Q<-1$. Then

$$(P-Q)U_{n-2}<U_n<(P-Q)U_{n-1},$$

$$(P-Q)V_{n-2}<V_n<(P-Q)V_{n-1}.$$

Proof. 

Using the recursive formula,

$$(P-Q)U_{n-2}<U_n=P{U}_{n-1}-Q{U}_{n-2}<(P-Q)U_{n-1}.$$

It is easy to check that the above inequality holds. Proceed as in the proof of $U_n$. We have a similar result of $V_n$. □

Lemma 2.

Let $P\ge 1,Q<-1$. For $n\ge 2k>0$.

$$U_n>{(-Q)}^k{U}_{n-2k},$$

$$V_n>{(-Q)}^k{V}_{n-2k}.$$

Proof. 

Proceed by induction on k. It is easy to check that the above inequality holds when $k=1$. Now, assume the inequality holds for $k=m$. By the induction method.

$$U_n>{(-Q)}^m{U}_{n-2m}.$$

Therefore, it follows that

$$U_n=P{U}_{n-1}-Q{U}_{n-2}>-Q{U}_{n-2}>{(-Q)}^{m+1}{U}_{n-2(m+1)}.$$

Proceed as in the proof of $U_n$. We have a similar result of $V_n$. □

Lemma 3.

Let n and k be positive integers. The following identities hold

• (a) $U_m{V}_k=U_{m+k}+Q^k{U}_{m-k}(m\ge k)$;
• (b) $U_m{V}_k=U_{m+k}-Q^m{U}_{k-m}(k\ge m)$;
• (c) $V_m{V}_k=V_{m+k}+Q^k{V}_{m-k}(m\ge k\ge 1)$;
• (d) $D{U}_m{U}_k=V_{m+k}+Q^k{V}_{m-k}(m\ge k\ge 1)$.

Proof.

• (a) Using (3),

  $$U_m{V}_k=\frac{{\alpha }^m-{\beta }^m}{\alpha -\beta }\times ({\alpha }^k+{\beta }^k)=\frac{{\alpha }^{m+k}-{\beta }^{m+k}}{\alpha -\beta }+{\left(\alpha \beta \right)}^k\frac{{\alpha }^{m-k}-{\beta }^{m-k}}{\alpha -\beta }.$$
• (b) It follows by the same method as in (a).
• (c) Using (3),

  $$V_m{V}_k=\left({\alpha }^m+{\beta }^m\right)\left({\alpha }^k+{\beta }^k\right)=\left({\alpha }^{m+k}+{\beta }^{m+k}\right)+{\left(\alpha \beta \right)}^k\left({\alpha }^{m-k}+{\beta }^{m-k}\right).$$
• (d) Using (3),

  $$D{U}_m{U}_k=\left({\alpha }^{m+k}+{\beta }^{m+k}\right)-{\left(\alpha \beta \right)}^k\left({\alpha }^{m-k}+{\beta }^{m-k}\right).$$

□

Corollary 1.

$V_k=U_{k+1}-Q{U}_{k-1}$.

Proof. 

It is easy to check Corollary 1 holds if one takes $m=1$ in formula (b) of Lemma 3. □

Lemma 4.

Let $n\ge k+1>0$ and $Q<-1$. Then

$$U_n=U_{k+1}{U}_{n-k}-Q{U}_k{U}_{n-k-1}$$

and

$$V_n=U_{k+1}{V}_{n-k}-Q{U}_k{V}_{n-k-1}.$$

Proof. 

Proceed by induction on k. It is easy to check that the above identities hold when $k=1$. Now, assume the equation holds for integer k. By the induction method.

$$\begin{array}{cc}\hfill U_n& =U_{k+1}{U}_{n-k}-Q{U}_k{U}_{n-k-1}\hfill \\ \hfill & =U_{k+1}(P{U}_{n-k-1}-Q{U}_{n-k-2})-Q{U}_k{U}_{n-k-1}\hfill \\ \hfill & =U_{k+2}{U}_{n-(k+1)}-Q{U}_{k+1}{U}_{n-(k+1)-1}.\hfill \end{array}$$

Proceed as in the proof of $U_n$. We have a similar result of $V_n$. □

Corollary 2.

For all $a,b,c,a_1,a_2,\dots ,a_n\in \mathbb{N}$. The following identities hold:

• (a)

  $$U_{a+b-1}=U_a{U}_b-Q{U}_{a-1}{U}_{b-1};$$

  (13)
• (b)

  $$U_{a+b-2}=\frac{1}{P}[U_a{U}_b-Q^2{U}_{a-2}{U}_{b-2}];$$

  (14)
• (c)

  $$U_{a+b+c-3}=\frac{1}{P}[U_a{U}_b{U}_c-PQ{U}_{a-1}{U}_{b-1}{U}_{c-1}+Q^3{U}_{a-2}{U}_{b-2}{U}_{c-2}];$$

  (15)
• (d) If $n\ge 3$, $P>1$, and $Q<-1$, then

  $$U_{a_1+\cdots +a_n-n}\ge \frac{1}{P}{U}_{a_1}{U}_{a_2}\cdots U_{a_n}.$$

  (16)

Proof.

• (a) Formula (13) follows easily from Lemma 4, if one takes $k+1=a$ and $n-k=b$.
• (b) Apply (13) to (14),

  $$\begin{array}{cc}\hfill U_{a+b-2}& =U_{a-1}{U}_b-Q{U}_{a-2}{U}_{b-1}\hfill \\ \hfill & =\frac{1}{P}[U_a{U}_b-Q{U}_{a-2}(-U_b+P{U}_{b-1})]\hfill \\ \hfill & =\frac{1}{P}[U_a{U}_b-Q^2{U}_{a-2}{U}_{b-2}].\hfill \end{array}$$
• (c) Combining Lemma 4 with (13) and (14),

  $$\begin{array}{cc}\hfill U_{a+b+c-3}& =U_{a+b-1}{U}_{c-1}-Q{U}_{a+b-2}{U}_{c-2}\hfill \\ \hfill & =\frac{1}{P}[U_a{U}_b{U}_c-PQ{U}_{a-1}{U}_{b-1}{U}_{c-1}+Q^3{U}_{a-2}{U}_{b-2}{U}_{c-2}].\hfill \end{array}$$
• (d) From (15), (16) holds. Then

  $$\begin{array}{cc}\hfill U_{a_1+a_2+\cdots +a_{n+1}-(n+1)}& =U_{\left[a_1+a_2+\cdots +a_n-n\right]+a_{n+1}-1}\hfill \\ \hfill & =U_{a_1+a_2+\cdots a_n-n}{U}_{a_{n+1}}-Q{U}_{a_1+a_2+\cdots +a_n-(n+1)}{U}_{a_{n+1}-1}\hfill \\ \hfill & \ge U_{a_1+a_2+\cdots +a_n-n}{U}_{a_{n+1}}\hfill \\ \hfill & \ge \frac{1}{P}{U}_{a_1}{U}_{a_2}\cdots U_{a_n}{U}_{a_{n+1}}.\hfill \end{array}$$

□

Lemma 5.

Assume $P>2$, $Q<-1$, and $P^2>-Q$. For all a, b, $c\in \mathbb{N}$. The following conditions hold

• (a)

  $$U_{a+b-2}<U_a{U}_b(a+b\ge 2);$$

  (17)
• (b)

  $$U_{a+b+c-3}<U_a{U}_b{U}_c(a+b+c\ge 3).$$

  (18)

Proof.

• (a) It is easy to check that (17) holds for a,$b<2$. Now, assume a, $b\ge 2$. By Lemma 2.

  $$U_a{U}_b>-Q^2{U}_{a-2}{U}_{b-2}.$$
  Combine with (14),

  $$P{U}_{a+b+c-2}<2U_a{U}_b$$

  implies $U_{a+b-2}<U_a{U}_b$.
• (b) Since

  $$\begin{array}{cc}\hfill U_a{U}_b{U}_c& =(P{U}_{a-1}-Q{U}_{a-2})(P{U}_{b-1}-Q{U}_{b-2})(P{U}_{b-1}-Q{U}_{b-2})\hfill \\ \hfill & >P^3{U}_{a-1}{U}_{b-1}{U}_{c-1}+Q^3{U}_{a-2}{U}_{b-2}\hfill \\ \hfill & >-PQ{U}_{a-1}{U}_{b-1}{U}_{c-1}+Q^3{U}_{a-2}{U}_{b-2}.\hfill \end{array}$$
  So

  $$P{U}_{a+b+c-3}<2U_a{U}_b{U}_c.$$
  Since $P>2$. It is easy to show that (18) holds.

□

Theorem 1

(Primitive divisor theorem of Carmichael [15]). If α and β are real and $n\ne 1,2,6$, then $U_n$ has a primitive divisor except when

$$n=12,\alpha +\beta =\pm 1,\alpha \beta =-1.$$

3. Main Theorems

Firstly, we begin this section by solving Equations (4)–(9) for $P>1$ and $Q<-1$. Then, we solve (10)–(12) for $P>2$, $Q<-1$, and $P^2>-Q$.

Theorem 2.

Let $n>m\ge k$. Equation (4) holds if and only if $n=2$, $m=k=1$ and $n=2m=2k$.

Proof. 

Equation (4) holds when $n=2$ and $m=k=1$. Thus, $m>2$ and $k>1$.

If $m=k$. Then $U_n=U_{2k}$ and $n=2m=2k$.

If $m-k=1$ and k is even. Then $n\ge 2k+2$. By Lemma 2.

$$\begin{array}{cc}\hfill U_{2k+1}+Q^k& =U_n=U_{n-1}-Q{U}_{n-2}\ge U_{2k+1}-Q{U}_{2k}\hfill \\ \hfill & >U_{2k+1}-QP{(-Q)}^{k-1}>U_{2k+1}+Q^k.\hfill \end{array}$$

If $m-k=1$ and k is odd. Then $2k+1\ge n+1$. By Lemma 2.

$$\begin{array}{c}\hfill U_n-Q^k=U_{2k+1}=U_{2k}-Q{U}_{2k-1}\ge U_n-Q{U}_{2k-1}>U_n-Q^k.\end{array}$$

If $k-m=1$ and m is odd. Then $n\ge 2m+2$. By Lemma 2.

$$\begin{array}{cc}\hfill U_{2m+1}-Q^m& =U_n=U_{n-1}-Q{U}_{n-2}\ge U_{2m+1}-Q{U}_{2m}\hfill \\ \hfill & >U_{2m+1}-Q{(-Q)}^{m-1}=U_{2m+1}-Q^m.\hfill \end{array}$$

If $k-m=1$ and m is even. Then $2m+1\ge n+1$. By Lemma 2.

$$\begin{array}{c}\hfill U_n+Q^m=U_{2m+1}=U_{2m}-Q{U}_{2m-1}\ge U_n-Q{U}_{2m-1}>U_n+Q^m.\end{array}$$

If $m-k>1$ and k is even. Then $n\ge m+k+1$. By Lemma 2.

$$\begin{array}{cc}\hfill U_{m+k}+Q^k{U}_{m-k}& =U_n=U_{n-1}-Q{U}_{n-2}\ge U_{m+k}-Q{U}_{m+k-1}\hfill \\ \hfill & >U_{m+k}+{(-Q)}^k{U}_{m-k+1}>U_{m+k}+Q^k{U}_{m-k}.\hfill \end{array}$$

If $m-k>1$ and k is odd. Then $m+k\ge n+1$. By Lemma 2.

$$\begin{array}{cc}\hfill U_n-Q^k{U}_{m-k}& =U_{m+k}=U_{m+k-1}-Q{U}_{m+k-2}\ge U_n-Q{U}_{m+k-1}\hfill \\ \hfill & >U_n+{(-Q)}^k{U}_{m-k+1}>U_n-Q^k{U}_{m-k}.\hfill \end{array}$$

If $k-m>1$ and m is odd. Then $n\ge m+k+1$. By Lemma 2.

$$\begin{array}{cc}\hfill U_{m+k}-Q^m{U}_{k-m}& =U_n=U_{n-1}-Q{U}_{n-2}\ge U_{m+k}-Q{U}_{m+k-1}\hfill \\ \hfill & >U_{m+k}+{(-Q)}^m{U}_{k-m+1}>U_{m+k}+{(-Q)}^m{U}_{k-m}.\hfill \end{array}$$

If $k-m>1$ and m is even. Then $m+k\ge n+1$. By Lemma 2.

$$\begin{array}{cc}\hfill U_n+Q^m{U}_{k-m}& =U_{m+k}=U_{m+k-1}-Q{U}_{m+k-2}\ge U_n-Q{U}_{m+k-2}\hfill \\ \hfill & >U_n+{(-Q)}^m{U}_{k-m}.\hfill \end{array}$$

□

Theorem 3.

Let $n>m\ge k$. Equation (5) possesses no solution.

Proof. 

If $U_k=U_3$. By Lemma 4.

$$U_3{U}_{n-2}<U_n=U_3{U}_{n-2}-Q{U}_{n-3}<U_3{U}_{n-1}.$$

Then $U_{n-2}<U_m<U_{n-1}$.

If $U_k=U_4$. By Lemma 4.

$$U_4{U}_{n-3}<U_n<U_4{U}_{n-2}.$$

Then $U_{n-3}<U_m<U_{n-2}$.

It follows from induction and Lemma 4 that

$$U_{n-1}{U}_2<U_n=U_3{U}_{n-2}-Q{U}_{n-3}<U_{n-1}{U}_3.$$

Then $U_2<U_m<U_3$. □

Theorem 4.

Let $n>m\ge k$. Equation (6) holds if and only if $n=4$, $m=2$, and $k=1$.

Proof. 

If $n=4$, $m=2$, and $k=1$. Equation (6) always holds. Suppose $m\ge k>1$.

By Lemma 3 (c) and Corollary 1. Equation (6) becomes

$$\begin{array}{cc}\hfill U_n& =U_{m+k+1}-Q{U}_{m+k-1}+Q^k{V}_{m-k}.\hfill \\ \hfill & =U_{m+k+1}-Q{U}_{m+k-1}+Q^k{U}_{m-k+1}-Q^{k+1}{U}_{m-k-1}.\hfill \end{array}$$

By Lemma 2

$$\begin{array}{cc}\hfill -Q{U}_{m+k-1}& =-PQ{U}_{m+k-2}+Q^2{U}_{m+k-1}\hfill \\ \hfill & >-Q^{k+1}{U}_{m-k-1}+Q^k{U}_{m-k+1}=Q^k{V}_{m-k}.\hfill \end{array}$$

Then $n\ge m+k+2$. By Lemma 2

$$P{Q}^2{U}_{m+k-3}>Q^k{U}_{m-k+1}$$

$$-Q^3{U}_{m+k-4}>-Q^{k+1}{U}_{m-k}>-Q^{k+1}{U}_{m-k-1}$$

$$\begin{array}{cc}\hfill U_{m+k+2}& =P{U}_{m+k+1}-PQ{U}_{m+k-1}+P{Q}^2{U}_{m+k-3}-Q^3{U}_{m+k-4}\hfill \\ \hfill & >U_{m+k+1}-Q{U}_{m+k-1}+Q^k{U}_{m-k+1}=Q^k{V}_{m-k}\hfill \end{array}$$

Then $n<m+k+2$. It is a contradiction. □

Theorem 5.

Let $n>m\ge k$. Equation (7) possesses no solution.

Proof. 

If $U_k=U_3$. By Lemma 4.

$$U_3{V}_{n-2}<V_n=U_3{V}_{n-2}-Q{V}_{n-3}<U_3{V}_{n-1}.$$

Then $V_{n-2}<V_k<V_{n-1}$.

If $U_k=U_4$. By Lemma 4.

$$U_4{U}_{n-3}<U_n<U_4{U}_{n-2}.$$

Then $V_{n-3}<V_k<V_{n-2}$. Thus, we obtain the contradiction.

It follows from induction and Lemma 4 that

$$U_{n-1}{V}_2<V_n<U_{n-1}{V}_3.$$

Then $V_2<V_k<V_3$. Thus, no integer k makes Equation (7) hold. □

Theorem 6.

Let $n>m\ge k$. Equation (8) possesses no solution.

Proof. 

Consider $V_n=V_{m+k}+Q^k{V}_{m-k}$ by Lemma 3 (c). If k is even. Then

$$n\ge m+k+1.$$

$$V_{m+k+1}\le V_{m+k}+Q^k{V}_{m-k}<P{V}_{m+k}+Q^k{V}_{m-k}$$

which is equal to

$$V_{m+k-1}<-Q^{k-1}{V}_{m-k}.$$

However,

$$V_{m+k-1}>{(-Q)}^{k-1}{V}_{m-k+1}>{(-Q)}^{k-1}{V}_{m-k},$$

a contradiction.

If k is odd. Then $n\le m+k-1$. By Lemma 2 and Lemma 3 (c).

$$V_n-Q^k{V}_{m-k}=V_{m+k}=P{V}_{m+k-1}-Q{V}_{m+k-2}>V_n-Q{V}_{m+k-2}>V_n-Q^k{V}_{m-k}.$$

□

Theorem 7.

Let $n>m\ge k$. Equation (9) possesses no solution.

Proof. 

By Lemma 3 (d). We can transform Equation (9) into

$$D{V}_n=V_{m+k}-Q^k{V}_{m-k}.$$

(19)

By Lemma 2. We have $D{V}_n<2V_{m+k}$. This implies that $n<m+k$.

Since

$$V_{m+k}=P{V}_{m+k-1}-Q{V}_{m+k-2}=(P^3-2PQ)V_{m+k-3}+Q^2-P^{2}Q{V}_{m+k-4}.$$

(20)

We plug (20) back into (19).

$$(P^2-4Q)V_n=(P^3-2PQ)V_{m+k-3}+(Q^2-P^{2}Q)V_{m+k-4}-Q^k{V}_{m-k}$$

By Lemma 2.

$$(Q^2-P^{2}Q)V_{m+k-4}>{(-Q)}^2{V}_{m+k-4}>-Q^k{V}_{m-k}.$$

This implies

$$(P^2-4Q)V_n>(P^3-2PQ)V_{m+k-3}.$$

We see that $n>m+k-3$. The proof falls into two conditions.

If $n=m+k-1$. Then

$$(P^2-4Q)V_{m+k-1}=V_{m+k}-Q^k{V}_{m-k}.$$

It can be deduced that

$$-Q{V}_{m+k-1}<{(-Q)}^k{V}_{m-k}.$$

However, we have

$$-Q{V}_{m+k-1}>{(-Q)}^k{V}_{m+k+1}$$

by Lemma 2.

If $n=m+k-2$. Then

$$(P^2-4Q)V_{m+k-1}=V_{ms+k}-Q^k{V}_{m-k},$$

which is equal to

$$-3Q{V}_{m+k-2}=-PQ{V}_{m+k-3}-Q^k{V}_{m-k}.$$

(21)

The left-hand side of (21) is equal to

$$-3Q{V}_{m+k-2}=-3PQ{V}_{m+k-3}+3Q^2{V}_{m+k-4}.$$

Thus, we obtain

$$-3PQ{V}_{m+k-3}+3Q^2{V}_{m+k-4}=-PQ{V}_{m+k-3}-Q^k{V}_{m-k}.$$

It is a contradiction by Lemma 2. Since

$$-3PQ{V}_{m+k-3}>-PQ{V}_{m+k-3}$$

$$3Q^2{V}_{m+k-4}>Q^2{V}_{m+k-4}>-Q^k{V}_{m-k}.$$

□

Theorem 8.

Let $n>m\ge k$. Equation (5) possesses no solution.

Proof. 

If triple $(a,b,c)$ is a solution of the equation. Then $b\mid c$ holds for $U_b\mid U_c$. Clearly, suppose $c=kb$ for $k\ge 2$.

$$\begin{array}{cc}\hfill U_a{U}_b& =U_{kb}\ge U_{2b}=U_b{U}_{b+1}-Q{U}_b{U}_{b-1}=P{U}_b^2-2Q{U}_b{U}_{b-1}>P{U}_b^2\ge U_a{U}_b,\hfill \end{array}$$

which is impossible. □

Theorem 9.

Let $a,b,c,d$ be natural numbers. Equation $U_a{U}_b=U_c{U}_d$ holds if and only if $U_a=U_c$ and $U_b=U_d$, $U_a=U_d$, and $U_b=U_c$.

Proof. 

If $a<b,c,d\le 3$. Then $U_b=U_c{U}_d$. Applying Theorem 8. We know that either $U_a=U_c=1$ and $U_b=U_d$ or $U_a=U_d=1$ and $U_b=U_c$. Next, we assume that $3\le a\le b,c,d$. Apply (14),

$$\begin{array}{cc}\hfill U_{a+b-2}& =\frac{1}{P}[U_a{U}_b+Q^2{U}_{a-2}{U}_{b-2}]\hfill \\ \hfill & <\frac{1}{P}{U}_a{U}_b\le U_a{U}_b=U_c{U}_d\hfill \\ \hfill & =U_{c+d-1}+Q{U}_{c-2}{U}_{d-2}\hfill \\ \hfill & <U_{c+d-1}.\hfill \end{array}$$

Thus,

$$U_{a+b-2}<U_a{U}_b<U_{c+d-1}.$$

It is clear that $a+b\le c+d$. The proof of $c+d\le a+b$ follows in a similar manner. Thus, we obtain $a+b=c+d$. Repeatedly using (13),

$$\begin{array}{cc}\hfill U_a{U}_b& =U_c{U}_d\hfill \\ \hfill \Rightarrow U_{a-1}{U}_{b-1}& =U_{c-1}{U}_{d-1}\hfill \\ \hfill & ⋮\hfill \\ \hfill \Rightarrow U_2{U}_{b-a+2}& =U_{c-a+2}{U}_{d-a+2}\hfill \\ \hfill \Rightarrow U_{b-a+2}& =U_{c-a+2}{U}_{c-a+2}.\hfill \end{array}$$

By Theorem 8. We have $U_{c-a+2}=1$ or $U_{c-a+2}=1$, which implies that either $a=c$ and $b=d$ or $a=d$ and $b=c$. □

Theorem 10.

Let $a,b,c,d$, and e be natural numbers. Equation $U_a{U}_b{U}_c=U_d{U}_e$ has no solution with $3\le a,b,c,d,e$.

Proof. 

If $a,b,c,d,e\in \mathbb{N}$. We assume that $(a,b,c;d,e)$ is a solution of the equation $U_a{U}_b{U}_c=U_d{U}_e$. Suppose $3\le a,b,c,d,e$. By (15),

$$\begin{array}{cc}\hfill U_{a+b+c-2}& =U_a{U}_{b+c-1}-Q{U}_{a-1}{U}_{b+c-2}\hfill \\ \hfill & =U_a{U}_b{U}_c-Q{U}_a{U}_{b-1}{U}_{c-1}-Q{U}_{a-1}{U}_{b+c-2}\hfill \\ \hfill & >U_a{U}_b{U}_c.\hfill \end{array}$$

By (17),

$$U_{d+e-2}<U_d{U}_e=U_a{U}_b{U}_c<U_{a+b+c-2}.$$

By (18),

$$U_{a+b+c-3}<U_a{U}_b{U}_c=U_d{U}_e<U_{d+e-1},$$

which implies that $a+b+c-3=d+e-2$.

$$\begin{array}{cc}\hfill U_{d+e-6}& <U_{d-2}{U}_{e-2}=U_{d+e-5}+Q{U}_{d-3}{U}_{e-2}\hfill \\ \hfill & =U_{a+b+c-6}+Q{U}_{d-3}{U}_{e-2}\hfill \\ \hfill & <U_{a+b+c-6}=\frac{1}{P}{U}_{a+b+c-5}+\frac{Q}{P}{U}_{a+b+c-7}\hfill \\ \hfill & <\frac{1}{P}{U}_{a+b+c-5}\hfill \\ \hfill & <U_{a+b+c-8}.\hfill \end{array}$$

Then, we have $d+e-6<a+b+c-8$, which is impossible. □

Let $P\ge 1$, $Q<0$, and $gcd(P,Q)=1$. Next, we solve Equations (10)–(12) using the primitive divisor theorem of Carmichael.

Theorem 11.

The only nontrivial solutions of Equation (10) with $a>1$, $b_i>1$. and $n\ge 2$ are

$$(2;2\cdots 2),(6;3^3,2\cdots 2),(6;3,2\cdots 2),(12;6,4^2,3,2\cdots 2),(12;4^2,3^4,2\cdots 2)$$

Here, nontrivial solution means that $n\ge 2$, $b_i\le 1$ for all $i=1,\cdots ,n$, and $a>1$.

Proof. 

If $a=12$, $P=1$, and $Q=-1$. Then, we obtain

$$U_{12}=144,U_6=8,U_4=3,U_3=2.$$

Note that

$$144=2^4{3}^2.$$

Then

$$U_{12}=U_6{U}_4^2{U}_3{U}_2^k.$$

$$U_{12}=U_4^2{U}_3^4{U}_2^k.$$

If $m\ge 7$ or $m=3,4,5$. By Theorem 1, there exists an odd primitive prime divisor p of $U_a$. We see p does not divide any generalized Fibonacci numbers $U_k$ with index less than a. Next, consider $m=2,6$. If $m=2$. Then

$$U_2=U_2{U}_2\cdots U_2.$$

holds if and only if $U_2=P=1.$

If $a=6$. Assume Equation (10) has a solution $(n_1,n_2,\cdots ,n_s)$ for $6>n_1\ge n_2\cdots \ge n_s$, $s\ge 2$, and $n_s>1$. Obviously, $n_i\ne 5$. If $n_1=4$. Since $gcd(U_6,U_4)=U_2$. It follows that $U_4=U_2$.

Then

$$U_4-U_2=P(P^2-2Q-1)=0.$$

Because $P\ge 1$ and $Q<0$, Equation (10) has no solution. If $n_1=n_2=3$. Since $U_2=P$.

$$U_3=P^2-Q|P^2-3Q.$$

Then

$$U_3=P^2-Q|-2Q.$$

It follows that $P^2=-Q$, $P^2-3Q=2U_3$ and

$$U_6=2U_3^2{U}_2.$$

If $U_3=P^2-Q=-2Q=2$. We obtain $P=1$$Q=-1$ and

$$U_6=U_3^3{U}_2^k.$$

If $U_2=P=2$ and $Q=-4$. It is contradict to $gcd(P,Q)=1$.

If $n_1=3$ and $n_2=2$. We have

$$U_6=U_3{U}_2^{l+1}.$$

and

$$U_2^l=P^l=P^2-3Q.(l>2)$$

However, it is contradict to $gcd(P,Q)=1$ if $P\ne 1$ and $P\ne 3$. If $P=1$.

$$P^2-3Q=1,$$

which has no solution with $P\ge 1$ and $Q<0$. If $P=3$.

$$81-36Q+3Q^2=3^{l+2}-3^{l}Q,$$

which has a solution with $P\ge 1$ and $Q<-6$.

If $n_1=2$. Then $U_2=P$ divides

$$\frac{U_6}{U_2}=P^4-4P^{2}Q+3Q^2.$$

Observe that $gcd(U_6,U_2)=U_2=P$, which is also contradict to P not divides

$$\frac{U_6}{U_2}=P^4-4P^{2}Q+3Q^2.$$

□

Theorem 12.

Equation (11) has a finite nontrivial solution. Here, nontrivial solution means that $n\ge 2$, $b_i\le 1$ for all $i=1,\cdots ,n$ and $a>1$.

Proof. 

Case 1 a is odd and $a\ge 6$. By Theorem 1. There exists an odd primitive divisor p of $U_a$. Therefore, p does not divide any generalized Lucas number with index less than a. Since $p|U_a$, we see that $U_a$ is not a product of generalized Lucas numbers.

Case 2 $a=2^{l}m$, $l\ge 1,m\ge 5$ and m is odd. By Theorem 1. There exists an odd primitive divisor p of $U_m$. Moreover, p does not divide any generalized Lucas number with index less than a. Since $p|U_m$ and $U_m|U_a$. We obtain $U_a$ is not a product of generalized Lucas numbers with index less than a.

Case 3 $a=3·2^l$

$$U_{3·2^l}=V_{3·2^{l-1}}{V}_{3·2^{l-2}}\cdots V_6{V}_3{U}_3$$

(22)

Since

$$U_3=P^2-Q>P=V_1$$

.

Equation (11) holds if and only if

$$U_3=P^2-Q=V_0=2.$$

Which contradict to $b_i\ge 2$.

Case 4 $a=2^l$

$$U_{2^l}=V_{2^{l-1}}{V}_{2^{l-2}}\cdots V_2{V}_1(l\ge 2).$$

(23)

We show that the representation of $U_{2^l}$ is unique for $l\ge 0$. It is easy to check that

$$U_2=V_1{U}_1.$$

Consider the equation

$$U_{2^l}=V_{a_1}{V}_{a_2}\cdots V_{a_k}=V_{2^{l-1}}{V}_{2^{l-2}}\cdots V_2{V}_1.$$

(24)

where $l\ge 2$, $a^1\ge a^2\cdots \ge a^k$. By the identity $U_{2m}=U_m{V}_m$. Transform (24) into

$$U_{2^{l-1}}{U}_{2a_1}{V}_{a_2}\cdots V_{a_k}=U_{a_1}{U}_{2^l}{V}_{2^{l-2}}\cdots V_2{V}_1.$$

(25)

If $2^l>2a_1$. By Theorem 1, there exists a prime p dividing $2^l$ but p does not divide any term on the left-hand side of Equation (25). It is a contradiction. Similarly, $2a_1>2^l$ leads to a contradiction. Therefore, $2a_1=2^l$. Equation (24) is reduced to

$$V_{a_2}\cdots V_{a^k}=V_{2^{l-2}}{V}_{2^{l-2}}\cdots V_2{V}_1.$$

Repeat the same process, we obtain

$$a_2=2^{l-2},a_3=2^{l-3},\cdots .$$

Equation (24) is reduced to

$$V_2{V}_1=V_{a_k}\cdots V_{a_i}.$$

It is obvious that $a_k=1,a_i=2$. □

By Theorem 11, Equation (12) has a nontrivial solution if and only if $P=1$ and $Q=-1$.

Theorem 13.

The only nontrivial solutions of Equation (12) with $1<a_1\le a_2\le \cdots \le a_m$, $1<b_1\le b_2\le \cdots \le b_n$ are

$$\begin{array}{ccc}\hfill (3,\dots ,3;6,\dots ,6)& ,& m=3n\hfill \\ \hfill (\stackrel{a}{\overbrace{3,\dots ,3}},\stackrel{b}{\overbrace{6,\dots ,6}},4,\dots ,4;12,\dots ,12)& ,& a+3b=4n\hfill \\ \hfill (\stackrel{a}{\overbrace{3,\dots ,3}},4,\dots ,4;\stackrel{b}{\overbrace{6,\dots ,6}},12,\dots ,12)& ,& a=3b+4n\hfill \\ \hfill (\stackrel{a}{\overbrace{6,\dots ,6}},4,\dots ,4;\stackrel{b}{\overbrace{3,\dots ,3}},12,\dots ,12)& ,& 3a=b+4n\hfill \\ \hfill (3,\dots ,3,4,\dots ,4;\stackrel{a}{\overbrace{12,\dots ,12}}\stackrel{b}{\overbrace{6,\dots ,6}})& ,& 3b+6a=m\hfill \end{array}$$

Here, nontrivial solution means that $a_i,b_j>1$ and $a_i\ne b_j$ for all $i=1,\cdots ,m$ and $j=1,\cdots ,n$.

4. Conclusions

In this paper, we mainly solve some Diophantine equations of the form $A_{n_1}\cdots A_{n_k}=B_{m_1}\cdots B_{m_r}{C}_{t_1}\cdots C_{t_s}$, where ($A_n$), ($B_m$), and ($C_t$) are generalized Fibonacci or Lucas numbers. Our theorems show that no generalized Fibonacci numbers can be expressed as the product of generalized Fibonacci or Lucas numbers except the trivial cases. In general, two different products of generalized Fibonacci numbers are not equal except the trivial cases.