# On some Diophantine equations

- Bahar Demirtürk Bitim
^{1}Email author and - Refik Keskin
^{1}

**2013**:162

https://doi.org/10.1186/1029-242X-2013-162

© Demirtürk Bitim and Keskin; licensee Springer. 2013

**Received: **14 December 2012

**Accepted: **22 March 2013

**Published: **9 April 2013

## Abstract

We consider the sequences $({u}_{n})$ and $({v}_{n})$ which are the generalizations of Fibonacci and Lucas sequences, respectively. Then we determine some identities involving these generalized sequences to present all solutions of the equations

and

for $p\ge 3$ and a square-free integer ${p}^{2}-4$. In addition to these, all solutions of some different Diophantine equations such as ${x}^{2}-{v}_{2n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$, ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4)$, ${x}^{2}-{v}_{n}xy+{y}^{2}=1$, ${x}^{2}-{v}_{2n}xy+{y}^{2}={u}_{n}^{2}$, ${x}^{2}-{v}_{2n}xy+{y}^{2}={v}_{n}^{2}$, ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=1$ are identified, by using divisibility rules of the sequences $({u}_{n})$ and $({v}_{n})$.

**MSC:**11B37, 11B39, 11C20, 11D09, 11D45.

## Keywords

## 1 Introduction

In this paper, we consider the generalized Fibonacci sequence $({u}_{n})$ and the generalized Lucas sequence $({v}_{n})$. Let $p\ge 3$ be an integer. For any $n\ge 2$, $({u}_{n})$ is defined by the recurrence relation ${u}_{n}=p{u}_{n-1}-{u}_{n-2}$ with the initial conditions ${u}_{0}=0$, ${u}_{1}=1$. The generalized Lucas sequence $({v}_{n})$ is defined by the recurrence relation $\phantom{\rule{0.25em}{0ex}}{v}_{n}=p{v}_{n-1}-{v}_{n-2}$ for any $n\ge 2$ with the initial conditions ${v}_{0}=2$ and ${v}_{1}=p$. The terms ${u}_{n}$ and ${v}_{n}$ are called the *n* th generalized Fibonacci and Lucas numbers, respectively.

Moreover, generalized Fibonacci and Lucas numbers can be extended to negative indices. In general, for all $n\in \mathbb{N}$, ${u}_{-n}=-{u}_{n}$ and ${v}_{-n}={v}_{n}$. Furthermore it is known that ${v}_{n}={u}_{n+1}-{u}_{n-1}$. For more detailed information about these sequences, one can consult [1–5] and [6].

In [3], McDaniel showed that the solutions of the equation ${x}^{2}-({p}^{2}-4){y}^{2}=4$ are given by $(x,y)=({v}_{n},{u}_{n})$ with $n\ge 1$. Moreover, in [7–9] and [10], Jones investigated whether the equations ${x}^{2}-({p}^{2}\mp 4){y}^{2}=\mp 4$, ${x}^{2}-({p}^{2}\mp 1){y}^{2}=\mp 4$, ${x}^{2}-({p}^{2}\mp 1){y}^{2}=\mp 1$ and ${x}^{2}-({p}^{2}\mp 4){y}^{2}=\mp 1$ have solutions or not. In his proofs, he used Fermat’s method of infinite descent.

*a*,

*b*,

*c*are generalized Fibonacci and generalized Lucas numbers. These equations can be listed as follows:

## 2 Divisibility rules of sequences $({u}_{n})$ and $({v}_{n})$

In this section, we recall some divisibility rules related to generalized Fibonacci and Lucas sequences $({u}_{n})$ and $({v}_{n})$. Since these rules are proved in [12–16], we omit their proofs. Using these divisibility rules, in the last section, we will find all solutions of Diophantine equations mentioned above.

**Theorem 1** *Let* $m,n\in \mathbb{N}$. *Then* ${v}_{n}|{u}_{m}$ *if and only if* $n|m$ *and* $m/n$ *is an even integer*.

**Theorem 2** *Let* $m,n\in \mathbb{N}$. *Then* ${u}_{n}|{u}_{m}$ *if and only if* $n|m$.

**Theorem 3** *Let* $m,n\in \mathbb{N}$. *Then* ${v}_{n}|{v}_{m}$ *if and only if* $n|m$ *and* $m/n$ *is an odd integer*.

**Theorem 4** *Let* $m,n\in \mathbb{N}$ *and* $n>1$. *Then* ${u}_{n}|{v}_{m}$ *if and only if* $n=2$ *and* *m* *is an odd integer*.

## 3 Some identities of the sequences $({u}_{n})$ and $({v}_{n})$

In this section, we obtain some identities by using special matrices including generalized Fibonacci and Lucas numbers.

for all $m,n\in \mathbb{Z}$.

**Theorem 5**

*for all* $m,n,t\in \mathbb{Z}$.

*Proof*If we consider identities (3.6) and (3.7), then the matrix multiplication

□

**Theorem 6**

*Let*$m,n,t\in \mathbb{Z}$

*and*$t\ne n$.

*Then*

*Proof*By using (3.6), we can consider the matrix multiplication

□

we can give the following theorem.

**Theorem 7**

*Let*$m,n,t\in \mathbb{Z}$

*and*$t\ne n$.

*Then*

## 4 Solutions of some Diophantine equations

In [4], Melham proved that all solutions of the equations ${y}^{2}-{v}_{m}xy+{x}^{2}=\mp {u}_{m}^{2}$ are $(x,y)=\mp ({u}_{n},{u}_{n+m})$ with $n\in \mathbb{Z}$. Moreover, he showed that if $m\in \mathbb{Z}$ and ${p}^{2}-4$ is a square-free integer, then all solutions of the equation ${y}^{2}-{v}_{m}xy+{x}^{2}=-({p}^{2}-4){u}_{m}^{2}$ are given by $(x,y)=\mp ({v}_{n},{v}_{n+m})$ with $n\in \mathbb{Z}$. These theorems of Melham are generalized forms of the theorems given in [3], by McDaniel. In [2], Kılıç and Ömür examined more general situations of the conics that McDaniel and Melham dealt in [3] and [4], respectively.

will be found.

Now we will remind some Diophantine equations with their solutions. The solutions of these equations are explored in [3] and [6]. We will use these equations for determining all solutions of other Diophantine equations.

**Theorem 8** *Let* $p>3$. *All solutions of the equation* ${x}^{2}-pxy+{y}^{2}=1$ *are given by* $(x,y)=\mp ({u}_{m},{u}_{m-1})$ *with* $m\in \mathbb{Z}$.

Since Corollary 1 can be seen from Theorem 8 and Corollary 2 is stated in [10], we will give them without proof.

**Corollary 1** *All solutions of the equation* ${x}^{2}-3xy+{y}^{2}=1$ *are given by* $(x,y)=\mp ({F}_{2m+2},{F}_{2m})$ *with* $m\in \mathbb{Z}$.

**Corollary 2** *Let* $p>3$. *All nonnegative solutions of the equation* ${u}^{2}-({p}^{2}-4){v}^{2}=4$ *are given by* $(u,v)=({v}_{m},{u}_{m})$ *with* $m\ge 1$.

Theorem 9 and Theorem 10 are stated in [21], so will give them without proof.

**Theorem 9** *Let* $p>3$. *Then the equation* ${x}^{2}-pxy+{y}^{2}=-1$ *has no solutions*.

**Theorem 10** *All solutions of the equation* ${x}^{2}-3xy+{y}^{2}=-1$ *are given by* $(x,y)=\mp ({F}_{2m+1},{F}_{2m-1})$ *with* $m\in \mathbb{Z}$.

From now on we will assume that *n* is an integer such that $n\ge 1$.

**Theorem 11** *If* $p\ge 3$, *then all solutions of the equation* ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}={v}_{n}^{2}$ *are given by* $(x,y)=\mp ({v}_{n+m},{u}_{m})$ *with* $m\in \mathbb{Z}$.

*Proof*Assume that ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}={v}_{n}^{2}$ for some integers

*x*and

*y*. Hence, we can write

Conversely, if $(x,y)=\mp ({v}_{n+m},{u}_{m})$ with $m\in \mathbb{Z}$, then it can be seen that ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}={v}_{n}^{2}$, by (3.12). □

Using Theorem 9 in the same manner with Theorem 11, the following corollary can be given.

**Corollary 3** *If* $p>3$, *then the equation* ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=-{v}_{n}^{2}$ *has no solutions*.

*Proof*Assume that ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=-{v}_{n}^{2}$ for some integers

*x*and

*y*. Similar with the proof of Theorem 11, taking $u=(x+{v}_{n-1}y)/{v}_{n}$ and $v=y$, it can be seen that

which is impossible by Theorem 9. Thus, it follows that the equation ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=-{v}_{n}^{2}$ has no integer solutions. □

The following corollary is a result of Theorem 11. Since it is proved in [11], we will give it without proof.

**Corollary 4** *All solutions of the equation* ${x}^{2}-5{F}_{2n}xy-5{y}^{2}=-{L}_{2n}^{2}$ *are given by* $(x,y)=\mp ({L}_{2n+2m+1},{F}_{2m+1})$ *with* $m\in \mathbb{Z}$.

Theorem 12 and Theorem 13 are stated by Melham, Kılıç and Ömür without proof in [4] and [2], respectively. Now we will prove them.

**Theorem 12** *Let* $p\ge 3$ *and* ${p}^{2}-4$ *be a square*-*free integer*. *Then all solutions of the equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$ *are given by* $(x,y)=\mp ({v}_{n+m},{v}_{m})$ *with* $m\in \mathbb{Z}$.

*Proof*Assume that ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$ for some integers

*x*and

*y*. Then multiplying both sides of this equation by 4 and using (3.10), we get ${(2x-{v}_{n}y)}^{2}-({p}^{2}-4){u}_{n}^{2}{y}^{2}=-4({p}^{2}-4){u}_{n}^{2}$. Since ${p}^{2}-4$ is square-free, it follows that ${u}_{n}|2x-{v}_{n}y$. Therefore, there is an integer

*z*such that $2x-{v}_{n}y={u}_{n}z$. From here we get ${({u}_{n}z)}^{2}-({p}^{2}-4){u}_{n}^{2}{y}^{2}=-4({p}^{2}-4){u}_{n}^{2}$, and then ${z}^{2}-({p}^{2}-4){y}^{2}=-4({p}^{2}-4)$. This implies that $({p}^{2}-4)|z$. Then there is an integer

*a*such that $z=({p}^{2}-4)a$, and we have $2x-{v}_{n}y=({p}^{2}-4){u}_{n}a$. Thus, it follows that

*y*and

*pa*have the same parity. Taking $u=(y+pa)/2$ and $v=a$, we obtain

Using the identities (3.6), (3.8) and (3.9) in (4.1), we get $(x,y)=\mp ({v}_{n+m},{v}_{m})$.

Conversely, if $(x,y)=\mp ({v}_{n+m},{v}_{m})$, then it follows that ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$, by (3.13). □

Using Theorem 9 in the same manner with Theorem 12, we can give the following corollary.

**Corollary 5** *Let* $p>3$ *and* ${p}^{2}-4$ *be a square*-*free integer*. *Then the equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=({p}^{2}-4){u}_{n}^{2}$ *has no solutions*.

We can give the following corollary from Corollary 5.

**Corollary 6** *Let* $p>3$ *and* ${p}^{2}-4$ *be a square*-*free integer*. *Then the equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=({p}^{2}-4)$ *has no solutions*.

When $p=3$, the equation ${x}^{2}-{v}_{n}xy+{y}^{2}=({p}^{2}-4){u}_{n}^{2}$ has solutions. In this case we have the equation ${x}^{2}-{L}_{2n}xy+{y}^{2}=5{F}_{2n}^{2}$. Now we can give all solutions of these equations in the following lemma. Since this lemma is proved in [11], we will give it without proof.

**Lemma 1** *All solutions of the equation* ${x}^{2}-{L}_{2n}xy+{y}^{2}=5{F}_{2n}^{2}$ *are given by* $(x,y)=\mp ({L}_{2n+2m+1},{L}_{2m+1})$ *with* $m\in \mathbb{Z}$.

**Theorem 13** *All solutions of the equation* ${x}^{2}-{v}_{n}xy+{y}^{2}={u}_{n}^{2}$ *are given by* $(x,y)=\mp ({u}_{n+m},{u}_{m})$ *with* $m\in \mathbb{Z}$.

*Proof*Suppose that ${x}^{2}-{v}_{n}xy+{y}^{2}={u}_{n}^{2}$ for some integers

*x*and

*y*. Completing the square gives ${(2x-{v}_{n}y)}^{2}-({p}^{2}-4){u}_{n}^{2}{y}^{2}=4{u}_{n}^{2}$, and it is seen that ${u}_{n}|2x-{v}_{n}y$. Thus, it follows that

Taking $u=(((2x-{v}_{n}y)/{u}_{n})+py)/2=(x+{u}_{n-1}y)/{u}_{n}$ and $v=y$, we have ${u}^{2}-puv+{v}^{2}=1$. Therefore, from Theorem 8, we get $(u,v)=\mp ({u}_{m+1},{u}_{m})$ with $m\in \mathbb{Z}$. From here, we obtain $(x,y)=\mp ({u}_{n}{u}_{m+1}-{u}_{n-1}{u}_{m},{u}_{m})$. Then by (3.5), it follows that $(x,y)=\mp ({u}_{n+m},{u}_{m})$.

Conversely, if $(x,y)=\mp ({u}_{n+m},{u}_{m})$, then it can be seen that ${x}^{2}-{v}_{n}xy+{y}^{2}={u}_{n}^{2}$, by (3.14). □

Using Theorem 9 in the same manner with Theorem 13, we can give the following corollaries.

**Corollary 7** *The equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=-{u}_{n}^{2}$ *has no solutions*.

The following corollary is a generalized form of Theorem 10. Since it is proved in [11], we will give it without proof.

**Corollary 8** *All solutions of the equation* ${x}^{2}-{L}_{2n}xy+{y}^{2}=-{F}_{2n}^{2}$ *are given by* $(x,y)=\mp ({F}_{2n+2m+1},{F}_{2m+1})$ *with* $m\in \mathbb{Z}$.

Now, let us examine all solutions of the following equations by using Diophantine equations given in Theorem 11, Theorem 12, Theorem 13 and the divisibility rules of the sequences $({u}_{n})$ and $({v}_{n})$.

**Theorem 14** *All solutions of the equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=1$ *are given by* $(x,y)=\mp ({u}_{(t+1)n}/{u}_{n},{u}_{tn}/{u}_{n})$ *with* $t\in \mathbb{Z}$.

*Proof*Assume that ${x}^{2}-{v}_{n}xy+{y}^{2}=1$ for some integers

*x*and

*y*. Multiplying both sides of this equation by ${u}_{n}^{2}$, we get

*m*. Hence, we get $x=\mp {u}_{n+m}/{u}_{n}$ and $y=\mp {u}_{m}/{u}_{n}$. Since

*x*and

*y*are integers, it is clear that $n|m$. Therefore, it follows that $m=tn$ for some $t\in \mathbb{Z}$. Then we obtain

Conversely, if $(x,y)=\mp ({u}_{(t+1)n}/{u}_{n},{u}_{tn}/{u}_{n})$ with $t\in \mathbb{Z}$, then it follows that ${x}^{2}-{v}_{n}xy+{y}^{2}=1$, by (3.14). □

Multiplying both sides of the equation ${x}^{2}-{v}_{n}xy+{y}^{2}=-1$ by ${u}_{n}^{2}$ and using Corollary 7, the following corollary can be given.

**Corollary 9** *The equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=-1$ *has no solutions*.

**Theorem 15** *If* $p\ge 3$, *then all solutions of the equation* ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=1$ *are given by* $(x,y)=\mp ({v}_{(2t+1)n}/{v}_{n},{u}_{2tn}/{v}_{n})$ *with* $t\in \mathbb{Z}$.

*Proof*Assume that ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=1$ for some integers

*x*and

*y*. Multiplying both sides of this equation by ${v}_{n}^{2}$, we get

Thus, it follows that ${v}_{n}x=\mp {v}_{n+m}$ and ${u}_{n}y=\mp {u}_{m}$ according to Theorem 11. Hence, we get $(x,y)=\mp ({v}_{n+m}/{v}_{n},{u}_{m}/{v}_{n})$. From Theorem 1 and Theorem 3, it can be seen that $n|m$ and $m/n$ is an even integer. This implies that $m=2tn$ for some $t\in \mathbb{Z}$. Therefore, we obtain $(x,y)=\mp ({v}_{(2t+1)n}/{v}_{n},{u}_{2tn}/{v}_{n})$.

Conversely, if $(x,y)=\mp ({v}_{(2t+1)n}/{v}_{n},{u}_{2tn}/{v}_{n})$ for some $t\in \mathbb{Z}$, then it follows that ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=1$, by (3.12). □

The following corollary can be given from Corollary 3.

**Corollary 10** *If* $p\ge 3$, *then the equation* ${x}^{2}-({p}^{2}-4){u}_{n}xy-({p}^{2}-4){y}^{2}=-1$ *has no solutions*.

**Theorem 16** *If* $p\ge 3$ *and* ${p}^{2}-4$ *is a square*-*free integer*, *then all solutions of the equation* ${x}^{2}-{v}_{2n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$ *are given by* $(x,y)=\mp ({v}_{(2t+3)n}/{v}_{n},{v}_{(2t+1)n}/{v}_{n})$ *with* $t\in \mathbb{Z}$.

*Proof*Suppose that ${x}^{2}-{v}_{2n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$ for some integers

*x*and

*y*. Multiplying both sides of this equation by ${v}_{n}^{2}$ and considering the fact that ${u}_{2n}={u}_{n}{v}_{n}$, we get

From Theorem 12, it follows that ${v}_{n}x=\mp {v}_{2n+m}$ and ${v}_{n}y=\mp {v}_{m}$. Hence we get $(x,y)=\mp ({v}_{2n+m}/{v}_{n},{v}_{m}/{v}_{n})$. Moreover, since *x* and *y* are integers, it follows that $n|m$ and $m/n$ is an odd integer from Theorem 3. Then there is an integer *t* such that $m=(2t+1)n$. Therefore, we obtain $(x,y)=\mp ({v}_{(2t+3)n}/{v}_{n},{v}_{(2t+1)n}/{v}_{n})$.

Conversely, if $(x,y)=\mp ({v}_{(2t+3)n}/{v}_{n},{v}_{(2t+1)n}/{v}_{n})$ for some $t\in \mathbb{Z}$, then it follows that ${x}^{2}-{v}_{2n}xy+{y}^{2}=-({p}^{2}-4){u}_{n}^{2}$, by (3.13). □

The following corollary can be proved similar to Theorem 16, by using Corollary 5. So, we omit its proof.

**Corollary 11** *If* $p>3$ *and* ${p}^{2}-4$ *is a square*-*free integer*, *then the equation* ${x}^{2}-{v}_{2n}xy+{y}^{2}=({p}^{2}-4){u}_{n}^{2}$ *has no solutions*.

**Theorem 17** *All solutions of the equations* ${x}^{2}-{v}_{2n}xy+{y}^{2}={u}_{n}^{2}$ *and* ${x}^{2}-{v}_{2n}xy+{y}^{2}={v}_{n}^{2}$ *are given by* $(x,y)=\mp ({u}_{(2t+2)n}/{v}_{n},{u}_{2tn}/{v}_{n})$ *and* $(x,y)=\mp ({u}_{(t+2)n}/{u}_{n},{u}_{tn}/{u}_{n})$ *with* $t\in \mathbb{Z}$, *respectively*.

*Proof*Assume that ${x}^{2}-{v}_{2n}xy+{y}^{2}={u}_{n}^{2}$ for some integers

*x*and

*y*. Multiplying both sides of this equation by ${v}_{n}^{2}$, we get

Then from Theorem 13, it follows that $(x,y)=\mp ({u}_{2n+m}/{v}_{n},{u}_{m}/{v}_{n})$ for some $m\in \mathbb{Z}$. Hence, using Theorem 1 it is seen that $n|m$ and $m/n$ is an even integer. Thus, we have $m=2tn$ for some $t\in \mathbb{Z}$. Therefore, $(x,y)=\mp ({u}_{(2t+2)n}/{v}_{n},{u}_{2tn}/{v}_{n})$.

Conversely, if $(x,y)=\mp ({u}_{(2t+2)n}/{v}_{n},{u}_{2tn}/{v}_{n})$ for some $t\in \mathbb{Z}$, then by (3.14) it follows that ${x}^{2}-{v}_{2n}xy+{y}^{2}={u}_{n}^{2}$.

Now suppose that ${x}^{2}-{v}_{2n}xy+{y}^{2}={v}_{n}^{2}$ for some integers *x* and *y*. Multiplying both sides of this equation by ${u}_{n}^{2}$ and considering Theorem 13, we get $(x,y)=\mp ({u}_{2n+m}/{u}_{n},{u}_{m}/{u}_{n})$ for some $m\in \mathbb{Z}$. Furthermore, since *x* and*y* are integers, it follows that $n|m$ from Theorem 2. Then we have $m=tn$ for some $t\in \mathbb{Z}$. Thus, we obtain $(x,y)=\mp ({u}_{(t+2)n}/{u}_{n},{u}_{tn}/{u}_{n})$.

Conversely, if $(x,y)=\mp ({u}_{(t+2)n}/{u}_{n},{u}_{tn}/{u}_{n})$, then by (3.14) it follows that ${x}^{2}-{v}_{2n}xy+{y}^{2}={v}_{n}^{2}$. □

The proof of the following corollary is similar to that of Theorem 17, by using Corollary 7. So, we omit its proof.

**Corollary 12** *The equations* ${x}^{2}-{L}_{4n}xy+{y}^{2}=-{F}_{2n}^{2}$ *and* ${x}^{2}-{L}_{4n}xy+{y}^{2}=-{L}_{2n}^{2}$ *have no solutions*.

**Theorem 18** *Let* $p\ge 3$, ${p}^{2}-4$ *be a square*-*free integer*. *If* $n>2$, *then the equation* ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4)$ *has no solutions and all solutions of the equation* ${x}^{2}-({p}^{2}-2)xy+{y}^{2}=-({p}^{2}-4)$ *are given by* $(x,y)=\mp (\frac{{v}_{2t+3}}{p},\frac{{v}_{2t+1}}{p})$ *with* $t\in \mathbb{Z}$.

*Proof*Assume that ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4)$ for some integers

*x*and

*y*. Multiplying both sides of this equation by ${u}_{n}^{2}$, we get

From Theorem 12, it follows that $(x,y)=\mp ({v}_{n+m}/{u}_{n},{v}_{m}/{u}_{n})$ with $m\in \mathbb{Z}$. If $n>2$, then ${u}_{n}\nmid {v}_{m}$ from Theorem 4. Therefore, the equation ${x}^{2}-{v}_{n}xy+{y}^{2}=-({p}^{2}-4)$ has no solutions. If $n=2$, then we get that all solutions of the equation ${x}^{2}-({p}^{2}-2)xy+{y}^{2}=-({p}^{2}-4)$ are given by $(x,y)=\mp ({v}_{m+2}/{u}_{2},{v}_{m}/{u}_{2})$ with $m\in \mathbb{Z}$. Hence, it is seen that *m* is an odd integer according to Theorem 4. Thus, $m=2t+1$ for some $t\in \mathbb{Z}$. Therefore, it follows that $(x,y)=\mp (\frac{{v}_{2t+3}}{p},\frac{{v}_{2t+1}}{p})$. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Authors’ Affiliations

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