Generalized Fibonacci numbers of the form $wx^{2}+1$

Abstract

Let $P\ge 3$ be an integer and let $(U_{n})$ and $(V_{n})$ denote generalized Fibonacci and Lucas sequences defined by $U_{0}=0,U_{1}=1$; $V_{0}=2,V_{1}=P,$ and $U_{n+1}=PU_{n}-U_{n-1}$, $V_{n+1}=PV_{n}-V_{n-1}$ for $n\ge 1.$ In this study, when P is odd, we solve the equation $U_{n}=wx^{2}+1$ for $w=1,2,3,5,6,7,10.$ After then, we solve some Diophantine equations utilizing solutions of these equations.

1 Introduction

Let P and Q be nonzero integers. Generalized Fibonacci sequence $(U_{n})$ and Lucas sequence $(V_{n})$ are defined by $U_{0}(P,Q)=0,U_{1}(P,Q)=1$; $V_{0}(P,Q)=2,V_{1}(P,Q)=P,$ and $U_{n+1}(P,Q)=PU_{n}(P,Q)+QU_{n-1}(P,Q)$, $V_{n+1}(P,Q)=PV_{n}(P,Q)+QV_{n-1}(P,Q)$ for $n\ge 1.$ $U_{n}(P,Q)$ and $V_{n}(P,Q)$ are called n-th generalized Fibonacci number and n-th generalized Lucas number, respectively. Generalized Fibonacci and Lucas numbers for negative subscripts are defined as $U_{-n}(P,Q)=-(-Q)^{-n}U_{n}(P,Q)$ and $V_{-n}(P,Q)=(-Q)^{-n}V_{n}(P,Q),$ respectively.

Since

$$\begin{aligned} U_{n}(-P,Q)=(-1)^{n-1}U_{n}(P,Q)\text { and }V_{n}(-P,Q)=(-1)^{n}V_{n}(P,Q), \end{aligned}$$

it will be assumed that $P\ge 1.$ Moreover, we will assume that $P^{2}+4Q>0.$ For $P=Q=1,$ we have classical Fibonacci and Lucas sequences $(F_{n})$ and $(L_{n}).$ For $P=2$ and $Q=1,$ we have Pell and Pell-Lucas sequences $(P_{n})$ and $(Q_{n}).$ For more information about generalized Fibonacci and Lucas sequences one can consult [1–4].

Generalized Fibonacci and Lucas numbers of the form $kx^{2}$ have been investigated since 1962. In [5], the authors solved $U_{n}=x^{2},V_{n}=x^{2},U_{n}=2x^{2},$ and $V_{n}=2x^{2}$ for odd relatively prime integers P and Q. The reader can consult [6] or [7] for a brief discussion of the subject.

In [8], the authors showed that when $a\ne 0$ and b are integers, then the equation $U_{n}(P,\pm 1)=ax^{2}+b$ has only a finite number of solutions n. In [9], Keskin solved the equations $V_{n}(P,-1)=wx^{2}+1$ and $V_{n}(P,-1)=wx^{2}-1$ for $w=1,2,3,6$ when P is odd. In [10], Karaatlı and Keskin solved the equations $V_{n}(P,-1)=5x^{2}\pm 1$ and $V_{n}(P,-1)=7x^{2}\pm 1$. Similar equations are tackled in [11] by using very different methods (see also [12–14]). In this study, we solve the equation $U_{n}(P,-1)=wx^{2}+1$ for $w=1,2,3,5,6,7,10.$

We will use the Jacobi symbol throughout this study. Our method is elementary and used by Cohn, Ribenboim and McDaniel in [15] and [16] , respectively.

2 Preliminaries

From now on, instead of $U_{n}(P,-1)$ and $V_{n}(P,-1),$ we sometimes write $U_{n}$ and $V_{n},$ respectively. Moreover, we will assume that $P\ge 3$.

The following lemmas can be proved by induction.

Lemma 2.1

If n is a positive integer, then $U_{2n}\equiv n(-1)^{n+1}P$ $( \mathrm{mod}P^{2})$ and $U_{2n+1}\equiv (-1)^{n}$ $(\mathrm{mod}P^{2}).$

Lemma 2.2

If n is a positive integer, then $V_{2n}\equiv 2(-1)^{n}$ $( \mathrm{mod}P^{2})$ and $V_{2n+1}\equiv (-1)^{n}(2n+1)P$ $(\mathrm{mod}P^{2}).$

The following theorems are given in [9].

Theorem 2.3

Let P be odd. If $V_{n}=kx^{2}$ for some k|P with $k>1,$ then $n=1.$

Theorem 2.4

Let P be odd. If $V_{n}=2kx^{2}$ for some k|P with $k>1,$ then $n=3.$

Theorem 2.5

Let P be odd. If $U_{n}=kx^{2}$ for some k|P with $k>1,$ then $n=2$ or $n=6$ and 3|P.

Theorem 2.6

Let P be odd. If k|P with $k>1,$ then the equation $U_{n}=2kx^{2}$ has no solutions.

Theorem 2.7

Let P be odd. If k|P with $k>1,$ then the equation $U_{n}=kx^{2}+1$ has only the solution $n=1.$

The following theorem is given in [10].

Theorem 2.8

Let P be odd. If $V_{n}=7kx^{2}$ for some k|P with $k>1,$ then $n=1.$

Now we give some known theorems from [5], which will be useful for solving the equation $U_{n}=wx^{2}+1.$ We use a theorem from [17] while solving $V_{n}=2x^{2}.$

Theorem 2.9

Let P be odd. If $V_{n}=x^{2}$ for some integer x,  then $n=1.$ If $V_{n}=2x^{2}$ for some integer x,  then $n=3,P=3,27.$

Theorem 2.10

Let P be odd. If $U_{n}=x^{2}$ for some integer x,  then $n=1$ or $n=2,$ $P=\square$ or $n=6,P=3.$ If $U_{n}=2x^{2}$ for some integer x,  then $n=3.$

The following lemma is a consequence of a theorem given in [18].

Lemma 2.11

All positive integer solutions of the equation $3x^{2}-2y^{2}=1$ are given by $(x,y)=(U_{n}(10,-1)-U_{n-1}(10,-1),U_{n}(10,-1)+U_{n-1}(10,-1))$ with $n\ge 1.$

The proof of the following lemma is easy and will be omitted.

Lemma 2.12

All positive integer solutions of the equation $x^{2}-7y^{2}=2$ are given by $(x,y)=\left( 3\left( U_{m+1}(16-1)-U_{m}(16,-1)\right) ,17U_{m}(16,-1)-U_{m-1}(16,-1)\right)$ with $m\ge 0.$

The following theorems are well known (see [19–22]).

Lemma 2.13

All positive integer solutions of the equation $x^{2}-(P^{2}-4)y^{2}=4$ are given by $(x,y)=\left( V_{n}(P,-1),U_{n}(P,-1)\right)$ with $n\ge 1.$

Lemma 2.14

All positive integer solutions of the equation $x^{2}-Pxy+y^{2}=1$ are given by $(x,y)=\left( U_{n}(P,-1),U_{n-1}(P,-1)\right)$ with $n\ge 2.$

The following two theorems are given in [23].

Theorem 2.15

Let $n\in \mathbb {N\cup }\left\{ 0\right\} ,$ m,  $r\in \mathbb {Z}$ and m be a nonzero integer. Then

$$\begin{aligned} U_{2mn+r}\equiv U_{r}\,\left( \mathrm{mod}U_{m}\right) . \end{aligned}$$

(2.1)

Theorem 2.16

Let $n\in \mathbb {N\cup }\left\{ 0\right\}$ and m,  $r\in \mathbb {Z}.$ Then

$$\begin{aligned} U_{2mn+r}\equiv \left( -1\right) ^{n}U_{r}\,\left( \mathrm{mod} V_{m}\right) . \end{aligned}$$

(2.2)

If $n=2\cdot 2^{k}a+r$ with a odd, then we get

$$\begin{aligned} U_{n}=U_{2\cdot 2^{k}a+r}\equiv -U_{r}\,(\mathrm{mod}V_{2^{k}}). \end{aligned}$$

(2.3)

by (2.2).

When P is odd, since $8|U_{3},$ using (2.1) we get

$$\begin{aligned} U_{6q+r}\equiv U_{r}\,\left( \mathrm{mod}U_{3}\right) \end{aligned}$$

(2.4)

and therefore

$$\begin{aligned} U_{6q+r}\equiv U_{r}\,\left( \mathrm{mod}8\right) . \end{aligned}$$

(2.5)

From Lemma 2.1 and Lemma 2.2, it follows that if q|P with $q>1,$ then

$$\begin{aligned}&q|V_{n}\Leftrightarrow n\text { is odd and }q|U_{n}\Leftrightarrow n\text { is even.} \end{aligned}$$

(2.6)

$$\begin{aligned}&\quad \text {If }P^{2}\equiv -1\,(\mathrm{mod}5),\quad \text { then }5|U_{5}. \end{aligned}$$

(2.7)

$$\begin{aligned}&\quad \text {If }P\text { is odd, then }2|V_{n}\Leftrightarrow 2|U_{n}\Leftrightarrow 3|n. \end{aligned}$$

(2.8)

Now we give some identities concerning generalized Fibonacci and Lucas numbers:

$$\begin{aligned} U_{-n}= & {} -U_{n}\text { and }V_{-n}=V_{n},\nonumber \\ U_{2n+1}-1= & {} U_{n}V_{n+1}, \end{aligned}$$

(2.9)

$$\begin{aligned} U_{2n}= & {} U_{n}V_{n}, \end{aligned}$$

(2.10)

$$\begin{aligned} V_{n}^{2}-(P^{2}-4)U_{n}^{2}= & {} 4, \end{aligned}$$

(2.11)

$$\begin{aligned} V_{2n}= & {} V_{n}^{2}-2. \end{aligned}$$

(2.12)

Let $m=2^{a}k,$ $n=2^{b}l,$ k and l odd, $a,b\ge 0,$ and $d=(m,n).$ Then (see [24])

$$\begin{aligned} \left( U_{m},V_{n}\right) =\left\{ \begin{array}{c} V_{d}~\quad \,\quad \text { if }a>b, \\ 1\text { or }2 \quad \text { if }a\le b. \end{array} \right. \end{aligned}$$

(2.13)

From (2.11) and Lemma 2.2, it follows that

$$\begin{aligned} 5|V_{n}\Leftrightarrow 5|P\text { and }n\text { is odd}. \end{aligned}$$

(2.14)

An induction method shows that

$$\begin{aligned} V_{2^{k}}\equiv 7\,(\mathrm{mod}8)\, \end{aligned}$$

and thus

$$\begin{aligned} \left( \frac{2}{V_{2^{k}}}\right) =1 \end{aligned}$$

(2.15)

and

$$\begin{aligned} \left( \frac{-1}{V_{2^{k}}}\right) =-1 \end{aligned}$$

(2.16)

for all $k\ge 1.$

Lemma 2.17

Let P be odd. Then

$$\begin{aligned} \left( \frac{P-1}{V_{2^{k}}}\right) =\left( \frac{P+1}{V_{2^{k}}}\right) =1 \end{aligned}$$

(2.17)

for all $k\ge 1.$ Moreover, if $3\not \mid P,$ then

$$\begin{aligned} \left( \frac{3}{V_{2^{k}}}\right) =1 \end{aligned}$$

(2.18)

for all $k\ge 1.$

Proof

If $3\not \mid P,$ then $P^{2}\equiv 1$ $(\mathrm{mod}3)$ and therefore $V_{2}=P^{2}-2\equiv -1$ $(\mathrm{mod}3).$ An induction method shows that $V_{2^{k}}\equiv -1$ $(\mathrm{mod}3)$ since $V_{2^{k}}=\left( V_{2^{k-1}}\right) ^{2}-2$ by (2.12). Thus

$$\begin{aligned} \left( \frac{3}{V_{2^{k}}}\right) =-\left( \frac{V_{2^{k}}}{3}\right) =-\left( \frac{-1}{3}\right) =1. \end{aligned}$$

Since $V_{2}=P^{2}-2\equiv -1$ $(\mathrm{mod}P^{2}-1),$ it follows that $V_{2^{k}}\equiv -1$ $(\mathrm{mod}P^{2}-1).$ Thus $V_{2^{k}}\equiv -1$ $(\mathrm{ mod}P-1)$ and $V_{2^{k}}\equiv -1$ $(\mathrm{mod}P+1).$ Let $P-1=2^{t}a$ with a odd. Then we get

$$\begin{aligned} \left( \frac{P-1}{V_{2^{k}}}\right) =\left( \frac{2^{t}a}{V_{2^{k}}}\right) =\left( \frac{2}{V_{2^{k}}}\right) ^{t}\left( \frac{a}{V_{2^{k}}}\right) =\left( \frac{a}{V_{2^{k}}}\right) =\left( -1\right) ^{\frac{a-1}{2}}\left( \frac{V_{2^{k}}}{a}\right) \end{aligned}$$

(2.19)

since $\left( \frac{2}{V_{2^{k}}}\right) =1$ by (2.15). By using the fact that $V_{2^{k}}\equiv -1$ $(\mathrm{mod}a),$ we get $\left( \frac{P-1}{ V_{2^{k}}}\right) =\left( -1\right) ^{\frac{a-1}{2}}\left( \frac{-1}{a} \right) =1$ by (2.19). Similarly, it is seen that $\left( \frac{P+1}{ V_{2^{k}}}\right) =1.$ $\square$

When P is odd, it can be shown that

$$\begin{aligned} \left( \frac{5}{V_{2^{k}}}\right) =\left\{ \begin{array}{c} \,-1\quad \text { if }P^{2}\equiv -1\,(\mathrm{mod}5), \\ 1\quad \text { if }P^{2}\equiv 1\,(\mathrm{mod}5) \end{array} \right. \end{aligned}$$

(2.20)

and

$$\begin{aligned} \left( \frac{7}{V_{2^{k}}}\right) =\left\{ \begin{array}{c} -1\quad \text {if }P^{2}\equiv 4\,(\mathrm{mod}7), \\ 1\quad \,\,\text {if }P^{2}\equiv 1\,(\mathrm{mod}7) \end{array} \right. \end{aligned}$$

(2.21)

for all $k\ge 1.$

3 Main theorems

From now on, we will assume that n is a positive integer and P is an odd integer.

Theorem 3.1

If $U_{n}=2kx^{2}+1$ with k|P and $k>1,$ then $n=1$ or $n=5.$

Proof

Assume that $U_{n}=2kx^{2}+1$ for some integer x. Then n is odd by Lemma 2.1. It is clear that $n=1$ is a solution. Assume that $n>1.$ Then we have $n=2m+1$ with $m\ge 1.$ Thus, we get $U_{m}V_{m+1}=U_{2m+1}-1=2kx^{2}$ by (2.9). It can be seen that m is even by (2.6). Thus, $(U_{m},V_{m+1})=P$ by (2.13). Then it follows that

$$\begin{aligned} U_{m}=k_{1}Pa^{2}\quad \text { and }\quad V_{m+1}=2k_{2}Pb^{2} \end{aligned}$$

or

$$\begin{aligned} U_{m}=2k_{1}Pa^{2}\quad \text { and }\quad V_{m+1}=k_{2}Pb^{2} \end{aligned}$$

for some natural numbers a and b with $k=k_{1}k_{2}.$ Thus, it is seen that

$$\begin{aligned} U_{m}=ut^{2}\text { and }V_{m+1}=2vs^{2} \end{aligned}$$

(3.1)

or

$$\begin{aligned} U_{m}=2ut^{2}\quad \text { and }\quad V_{m+1}=vs^{2} \end{aligned}$$

(3.2)

for some natural numbers u, v, s, t with u|P and v|P. Assume that (3.1) is satisfied. By using Theorems 2.4 and 2.10, it is seen that $m=2.$ Therefore $n=5.$ The identity (3.2) is impossible by Theorems 2.6 and 2.10.

Theorem 3.2

Let $w=1,2,3,6.$ If $U_{n}=wx^{2}+1$ for some integer x, then $(w,n)=(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(6,1),(6,2),(6,5).$

Proof

Assume that $U_{n}=wx^{2}+1$ for some integer x. Let $n>3$. Then $n=4q+r$ for some $q>0$ with $0\le r\le 3$. Then $n=2\cdot 2^{k}a+r$ with a odd and $k\ge 1.$ Thus,

$$\begin{aligned} wx^{2}=-1+U_{n}\equiv -1-U_{r}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3). This shows that

$$\begin{aligned} wx^{2}\equiv -1,-2,-(P+1),-P^{2}\,(\mathrm{mod}V_{2^{k}}). \end{aligned}$$

Since $\left( \dfrac{2}{V_{2^{k}}}\right) =1$, $\left( \dfrac{-1}{V_{2^{k}}} \right) =-1,$ and $\left( \dfrac{P+1}{V_{2^{k}}}\right) =1$ by (2.15), (2.16), and (2.17), respectively, we get

$$\begin{aligned} \left( \dfrac{w}{V_{2^{k}}}\right) =-1. \end{aligned}$$

(3.3)

If $w=1,2,$ then (3.3) is impossible. Let $w=3,6.$ If $3\not \mid P$, then again (3.3) is impossible since $\left( \dfrac{3}{V_{2^{k}}} \right) =1$ by (2.18). Therefore $n\le 3$ in case $3\not \mid P$ and $w=3,6.$ But $n=3$ is not a solution in this case. If $w=6$ and 3|P, then by Theorem 3.1, we get $n=1$ or $n=5.$ Thus, $n=1,5$ for the case $w=6$ and 3|P. If $w=3$ and 3|P, then by Theorem 2.7, we get $n=1.$

$\square$

Theorem 3.3

If $U_{n}=5x^{2}+1$ for some integer x,  then $n=1$ or $n=2.$

Proof

Assume that $U_{n}=5x^{2}$ $+1$ for some integer x. If 5|P,  then by Theorem 2.7, $n=1.$ Assume that $5\not \mid P$. Let $n>2$ and n be even. Now we divide the proof into two cases.

Case I. Let $\ P^{2}\equiv 1$ $(\mathrm{mod}5).$ Since n is even, $n=4q+r$ for some positive integer q with $r=0,2.$ Thus, $n=2\cdot 2^{k}a+r$ with a odd and $k\ge 1.$ Then

$$\begin{aligned} 5x^{2}=-1+U_{n}\equiv -1-U_{r}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3). This shows that

$$\begin{aligned} 5x^{2}\equiv -1,-(P+1)\,(\mathrm{mod}V_{2^{k}}), \end{aligned}$$

which is impossible since $\left( \dfrac{-1}{V_{2^{k}}}\right) =-1$, $\left( \dfrac{P+1}{V_{2^{k}}}\right) =1$, and $\left( \dfrac{5}{V_{2^{k}}}\right) =1$ by (2.16), (2.17), and (2.20), respectively.

Case II. Let $\ P^{2}\equiv -1$ $(\mathrm{mod}5)$. We get $5x^{2}\equiv -1$ $(\mathrm{mod}P)$ since $P|U_{n}$ when n is even. This shows that

$$\begin{aligned} -1=\left( \dfrac{P}{5}\right) =\left( \dfrac{5}{P}\right) =\left( \dfrac{-1}{ P}\right) , \end{aligned}$$

which implies that $P\equiv 3,7$ $(\mathrm{mod}8)$. Since n is even, we get $n=6q+r$, $r=0,2,4.$ Then $5x^{2}+1$ $\equiv U_{r}$ $(\mathrm{mod}8)$ by (2.5). If $r=0,$ then we get $5x^{2}$ $\equiv -1$ $(\mathrm{mod}8),$ which is impossible. Let $r=2$. Then $5x^{2}$ $+1\equiv U_{2}$ $(\mathrm{mod}8)$ by (2.5), which shows that $5x^{2}$ $+1\equiv P$ $(\mathrm{mod}8).$ But this is impossible since $P\equiv 3,7$ $(\mathrm{mod}8).$ Let $r=4$. Then $n=12t+4$ or $n=12t+10$ for some integer t. Let $n=12t+10$. Then $n=12q_{1}-2$ with $q_{1}>0.$ Thus, $n=2\cdot 2^{k}a-2$ with a odd and $k\ge 1.$ Then it follows that

$$\begin{aligned} 5x^{2}=-1+U_{n}\equiv -1-U_{-2}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3), which implies that

$$\begin{aligned} 5x^{2}\equiv P-1\,(\mathrm{mod}V_{2^{k}}). \end{aligned}$$

This is impossible since $\left( \dfrac{P-1}{V_{2^{k}}}\right) =1$ and $\left( \dfrac{5}{V_{2^{k}}}\right) =-1$ by (2.17) and (2.20), respectively. Let $n=12t+4.$ Since $16|U_{6},$ we get $5x^{2}+1=U_{n}\equiv U_{4}$ $(\mathrm{mod}16)$ by (2.1). A simple computation shows that $5x^{2}+1\equiv 1,5,6,14$ $(\mathrm{mod}16)$ and therefore $U_{4}\equiv 1,5,6,14$ $(\mathrm{mod}16)$. Moreover, we have $5x^{2}+1=U_{n}\equiv U_{4}\equiv -P$ $(\mathrm{mod}8)$ by (2.5). Using the fact that $5x^{2}+1\equiv 1,5,6,14$ $(\mathrm{mod}8),$ we see that $P\equiv 3,7$ $(\mathrm{mod}8).$ Since $P\equiv 3,7$ $\ (\mathrm{mod}8)$ and $P^{3}-2P=U_{4}\equiv 1,5,6,14$ $(\mathrm{mod}16),$ it is seen that $P\equiv 3,15$ $(\mathrm{mod}16).$ Let $P\equiv 3$ $(\mathrm{mod} 16)$ and $P\equiv 3$ $(\mathrm{mod}5).$ Since n is even, $\ n=10q+r,$ $r\in \{0,2,4,6,8\}.$ Using $5|U_{5},$ we get $5x^{2}+1=U_{n}\equiv U_{r}$ $(\mathrm{ mod}5)$ by (2.1). A simple computation shows that $r=4.$ Since $n=10q+4$ and $n=12t+4,$ we get $\ n=60k+4$ for some natural number k. Thus, by using (2.2), it is seen that

$$\begin{aligned} U_{n}=U_{60k+4}\equiv U_{4}\,(\mathrm{mod}V_{5}), \end{aligned}$$

which implies that

$$\begin{aligned} 5x^{2}\equiv P^{3}-2P-1\,(\mathrm{mod}P^{4}-5P^{2}+5) \end{aligned}$$

since $V_{5}=P(P^{4}-5P^{2}+5).$ This shows that

$$\begin{aligned} \left( \dfrac{5}{P^{4}-5P^{2}+5}\right) =\left( \dfrac{P^{3}-2P-1}{ P^{4}-5P^{2}+5}\right) =\left( \dfrac{(P^{3}-2P-1)/4}{P^{4}-5P^{2}+5}\right) . \end{aligned}$$

By using the facts that $(P^{3}-2P-1)/4\equiv 1$ $(\mathrm{mod}4),$ $P^{4}-5P^{2}+5\equiv 1$ $(\mathrm{mod}5),$ $P^{4}-5P^{2}+5\equiv 9$ $(\mathrm{mod }16),$ and $-3P^{2}+P+5\equiv 13$ $(\mathrm{mod}16),$ we get

$$\begin{aligned} 1= & {} \left( \dfrac{5}{P^{4}-5P^{2}+5}\right) =\left( \dfrac{P^{4}-5P^{2}+5}{ (P^{3}-2P-1)/4}\right) =\left( \dfrac{-3P^{2}+P+5}{(P^{3}-2P-1)/4}\right) \\= & {} \left( \dfrac{(P^{3}-2P-1)/4}{-3P^{2}+P+5}\right) =\left( \dfrac{ P^{3}-2P-1}{-3P^{2}+P+5}\right) =\left( \dfrac{9(P^{3}-2P-1)}{-3P^{2}+P+5} \right) \\= & {} \left( \dfrac{-2(P+2)}{-3P^{2}+P+5}\right) =\left( \dfrac{-2}{-3P^{2}+P+5} \right) \left( \dfrac{P+2}{-3P^{2}+P+5}\right) \\= & {} -\left( \dfrac{P+2}{-3P^{2}+P+5}\right) =-\left( \dfrac{-3P^{2}+P+5}{P+2} \right) =-\left( \dfrac{-1}{P+2}\right) =-1, \end{aligned}$$

a contradiction. Let $P\equiv 3$ $(\mathrm{mod}16)$ and $P\equiv 2$ $(\mathrm{mod }5).$ Then $(P-1)/2\equiv 3$ $(\mathrm{mod}5)$ and $(P-1)/2\equiv 1$ $(\mathrm{ mod}8)$ and therefore

$$\begin{aligned} \left( \dfrac{5}{(P-1)/2}\right) =-1\text { and }\left( \dfrac{-2}{(P-1)/2} \right) =1. \end{aligned}$$

(3.4)

Moreover,

$$\begin{aligned} 5x^{2}=-1+U_{n}=-1+U_{12t+4}\equiv -1+U_{4}\equiv P^{3}-2P-1\,(\mathrm{ mod}U_{3}) \end{aligned}$$

by (2.1). This implies that

$$\begin{aligned} 5x^{2}\equiv P^{3}-2P-1\,(\mathrm{mod}P-1). \end{aligned}$$

This shows that

$$\begin{aligned} 5x^{2}\equiv -2\,(\mathrm{mod}(P-1)/2), \end{aligned}$$

which is impossible by (3.4). Let $P\equiv 15$ $(\mathrm{mod}16).$ Then $(P^{2}-3)/2\equiv 3$ $(\mathrm{mod}5)$ and $(P^{2}-3)/2\equiv 7$ $(\mathrm{mod} 8).$ Moreover, we get

$$\begin{aligned} 5x^{2}=-1+U_{n}=-1+U_{12t+4}\equiv -1+U_{4}\equiv P^{3}-2P-1\,(\mathrm{ mod}V_{3}) \end{aligned}$$

by (2.2). This shows that

$$\begin{aligned} 5x^{2}\equiv P-1\,(\mathrm{mod}(P^{2}-3)/2) \end{aligned}$$

and therefore

$$\begin{aligned} \left( \dfrac{5}{(P^{2}-3)/2}\right) =\left( \dfrac{P-1}{(P^{2}-3)/2}\right) . \end{aligned}$$

This is impossible since

$$\begin{aligned} -1= & {} \left( \dfrac{(P^{2}-3)/2}{5}\right) =\left( \dfrac{5}{(P^{2}-3)/2} \right) =\left( \dfrac{(P-1)/2}{(P^{2}-3)/2}\right) \left( \dfrac{2}{ (P^{2}-3)/2}\right) \\= & {} \left( \dfrac{(P-1)/2}{(P^{2}-3)/2}\right) =-\left( \dfrac{(P^{2}-3)/2}{ (P-1)/2}\right) =-\left( \dfrac{-1}{(P-1)/2}\right) =1. \end{aligned}$$

Now assume that $n>3$ and n is odd. Then $n=2m+1$ with $m>1.$ Therefore $U_{2m+1}=5x^{2}+1,$ which implies that $5x^{2}=U_{2m+1}-1=U_{m}V_{m+1}$ by ( 2.9). Let m be odd. Then $(U_{m},V_{m+1})=1$ by (2.13) and ( 2.8). Thus,

$$\begin{aligned} U_{m}=a^{2} \quad \text { and }\quad V_{m+1}=5b^{2} \end{aligned}$$

(3.5)

or

$$\begin{aligned} U_{m}=5a^{2}\quad \text { and }\quad V_{m+1}=b^{2} \end{aligned}$$

(3.6)

for some integers a and b. The identities (3.5) and (3.6) are impossible by (2.14) and Theorem 2.9, respectively. Let m be even. Then $(U_{m},V_{m+1})=P$ by (2.13). Thus,

$$\begin{aligned} U_{m}=Pa^{2}\text { and }V_{m+1}=5Pb^{2} \end{aligned}$$

(3.7)

or

$$\begin{aligned} U_{m}=5Pa^{2}\text { and }V_{m+1}=Pb^{2} \end{aligned}$$

(3.8)

for some integers a and b. The identities (3.7) and (3.8) are impossible by (2.14) and Theorem 2.3, respectively. Therefore $n\le 3$. If $n=3,$ we get $P^{2}-1=U_{3}=5x^{2}+1,$ which implies that $P^{2}\equiv 2$ $(\mathrm{mod}5).$ This is impossible. Thus, $n=1$ or $n=2.$ $\square$

Theorem 3.4

If $U_{n}=7x^{2}+1$ for some integer x,  then $n=1,2,3.$

Proof

Assume that $U_{n}=7x^{2}$ $+1$ for some integer x. If $7|P,\$ then by Theorem 2.7, $n=1.$ Assume that $7\not \mid P$. Let $n>2$ and n be even. Then $7x^{2}+1\equiv 0$ $(\mathrm{mod}P).$ This shows that $\left( \dfrac{7}{P} \right) =\left( \dfrac{-1}{P}\right) ,$ which implies that $\left( \dfrac{P}{ 7}\right) =1$. Therefore $P\equiv 1,2,4$ $(\mathrm{mod}7).$ Now we distinguish three cases.

Case I. Let $P\equiv 1$ $(\mathrm{mod}7).$ Since n is even, $n=4q+r$ for some $q>0$ with $r=0,2.$ Thus, $n=2\cdot 2^{k}a+r$ with a odd and $k\ge 1.$ Then we get

$$\begin{aligned} 7x^{2}=-1+U_{n}\equiv -1-U_{r}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3), which implies that

$$\begin{aligned} 7x^{2}\equiv -1,-(P+1)\,(\mathrm{mod}V_{2^{k}}). \end{aligned}$$

This is impossible since $\left( \dfrac{-1}{V_{2^{k}}}\right) =-1$, $\left( \dfrac{P+1}{V_{2^{k}}}\right) =1,$ and $\left( \dfrac{7}{V_{2^{k}}}\right) =1$ by (2.16), (2.17), and (2.21), respectively.

Case II. Let $P\equiv 4$ $(\mathrm{mod}7).$ Then $7|V_{2}$ and

$$\begin{aligned} 7x^{2}=-1+U_{n}=-1+U_{4q+r}\equiv -1\pm U_{r}\,(\mathrm{mod}V_{2}) \end{aligned}$$

by (2.2). This is impossible since $7\not \mid (-1\pm U_{r})$ for $r=0,2.$

Case III. Let $P\equiv 2$ $(\mathrm{mod}7).$ If $n=4q+2,$ then $n=4(q+1)-2=2\cdot 2^{k}a-2$ with a odd and $k\ge 1.$ Thus, we get

$$\begin{aligned} 7x^{2}=-1+U_{n}\equiv -1-U_{-2}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3), which implies that

$$\begin{aligned} 7x^{2}\equiv P-1\,(\mathrm{mod}V_{2^{k}}). \end{aligned}$$

But this is impossible since $\left( \dfrac{P-1}{V_{2^{k}}}\right) =1$ and $\left( \dfrac{7}{V_{2^{k}}}\right) =-1$ by (2.17) and (2.21), respectively. Let $n=4q.$ Then $n=12t+r$ with $r=0,4,8.$ Assume that $n=12t.$ Since P is odd, we can write $P^{2}-1=2^{m}a$ with a odd. Thus,

$$\begin{aligned} 7x^{2}=-1+U_{n}\equiv -1+U_{0}\,(\mathrm{mod}U_{3}) \end{aligned}$$

by (2.1), which implies that

$$\begin{aligned} 7x^{2}\equiv -1\,(\mathrm{mod}a). \end{aligned}$$

This shows that $\left( \dfrac{7}{a}\right) =\left( \dfrac{-1}{a}\right)$ and therefore $\left( \dfrac{a}{7}\right) =1.$ Thus,

$$\begin{aligned} 1=\left( \dfrac{a}{7}\right) =\left( \dfrac{2^{m}a}{7}\right) =\left( \dfrac{ P^{2}-1}{7}\right) =\left( \dfrac{3}{7}\right) =-1, \end{aligned}$$

a contradiction. Assume that $n=12t+4.$ Since $16|U_{6},$ we get $U_{n}\equiv U_{4}$ $(\mathrm{mod}16)$ by (2.1). This shows that $7x^{2}+1\equiv P^{3}-2P$ $(\mathrm{mod}16).$ Since $7x^{2}+1\equiv 0,1,8,13$ $( \mathrm{mod}16)$, a simple computation shows that $P\equiv 11,15$ $(\mathrm{mod} 16).$ Let $P\equiv 11$ $(\mathrm{mod}16).$ Then

$$\begin{aligned} 7x^{2}=-1+U_{n}=-1+U_{12t+4}\equiv -1+U_{4}\equiv P^{3}-2P-1\,(\mathrm{ mod}U_{3}) \end{aligned}$$

by (2.1), which shows that $7x^{2}\equiv -2$ $(\mathrm{mod}P-1).$ Thus, we get $\left( \dfrac{7}{(P-1)/2}\right) =$ $\left( \dfrac{-2}{(P-1)/2} \right)$ and therefore $\left( \dfrac{(P-1)/2}{7}\right) =$ $\left( \dfrac{2 }{(P-1)/2}\right) .$ But this is impossible since $(P-1)/2\equiv 5$ $(\mathrm{ mod}8)$ and $(P-1)/2\equiv 4$ $(\mathrm{mod}7).$ Let $P\equiv 15$ $(\mathrm{mod} 16).$ By using a similar argument, it is seen that $7x^{2}\equiv P-1$ $( \mathrm{mod}P^{2}-3).$ This shows that

$$\begin{aligned} \left( \dfrac{7}{(P^{2}-3)/2}\right) =\left( \dfrac{(P-1)/2}{(P^{2}-3)/2} \right) \left( \dfrac{2}{(P^{2}-3)/2}\right) . \end{aligned}$$

Since $(P^{2}-3)/2\equiv 4$ $(\mathrm{mod}7)$, $(P^{2}-3)/2\equiv 7$ $(\mathrm{ mod}8),$ and $(P-1)/2\equiv 7$ $(\mathrm{mod}8),$ we get

$$\begin{aligned} -1=\left( \dfrac{7}{(P^{2}-3)/2}\right) =\left( \dfrac{(P-1)/2}{(P^{2}-3)/2} \right) =-\left( \dfrac{(P^{2}-3)/2}{(P-1)/2}\right) =-\left( \dfrac{-1}{ (P-1)/2}\right) =1, \end{aligned}$$

a contradiction. Assume that $n=12t+8.$ Then we can write $n=12m-4.$ A simple computation shows that $P\equiv 1,5$ $(\mathrm{mod}16)$ in this case. Let $P\equiv 1$ $(\mathrm{mod}16).$ Then

$$\begin{aligned} 7x^{2}=-1+U_{n}=-1+U_{12m-4}\equiv -1+U_{-4}\equiv -(P^{3}-2P+1)\,( \mathrm{mod}U_{3}), \end{aligned}$$

which implies that $7x^{2}\equiv -2$ $(\mathrm{mod}P+1).$ Thus, we get

$$\begin{aligned} \left( \dfrac{7}{(P+1)/2}\right) =\left( \dfrac{-2}{(P+1)/2}\right) . \end{aligned}$$

Therefore by using the facts that $(P+1)/2\equiv 1$ $(\mathrm{mod}8)$ and $(P+1)/2\equiv 5$ $(\mathrm{mod}7),$ we get

$$\begin{aligned} -1=\left( \dfrac{(P+1)/2}{7}\right) =\left( \dfrac{2}{(P+1)/2}\right) =1, \end{aligned}$$

a contradiction. Let $P\equiv 5$ $(\mathrm{mod}16).$ Then

$$\begin{aligned} 7x^{2}=-1+U_{n}=-1+U_{12m-4}\equiv -1+U_{-4}\equiv -(P^{3}-2P+1)\,( \mathrm{mod}V_{3}) \end{aligned}$$

by (2.2), which implies that $7x^{2}\equiv -(P+1)$ $(\mathrm{mod} P^{2}-3).$ By using the facts that $(P^{2}-3)/2\equiv 4$ $(\mathrm{mod}7)$, $(P^{2}-3)/2\equiv 3$ $(\mathrm{mod}8),$ and $(P+1)/2\equiv 3$ $(\mathrm{mod}8),$ we get

$$\begin{aligned} 1= & {} \left( \dfrac{7}{(P^{2}-3)/2}\right) \left( \dfrac{-1}{(P^{2}-3)/2} \right) \left( \dfrac{(P+1)/2}{(P^{2}-3)/2}\right) \left( \dfrac{2}{ (P^{2}-3)/2}\right) \\= & {} -\left( \dfrac{(P^{2}-3)/2}{7}\right) \left( \dfrac{(P+1)/2}{(P^{2}-3)/2} \right) =-\left( \dfrac{(P+1)/2}{(P^{2}-3)/2}\right) \\= & {} \left( \dfrac{(P^{2}-3)/2}{(P+1)/2}\right) =\left( \dfrac{-1}{(P+1)/2} \right) =-1, \end{aligned}$$

a contradiction. Thus, we conclude that $n\le 2.$ Now assume that n is odd. Then $n=2m+1$ with $m\ge 0.$ Thus, $U_{2m+1}=7x^{2}+1,$ which implies that $7x^{2}=U_{2m+1}-1=U_{m}V_{m+1}$ by (2.9). Let m be odd. Then $(U_{m},V_{m+1})=1$ by (2.13) and (2.8). Thus,

$$\begin{aligned} U_{m}=a^{2}\quad \text { and }\quad V_{m+1}=7b^{2} \end{aligned}$$

(3.9)

or

$$\begin{aligned} U_{m}=7a^{2}\quad \text { and }\quad V_{m+1}=b^{2} \end{aligned}$$

(3.10)

for some integers a and b. Assume that (3.9) is satisfied. Then by Theorem 2.10, we get $m=1$ and therefore $n=3.$ The identity (3.10) is impossible by Theorem 2.9. Let m be even. Then $(U_{m},V_{m+1})=P$ by (2.13). This implies that

$$\begin{aligned} U_{m}=Pa^{2}\text { and }V_{m+1}=7Pb^{2} \end{aligned}$$

(3.11)

or

$$\begin{aligned} U_{m}=7Pa^{2}\text { and }V_{m+1}=Pb^{2}. \end{aligned}$$

(3.12)

for some integers a and b. By using Theorems 2.3 and 2.8, we have in both cases that $m+1=1$ and therefore $n=1.$ Consequently, we have $n=1,2,3.$ $\square$

Theorem 3.5

If $U_{n}=10x^{2}+1$ for some integer x,  then $n=1,2.$

Proof

If 5|P,  then by Theorem 3.1, $n=1$ or $n=5.$ Assume that $5\not \mid P$ . Let $n>2$ and n be even. Then $10x^{2}+1\equiv 0$ $(\mathrm{mod}P)$ since $P|U_{n}$ when n is even. Therefore

$$\begin{aligned} \left( \dfrac{5}{P}\right) =\left( \dfrac{-2}{P}\right) . \end{aligned}$$

If $P\equiv \pm 1$ $(\mathrm{mod}5),$ then $P\equiv 1,3$ $(\mathrm{mod}8).$ If $P\equiv \pm 2$ $(\mathrm{mod}5),$ then $P\equiv 5,7$ $(\mathrm{mod}8).$ The remainder of the proof is split into two cases.

Case I. Let $P\equiv \pm 1$ $(\mathrm{mod}5).$ Since n is even, we get $n=4q+r$ for some positive integer q with $r=0,2.$ Thus, $n=2\cdot 2^{k}a+r$ with a odd and $k\ge 1.$ Then

$$\begin{aligned} 10x^{2}=-1+U_{n}\equiv -1-U_{r}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3). This shows that

$$\begin{aligned} 10x^{2}\equiv -1,-(P+1)\,(\mathrm{mod}V_{2^{k}}), \end{aligned}$$

which is impossible since $\left( \dfrac{2}{V_{2^{k}}}\right) =1$, $\left( \dfrac{-1}{V_{2^{k}}}\right) =-1$, $\left( \dfrac{P+1}{V_{2^{k}}}\right) =1,$ and $\left( \dfrac{5}{V_{2^{k}}}\right) =1$ by (2.15), (2.16), ( 2.17), and (2.20), respectively.

Case II. Let $P\equiv \pm 2$ $(\mathrm{mod}5).$ Since n is even, we get $n=6q+r$ with $r=0,2,4.$ Then $10x^{2}+1$ $\equiv U_{r}$ $(\mathrm{mod}8)$ by (2.5). If $r=0,$ then we get $10x^{2}$ $\equiv -1$ $(\mathrm{mod}8),$ which is impossible. Let $r=2$. Then $10x^{2}$ $+1\equiv U_{2}$ $(\mathrm{mod} 8)$, which shows that $10x^{2}$ $+1\equiv P$ $(\mathrm{mod}8),$ which is impossible since $P\equiv 5,7$ $(\mathrm{mod}8).$ Let $r=4.$ Then either $n=12t+10$ or $n=12t+4$ for some nonnegative integer t. Assume that $n=12t+10.$ Then $n=12q_{1}-2$ with $q_{1}>0.$ Thus, $n=2\cdot 2^{k}a-2$ with a odd and $k\ge 1.$ This shows that

$$\begin{aligned} 10x^{2}\equiv -1+U_{n}\equiv -1-U_{-2}\,(\mathrm{mod}V_{2^{k}}) \end{aligned}$$

by (2.3), which shows that

$$\begin{aligned} 10x^{2}\equiv P-1\,(\mathrm{mod}V_{2^{k}}). \end{aligned}$$

This is impossible since $\left( \dfrac{2}{V_{2^{k}}}\right) =1$, $\left( \dfrac{P-1}{V_{2^{k}}}\right) =1,$ and $\left( \dfrac{5}{V_{2^{k}}}\right) =-1$ by (2.15), (2.17), and (2.20), respectively. Assume that $n=12t+4.$ It can be seen that $U_{n}=10x^{2}+1$ $\equiv 1,9,11$ $( \mathrm{mod}16).$ Moreover, we get $U_{n}\equiv U_{4}$ $(\mathrm{mod}16)$ by ( 2.1) since $16|U_{6}.$ A simple computation shows that $P\equiv 7,13,15$ $(\mathrm{mod}16)$ since $P\equiv 5,7$ $(\mathrm{mod}8)$ and $U_{4}=P^{3}-2P.$ Let $P\equiv 7,15$ $(\mathrm{mod}16).$ Then $(P^{2}-3)/2\equiv 3$ $(\mathrm{mod} 5)$ and $(P^{2}-3)/2\equiv 7$ $(\mathrm{mod}8).$ Moreover, we get

$$\begin{aligned} 10x^{2}=-1+U_{n}=-1+U_{12t+4}\equiv -1+U_{4}\equiv P^{3}-2P-1\,(\mathrm{ mod}V_{3}) \end{aligned}$$

by (2.2). This shows that

$$\begin{aligned} 10x^{2}\equiv P-1\,(\mathrm{mod}(P^{2}-3)) \end{aligned}$$

and therefore

$$\begin{aligned} 5x^{2}\equiv (P-1)/2\,(\mathrm{mod}(P^{2}-3)/2). \end{aligned}$$

Then we get

$$\begin{aligned} \left( \dfrac{5}{(P^{2}-3)/2}\right) =\left( \dfrac{(P-1)/2}{(P^{2}-3)/2} \right) . \end{aligned}$$

This is impossible since

$$\begin{aligned} -1= & {} \left( \dfrac{(P^{2}-3)/2}{5}\right) =\left( \dfrac{5}{(P^{2}-3)/2} \right) =\left( \dfrac{(P-1)/2}{(P^{2}-3)/2}\right) \\= & {} -\left( \dfrac{(P^{2}-3)/2}{(P-1)/2}\right) =-\left( \dfrac{-1}{(P-1)/2} \right) =1. \end{aligned}$$

Let $P\equiv 13$ $(\mathrm{mod}16)$ and $P\equiv 2$ $(\mathrm{mod}5).$ Then

$$\begin{aligned} (P-1)/4\equiv 4\,(\mathrm{mod}5)\text { and }(P-1)/4\equiv 3\,( \mathrm{mod}4) \end{aligned}$$

and therefore

$$\begin{aligned} \left( \dfrac{5}{(P-1)/4}\right) =1\text { and }\left( \dfrac{-1}{(P-1)/4} \right) =-1. \end{aligned}$$

(3.13)

Moreover,

$$\begin{aligned} 10x^{2}=-1+U_{n}=-1+U_{12t+4}\equiv -1+U_{4}\equiv P^{3}-2P-1\,(\mathrm{ mod}U_{3}) \end{aligned}$$

by (2.1). This implies that

$$\begin{aligned} 10x^{2}\equiv P^{3}-2P-1\,(\mathrm{mod}P-1). \end{aligned}$$

This shows that

$$\begin{aligned} 10x^{2}\equiv -2\,(\mathrm{mod}(P-1)/2) \end{aligned}$$

and therefore

$$\begin{aligned} 5x^{2}\equiv -1\,(\mathrm{mod}(P-1)/4), \end{aligned}$$

which is impossible by (3.13). Let $P\equiv 13$ $(\mathrm{mod}16)$ and $P\equiv 3$ $(\mathrm{mod}5).$ Since n is even, $\ n=10q+r$ with $r\in \{0,2,4,6,8\}.$ Since $5|U_{5}$ by (2.7), we get $10x^{2}+1=U_{n} \equiv U_{r}$ $(\mathrm{mod}5)$ by (2.1). A simple computation shows that $r=4.$ Since $n=10q+4$ and $n=12t+4,$ we get $\ n=60k+4$ for some natural number k. Thus, by using (2.2), it is seen that

$$\begin{aligned} U_{n}=U_{60k+4}\equiv U_{4}\,(\mathrm{mod}V_{5}), \end{aligned}$$

which implies that

$$\begin{aligned} 10x^{2}\equiv P^{3}-2P-1\,(\mathrm{mod}P^{4}-5P^{2}+5) \end{aligned}$$

since $V_{5}=P(P^{4}-5P^{2}+5).$ This shows that

$$\begin{aligned} 5x^{2}\equiv (P^{3}-2P-1)/2\,(\mathrm{mod}P^{4}-5P^{2}+5) \end{aligned}$$

and therefore

$$\begin{aligned} \left( \dfrac{5}{P^{4}-5P^{2}+5}\right) =\left( \dfrac{(P^{3}-2P-1)/2}{ P^{4}-5P^{2}+5}\right) . \end{aligned}$$

Since $(P^{3}-2P-1)/2\equiv 5$ $(\mathrm{mod}8),$ $P^{4}-5P^{2}+5\equiv 1$ $( \mathrm{mod}5),$ $P^{4}-5P^{2}+5\equiv 9$ $(\mathrm{mod}16),$ and $-3P^{2}+P+5\equiv 7$ $(\mathrm{mod}16),$ we get

$$\begin{aligned} 1= & {} \left( \dfrac{P^{4}-5P^{2}+5}{5}\right) =\left( \dfrac{5}{P^{4}-5P^{2}+5 }\right) =\left( \dfrac{(P^{3}-2P-1)/2}{P^{4}-5P^{2}+5}\right) \\= & {} \left( \dfrac{P^{4}-5P^{2}+5}{(P^{3}-2P-1)/2}\right) =\left( \dfrac{ -3P^{2}+P+5}{(P^{3}-2P-1)/2}\right) =\left( \dfrac{(P^{3}-2P-1)/2}{ -3P^{2}+P+5}\right) \\= & {} \left( \dfrac{P^{3}-2P-1}{-3P^{2}+P+5}\right) \left( \dfrac{2}{-3P^{2}+P+5 }\right) =\left( \dfrac{P^{3}-2P-1}{-3P^{2}+P+5}\right) \\= & {} \left( \dfrac{9(P^{3}-2P-1)}{-3P^{2}+P+5}\right) =\left( \dfrac{-2(P+2)}{ -3P^{2}+P+5}\right) \\= & {} \left( \dfrac{-2}{-3P^{2}+P+5}\right) \left( \dfrac{P+2}{-3P^{2}+P+5} \right) \\= & {} -\left( \dfrac{P+2}{-3P^{2}+P+5}\right) =\left( \dfrac{-3P^{2}+P+5}{P+2} \right) =\left( \dfrac{-1}{P+2}\right) =-1, \end{aligned}$$

a contradiction. Now assume that $n>1$ and n is odd. Then $n=2m+1$ with $m\ge 1.$ Therefore $U_{2m+1}=10x^{2}+1,$ which implies that $10x^{2}=U_{2m+1}-1=U_{m}V_{m+1}$ by (2.9). Let m be odd. Then $(U_{m},V_{m+1})=1$ by (2.13) and (2.8). Thus,

$$\begin{aligned} U_{m}= & {} a^{2}\text { and }V_{m+1}=10b^{2}, \end{aligned}$$

(3.14)

$$\begin{aligned} U_{m}= & {} 10a^{2}\text { and }V_{m+1}=b^{2}, \end{aligned}$$

(3.15)

$$\begin{aligned} U_{m}= & {} 2a^{2}\text { and }V_{m+1}=5b^{2}, \end{aligned}$$

(3.16)

or

$$\begin{aligned} U_{m}=5a^{2}\text { and }V_{m+1}=2b^{2} \end{aligned}$$

(3.17)

for some integers a and b. The identity (3.15) is impossible by Theorem 2.9. The identities (3.14) and (3.16) are impossible by (2.14), and (3.17) is impossible by Theorem . Let m be even. Then $(U_{m},V_{m+1})=P$ by (2.13). Thus,

$$\begin{aligned} U_{m}= & {} Pa^{2}\text { and }V_{m+1}=10Pb^{2}, \end{aligned}$$

(3.18)

$$\begin{aligned} U_{m}= & {} 10Pa^{2}\text { and }V_{m+1}=Pb^{2}, \end{aligned}$$

(3.19)

$$\begin{aligned} U_{m}= & {} 2Pa^{2}\text { and }V_{m+1}=5Pb^{2}, \end{aligned}$$

(3.20)

or

$$\begin{aligned} U_{m}=5Pa^{2}\text { and }V_{m+1}=2Pb^{2} \end{aligned}$$

(3.21)

for some integers a and b. The identities (3.18) and (3.20) are impossible by (2.14), and (3.19) is impossible by Theorem 2.3. Assume that (3.21) is satisfied. Then by Theorem 2.4, we get $m=2$ and therefore $n=5.$ Consequently, we have $n=1,2,5$. But it can be seen that 5 is not a solution and therefore $n=1,2.$ $\square$

By using MAGMA [25], it can be shown that the equation $2Px^{2}+1=U_{5}=P^{4}-3P^{2}+1$ has only the solution $P=3.$ Therefore we can give the following corollary by using Theorem 3.1 and Lemmas 2.13 and 2.14.

Corollary 3.6

The equations $x^{2}-(P^{2}-4)(2Py^{2}+1)^{2}=4$ and $(2Px^{2}+1)^{2}-P(2Px^{2}+1)y+y^{2}=1$ have positive integer solutions only when $P=3$. The solutions are given by $(x,y)=(123,3)$ and $(x,y)=(3,21),$ respectively.

Corollary 3.7

Let $k=1,2,3,5,10.$ The equations $x^{2}-(P^{2}-4)(ky^{2}+1)^{2}=4$ and $(kx^{2}+1)^{2}-P(kx^{2}+1)y+y^{2}=1$ have positive integer solutions only when $P=ka^{2}+1$ for some integer a.

Corollary 3.8

The equations $x^{2}-(P^{2}-4)(6y^{2}+1)^{2}=4$ and $(6x^{2}+1)^{2}-P(6x^{2}+1)y+y^{2}=1$ have positive integer solutions only when $P=6a^{2}+1$ for some integer a or $P=3(U_{m}(10,-1)+U_{m-1}(10,-1))$ for some $m\ge 1$ and there is only one solution in each case.

Proof

In order to prove the corollary we must solve the equation $6x^{2}+1=U_{5}=P^{4}-3P^{2}+1.$ Since $6x^{2}+1=P^{4}-3P^{2}+1,$ it is seen that $P=3a$ and $x=3b$ for some integers a and b. Then we get $a^{2}(3a^{2}-1)=2b^{2},$ which implies that $3a^{2}-1=2v^{2}.$ This shows that $3a^{2}-2v^{2}=1.$ Thus, by Lemma 2.11, we get $a=$ $U_{m}(10,-1)-U_{m-1}(10,-1)$ for some $m\ge 1.$ Since $P=3a,$ the proof follows. $\square$

From Theorem 3.4 and Lemma 2.12, we can give the following corollary easily.

Corollary 3.9

The equations $x^{2}-(P^{2}-4)(7y^{2}+1)^{2}=4$ and $(7x^{2}+1)^{2}-P(7x^{2}+1)y+y^{2}=1$ have positive integer solutions only when $P=7a^{2}+1$ for some integer a or $P=3\left( U_{m+1}(16-1)-U_{m}(16,-1)\right)$ for some $m\ge 1$ and there is only one solution in each case.