On The Lucas Sequence Equations $V_{n}=7\square$ and $V_{n}=7V_{m}\square$

Abstract

Let P be a nonzero integer and let $(U_{n})$ and $(V_{n})$ denote Lucas sequences of first and second kind 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 show that the equation $U_{n} =7\square$ has only the solution $(n,P)=(2,7\square )$ when 7|P and the equation $V_{n}=7\square$ has only the solution $(n,P)=(1,7\square )$ when 7|P or $(n,P)=(4,1)$ when $P^{2}\equiv 1( \text{ mod } 7).$ In addition, we show that the equation $V_{n}=7V_{m}\square$ has a solution if and only if $P^{2}=-3+7\square$ and $(n,m)=(3,1).$ Moreover, we show that the equation $U_{n}=7U_{m} \square$ has only the solution $(n,m,P,\square )=(8,4,1,1)$ when P is odd.

1 Introduction

Let P and Q be nonzero integers. Lucas sequence of first kind $\left( U_{n}\left( P,Q\right) \right)$ and Lucas sequence of second kind $\left( V_{n}\left( P,Q\right) \right) ,$ with parameters P and Q,  are defined as follows:

$$\begin{aligned} U_{0}\left( P,Q\right)= & {} 0,\text { }U_{1}\left( P,Q\right) =1,\text { } U_{n+1}\left( P,Q\right) =PU_{n}\left( P,Q\right) \\&+QU_{n-1}\left( P,Q\right) \text { }(n\ge 1) \end{aligned}$$

and

$$\begin{aligned} V_{0}\left( P,Q\right)= & {} 2,\text { }V_{1}\left( P,Q\right) =P,\text { } V_{n+1}\left( P,Q\right) \!=\!PV_{n}\left( P,Q\right) \\&+ QV_{n-1}\left( P,Q\right) \text { }(n\ge 1). \end{aligned}$$

Lucas sequences of first and second kind for negative subscripts are defined as

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

respectively. Assume that $P^{2}+4Q\ne 0.$ Explicitly, if $\alpha =\left( P+ \sqrt{P^{2}+4Q}\right) /2$ and $\beta =\left( P-\sqrt{P^{2}+4Q}\right) /2$ are the roots of $x^{2}-Px-Q=0,$ then we have the following well-known expressions named Binet’s formula

$$\begin{aligned} U_{n}=U_{n}\left( P,Q\right) =(\alpha ^{n}-\beta ^{n})/(\alpha -\beta )\text { and }V_{n}=V_{n}\left( P,Q\right) =\alpha ^{n}+\beta ^{n} \end{aligned}$$

for all $n\in \mathbb {Z} .$ Special cases of the sequences $(U_{n})$ and $(V_{n})$ are known under different names. The names under which they are known are Fibonacci, Pell, etc. When $Q=-1,$ we present $(U_{n})$ and $(V_{n})$ by $\left( U_{n}(P,-1)\right)$ and $\left( V_{n}(P,-1)\right) ,$ respectively. For more information about Lucas sequences of first and second kind, see [1].

Investigations of the properties of second-order linear sequences have a very long history and literature. These investigations have given rise to questions concerning whether, for certain pairs $\left( P,Q\right) , U_{n}$, or $V_{n}$ is a square $\left( =\square \right)$ or $k\square .$ There are several results when the terms of these sequences are square. Now we summarize briefly the relevant known facts. In [2, 3], Cohn determined the square and twice the squares in $(U_{n})$ and $(V_{n})$ when P is odd and $Q=\pm 1.$ Ribenboim and McDaniel [4] determined all indices n such that $U_{n}=\square ,2U_{n}=\square , V_{n}=\square ,$ and $2V_{n}=\square$ for all odd relatively prime integers P and Q. In [5], the same authors solved the equations $V_{n}=3\square$ for $P\equiv 1,3(\text{ mod } 8),Q\equiv 3(\text{ mod } 4), (P,Q)=1$ and solved the equation $U_{n}=3\square$ for all odd relatively prime integers P and Q. Bremner and Tzanakis [6] extended the result of the equation $U_{n}=\square$ by determining all sequence $(U_{n})$ with $U_{12}=\square ,$ subject only to the restriction that $(P,Q)=1.$ In a later paper, the same authors [7] showed that for $n=2,\ldots ,7$, $U_{n}$ is a square for infinitely many coprime P, Q and determined all sequences $(U_{n})$ with $U_{n}=\square , n=8,10,11.$ And also in [8], they discussed the more general problem of finding all integers n, P, Q for which $U_{n}=k\square$ for a given integer k. Cohn [9] dealt with the equations $V_{n}=V_{m}\square$ and $V_{n}=2V_{m}\square$ when P is odd and $Q=1.$ McDaniel and Ribenboim [10] dealt with the equations $V_{n}=V_{m}\square$ and $U_{n}=U_{m}\square$ for general odd Q with $(P,Q)=1.$ Şiar and Keskin [11], assuming $Q = 1$, showed that there are no integer solutions to the equations $V_{n}=3\square$ and $V_{n}=6\square$ when P is odd and they also gave solutions to the equations $V_{n}=3V_{m}\square$ and $V_{n}=6V_{m}\square .$ In [12], when $Q=1,$ Keskin and Karaatlı solved the equations $U_{n}=5\square$ and $U_{n}=5U_{m}\square$ under some assumptions on P. They solved the equations $V_{n}=5\square$ with P odd and $Q=1$ and they also showed that the equation $V_{n}=5V_{m}\square$ has no solutions. Interested readers can also consult [11] or [12] to see the survey of such results.

In this study, using congruences, with extensive reliance upon the Jacobi symbol, we will solve the equations $U_{n}=7\square ,V_{n}=7\square , U_{n}=7U_{m}\square ,$ and $V_{n}=7V_{m}\square$ under the conditions that P is odd and $Q=1.$ Our method of proof is similar to that presented by Cohn, McDaniel, and Ribenboim [2–4, 9].

2 Preliminary Facts, Some Lemmas, and Theorems

In this section, we state the necessary identities, lemmas, and theorems. Throughout the paper, unless otherwise stated, $Q=1$ and m, n are positive integers. Also, $\left( \frac{*}{*}\right)$ will denote the Jacobi symbol. The proofs of the following four theorems can be found in [13].

Theorem 1

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

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

(2.1)

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

(2.2)

Theorem 2

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

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

(2.3)

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

(2.4)

The following congruences are valid for the numbers $U_{n}=U_{n}(P,-1)$ and $V_{n}=V_{n}(P,-1).$

Theorem 3

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

$$\begin{aligned} U_{2mn+r}\equiv U_{r}\text { }(\text{ mod }\text { }U_{m}), \end{aligned}$$

(2.5)

$$\begin{aligned} V_{2mn+r}\equiv V_{r}\text { }(\text{ mod }\text { }U_{m}). \end{aligned}$$

(2.6)

Theorem 4

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

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

(2.7)

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

(2.8)

The following lemmas can be proved by using Theorems 1 and 2.

Lemma 5

$$\begin{aligned} 3|V_{n}\Leftrightarrow \left\{ \begin{array}{ll} n\equiv 1\text { }(\text{ mod }\text { }2)&{}\quad \text { if }3|P, \\ n\equiv 2\text { }(\text{ mod }\text { }4)&{}\quad \text { if }3\not \mid P. \end{array} \right. \end{aligned}$$

Lemma 6

$$\begin{aligned} 7|V_{n}\Leftrightarrow \left\{ \begin{array}{ll} 2\not \mid n&{}\quad {\text {if}}\;7|P, \\ 4|n\;{\text {and}}\;n/4\;{\text {{is}\; odd}}&{}\quad {\text {if}}\;P^{2}\equiv 1(\text{ mod } 7), \\ 4|n\;{\text {and}}\;n/4\;{\text {is\; odd}}&{}\quad {\text {if}}\;P^{2}\equiv 2(\text{ mod } 7), \\ 3|n\;{\text {and}}\;n\;{\text {is\; odd}}&{}\quad {\text {if}}\;P^{2}\equiv 4(\text{ mod } 7). \end{array} \right. \end{aligned}$$

Lemma 7

$$\begin{aligned} 7|U_{n}\Leftrightarrow \left\{ \begin{array}{ll} 2|n &{}\quad \text { if }7|P, \\ 8|n &{}\quad \text { if }P^{2}\equiv 1(\text{ mod } \text { }7), \\ 16|n &{}\quad \text { if }P^{2}\equiv 2(\text{ mod } \text { }7), \\ 6|n &{}\quad \text { if }P^{2}\equiv 4(\text{ mod } \text { }7). \end{array} \right. \end{aligned}$$

We state the following lemma without proof.

Lemma 8

All positive integer solutions to the equation $x^{2}-7y^{2}=-3$ are given by $(x,y)=\left( 2U_{m+1}(16,-1)+5U_{m}(16,-1),U_{m+1}(16,-1)-2U_{m}(16,-1)\right)$ with $m\ge 0$ or $(x,y)=\left( 5U_{m}(16,-1)+2U_{m-1}(16,-1),2U_{m}(16,-1)-U_{m-1}(16,-1)\right)$ with $m\ge 1.$

We omit the proofs of the following lemma, as it is based a straightforward induction.

Lemma 9

If n is even, then $V_{n}\equiv 2 (\text{ mod } P^{2})$ and if n is odd, then $V_{n}\equiv nP (\text{ mod } P^{2}).$

The following theorem can be obtained from Theorem 11 given in [9].

Theorem 10

Let P be an odd integer, $m\ge 1$ be an integer, and $V_{n}=V_{m}x^{2}$ for some integer x. Then $n=m$ or $n=3, m=1, P=1.$

Now we begin by listing the properties which will be used.

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

(2.9)

$$\begin{aligned} V_{2n}= & {} V_{n}^{2}-2(-1)^{n}, \end{aligned}$$

(2.10)

$$\begin{aligned} V_{3n}= & {} V_{n}\left( V_{n}^{2}-3(-1)^{n}\right) , \end{aligned}$$

(2.11)

$$\begin{aligned} \text {If }P\text { is odd and }n\ge & {} 1,\text { then }2|V_{n}\Leftrightarrow 2|U_{n}\Leftrightarrow 3|n, \end{aligned}$$

(2.12)

$$\begin{aligned} \text {If }3|P,\text { then }V_{2^{r}}\equiv & {} 2\text { }(\text{ mod }\text { }3) \text { for all positive integer }r. \end{aligned}$$

(2.13)

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

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

(2.14)

$$\begin{aligned} \text {If }r>1,\text { then }\left( \frac{U_{3}}{V_{2^{r}}}\right)= & {} 1, \end{aligned}$$

(2.15)

$$\begin{aligned} \text {If }V_{m}\ne & {} 1,\text { then }V_{m}|V_{n}\text { iff }m|n\text { and }n/m \text { is odd,} \end{aligned}$$

(2.16)

$$\begin{aligned} \text {If }U_{m}\ne & {} 1,\text { then }U_{m}|U_{n}\text { iff }m|n, \end{aligned}$$

(2.17)

$$\begin{aligned} V_{7n}= & {} \left\{ \begin{array}{ll} V_{n}\left( V_{2n}^{3}-V_{2n}^{2}-2V_{2n}+1\right) &{}\quad \text { if }n\text { is even,} \\ V_{n}\left( V_{2n}^{3}+V_{2n}^{2}-2V_{2n}-1\right) &{}\quad \text { if }n\text { is odd. } \end{array} \right. \end{aligned}$$

(2.18)

By using (2.10), we readily obtain from (2.18) that

$$\begin{aligned} V_{7n}=\left\{ \begin{array}{ll} V_{n}\left( V_{n}^{6}-7V_{n}^{4}+14V_{n}^{2}-7\right) &{}\quad \text { if }n\text { is even,} \\ V_{n}\left( V_{n}^{6}+7V_{n}^{4}+14V_{n}^{2}+7\right) &{}\quad \text { if }n\text { is odd.} \end{array} \right. \end{aligned}$$

(2.19)

If 7|P and n is odd, then $7|V_{n}$ from Lemma 6 and therefore from (2.19), it follows that

$$\begin{aligned} V_{7n}=7V_{n}\left( 7a+1\right) . \end{aligned}$$

(2.20)

If P is odd, then $V_{2^{r}}\equiv 3 (\text{ mod } 8)$ if $r=1$ and $V_{2^{r}}\equiv 7(\text{ mod }8)$ if $r>1,$ which can be found in [4]. And so, it is clear that

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

(2.21)

for $r\ge 1.$ Moreover,

$$\begin{aligned} V_{n}^{2}(P,-1)-(P^{2}-4)U_{n}^{2}(P,-1)=4. \end{aligned}$$

(2.22)

If P is even, then

$$\begin{aligned} \left. \begin{array}{c} U_{n}(P,-1)\text { is even}\Leftrightarrow n\text { is even,} \\ U_{n}(P,-1)\text { is odd}\Leftrightarrow n\text { is odd,} \\ V_{n}(P,-1)\text { is even for all }n\in \mathbb {N} .\text { } \end{array} \right. \end{aligned}$$

(2.23)

Properties between (2.11)–(2.18) can be found in [5, 14, 15]. The others are well known.

3 Main Theorems

In this section, we shall establish some theorems and lemmas which will be required later.

Theorem 11

The equation $x^{4}-7y^{2}=-3$ has no positive integer solution.

Proof

Assume that $x^{4}-7y^{2}=-3$ for some $x,y>0.$ If y is odd, then it follows that $x^{4}\equiv 4\left( \text{ mod }\text { }8\right) ,$ which is impossible. Thus, y is even and therefore x is odd. Note that the equation $x^{4}-7y^{2}=-3$ implies that

$$\begin{aligned} (x^{2})^{2}-7y^{2}=-3. \end{aligned}$$

By Lemma 8, we get $x^{2}=2U_{m+1}(16,-1)+5U_{m}(16,-1)$ or $x^{2}=5U_{m}(16,-1)+2U_{m-1}(16,-1).$ Assume that

$$\begin{aligned} x^{2}=2U_{m+1}(16,-1)+5U_{m}(16,-1). \end{aligned}$$

Since x is odd, it is seen from (2.23) that m is odd. Besides,

$$\begin{aligned} x^{2}\equiv 2U_{m+1}(16,-1)\text { }(\text{ mod }\text { }5), \end{aligned}$$

which implies that $5|U_{m+1}.$ It can be easily shown that $5|U_{m}(16,-1)$ iff 3|m. Thus, we get $m+1=3k$ for some $k>0.$ Since m is odd, k is even and therefore $k=2q.$ Hence, we have $m=6q-1$ with $q\ge 0.$ And so, by (2.5), we get

$$\begin{aligned} x^{2}&=2U_{2.3q}(16,-1)+5U_{2.3q-1}(16,-1)\\&\equiv 2U_{0}(16,-1)+5U_{-1}(16,-1) \text { }\left( \text{ mod }\text { }U_{3}(16,-1)\right) , \end{aligned}$$

implying that $x^{2}\equiv -5 \left( \text{ mod }\text { }17\right) ,$ because $17|U_{3}(16,-1).$ But this is impossible since $\left( \frac{-5}{17} \right) =-1.$

Similarly, we prove that the equation $x^{2}=5U_{m}(16-1)+2U_{m-1}(16,-1)$ is impossible. This completes the proof. $\square$

Lemma 12

The equation $9x^{4}-21y^{2}=-3$ has no positive integer solution.

Proof

Dividing both sides of the equation by 3 gives $7y^{2}-3x^{4}=1.$ Now let us consider the equation

$$\begin{aligned} 7u^{2}-3v^{2}=1. \end{aligned}$$

(3.1)

Since the fundamental solution to (3.1) is $2\sqrt{7}+3\sqrt{3},$ it can be shown that all positive integer solutions to (3.1) are given by $(u,v)=(2(U_{n+1}-U_{n}),3(U_{n+1}+U_{n})),$ where $U_{n}=U_{n}(150,-1).$ Therefore, we have $x^{2}=3(U_{n+1}+U_{n}).$ It can be easily seen that

$$\begin{aligned} U_{n}\equiv \left\{ \begin{array}{ll} n\text { }(\text{ mod }\text { }8)&{}\quad \text { if }n\text { is odd,} \\ -n\text { }(\text{ mod }\text { }8)&{}\quad \text { if }n\text { is even.} \end{array} \right. \end{aligned}$$

(3.2)

Hence, if n is odd, then $x^{2}\equiv -3n-3+3n\equiv -3(\text{ mod } 8),$ a contradiction. If n is even, then $x^{2}\equiv 3n+3-3n\equiv 3(\text{ mod } 8),$ a contradiction. This completes the proof. $\square$

We begin with the following theorem. This result will be used in the solution to the equation $U_{n}=7\square .$

Theorem 13

If P is an odd integer, then there is no integer x such that $V_{n}=14x^{2}.$

Proof

Assume $V_{n}=14x^{2}$ and P is odd. Since $2|V_{n}, 3|n$ by (2.12 ). The remainder of the proof is split into two cases.

Case I: Assume 7|P or $P^{2}\equiv 4 (\text{ mod } 7).$ Since $7|V_{n},$ it is seen from Lemma 6 that $2\not \mid n.$ Since 3|n,  we get $n=3t$ and therefore $2\not \mid t.$ Thus we can write $n=12q\pm 3.$ And so, by (2.2), we obtain $V_{n}=V_{12q\pm 3}\equiv \pm V_{3} (\text{ mod } U_{6}),$ which implies that $14x^{2}\equiv \pm 4P\equiv 4 (\text{ mod } 8).$ This shows that $x^{2}\equiv 2 (\text{ mod } 4),$ a contradiction.

Case II: Assume $P^{2}\equiv 1 (\text{ mod } 7)$ or $P^{2}\equiv 2 ( \text{ mod } 7).$ Since $7|V_{n},$ it follows that $n=4t$ for some odd t by Lemma 6. Since 3|n,  we see that 3|t and therefore $t=6q+3.$ Thus we can write $n=24q+12.$ Let $P^{2}\equiv 1 (\text{ mod } 7).$ And so, by (2.4), we get

$$\begin{aligned} V_{n}=V_{24q+12}=V_{2\cdot 2\cdot 6q+12}\equiv V_{12}\equiv -V_{0}\equiv -2 \text { }(\text{ mod } V_{2}), \end{aligned}$$

which implies that $14x^{2}\equiv -2 (\text{ mod } P^{2}+2).$ Hence, we obtain $1=\left( \frac{-7}{P^{2}+2}\right)$ . But this is impossible since

$$\begin{aligned} \left( \frac{-7}{P^{2}+2}\right) =(-1)^{\frac{P^{2}+1}{2}}(-1)^{\frac{P^{2}+1 }{2}}\left( \frac{P^{2}+2}{7}\right) =\left( \frac{3}{7}\right) =\left( \frac{-4}{7}\right) =-1. \end{aligned}$$

Now let $P^{2}\equiv 2(\text{ mod } 7).$ Since $n=24q+12,$ it follows from ( 2.2) that

$$\begin{aligned} V_{n}=V_{24q+12}=V_{2\cdot 3(4q+2)}\equiv V_{0}\equiv 2\text { }(\text{ mod } U_{3}), \end{aligned}$$

which implies that $14x^{2}\equiv 2 (\text{ mod } P^{2}+1),$ i.e., $7x^{2}\equiv 1 (\text{ mod }(P^{2}+1)/2).$ But this is also impossible since

$$\begin{aligned} \left( \frac{7}{(P^{2}+1)/2}\right) =(-1)^{\frac{P^{2}-1}{4}}\left( \frac{ (P^{2}+1)/2}{7}\right) =\left( \frac{5}{7}\right) =\left( \frac{-2}{7} \right) =-1. \end{aligned}$$

This completes the proof. $\square$

Theorem 14

Let P be odd. If 7|P,  then $V_{n}=7x^{2}$ is possible if and only if $(P,n)=(7\square ,1).$ If $7\not \mid P,$ then $V_{n}=7x^{2}$ is impossible, except for the case $(P,n)=(1,4).$

Proof

Assume that $V_{n}=7x^{2}, 7|P$ and P is odd. Then by Theorem 3.4 given in [11], we get $n=1$ or $n=3.$ If $n=1,$ then $V_{1}=P=7x^{2}$ is a solution. If $n=3,$ then $V_{3}=P(P^{2}+3)=7x^{2}.$ Since 7|P,  it follows that $\left( P/7\right) \left( P^{2}+3\right) =x^{2}.$ Clearly, $d=\left( P/7,P^{2}+3\right) =1$ or 3.

Let $d=1.$ Then, $P^{2}+3=b^{2}$ gives $3=(b-P)(b+P)$ and the only positive integer solution is $(b,P)=(2,1),$ which is not convenient since we must have 7|P.

Let $d=3.$ Then we have

$$\begin{aligned} P=21a^{2}\text { and }P^{2}+3=3b^{2}. \end{aligned}$$

(3.3)

It is seen from (3.3) that 3|P and therefore

$$\begin{aligned} P=3c \end{aligned}$$

(3.4)

for some $c>0.~$Substituting (3.4) into (3.3), we immediately have the Pell equation $b^{2}-3c^{2}=1.$ It is well known that all positive integer solutions to the equation $b^{2}-3c^{2}=1$ are given by $\left( b,c\right) =\left( V_{m}\left( 4,-1\right) /2,U_{m}\left( 4,-1\right) \right)$ with $m\ge 1.$ On the other hand, since $3c=21a^{2},$ we get $c=7a^{2}.$ So, we are looking for the solutions $U_{m}\left( 4,-1\right) =7\square .$ Since $7|U_{4}(4,-1),$ it can be easily shown that $7|U_{m}(4,-1)$ iff $m=4k$ for some $k\ge 1.$ Then by (2.9), it follows that

$$\begin{aligned} 7\square =U_{4k}\left( 4,-1\right) =U_{2k}\left( 4,-1\right) V_{2k}\left( 4,-1\right) . \end{aligned}$$

From (2.14) and (2.23), it is seen that $\left( U_{2k}\left( 4,-1\right) ,V_{2k}\left( 4,-1\right) \right) =2.$ Then either

$$\begin{aligned} U_{2k}\left( 4,-1\right) =2u^{2}\text { and }V_{2k}\left( 4,-1\right) =14v^{2} \end{aligned}$$

(3.5)

or

$$\begin{aligned} U_{2k}\left( 4,-1\right) =14u^{2}\text { and }V_{2k}\left( 4,-1\right) =2v^{2}. \end{aligned}$$

(3.6)

From now on and until the end of this paragraph, instead of $U_{n}(4,-1)$ and $V_{n}(4,-1),$ we will write $U_{n}$ and $V_{n},$ respectively. Suppose ( 3.5) is satisfied. Clearly, $7|V_{2k}.$ Since $7|V_{2},$ it can be easily shown that k is odd. Let $k=4q\pm 1.$ By (2.5), we get

$$\begin{aligned} 2u^{2}=U_{2(4q\pm 1)}\equiv U_{\pm 2}\text { }(\text{ mod }U_{4}). \end{aligned}$$

Since $8|U_{4},$ the previous congruence becomes $2u^{2}\equiv \pm 4 ( \text{ mod } 8),$ which is impossible. Suppose (3.6) is satisfied. We show that if $V_{n}=2v^{2},$ then 3|n. Let $n=6q+r, 0\le r\le 5.$ Then by (2.6), it follows that $V_{n}\equiv V_{r} (\text{ mod } U_{3}),$ implying that $2v^{2}\equiv V_{r} (\text{ mod } 5)$ since $5|U_{3}.$ From this, it follows that $r=0$ or 3. This shows that 3|n. Returning to the equation $V_{2k}=2v^{2},$ we have $k=3r.$ Thus $V_{6r}=V_{3\cdot 2r}=V_{2r}(V_{2r}^{2}-3)=2v^{2}$ by (2.11). This implies that $v^{2}= \frac{V_{2r}}{2}(V_{2r}^{2}-3).$ On the other hand, since $V_{n}^{2}-12U_{n}^{2}=4$ by (2.22), we see that $3\not \mid V_{n}.$ Thus, $\left( \frac{V_{2r}}{2},V_{2r}^{2}-3\right) =1.$ Then, we have $V_{2r}^{2}-3=a^{2},$ which is impossible.

Now we consider the case $P^{2}\equiv 1 (\text{ mod } 7).$ Since $7|V_{n},$ it follows from Lemma 6 that $n=4t$ for some odd integer t. Let $t>1.$ We can write $t=4q\pm 1$ with $q\ge 1$ and therefore $n=4t=2\cdot 2^{r}a\pm 4,$ with a odd and $r\ge 3.$ Thus by (2.4), we get $V_{n}\equiv -V_{4} (\text{ mod } V_{2^{r}}).$ If $r=3,$ then

$$\begin{aligned} 7x^{2}\equiv -V_{4}\text { }(\text{ mod }\text { }V_{8}) \end{aligned}$$

(3.7)

and if $r>3,$ then

$$\begin{aligned} 7x^{2}\equiv -V_{4}\text { }(\text{ mod }\text { }V_{2^{r}}). \end{aligned}$$

(3.8)

Since $V_{8}=V_{4}^{2}-2$ by (2.10), it follows that $V_{8}\equiv -2 (\text{ mod } V_{4})$ and therefore $V_{2^{r}}\equiv 2 (\text{ mod } V_{4}).$ Note that $V_{4}=P^{4}+4P^{2}+2.$ Since $P^{2}\equiv 1 (\text{ mod } 7),$ we see that $7|V_{4}$ and therefore by (2.10), we have $V_{8}\equiv -2 (\text{ mod } 7)$ and $V_{2^{r}}\equiv 2 (\text{ mod } 7).$ Besides, since P is odd, it follows that $V_{4}\equiv 7 (\text{ mod } 8)$ and $V_{8}\equiv 7 (\text{ mod } 8).$ Also, $\left( \frac{-1}{ V_{2^{r}}}\right) =-1$ by (2.21). Assume that $r=3,$ so that, by (3.7 ), we have $\left( \frac{7}{V_{8}}\right) =\left( \frac{-1}{V_{8}}\right) \left( \frac{V_{4}}{V_{8}}\right) .$ But this is impossible since

$$\begin{aligned} \left( \frac{7}{V_{8}}\right) =(-1)\left( \frac{V_{8}}{7}\right) =(-1)(\frac{ -2}{7})=1,\left( \frac{-1}{V_{8}}\right) =-1 \end{aligned}$$

and

$$\begin{aligned} \left( \frac{V_{4}}{V_{8}}\right) =(-1)\left( \frac{V_{8}}{V_{4}}\right) =(-1)\left( \frac{-2}{V_{4}}\right) =\left( \frac{2}{V_{4}}\right) =1, \end{aligned}$$

because $V_{4}+2=(P^{2}+2)^{2}.$

Now assume that $r>3,$ so that, (3.8) is satisfied. Then, it follows that $\left( \frac{7}{V_{2^{r}}}\right) =\left( \frac{-1}{V_{2^{r}}}\right) \left( \frac{V_{4}}{V_{2^{r}}}\right) .$ But this is also impossible since

$$\begin{aligned} \left( \frac{7}{V_{2^{r}}}\right) =(-1)\left( \frac{V_{2^{r}}}{7}\right) =(-1)\left( \frac{2}{7}\right) =-1,\left( \frac{-1}{V_{2^{r}}}\right) =-1 \end{aligned}$$

and

$$\begin{aligned} \left( \frac{V_{4}}{V_{2^{r}}}\right) =(-1)\left( \frac{V_{2^{r}}}{V_{4}} \right) =(-1)\left( \frac{2}{V_{4}}\right) =-1. \end{aligned}$$

Hence, we conclude that $t=1.$ Then $n=4$ and therefore $V_{4}=(P^{2}+2)^{2}-2=V_{n}=7x^{2}.$ Now, we consider the equation $u^{2}-7v^{2}=2$ with $u=P^{2}+2.$ It can be easily shown that all positive integer solutions to the equation $u^{2}-7v^{2}=2$ are given by

$$\begin{aligned} (u,v)=\left( 3(U_{m+1}(16,-1)-U_{m}(16,-1)),\text { } 17U_{m}(16,-1)-U_{m-1}(16,-1)\right) \end{aligned}$$

with $m\ge 0.$ And so, we get $P^{2}+2=3(U_{m+1}(16,-1)-U_{m}(16,-1)).$ From now on and until the end of the case $P^{2}\equiv 1 (\text{ mod } 7),$ instead of $U_{m}(16,-1),$ we will write $U_{m}.$ Let $m=4q+r, 0\le r\le 3.$ Then by (2.5), it follows that $U_{4q+r}\equiv U_{r} (\text{ mod } U_{2}),$ implying that $P^{2}+2\equiv 3(U_{r+1}-U_{r}) (\text{ mod } 16)$ since $16|U_{2}.$ A simple calculation shows that $r=0$ and therefore 4|m. So, we can write $m=12q, 12q+4$ or $12q+8.$ If $m=12q+4,$ then we obtain $P^{2}+2\equiv 3(U_{12q+5}-U_{12q+4})\equiv 3(U_{5}-U_{4}) (\text{ mod } U_{3})$ by (2.5). Since $5|U_{3},$ we immediately have $P^{2}+2\equiv 0 (\text{ mod } 5),$ which is impossible. Now the remainder of the proof is split into two cases.

Case I: Let $m=12q$ with $q\ge 0.$ If $q>0,$ then we can write $m=12q=2\cdot 2^{r}\cdot 3a,$ with a odd and $r\ge 1.$ Thus by (2.7 ), we get

$$\begin{aligned} P^{2}+2=3(U_{2\cdot 2^{r}\cdot 3a+1}-U_{2\cdot 2^{r}\cdot 3a}) \equiv -3\text { }(\text{ mod } \text { }V_{2^{r}}), \end{aligned}$$

leading to

$$\begin{aligned} P^{2}\equiv -5\text { }(\text{ mod } \text { }V_{2^{r}}/2). \end{aligned}$$

(3.9)

If $r\ge 2,$ then a simple calculation shows that $V_{2^{r}}\equiv 2 ( \text{ mod } 8)$ and $V_{2^{r}}\equiv -1 (\text{ mod } 5).$ Thus, $V_{2^{r}}/2\equiv 1 (\text{ mod } 4)$ and $V_{2^{r}}/2\equiv 2 (\text{ mod } 5).$ From (3.9), it is seen that $1=\left( \frac{-1}{V_{2^{r}}/2} \right) \left( \frac{5}{V_{2^{r}}/2}\right) .$ But this is impossible since

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

and

$$\begin{aligned} \left( \frac{5}{V_{2^{r}}/2}\right) =\left( \frac{V_{2^{r}}/2}{5}\right) =\left( \frac{2}{5}\right) =-1. \end{aligned}$$

Hence, we get $r=1.$ By (2.7), it follows that $P^{2}+2=3(U_{2\cdot 6a+1}-U_{2\cdot 6a}) \equiv -3 (\text{ mod } V_{6}),$ i.e., $P^{2}\equiv -5 (\text{ mod } V_{6}/2).$ This implies that $1=\left( \frac{ -1}{V_{6}/2}\right) \left( \frac{5}{V_{6}/2}\right) .$ Using the fact that $V_{6}/2\equiv 1 (\text{ mod } 5)$ and $V_{6}/2\equiv 3 (\text{ mod } 4),$ we readily obtain

$$\begin{aligned} 1=\left( \frac{-1}{V_{6}/2}\right) \left( \frac{5}{V_{6}/2}\right) =(-1)\left( \frac{V_{6}/2}{5}\right) =(-1)\left( \frac{1}{5}\right) =-1, \end{aligned}$$

a contradiction. Thus, we get $q=0.$ Then $m=0$ and therefore $P^{2}+2=3.$ This gives that $P=1.$

Case II: Let $m=12q+8$ with $q\ge 0.$ This implies that $m=12u-4.$ Then by (2.7), we get

$$\begin{aligned} P^{2}+2=3(U_{2\cdot 3\cdot 2q-3}-U_{2\cdot 3\cdot 2q-4})\equiv 3(U_{-3}-U_{-4})\equiv 3(U_{4}-U_{3})(\text{ mod } \text { }V_{3}). \end{aligned}$$

A simple calculation shows that $11|V_{3}, U_{4}\equiv 5 (\text{ mod } 11),$ and $U_{3}\equiv 2 (\text{ mod } 11).$ Thus, it is seen that $P^{2}\equiv 7 (\text{ mod } 11),$ which is impossible since $\left( \frac{7 }{11}\right) =\left( \frac{-4}{11}\right) =-1.$

Assume that $P^{2}\equiv 2 (\text{ mod } 7).$ Since $7|V_{n},$ it follows from Lemma 6 that $n=4t$ for some odd integer t. Similar arguments used for the case when $P^{2}\equiv 1 (\text{ mod } 7)$ show that $P=1.$ But this is impossible since $P^{2}\equiv 2 (\text{ mod } 7).$

Assume that $P^{2}\equiv 4 (\text{ mod } 7).$ Since $7|V_{n},$ it follows that $n=3t$ for some odd integer t by Lemma 6. Hence, $V_{n}=V_{3t}=V_{t}(V_{t}^{2}+3)$ from (2.11). Clearly, $\left( V_{t},V_{t}^{2}+3\right) =1$ or 3.

Let $\left( V_{t},V_{t}^{2}+3\right) =1.$ Then either

$$\begin{aligned} V_{t}=a^{2},\text { }V_{t}^{2}+3=7b^{2} \end{aligned}$$

(3.10)

or

$$\begin{aligned} V_{t}=7a^{2},\text { }V_{t}^{2}+3=b^{2} \end{aligned}$$

(3.11)

for some positive integers a and b. But the two relations (3.11) lead to $b^{2}\equiv 3 (\text{ mod } 7)$ which is impossible, hence (3.10) is satisfied. Solving the systems of equations $V_{t}=a^{2}, V_{t}^{2}+3=7b^{2}$ gives $a^{4}-7b^{2}=-3,$ which has no positive integer solution by Theorem 11.

Let $\left( V_{t},V_{t}^{2}+3\right) =3.$ This implies that either

$$\begin{aligned} V_{t}=3a^{2},\text { }V_{t}^{2}+3=21b^{2} \end{aligned}$$

(3.12)

or

$$\begin{aligned} V_{t}=21a^{2},\text { }V_{t}^{2}+3=3b^{2} \end{aligned}$$

(3.13)

for some $a,b>0.$ Assume that (3.12) is satisfied. Then we get $9a^{4}-21b^{2}=-3,$ which has no positive integer solution by Lemma 12. Now assume that (3.13) is satisfied. Since $3|V_{t}$ and t is odd, it follows from Lemma 5 that 3|P. On the other hand, it is seen that $V_{t}^{2}=V_{2t}-2$ by (2.10). Combining the equation $V_{t}^{2}=V_{2t}-2$ with $V_{t}^{2}+3=3b^{2}$ gives $V_{2t}=3b^{2}-1.$ Let $t>1.$ Then we can write $t=4q\pm 1=2^{r}z\pm 1$ with z odd and $r\ge 2.$ And so by (2.4), we get

$$\begin{aligned} V_{2t}=V_{2\cdot 2^{r}z\pm 2}\equiv -V_{2}\text { }\left( \text{ mod } \text { } V_{2^{r}}\right) , \end{aligned}$$

implying that

$$\begin{aligned} 3b^{2}\equiv -\left( P^{2}+2-1\right) \equiv -U_{3}\text { }\left( \text{ mod } \text { }V_{2^{r}}\right) . \end{aligned}$$

This means that

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

We have $\left( \frac{-1}{V_{2^{r}}}\right) =-1$ and $\left( \frac{U_{3}}{ V_{2^{r}}}\right) =1$ by (2.21) and (2.15), respectively. On the other hand, $V_{2^{r}}\equiv 2 \left( \text{ mod } \text { }3\right)$ by (2.13), leading to

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

Therefore,

$$\begin{aligned} \left( \frac{-3U_{3}}{V_{2^{r}}}\right) =\left( \frac{-1}{V_{2^{r}}}\right) \left( \frac{3}{V_{2^{r}}}\right) \left( \frac{U_{3}}{V_{2^{r}}}\right) =\left( -1\right) \left( 1\right) \left( 1\right) =-1, \end{aligned}$$

which contradicts the displayed relation a few lines above. Hence, $t=1$ and therefore $V_{1}=P=21\square .$ But this contradicts the fact that $P^{2}\equiv 4 \left( \text{ mod } \text { }7\right) .$ This completes the proof. $\square$

Theorem 15

If P is odd, then, a relation of the form $V_{n}=7V_{m}x^{2}$ with $V_{m}\ne 1,$ is possible if and only if $P^{2}=-3+7\square ,$ (hence P is given by Lemma 8) and $(n,m)=(3,1).$ If P is even and $P^{2}\equiv 2(\text{ mod } 7),$ then a relation of the form $V_{n}=7V_{m}x^{2}$ with $V_{m}\ne 1,$ is impossible.

Proof

The strategy of the proof is as follows. Let $V_{n}=7V_{m}x^{2},$ where $V_{m}\ne 1.$ Under the assumption that, either $P\equiv 0,\pm 1(\text{ mod } 7)$ and P is odd, or $P^{2}\equiv 2(\text{ mod } 7)$ (independently of the parity of P), we will arrive at a contradiction. Finally, if $P^{2}\equiv 4(\text{ mod } 7)$ and P is odd, we will prove that $(n,m)=(3,1).$ In the last case, $V_{3}=7V_{1}x^{2},$ which is equivalent to $P^{2}-7x^{2}=-3;$ hence P is obtained by applying Lemma 8.

Case I: Assume that 7|P and $V_{n}=7V_{m}x^{2}.$ Since $7|V_{n},$ it follows from Lemma 6 that $n\ge 3$ is odd. Furthermore, since $V_{m}|V_{n},$ there exists an odd integer t such that $n=mt$ by (2.16 ). Thus, m is odd. Therefore, we have $V_{n}\equiv nP \left( \text{ mod } \text { }P^{2}\right)$ and $V_{m}\equiv mP \left( \text{ mod } \text { } P^{2}\right)$ by Lemma 9. This shows that $nP\equiv 7mPx^{2} \left( \text{ mod } \text { }P^{2}\right) ,$ i.e., $n\equiv 7mx^{2} \left( \text{ mod } \text { }P\right) .$ Since 7|P,  it is obvious that 7|n. Since 7|n and $n=mt,$ it is seen that 7|mt.

Assume that 7|t. Then $t=7s$ for some odd positive integer s and therefore $n=mt=7ms.$ By (2.18), we immediately have

$$\begin{aligned} 7V_{m}x^{2}= & {} V_{n}=V_{7ms}=V_{ms}\left( V_{2ms}^{3}+V_{2ms}^{2}-2V_{2ms}-1\right) \\= & {} V_{ms}(V_{ms}^{6}+7V_{ms}^{4}+14V_{ms}^{2}+7), \end{aligned}$$

by (2.18). This implies that 7 divides the parenthesis, i.e.,

$$\begin{aligned} 7|\left( V_{2ms}^{3}+V_{2ms}^{2}-2V_{2ms}-1\right) . \end{aligned}$$

Hence, we get $x^{2}=\frac{V_{ms}}{V_{m}}\left( \frac{ V_{2ms}^{3}+V_{2ms}^{2}-2V_{2ms}-1}{7}\right) .$ We have,

$$\begin{aligned} \left( \frac{V_{ms}}{V_{m}},\frac{V_{2ms}^{3}\!+\!V_{2ms}^{2}-2V_{2ms}-1}{7} \right) \!=\!\left( \frac{V_{ms}}{V_{m}} ,(V_{ms}^{6}\!+\!7V_{ms}^{4}\!+\!14V_{ms}^{2}+7)/7)\right) =1. \end{aligned}$$

Then

$$\begin{aligned} V_{ms}=V_{m}a^{2}\text { and }V_{2ms}^{3}+V_{2ms}^{2}-2V_{2ms}-1=7b^{2} \end{aligned}$$

for some $a,b>0.$ By Theorem 10, we have $ms=3, m=1,$ and $P=1$ or $ms=m.$ If $m=1$ and $P=1,$ then, we see that $V_{m}=V_{1}=P=1,$ which is impossible since $V_{m}\ne 1.$ If $ms=m,$ then, $s=1.$ Since $n=7ms,$ we get $n=7m.$ By Theorem 2, it follows that

$$\begin{aligned} V_{n}=V_{7m}=V_{8m-m}=V_{2\cdot 4m-m}\equiv -V_{-m}\text { }(\text{ mod } V_{4}), \end{aligned}$$

implying that

$$\begin{aligned} 7V_{m}x^{2}\equiv V_{m}\text { }(\text{ mod } V_{4}). \end{aligned}$$

Since $V_{4}$ is odd, it follows by (2.14) that $(U_{2m},V_{4})=1.$ But $U_{2m}=U_{m}V_{m},$ by (2.9), hence $(V_{m},V_{4})=1.$ Therefore, the congruence becomes $7x^{2}\equiv 1 (\text{ mod } V_{4}).$ Using the fact that 7|P,  we have

$$\begin{aligned} 1=\left( \frac{7}{V_{4}}\right) =(-1)^{\frac{V_{4}-1}{2}}\left( \frac{V_{4}}{ 7}\right) =(-1)\left( \frac{2}{7}\right) =-1, \end{aligned}$$

a contradiction.

Now assume that $7\not \mid t,$ so that 7|m. So, we can write $m=7^{r}a$ with $7\not \mid a$ and $r\ge 1.$ By (2.20), we get $V_{m}=V_{7^{r}a}=7V_{7^{r-1}a}\left( 7a_{1}+1\right)$ for some positive integer $a_{1}.$ Thus, we conclude that

$$\begin{aligned} V_{m}=V_{7^{r}a}=7^{r}V_{a}\left( 7a_{1}+1\right) \left( 7a_{2}+1\right) \cdot \cdot \cdot \left( 7a_{r}+1\right) \end{aligned}$$

for some $a_{i}>0$ with $1\le i\le r.$ Let $A=\left( 7a_{1}+1\right) \left( 7a_{2}+1\right) \cdot \cdot \cdot \left( 7a_{r}+1\right) .$ As a consequence, $V_{m}=7^{r}V_{a}A.$ It is clear that $7\not \mid A.$ In a similar way, we see that

$$\begin{aligned} V_{n}=V_{7^{r}at}=7^{r}V_{at}\left( 7b_{1}+1\right) \left( 7b_{2}+1\right) \cdot \cdot \cdot \left( 7b_{r}+1\right) \end{aligned}$$

for some $b_{j}>0$ with $1\le j\le r.$ Thus, we have $V_{n}=7^{r}V_{at}B,$ where $B=\left( 7b_{1}+1\right) \left( 7b_{2}+1\right) \cdot \cdot \cdot \left( 7b_{r}+1\right) .$ Clearly, $7\not \mid B.$ Upon substituting the values of $V_{n}$ and $V_{m}$ into $V_{n}=7V_{m}x^{2},$ we obtain

$$\begin{aligned} 7^{r}V_{at}B=7\cdot 7^{r}V_{a}Ax^{2}, \end{aligned}$$

implying that $V_{at}B=7V_{a}Ax^{2}.$ By Lemma 9, it is seen that $atPB\equiv 7aPAx^{2} \left( \text{ mod }\text { }P^{2}\right) ,$ which gives that $atB\equiv 7aAx^{2} \left( \text{ mod } \text { }P\right) .$ Since 7|P,  it follows that 7|atB. But this is impossible since $7\not \mid a, 7\not \mid t,$ and $7\not \mid B.$

Case II:  Assume that $P^{2}\equiv 1 (\text{ mod } 7).$ From Lemma 6, it is seen that $n=4t$ for some odd positive integer t. Therefore, we can write $n=12q$ for some odd q or $n=12q\pm 4$ for some even q. Firstly, let $n=12q$. And so by (2.2), we get

$$\begin{aligned} V_{n}=V_{12q}=V_{2\cdot 6q}\equiv V_{0}\text { }\left( \text{ mod } \text { } U_{6}\right) . \end{aligned}$$

Since $U_{6}=P^{5}+4P^{3}+3P$ and P is odd, it is easily seen that $8|U_{6}.$ Hence, we have $V_{n}\equiv 2\left( \text{ mod } \text { }8\right) .$ Secondly, let $n=12q\pm 4.$ Then we immediately have from (2.2) that

$$\begin{aligned} V_{n}=V_{12q\pm 4}=V_{2\cdot 6q\pm 4}\equiv V_{\pm 4}\text { }\left( \text{ mod } \text { }U_{6}\right) , \end{aligned}$$

implying that $V_{n}\equiv V_{4} \left( \text{ mod } \text { }8\right) .$ Using the fact that $V_{4}=P^{4}+4P^{2}+2$ and P is odd, we obtain $V_{4}\equiv 7 \left( \text{ mod } \text { }8\right)$ in this case. Hence, we conclude that $V_{n}\equiv 2,7 \left( \text{ mod } \text { }8\right) .$ On the other hand, since $V_{m}|V_{n},$ we get $n=ms$ for some odd s by (2.16). It is known that 4|n and s is odd. Hence, we see that 4|m and therefore $m=4u$ for some odd u. And so, with arguments similar to those a few lines above, we have $V_{m}\equiv 2,7 \left( \text{ mod } \text { } 8\right) .$ Thus, $7V_{m}\equiv 14,49\equiv 6,1 \left( \text{ mod } \text { } 8\right) .$ As a consequence,

$$\begin{aligned} V_{n}=7V_{m}x^{2}\equiv 1,6\left\{ \begin{array}{c} 0 \\ 1 \\ 4 \end{array} \right\} \equiv 0,1,4,6\text { }\left( \text{ mod } \text { }8\right) . \end{aligned}$$

But this contradicts the fact that $V_{n}\equiv 2,7 \left( \text{ mod } \text { }8\right) .$

Case III: Assume that $P^{2}\equiv 2 \left( \text{ mod } \text { }7\right) .$ Since $7|V_{n},$ it follows from Lemma 6 that $n=4t$ for some odd t. Furthermore, since $V_{m}|V_{n},$ there exists an odd integer $s(>1)$ such that $n=ms$ by (2.16). Thus, we can write $s=4q\pm 1$ for some $q\ge 1.$ Since 4|n and s is odd, it is seen that m is even and also 4|m. Upon substituting $n=ms$ and $s=4q\pm 1$ into $V_{n},$ we get

$$\begin{aligned} V_{n}=V_{ms}=V_{m(4q\pm 1)}=V_{2\cdot 2mq\pm m}\equiv V_{m}\text { }\left( \text{ mod } \text { }U_{2m}\right) \end{aligned}$$

by (2.2). This implies that $7V_{m}x^{2}\equiv V_{m} \left( \text{ mod } \text { }U_{m}V_{m}\right)$ by (2.9). Dividing both sides of the congruence by $V_{m}$ gives $7x^{2}\equiv 1\left( \text{ mod } \text {~} U_{m}\right) .$ Since 4|m,  it is clear from (2.17) that $U_{4}|U_{m}.$ Since $U_{4}=U_{2}V_{2}$ by (2.9), the preceding congruence becomes $7x^{2}\equiv 1 \left( \text{ mod } \text { }V_{2}\right) ,$ i.e.,

$$\begin{aligned} 7x^{2}\equiv 1\text { }\left( \text{ mod } \text { }P^{2}+2\right) . \end{aligned}$$

(3.14)

Assume that P is odd. Using $P^{2}\equiv 2 \left( \text{ mod } \text { } 7\right) ,$ we get from (3.14) that

$$\begin{aligned} 1=\left( \frac{7}{P^{2}+2}\right) =\left( -1\right) ^{\frac{P^{2}+1}{2} }\left( \frac{P^{2}+2}{7}\right) =\left( -1\right) \left( \frac{4}{7}\right) =-1, \end{aligned}$$

a contradiction. Now assume that P is even. If 2|P and $4\not \mid P,$ then from (3.14), we obtain $7x^{2}\equiv 1 \left( \text{ mod } \text { }\frac{P^{2}+2}{2}\right) ,$ implying that $1=\left( \frac{7}{\left( P^{2}+2\right) /2}\right) .$ Since $P^{2}\equiv 2 \left( \text{ mod } \text { } 7\right) ,$ it can be seen that $\left( P^{2}+2\right) /2\equiv 2 \left( \text{ mod } \text { }7\right) .$ Hence, we get

$$\begin{aligned} 1=\left( \frac{7}{\left( P^{2}+2\right) /2}\right) =\left( -1\right) ^{\frac{ P^{2}}{4}}\left( \frac{\left( P^{2}+2\right) /2}{7}\right) \left( -1\right) \left( \frac{2}{7}\right) =-1, \end{aligned}$$

a contradiction. For the case when 4|P,  we turn back again to the congruence $7x^{2}\equiv 1 \left( \text{ mod } \text { }U_{4}\right) .$ Since $P|U_{4}$ we have $7x^{2}\equiv 1 \left( \text{ mod } \text { }P\right) ,$ which implies that $7x^{2}\equiv 1 \left( \text{ od } \text { }4\right) ,$ a contradiction.

Case IV: Assume that $P^{2}\equiv 4 \left( \text{ mod } \text { }7\right) .$ Since $7|V_{n},$ it follows from Lemma 6 that $n=3t$ for some odd positive integer t. Moreover, since $V_{m}|V_{n},$ it is obvious that $n=ms$ for some odd $s(>1)$ by (2.16). And so, we can write $s=4q\pm 1$ with $q\ge 1.$ Thus, we get $n=ms=4qm\pm m.$ From now on, we divide the proof into two subcases.

Subcase i Let 3|m. Then by (2.17), $U_{3}|U_{m}.$ Substituting $n=4qm\pm m$ into $V_{n}$ and using (2.2) and (2.9), we obtain $V_{n}=V_{4qm\pm m}=V_{2\cdot 2mq\pm m}\equiv V_{\pm m}\left( \text{ mod } \text { }U_{m}V_{m}\right) ,$ i.e., $7x^{2}\equiv \pm 1 \left( \text{ mod } \text { } U_{m}\right) .$ Since $U_{3}|U_{m}$ and $U_{3}=P^{2}+1,$ we conclude that

$$\begin{aligned} 7x^{2}\equiv \pm 1\text { }\left( \text{ mod } \text { }P^{2}+1\right) . \end{aligned}$$

(3.15)

It is clear from (3.15) that $7x^{2}\equiv \pm 1 \left( \text{ mod } \text { }\frac{P^{2}+1}{2}\right) .$ Let $7x^{2}\equiv 1 \left( \text{ mod } \text { } \frac{P^{2}+1}{2}\right) .$ This shows that $1=\left( \frac{7}{\left( P^{2}+1\right) /2}\right) .$ Since $P^{2}\equiv 4 \left( \text{ mod } \text { } 7\right) ,$ it follows that $\frac{P^{2}+1}{2}\equiv -1 \left( \text{ mod } \text { }7\right) .$ Hence, we get

$$\begin{aligned} 1=\left( \frac{7}{\left( P^{2}+1\right) /2}\right) =\left( -1\right) ^{\frac{ P^{2}-1}{4}}\left( \frac{\left( P^{2}+1\right) /2}{7}\right) =\left( \frac{-1 }{7}\right) =-1, \end{aligned}$$

a contradiction. Similarly $7x^{2}\equiv -1 \left( \text{ mod } \text { }\frac{ P^{2}+1}{2}\right)$ leads to a contradiction.

Subcase ii Let $3\not \mid m.$ Since 3|n and $n=ms,$ it follows that 3|s and therefore $s=3k$ for some odd k. Thus, we get $n=ms=3mk.$ Substituting this into $V_{n}$ gives $V_{n}=V_{3mk}=V_{mk}\left( V_{mk}^{2}+3\right)$ by (2.11). This implies that $7V_{m}x^{2}=V_{mk}\left( V_{mk}^{2}+3\right) ,$ i.e., $7x^{2}=\frac{V_{mk}}{V_{m}}\left( V_{mk}^{2}+3\right) .$ Clearly, $d=\left( \frac{V_{mk}}{V_{m}} ,V_{mk}^{2}+3\right) =1$ or 3.

Let $d=1.$ Then either

$$\begin{aligned} V_{mk}=V_{m}a^{2},\text { }V_{mk}^{2}+3=7b^{2} \end{aligned}$$

(3.16)

or

$$\begin{aligned} V_{mk}=7V_{m}a^{2},\text { }V_{mk}^{2}+3=b^{2} \end{aligned}$$

(3.17)

for some $a,b>0.$ We immediately see that (3.17) does not hold. Because the only possible value of $V_{mk}$ for which $V_{mk}^{2}+3=b^{2}$ is $V_{mk}=1,$ which is impossible. Assume that (3.16) holds. Then by Theorem 10, we obtain $mk=3, m=1,$ and $P=1$ or $mk=m.$ If $mk=3$ and $P=1,$ then, $V_{mk}^{2}+3=V_{3}^{2}+3=(P^{3}+3P)^{2}+3 =19=7b^{2},$ which is impossible. If $mk=m,$ then, $k=1.$ So, it is sufficient to consider the equation $V_{m}^{2}+3=7b^{2}.$ From (2.10), it follows that $V_{2m}+1=7b^{2}.$ Assume that $m>1.$ Since m is odd, we can write $2m=2\left( 4q\pm 1\right) =2\left( 2^{r}z\right) \pm 2$ with z odd and $r\ge 2.$ Then by (2.4),

$$\begin{aligned} V_{2m}=V_{2\cdot 2^{r}z\pm 2}\equiv -V_{2}\equiv -\left( P^{2}+2\right) \text { }\left( \text{ mod } \text { }V_{2^{r}}\right) , \end{aligned}$$

implying that

$$\begin{aligned} 7b^{2}\equiv -\left( P^{2}+2-1\right) \equiv -U_{3}\text { }\left( \text{ mod } \text { }V_{2^{r}}\right) . \end{aligned}$$

This means that $1=\left( \frac{-7U_{3}}{V_{2^{r}}}\right) .$ We have $\left( \frac{-1}{V_{2^{r}}}\right) =-1$ and $\left( \frac{U_{3}}{V_{2^{r}}} \right) =1$ by (2.21) and (2.15), respectively. On the other hand, it is easy to see that $V_{2^{r}}\equiv 6 \left( \text{ mod } \text { }7\right)$ when $P^{2}\equiv 4 \left( \text{ mod } \text { }7\right) .$ Thus, we get

$$\begin{aligned} \left( \frac{7}{V_{2^{r}}}\right) =\left( -1\right) ^{\frac{V_{2^{r}}-1}{2} }\left( \frac{V_{2^{r}}}{7}\right) =\left( -1\right) \left( \frac{6}{7} \right) =-1\left( \frac{-1}{7}\right) =1. \end{aligned}$$

This concludes that $1=\left( \frac{-7U_{3}}{V_{2^{r}}}\right) =\left( \frac{ -1}{V_{2^{r}}}\right) \left( \frac{7}{V_{2^{r}}}\right) \left( \frac{U_{3}}{ V_{2^{r}}}\right) =(-1)(1)(1)=-1,$ a contradiction. Hence, $m=1$ and therefore $n=3m=3.$ It is clear that $m=1$ and $n=3$ is a solution by Lemma 8.

Let $d=3.$ Then, we obtain

$$\begin{aligned} V_{mk}=3V_{m}a^{2},\text { }V_{mk}^{2}+3=21b^{2} \end{aligned}$$

(3.18)

or

$$\begin{aligned} V_{mk}=21V_{m}a^{2},\text { }V_{mk}^{2}+3=3b^{2} \end{aligned}$$

(3.19)

for some $a,b>0.$ Assume that (3.18) is satisfied. Then, by Theorem 3.7 given in [11], the only possible values of mk and m for which $V_{mk}=3V_{m}a^{2}$ are $mk=3$ and $m=1.$ This implies that $V_{3}^{2}+3=21b^{2}.$ Thus, we get $V_{3}^{2}\equiv 4 \left( \text{ mod } \text { }7\right) .$ This is impossible since $7|V_{3}.$ Now assume that (3.19) is satisfied. Since $3|V_{mk}$ and mk is odd, it is seen that 3|P by Lemma 5. On the other hand, $V_{mk}^{2}=V_{2mk}-2$ by (2.10). Combining the equations $V_{mk}^{2}=V_{2mk}-2$ and $V_{mk}^{2}+3=3b^{2},$ we get $V_{2mk}=3b^{2}-1.$ Let $mk=4q\pm 1=2^{r}z\pm 1$ with odd z and $r\ge 2.$ And so by (2.4), we obtain

$$\begin{aligned} V_{2mk}=V_{2\cdot 2^{r}z\pm 2}\equiv -V_{2}\text { }( \text{ mod } V_{2^{r}}), \end{aligned}$$

implying that

$$\begin{aligned} 3b^{2}\equiv -\left( P^{2}+2-1\right) \equiv -U_{3}\text { }\left( \text{ mod } \text { }V_{2^{r}}\right) . \end{aligned}$$

This shows that $1=\left( \frac{-3U_{3}}{V_{2^{r}}}\right) .$ We have $\left( \frac{-1}{V_{2^{r}}}\right) =-1$ by (2.21), $\left( \frac{U_{3}}{ V_{2^{r}}}\right) =1$ by (2.15), and $V_{2^{r}}\equiv 2 \left( \text{ mod }\text { }3\right)$ by (2.13). Thus,

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

Combining the above, we see that

$$\begin{aligned} 1=\left( \frac{-3U_{3}}{V_{2^{r}}}\right) =\left( \frac{-1}{V_{2^{r}}} \right) \left( \frac{3}{V_{2^{r}}}\right) \left( \frac{U_{3}}{V_{2^{r}}} \right) =\left( -1\right) \left( 1\right) \left( 1\right) =-1, \end{aligned}$$

a contradiction. This completes the proof. $\square$

Theorem 16

If P is odd, then $U_{n}=7x^{2}$ is possible if and only if $P=7\square$ and $n=2.$

Proof

Assume that $U_{n}=7x^{2}$ for some $x>0.$ Since $7|U_{n}, n=2t$ for some positive integer t by Lemma 7. And so by (2.9), we get $U_{n}=U_{2t}=U_{t}V_{t}=7x^{2}.$ Clearly, $\left( U_{t},V_{t}\right) =1$ or 2 by (2.14).

Let $\left( U_{t},V_{t}\right) =1.$ Then either

$$\begin{aligned} U_{t}=a^{2},\text { }V_{t}=7b^{2} \end{aligned}$$

(3.20)

or

$$\begin{aligned} U_{t}=7a^{2},\text { }V_{t}=b^{2} \end{aligned}$$

(3.21)

for some $a,b>0.$ Assume that (3.20) is satisfied. Then by Theorem 14, the possible values of t for which $V_{t}=7x^{2}$ are $t=1$ when 7|P and $t=4,P=1$ when $P^{2}\equiv 1 ( \text{ mod } 7).$ If $t=1,$ then $n=2$ and therefore $P=7\square$ is a solution. If $t=4$ and $P=1,$ then $U_{4}=P^{3}+2P=3=a^{2},$ which is impossible in integers. Now assume that ( 3.21) is satisfied. Since $7|U_{t},$ it is seen from Lemma 7 that t is even. Let $t=2m.$ Then by (2.10), we see that $b^{2}=V_{t}=V_{2m}=V_{m}^{2}\pm 2,$ which is impossible.

Let $\left( U_{t},V_{t}\right) =2.$ Then either

$$\begin{aligned} U_{t}=2a^{2},\text { }V_{t}=14b^{2} \end{aligned}$$

(3.22)

or

$$\begin{aligned} U_{t}=14a^{2},\text { }V_{t}=2b^{2} \end{aligned}$$

(3.23)

for some $a,b>0.$ According to Theorem 13, (3.22) cannot hold. Assume that (3.23) is satisfied. Then by Theorem 2 given in [2] and [4], we have $t=6$ and $P=1,5.$ But this is also impossible. For, otherwise we would have $14a^{2}=U_{6}=U_{3}V_{3}=\left( P^{2}+1\right) \left( P^{3}+3P\right)$ which is impossible for $P=1,5.$ This completes the proof. $\square$

Theorem 17

Let P be odd, $m>1$ and $U_{m}\ne 1.$ The equation $U_{n}=7U_{m}x^{2}$ has solution only when $P^{2}\equiv 1 ( \text{ mod } 7),$ in which case, the only solution is given by $(n,m,P,x)=(8,4,1,1).$

Proof

Assume that $U_{n}=7U_{m}x^{2}$ with $m>1.$ Since $U_{m}|U_{n},$ it follows from (2.17) that m|n. Thus, $n=mt$ for some positive integer t. It is easy to see that $n\ne m.$ Then, we have $t>1.$ On the other hand, since $7|U_{n},$ it is seen that n is even by Lemma 7. Since n is even and $n=mt,$ either m or t is even.

Case I: t is even. Then $t=2s$ for some $s>0.$ By (2.9), we have $U_{n}=U_{2ms}=U_{ms}V_{ms}=7U_{m}x^{2}.$ This yields that $\left( U_{ms}/U_{m}\right) V_{ms}=7x^{2}.$ Clearly, $\left( U_{ms}/U_{m},V_{ms}\right) =1$ or 2 by (2.14). If $\left( U_{ms}/U_{m},V_{ms}\right) =1,$ then either

$$\begin{aligned} U_{ms}=U_{m}a^{2},\text { }V_{ms}=7b^{2} \end{aligned}$$

(3.24)

or

$$\begin{aligned} U_{ms}=7U_{m}a^{2},\text { }V_{ms}=b^{2} \end{aligned}$$

(3.25)

for some positive integers a and b. By Theorem 14, (3.24) is impossible when $P^{2}\equiv 2 ( \text{ mod } 7)$ or $P^{2}\equiv 4 ( \text{ mod } 7).$ If 7|P,  then by Theorem 14, we have $ms=1.$ But this contradicts the fact that $m>1.$ If $P^{2}\equiv 1 ( \text{ mod } 7),$ then by Theorem 14, it follows that $ms=4$ and $P=1.$ Since $m>1,$ we get $m=4, s=1$ or $m=2, s=2.$ Let $m=4, s=1.$ Since $t=2s$ and $n=mt,$ we get $n=8.$ Hence, $U_{8}=7U_{4}x^{2},$ implying by (2.9) that $V_{4}=7x^{2}.$ Since $P=1,$ we obtain $x=1.$ So, $(n,m,P,x)=(8,4,1,1)$ is a solution. Now, let $m=2, s=2.$ Then we readily obtain $n=8$ and therefore $U_{8}=7U_{2}x^{2}.$ By (2.9), it follows that $V_{2}V_{4}=7x^{2}.$ Since $7|V_{4},$ we get $V_{2}\frac{V_{4}}{7}=x^{2}.$ Clearly, $\left( V_{2},\frac{V_{4}}{7}\right) =1$ by (2.14) and (2.12). Then $V_{2}=a^{2}, V_{4}=7b^{2}$ for some $a,b>0.$ Since $P=1,$ there is no integer a such that $V_{2}=P^{2}+2=3=a^{2}.$

If (3.25 ) holds, then by Theorem 1 given in [2] and [4], we have $ms=3$ and $P=1$ or 3. Since $m>1$ and $ms=3,$ it follows that $m=3.$ This implies that $U_{3}=7U_{3}x^{2},$ which is impossible.

If $\left( U_{ms}/U_{m},V_{ms}\right) =2,$ then either

$$\begin{aligned} U_{ms}=2U_{m}a^{2},\text { }V_{ms}=14b^{2} \end{aligned}$$

(3.26)

or

$$\begin{aligned} U_{ms}=14U_{m}a^{2},\text { }V_{ms}=2b^{2}. \end{aligned}$$

(3.27)

Clearly, (3.26) is excluded by Theorem 13. Suppose (3.27) is satisfied. Then by Theorem 2 given in [2] and [4], we have $ms=6$ and $P=1,5.$ Since $m>1,$ it follows that $m=2,3$ or 6. If $m=2,$ then $U_{6}=14U_{2}a^{2},$ implying that $\left( P^{2}+1\right) \left( P^{2}+3\right) =14a^{2}$ which is impossible in integers for the case when $P=1,5.$ If $m=3,$ then $U_{6}=14U_{3}a^{2},$ implying that $\left( P^{3}+3P\right) =14a^{2},$ which is impossible. Lastly, if $m=6,$ then $U_{6}=14U_{6}a^{2},$ implying that $1=14a^{2},$ which is also impossible.

Case II: t is odd. Since $n=mt$ and n is even, it follows that m is even. Let $m=2s.$ Then $n=2st$ and so by (2.9), we get $U_{n}=U_{2st}=U_{st}V_{st}=7U_{2st}x^{2}=7U_{st}V_{st}x^{2}.$ This implies that $\frac{U_{st}}{U_{s}}\frac{V_{st}}{V_{s}}=7x^{2}.$ Clearly, $d=\left( \frac{U_{st}}{U_{s}},\frac{V_{st}}{V_{s}}\right) =1$ or 2.

Let $d=1.$ Then either

$$\begin{aligned} U_{st}=U_{s}a^{2},\text { }V_{st}=7V_{s}b^{2} \end{aligned}$$

(3.28)

or

$$\begin{aligned} U_{st}=7U_{s}a^{2},\text { }V_{st}=V_{s}b^{2}. \end{aligned}$$

(3.29)

Suppose (3.28) is satisfied. Then by Theorem 15, we get $s=1$ and $st=3.$ This implies that $U_{3}=U_{1}a^{2},$ that is $P^{2}+1=a^{2},$ which is impossible. Suppose (3.29) is satisfied. Then by Theorem 10, we obtain $st=3, s=1,$ and $P=1$ or $st=s.$ If $st=3, s=1,$ and $P=1,$ then from $U_{st}=7U_{s}a^{2},$ we have $U_{3}=7U_{1}a^{2},$ leading to $2=7a^{2},$ which is impossible. If $st=s,$ then again from $U_{st}=7U_{s}a^{2},$ we have $1=7a^{2},$ which is impossible.

Let $d=2.$ Then either

$$\begin{aligned} U_{st}=2U_{s}a^{2},\text { }V_{st}=14V_{s}b^{2} \end{aligned}$$

(3.30)

or

$$\begin{aligned} U_{st}=14U_{s}a^{2},\text { }V_{st}=2V_{s}b^{2}. \end{aligned}$$

(3.31)

Assume that (3.30) is satisfied. Then, by Theorem 6 given in [9], the only possible values of P, st,  and s for which $U_{st}=2U_{s}a^{2}$ are $st=3, s=2, P=1; st=6, s=2, P=1; st=12, s=3, P=1; st=12, s=6, P=1;$ or $st=12, s=6, P=5.$ A simple computation shows that $V_{st}=14V_{s}b^{2}$ is impossible under all the conditions that when $P=1.$ If $P=5,$ then, this is impossible for the case when 7|P or if $P^{2}\equiv 1,2 \left( \text{ mod } \text { }7\right) .$ On the other hand, since $7|V_{st},$ it follows from Lemma 6 that $st=3r$ with r odd for the case when $P^{2}\equiv 4 \left( \text{ mod } \text { }7\right) .$ This means that st is odd. But this contradicts the fact that $st=12$ is even. Assume that (3.31) is satisfied. Then by Theorem 12 given in [9], we get $s=1$ and $P=1.$ Since $m=2s,$ it follows that $m=2.$ Substituting this value of m into $U_{n}=7U_{m}x^{2}$ gives $U_{n}=7U_{2}x^{2}=7x^{2}.$ By Theorem 16, the equation $U_{n}=7x^{2}$ is possible if and only if $n=2.$ As a consequence $m=2$ and $n=2.$ But this is impossible since $n\ne m.$ This completes the proof. $\square$