On Fibonacci and Lucas Numbers of the Form cx^{2}

Refik Keskin and Zafer Yosma Department of Mathematics Faculty of Science and Arts Sakarya University Sakarya Turkey

                    rkeskin@sakarya.edu.tr
                   d085016004@sakarya.edu.tr

Abstract

In this paper, by using some congruences concerning with Fibonacci and Lucas numbers, we completely solve the Diophantine equations L_{n} = 2L_{m}x^{2}, F_{n} = 2F_{m}x^{2}, L_{n} = 6L_{m}x^{2}, F_{n} = 3F_{m}x^{2}, and F_{n} = 6F_{m}x^{2}.

1 Introduction

Fibonacci and Lucas sequences are defined as follows; F_{0} = 0, F_{1} = 1, F_{n} = F_{n−1} + F_{n−2} for n ≥ 2 and L_{0} = 2, L_{1} = 1, L_{n} = L_{n−1} + L_{n−2} for n ≥ 2, respectively. F_{n} is called the n-th Fibonacci number and L_{n} is called the n-th Lucas number. Fibonacci and Lucas numbers for negative subscripts are given by F_{−n} = (−1)^{n+1} F_{n} for n ≥ 1 and L_{−n} = (−1)^{n} L_{n} for n ≥ 1. It can be seen that L_{n} = F_{n−1} + F_{n+1} and L_{n−1} + L_{n+1} = 5F_{n} for every n ∈ ℤ. For more information about Fibonacci and Lucas sequences, one can consult [9], [18].

Let α and β denote the roots of the equation x^{2} − x − 1 = 0. Then α = (1 + \sqrt{5})/2 and β = (1 − \sqrt{5})/2. It can be seen that αβ = −1 and α + β = 1. Moreover it is well known and easy to show that

F_{n} = (α^{n} − β^{n}) /\sqrt{5}

and

L_{n} = α^{n} + β^{n}

for every n ∈ ℤ.

In the following section, we will give some congruences concerning with Fibonacci and Lucas numbers. By using these congruences, we may prove many properties known before.

2 Preliminaries

The problem of characterizing the square Fibonacci numbers was first introduced in the book by Ogilvy [12, p. 100]. In 1963, both Moser and Carlitz [10], and Rollet [17] proposed this problem. In 1964, the square conjecture was proved by Cohn [4] and independently by Wyler [19]. Later the problem of characterizing the square Lucas numbers was solved by Cohn [6] and by Alfred [1]. Moreover in 1965, Cohn solved the Diophantine equations F_{n} = 2x^{2} and L_{n} = 2x^{2} in [6].

We give the following theorem from [5].

Theorem 1. If F_{n} = x^{2}, then n = 1, 2, 12. If F_{n} = 2x^{2}, then n = 3, 6. If L_{n} = x^{2}, then n = 1, 3 and if L_{n} = 2x^{2}, then n = 6.

The proofs of the following two theorems are given in [8].

Theorem 2. Let n ∈ ℕ ∪{0} and k, m ∈ ℤ. Then

F_{2mn+k} ≡ (−1)^{mn} F_{k} (mod F_{m})  (1)

and

L_{2mn+k} ≡ (−1)^{mn} L_{k} (mod F_{m}) .  (2)

Theorem 3. Let n ∈ ℕ∪{0} and k, m ∈ ℤ. Then

L_{2mn+k} ≡ (−1)^{(m+1)n} L_{k} (mod L_{m})  (3)

and

F_{2mn+k} ≡ (−1)^{(m+1)n} F_{k} (mod L_{m}) .  (4)

From the identity (2), it follows that 8 ∤ L_{n} for any natural number n.

Now we give two lemmas and a corollary, which will be needed later. The proofs of the lemmas can be achieved by induction. For the proof of the corollary, one can consult [2] or [11].

Lemma 4. L_{2^{k}} ≡ 3 (mod 4) for the all positive integers k with k ≥ 1.

Lemma 5. If r ≥ 3, then L_{2^{r}} ≡ 2 (mod 3).

Corollary 6. If k ≥ 1, then there is no integer x such that x^{2} ≡ −1 (mod L_{2^{k}}).

The following lemma can be proved by induction.

Lemma 7. If r ≥ 2, then L_{2^{r}} ≡ 7 (mod 8) .

The proofs of the following theorems can be found in [3], [18] or [8].

Theorem 8. Let m, n ∈ ℕ and m ≥ 2. Then L_{m}|L_{n} if and only if m|n and \frac{n}{m} is an odd integer.

Theorem 9. Let m, n ∈ ℕ and m ≥ 3. Then F_{m}|F_{n} if and only if m|n.

Theorem 10. Let m, n ∈ ℕ and m ≥ 2. Then L_{m}|F_{n} if and only if m|n and \frac{n}{m} is an even integer.

Also we give some identities about Fibonacci and Lucas numbers which will be needed in the sequel:

L_{2n} = L_{n}^{2} − 2(−1)^{n}  (5)
L_{3n} = L_{n}(L_{n}^{2} − 3(−1)^{n})  (6)
F_{2n} = F_{n}L_{n}  (7)
F_{3n} = F_{n}(5F_{n}^{2} + 3(−1)^{n})  (8)
L_{n}^{2} − 5F_{n}^{2} = 4(−1)^{n}  (9)
2|F_{n} ⇔ 2|L_{n} ⇔ 3|n  (10)
(F_{n}, L_{n}) = 1 or (F_{n}, L_{n}) = 2  (11)

Let (\frac{a}{p}) represent the Legendre symbol. Then we have

(\frac{2}{p}) = 1 if and only if p ≡ ±1 (mod 8)  (12)

and

(\frac{−2}{p}) = 1 if and only if p ≡ 1, 3 (mod 8) .  (13)

For the proof of (12) and (13), one can consult [2] or [11].

3 Main Theorems

Many authors investigated Fibonacci and Lucas numbers of the form cx^{2}. In [5], Cohn solved F_{n} = cx^{2} and L_{n} = cx^{2} for c = 1, 2. In [14], Robbins considered Fibonacci numbers of the form px^{2}. Robbins solved the equation F_{n} = px^{2} for all p such that p ≡ 3 (mod 4) or p < 10000. Later, in [15] Robbins considered Fibonacci numbers of the form cx^{2}. The author obtained all solutions of F_{n} = cx^{2} for composite values of c ≤ 1000. After that, in [16], the same author solved L_{n} = px^{2}, where p is an odd prime and p < 1000. Moreover, in [20], Zhou dealt with Lucas numbers of the form L_{n} = px^{2}, where p is a prime number, and he gave solutions for 1000 < p < 60000. In this section, we consider the equations L_{n} = 2L_{m}x^{2}, F_{n} = 2F_{m}x^{2}, L_{n} = 6L_{m}x^{2}, F_{n} = 3F_{m}x^{2}, and F_{n} = 6F_{m}x^{2}.

In [13], Ribenboim considers square-classes of Fibonacci numbers. F_{m}, F_{n} are in the same square-class if there exist non-zero integers x, y such that F_{m}x^{2} = F_{n}y^{2}; or equivalently, when F_{m}F_{n} is a square. In a similar way, he considers square-classes of Lucas numbers. A square-class will be called trivial if it consists of only one number. Ribenboim showed that the square-class of L_{m} is trivial when m ≠ 0, 1, 3, and 6. He also showed that the squareclass of F_{m} is trivial when m ≠ 1, 2, 3, 6, 12. Now, we can give following two theorems, which can be obtained from Proposition 1 and Proposition 2 given in [13].

From now on, we will assume that n and m are positive integers.

Theorem 11. Let m > 3 be an integer and F_{n} = F_{m}x^{2} for some x ∈ ℤ. Then n = m.

Theorem 12. Let m ≥ 2 be an integer and L_{n} = L_{m}x^{2} for some x ∈ ℤ. Then n = m.

The proofs of the following two theorems can be obtained from Theorem 6 and Theorem 12 given in [7], but we will give a different proof.

Theorem 13. There is no integer x such that L_{n} = 2L_{m}x^{2} for m > 1.

Proof. Assume that L_{n} = 2L_{m}x^{2}. Then L_{m}|L_{n} and therefore n = mk for some odd natural number k by Theorem 8. Firstly assume that m is an odd integer. Since 2|L_{n}, we get 3|n by (10). Thus we see that 3 ∤ m. For if 3|m, then L_{3}|L_{m}, i.e., 4|L_{m} by Theorem 8. This implies that 8|L_{n}, which is impossible. Since 3 ∤ m, it follows that 3|k. That is, k = 3t for some odd positive integer t. Thus n = mk = 3mt and mt is an odd integer. Therefore, since 3|n, it follows that L_{3}|L_{n}, i.e., 4|2L_{m}x^{2} by Theorem 8. Since 3 ∤ m, L_{m} is an odd integer. Therefore 2|x^{2}, i.e., x is an even integer. This implies that 8|L_{n}, which is impossible.

Now assume that m is an even integer. If x is an even integer, then we see that 8|L_{n}, which is impossible. Therefore x is an odd integer. Assume that 3|m. Then L_{m} is an even integer. Therefore L_{3}|L_{n} by Theorem 8. It follows that n = 3b for some odd integer b by Theorem 8. That is, n is an odd integer. But this is impossible. Because since m is an even integer, n is also an even integer. Assume that 3 ∤ m. Then since n = mk and 3|n, we get 3|k, i.e., k = 3t for some odd integer t. Since t is an odd integer, t = 4q ± 1 for some nonnegative integer t. Thus n = mk = 3m(4q ± 1) = 2 · 6mq ± 3m. Then

L_{n} = L_{2·6mq±3m} ≡ L_{±3m} (mod F_{6})

and therefore

2L_{m}x^{2} ≡ L_{3m} (mod 8)

by (2). Since x^{2} ≡ 1 (mod 8) and m is even integer, we get

2L_{m} ≡ L_{m}(L_{m}^{2} − 3) (mod 8)

by (6). Moreover, since 3 ∤ m, L_{m} is odd integer. Therefore we get

2 ≡ L_{m}^{2} − 3 (mod 8).

Thus

2 ≡ −2 (mod 8),

which is impossible. This completes the proof.

In [5], it is shown that, for m = 1, 2, the equation F_{n} = 2F_{m}x^{2} = 2x^{2} has solution only for n = 3, 6. More generally, we can give the following theorem.

Theorem 14. If F_{n} = 2F_{m}x^{2} and m ≥ 3, then m = 3, x^{2} = 36, and n = 12 or m = 6, x^{2} = 9, and n = 12.

Proof. If m = 3, then F_{n} = 2F_{3}x^{2} = (2x)^{2}. Thus it can be seen that n = 12, x^{2} = 36 by Theorem 1. Assume that m > 3 and F_{n} = 2F_{m}x^{2}. Then F_{m}|F_{n} and therefore n = mk for some natural number k by Theorem 9.

Firstly, assume that k is an even integer. Then k = 2t for some integer t. Therefore n = mk = 2mt. Thus

F_{n} = F_{2mt} = F_{mt}L_{mt} = 2F_{m}x^{2}

by (7). This shows that (F_{mt}/F_{m}) L_{mt} = 2x^{2}. It can be easily seen that if (F_{mt}/F_{m}, L_{mt}) = d, then d = 1 or d = 2 by (11). Thus we have the following equations:

\frac{F_{mt}}{F_{m}} = u^{2}, L_{mt} = 2v^{2},  (14)

\frac{F_{mt}}{F_{m}} = 2u^{2}, L_{mt} = v^{2},  (15)

\frac{F_{mt}}{F_{m}} = 2u^{2}, L_{mt} = (2v)^{2},  (16)

and

\frac{F_{mt}}{F_{m}} = (2u)^{2}, L_{mt} = 2v^{2}.  (17)

Assume that (14) is satisfied. Then mt = m, i.e., t = 1 by Theorem 11. Therefore L_{m} = 2v^{2} and this implies that m = 6 by Theorem 1. Thus we get m = 6, x^{2} = 9, and n = 12. By using Theorem 1 and Theorem 11, it can be seen that the other three cases are impossible.

Secondly, assume that k is an odd integer. Suppose that m is an even integer, i.e., m = 2r for some natural number r. Then we can write n = mk = 2kr. Thus

F_{n} = F_{2kr} = F_{kr}L_{kr} = 2F_{r}L_{r}x^{2}

by (7). This shows that (F_{kr}/F_{r}) (L_{kr}/L_{r}) = 2x^{2}. A similar argument shows that the equation (F_{kr}/F_{r}) (L_{kr}/L_{r}) = 2x^{2} has no solution. Now assume that m is an odd integer. Firstly, suppose that 3 ∤ k. Since k is an odd integer, we can write k = 6q ± 1 for some nonnegative integer q. Therefore n = mk = m(6q ± 1) = 2 · 3mq ± m. Thus we get

F_{n} = F_{2·3mq±m} ≡ F_{±m} (mod L_{3}),

i.e.,

F_{n} ≡ F_{m} (mod 4)

by (4). Since F_{n} is even integer, F_{m} is also an even integer. Thus 3|m, and therefore m = 3a for some integer a by (10). On the other hand, since F_{m} is even integer, 4|F_{n}, and thus 6|n by Theorem 9. Since n = mk = 3ak, we get 6|3ak, i.e., 2|ak. Moreover, since k is odd integer, it is seen that 2|a. This implies that 2|m, which is impossible. Because m is an odd integer. Assume that 3|k. Then k = 3s for some odd integer s. Therefore n = mk = 3ms. Thus since ms is odd integer, we get

F_{n} = F_{3ms} = F_{ms}(5F_{ms}^{2} − 3) = 2F_{m}x^{2}

by (8). This shows that (F_{ms}/F_{m}) (5F_{ms}^{2} − 3) = 2x^{2}. It can be easily seen that if d = (F_{ms}/F_{m}, 5F_{ms}^{2} − 3) , then d = 1 or d = 3. Assume that d = 3. Then 3|F_{ms}, and thus 4|ms by Theorem 9. But this is impossible, since ms is odd integer. Therefore d = 1. Then we get

\frac{F_{ms}}{F_{m}} = u^{2}, 5F_{ms}^{2} − 3 = 2v^{2}  (18)

or

\frac{F_{ms}}{F_{m}} = 2u^{2}, 5F_{ms}^{2} − 3 = v^{2}  (19)

for some integers u and v. Assume that (18) is satisfied. Then ms = m, i.e., s = 1 by Theorem 11. Therefore 5F_{m}^{2}−3 = 2v^{2} and this shows that 2v^{2} = 5F_{m}^{2}−3 = L_{m}^{2}+1 = L_{2m}−1 by (9) and (5). This implies that L_{2m} = 2v^{2} + 1. Since L_{2m} = 2v^{2} + 1, we get 3 ∤ m. Thus we can write m = 6q ± 1 = 3 · 2^{r+1}b ± 1, where q = 2^{r}b for some odd integer b with r ≥ 0. This shows that

L_{2m} = L_{2·2^{r+1}3b±2} ≡ −L_{±2} (mod L_{2^{r+1}})

and therefore

2v^{2} + 1 ≡ −3 (mod L_{2^{r+1}}),

i.e.,

2v^{2} ≡ −4 (mod L_{2^{r+1}})

by (3). Since L_{2^{r+1}} is an odd integer, we get

v^{2} ≡ −2 (mod L_{2^{r+1}}).

This shows that (\frac{−2}{p}) = 1 for every prime divisor of L_{2^{r+1}}. Then it follows that

p ≡ 1, 3 (mod 8)

by (13) and therefore

L_{2^{r+1}} ≡ 1, 3 (mod 8).

This shows that r = 0 by Lemma 7. Consequently, q is an odd integer. Therefore it can be easily seen that m = 12c + 5 or m = 12c + 7 for some integer c. Thus we get

L_{m} ≡ 3 (mod 8)

or

L_{m} ≡ 5 (mod 8)

by (2). On the other hand, since

2v^{2} = L_{m}^{2} + 1,

we get

2v^{2} ≡ 1 (mod L_{m}),

and therefore

(2v)^{2} ≡ 2 (mod L_{m}).

This shows that (\frac{2}{p}) = 1 for every prime divisor p of L_{m}. Then it follows that

p ≡ ±1 (mod 8)

by (12) and therefore

L_{m} ≡ ±1 (mod 8).

But this contradicts with the fact that L_{m} ≡ 3, 5 (mod 8). Assume that (19) is satisfied. Then we get v^{2} = 5F_{ms}^{2} −3 = L_{ms}^{2} +1 by (9). This implies that L_{ms} = 0, which is impossible. This completes the proof.

Theorem 15. If L_{n} = 6L_{m}x^{2} and m ≥ 1, then m = 2, x^{2} = 1, and n = 6.

Proof. Assume that L_{n} = 6L_{m}x^{2} for some integer x. Then 3|L_{n} and therefore n = 2k_{0} for some odd integer k_{0} by Theorem 8. Moreover, since 2|L_{n}, we get 3|n by (10). This shows that 3|k_{0} and then k_{0} = 3t for some odd integer t. Thus n = 6t = 6(2u + 1) = 12u + 6. Therefore

L_{n} = L_{12u+6} ≡ L_{6} (mod 8)

by (2). That is,

L_{n} ≡ 2 (mod 8).

Since 8 ∤ L_{n}, it can be seen that x is an odd integer. Therefore

x^{2} ≡ 1 (mod 8),

which implies that

6L_{m}x^{2} ≡ 6L_{m} (mod 8).

This shows that

6L_{m} ≡ 2 (mod 8),

which implies that m ≠ 1. Now assume that m > 2. Since L_{m}|L_{n}, there exists an odd integer k such that n = mk by Theorem 8. On the other hand, since 2|n, it is seen that 2|m. Therefore m = 2r for some odd integer r. If r = 6q +3, then m = 2r = 12q +6 and therefore

L_{m} ≡ L_{6} (mod 8)

by (2). That is,

L_{m} ≡ 2 (mod 8),

which is impossible since

6L_{m} ≡ 2 (mod 8).

Therefore 3 ∤ r. Since n = mk, m = 2r and 3 ∤ r, it follows that 3|k and thus k = 3s for some odd integer s. Then

L_{n} = L_{mk} = L_{3ms} = L_{ms}(L_{ms}^{2} − 3) = 6L_{m}x^{2}

by (6). It can be seen that (L_{ms}, L_{ms}^{2} − 3) = 3. Thus (L_{ms}, \frac{L_{ms}^{2} − 3}{3}) = 1. Then we get

\frac{L_{ms}}{L_{m}} (\frac{L_{ms}^{2} − 3}{3}) = 2x^{2}.

This shows that

\frac{L_{ms}}{L_{m}} = 2u^{2} and \frac{L_{ms}^{2} − 3}{3} = v^{2}  (20)

or

\frac{L_{ms}}{L_{m}} = u^{2} and \frac{L_{ms}^{2} − 3}{3} = 2v^{2}  (21)

for some integers u and v. Assume that (20) is satisfied. Then 3 (\frac{L_{ms}}{3} )^{2} − 1 = v^{2} and

therefore

v^{2} ≡ −1 (mod 3),

which is a contradiction. Now assume that (21) is satisfied. Then L_{ms} = L_{m}u^{2}, which implies that ms = m by Theorem 12. That is, s = 1. Thus L_{m}^{2} − 3 = 6v^{2}. Since L_{m}^{2} = L_{2m} + 2 by (5), we see that L_{2m} − 1 = 6v^{2}. Moreover, since m = 2r, it follows that L_{4r} − 1 = 6v^{2}. On the other hand, we can write 4r as 4r = 4(4u ± 1) = 16u ± 4 = 2 · 2^{k}a ± 4 for some odd integer a with k ≥ 3. This shows that

L_{4r} = L_{2·2^{k}a±4} ≡ −L_{±4} (mod L_{2^{k}})

by (3) and therefore

1 + 6v^{2} ≡ −7 (mod L_{2^{k}}).

Then we get

6v^{2} ≡ −8 (mod L_{2^{k}}).

That is,

3v^{2} ≡ −4 (mod L_{2^{k}}).

Thus

(3v)^{2} ≡ −12 (mod L_{2^{k}}).

This shows that (\frac{−12}{p}) = 1 for every prime divisor p of L_{2^{k}}. Then it follows that

p ≡ 1 (mod 3)

and therefore

L_{2^{k}} ≡ 1 (mod 3).

But this contradicts with Lemma 5. This completes the proof.

In [8], the authors showed that L_{n} = 3L_{m}x^{2} has no solution if m > 1. Now we give a similar result for Fibonacci numbers.

Theorem 16. Let m ≥ 3 be an integer and F_{n} = 3F_{m}x^{2} for some integer x. Then m = 4, x^{2} = 16, and n = 12.

Proof. Assume that m ≥ 3 and F_{n} = 3F_{m}x^{2} for some integer x. Then F_{m}|F_{n} and therefore n = mk for some integer k by Theorem 9.

Firstly, assume that k is an even integer. Then k = 2s for some s ∈ ℕ. Therefore n = mk = 2ms. Thus

F_{n} = F_{2ms} = F_{ms}L_{ms} = 3F_{m}x^{2}

by (7). This shows that

(F_{ms}/F_{m})L_{ms} = 3x^{2}.

By using Theorem 1, Theorem 12, and Theorem 15, it can be shown that the equation (F_{ms}/F_{m})L_{ms} = 3x^{2} has no solution.

Now assume that k is an odd integer. Since F_{n} = 3F_{m}x^{2}, we get 4|n by Theorem 9. Moreover, since n = mk and k is odd, we get 4|m. Assume that x is an even integer. Then 4|F_{n}. Thus L_{3}|F_{n} and 3|n by Theorem 10. Therefore since 4|n and 3|n, we get 12|n. That is, n = 12t for some t ∈ ℕ. On the other hand since 4|m, we get m = 4r for some r ∈ ℕ. Therefore 12t = n = mk = 4rk. Then it follows that 3t = rk. Thus

F_{n} = F_{12t} = F_{6t}L_{6t} = 3F_{2r}L_{2r}x^{2}

by (7). Since (6t/2r) = k and k is odd, we can write

\frac{F_{6t}}{F_{2r}} .\frac{L_{6t}}{L_{2r}} = 3x^{2}.

Assume that 3|r. Then, it can be seen that (\frac{F_{6t}}{F_{2r}} , \frac{L_{6t}}{L_{2r}} ) = 1. Therefore

\frac{F_{6t}}{F_{2r}} = u^{2}, \frac{L_{6t}}{L_{2r}} = 3v^{2}  (22)

or

\frac{F_{6t}}{F_{2r}} = 3u^{2}, \frac{L_{6t}}{L_{2r}} = v^{2}  (23)

for some integers u and v. A similar argument shows that (22) and (23) are impossible. Now assume that 3 ∤ r. Then since 3t = rk, it follows that 3|k. Thus k = 3s for some s ∈ ℕ. Then

3t = rk = 3rs and therefore t = rs. Also since 3 ∤ r, it can be seen that (\frac{F_{6t}}{F_{2r}} , \frac{L_{6t}}{L_{2r}} ) = 2.

Therefore

\frac{F_{6t}}{F_{2r}} = 2u^{2}, \frac{L_{6t}}{L_{2r}} = 6v^{2}  (24)

or

\frac{F_{6t}}{F_{2r}} = 6u^{2}, \frac{L_{6t}}{L_{2r}} = 2v^{2}  (25)

for some integers u and v. Assume that (24) is satisfied. Then 2r = 2 by Theorem 15. This shows that r = 1 and t = s. Thus L_{6t} = 6L_{2}v^{2} = L_{6}v^{2} and this implies that 6t = 6, i.e., t = 1 by Theorem 12. Therefore k = 3s = 3t = 3 and m = 4r = 4. Therefore n = 12 and x^{2} = 16.

Now assume that (25) is satisfied. Then it follows that

L_{6t} = 2L_{2r}v^{2},

which is impossible by Theorem 13 and Theorem 15.

Now assume that x is an odd integer. Then

F_{n} ≡ 3F_{m} (mod 8).

Since 4|m, it follows that m = 12q or m = 12q ±4 for some integer q. If m = 12q ±4, then

F_{m} ≡ F_{12q±4} ≡ F_{±4} ≡ ±3 (mod 8)

by (1). Therefore

F_{n} ≡ ±1 (mod 8),

which is impossible since 4|n. Because if 4|n, then n = 12r ± 4 or n = 12r for some integer r, and therefore F_{n} ≡ ±3, 0 (mod 8) by (1). If m = 12q, then n = mk = 12qk. This shows that 6qk/6q is an odd integer. Then from the identity

F_{n} = F_{12qk} = F_{6qk}L_{6qk} = 3F_{m}x^{2} = 3F_{6q}L_{6q}x^{2},

it follows that

\frac{F_{6qk}}{F_{6q}} .\frac{L_{6qk}}{L_{6q}} = 3x^{2}.

Since (\frac{F_{6qk}}{F_{6q}} , \frac{L_{6qk}}{L_{6q}} ) = 1, we get

\frac{F_{6qk}}{F_{6q}} = u^{2}, \frac{L_{6qk}}{L_{6q}} = 3v^{2}  (26)

or

\frac{F_{6qk}}{F_{6q}} = 3u^{2}, \frac{L_{6qk}}{L_{6q}} = v^{2}  (27)

for some integers u and v. Similarly, it can be seen that (26) and (27) are impossible. This completes the proof.

Lastly, we can give the following theorem without proof since its proof is similar to that of Theorem 16. Theorem 17. There is no integer x such that F_{n} = 6F_{m}x^{2}.

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[18] S. Vajda, Fibonacci and Lucas Numbers and The Golden Section, Ellis Horwood, 1989.

[19] O. Wyler, In the Fibonacci series F_{1} = 1, F_{2} = 1, F_{n+1} = F_{n} + F_{n−1} the first, second and twelfth terms are squares, Amer. Math. Monthly 71 (1964), 221–222.

[20] C. Zhou, A general conclusion on Lucas numbers of the form px^{2}, where p is prime, Fibonacci Quart. 37 (1999), 39–45.

2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39. Keywords: Fibonacci numbers, Lucas numbers, congruences.

(Concerned with sequence \underline{A000045}.)

Received April 29 2011; revised version received August 18 2011; September 25 2011. Published in Journal of Integer Sequences, October 16 2011.

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