PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 100(114) (2016), 259–269 DOI: 10.2298/PIM1614259I

PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES

Nurettin Irmak and Murat Alp

Abstract. We introduce a novel fourth order linear recurrence sequence {S_{n} } using the two periodic binary recurrence. We call it “pellans sequence” and then we solve the system

ab + 1 = S_{x}, ac + 1 = S_{y} bc + 1 = S_{z} where a < b < c are positive integers. Therefore, we extend the order of recurrence sequence for this variant diophantine equations by means of pellans sequence.

1. Introduction

A Diophantine m-tuple is a set of {a_{1}, a_{2}, …, a_{n}} of positive rational numbers or integers such that a_{i}a_{j} +1 = □ for all 1 ⩽ i, j ⩽ n. Diophantus investigated first the problem of finding rational quadruples, and found the example \{\frac{1}{16} , \frac{33}{16} , \frac{68}{16} , \frac{105}{16} \}. Then, Fermat found the first integer quadruples as {1, 3, 8, 120}. In [6], the general form of this set was found by Hoggatt and Bergum as follows

{F_{2k}, F_{2k+2} , F_{2k+4} , 4F_{2k+1} F_{2k+2} F_{2k+3} },

where {F_{n}}_{n⩾0} denotes the Fibonacci sequence. Therefore, there exist infinitely many quadruples. The famous theorem of Dujella [4] states that there are only finitely many quintuples.

A variant of the problem is obtained if one replaces the squares by the terms of given binary recurrences. This type of problem was started by Luca and Szalay. Luca and Szalay replaced the squares by the terms of Fibonacci and Lucas sequence and found that there is no Fibonacci diophantine triple and the only Lucas diophantine triple is {1, 2, 3}. Similarly, Alp, Irmak and Szalay put the terms of balancing sequences instead of the squares and they did not found any triples. For details, see [1, 8, 9]. Moreover, Fuchs, Luca and Szalay [5] investigated the general case for binary sequence and they gave sufficient and necessary conditions to have finitely many triples. For an integer A ⩾ 3, Irmak and Szalay [7] showed that there is no diophantine triples for the sequence {u_{n}} where {u_{n}} satisfies the relation u_{n} = Au_{n−1} − u_{n−2} with the initial conditions u_{0} = 0 and u_{1} = 1.

2010 Mathematics Subject Classification: Primary 11D09; Secondary 11B39.

Key words and phrases: Diophantine triples, pell numbers, balancing numbers, pellans sequence.

Communicated by Žarko Mijajlović.

Up to now, the authors have studied for the special cases of binary recurrence. One way for this type diophantine problems is to extend the problem to recurrent sequences of larger orders. In this paper, we define a fourth order recurrence sequence which we call pellans sequence since odd terms of the sequence are Pell numbers and even terms of the sequence are balancing numbers or half of the terms of even Pell numbers. Afterwards, we investigate its diophantine triples. Now, it is suitable to give the definitions of Pell, Pell–Lucas and Balancing numbers. The terms of Pell sequence {P_{n}}_{n⩾0} satisfy the recurrence relation P_{n} = 2P_{n−1} + P_{n−2} with initial conditions P_{0} = 0 and P_{1} = 1. The terms of Pell–Lucas sequence {Q_{n}}_{n⩾0} satisfy the same recurrence relation with Pell sequence together with initial conditions Q_{0} = 2 and Q_{1} = 2. The terms of Balancing sequence {B_{n}}_{n⩾0} are defined by B_{n} = 6B_{n−1} − B_{n−2} together with the initial conditions B_{0} = 0 and B_{1} = 1.

Definition 1.1. For a nonnegative integer n, the pellans sequence {S_{n}}_{n⩾0} is defined by S_{2n} = \frac{1}{2}P_{2n} and S_{2n+1} = P_{2n+1} .

The first few terms of the pellans sequence are 0, 1, 1, 5, 6, 29, 35, . . .. It is obvious that the terms of pellans sequence {S_{n}}_{n⩾0} satisfy S_{n} = 6S_{n−2} − S_{n−4} with S_{0} = 0, S_{1} = 1, S_{2} = 1 and S_{3} = 5.

Our main result is the following

Theorem 1.1. There is no integer solution for the system

(1.1)  ab + 1 = S_{x}, ac + 1 = S_{y}, bc + 1 = S_{z}

where 0 < a < b < c are integers and the sequence {S_{n}} is the pellans sequence.

2. Preliminaries

Now, we present required properties to prove Theorem 1.1.

Lemma 2.1. Let m and n are positive integers. Then

(1) gcd(S_{n}, S_{m}) = S_{gcd (n,m)} .

(2) P_{2n+1} − 1 = \{ \begin{matrix}P_{n}Q_{n+1} , & n is even \\ P_{n+1} Q_{n}, & n is odd \end{matrix} and P_{2n} − 2 = \{ \begin{matrix}P_{n+1} Q_{n−1} & n is even \\ P_{n−1} Q_{n+1} & n is odd \end{matrix}

where {Q_{n}} is the Pell–Lucas sequence.

(3) 2B_{n} = P_{2n} = P_{n}Q_{n}.

(4) (B_{n} − 1)(B_{n} + 1) = B_{n−1} B_{n+1} .

(5) B_{2n+1} − 1 = B_{n}C_{n+1} where {C_{n}} is associated sequence of {B_{n}}.

(6) C_{n} = Q_{2n}.

(7) If n is odd integer, then B_{n} − 1 = \frac{1}{2}P_{n−1} Q_{n+1} .

Proof. The third and sixth identities can be found in [10]. For the fourth one, we refer to [2]. Since the terms of the sequence {S_{n}} are balancing and Pell numbers and special cases of the Lucas sequence, we can write easily the first identity by means of [3]. In order to prove the seventh one, we use the Binet formulas for the Pell and associated Pell sequences. Since n is an odd integer, we get

\frac{1}{2}P_{n−1} Q_{n+1} = \frac{α^{n−1} − β^{n−1}}{α − β} (α^{n+1} + β^{n+1} )

= \frac{1}{2}( \frac{α^{2n} − β^{2n}}{α − β} + 2) = \frac{κ^{n} − τ^{n}}{κ − τ} + 1 = B_{n} + 1

where α and β are the roots of the equation x^{2} − 2x − 1 = 0 and we use the facts κ = α^{2} and τ = β^{2}. The remaining identities can be proven by using Binet formulas. □

Lemma 2.2. For all integers n ⩾ 3, the following inequalities hold

α^{n−1.19} < P_{n} < α^{n−1.16}
α^{n−0.1} < Q_{n} < α^{n+0.1}
α^{2n−1.97} < B_{n} < α^{2n−1.96}

where α is the dominant root of the equation x^{2} − 2x − 1 = 0.

Proof. The third inequality is from [1]. For n ⩾ 3, we have

P_{n} ⩾ \frac{α^{n} − |β^{3} |}{α − β} ⩾ α^{n} \frac{1 − (|β|/α)^{3}}{α − β} ⩾ α^{n−1.19} ,

which gives the lower bound for the sequence {P_{n}}. Similarly,

P_{n} ⩽ \frac{α^{n} + |β^{3} |}{α − β} ⩽ α^{n} \frac{1 + (|β|/α)^{3}}{α − β} ⩽ α^{n−1.16}

gives the upper bound. The bounds for the sequence {Q_{n}} can be proven in a similar way. □

3. Proof of Theorem 1.1

Since 0 < a < b < c, then 1 · 2 + 1 ⩽ ab + 1 = S_{x}. We get that 3 ⩽ x. From now on, the proof is split into two parts.

Case 1. Assume that z ⩽ 280.

In this case, we run a computer search to detect system (1.1) for each case. Note that the balancing case has been already solved in [1]. Observe that we have

a = \sqrt{\frac{(S_{x} − 1)(S_{y} − 1)}{S_{z} − 1}} , 3 ⩽ x < y < z ⩽ 280.

Going through all the eligible values for x, y and z, and we found no integer solution.

Case 2. Assume that z > 280.

In this case, we distinguish four main cases depending on the integers y and z.

(1) Both y and z are even

If x is an even integer, then the case corresponds to [1] since the pellans sequence turns to balancing sequence. Assume that x is odd. The proof mainly depends on the indices y and z apart from the case k = 2 and l = 1 in [1]. If we take S_{x} = P_{2x−1} in the case k = 2, l = 1 in [1], then we can follow the similar way to complete the proof of theorem. Hence, we omit this case.

(2) Both y and z are odd

Since both y and z are odd integers, then the terms of pellans sequence turn to Pell numbers. Now, we give a lemma which gives a relation between the indices y and z.

Lemma 3.1. The system

(3.1)  ab + 1 = S_{x}, ac + 1 = P_{y}, bc + 1 = P_{z}

satisfy z ⩽ 2y − 2.

Proof. The last two equations of (1.1) give \sqrt{P_{z}} < c < P_{y}. By Lemma 2.2, we have

α^{(z−1.19)/2} < \sqrt{P_{z}} < c < P_{y} < α^{y−1.16}

which yields z ⩽ 2y − 2. □

Put q_{1} = gcd(S_{y} − 1, S_{z} − 1). Since y and z are odd integers, we replace the pellans sequence by Pell sequence according to the definition of pellans sequence. Applying the second and fourth identities of Lemma 2.1, we get

q_{1} = gcd(P_{y} − 1, P_{z} − 1) = gcd(P_{\frac{y−i}{2}} Q_{\frac{y+i}{2}} , P_{\frac{z−j}{2}} Q_{\frac{z+j}{2}} )
⩽ P_{gcd}_{(\frac{y−i}{2} , \frac{z−j}{2} )}Q_{gcd}_{(\frac{y−i}{2} , \frac{z+j}{2} )}Q_{gcd}_{(\frac{y+i}{2} , \frac{z−j}{2} )}Q_{gcd}_{(\frac{y+i}{2} , \frac{z+j}{2} )}

where i, j ∈ \{±1\}. Let gcd( \frac{z+μ_{1} j}{2} , \frac{y+μ_{2} i}{2} ) = \frac{z+μ_{1} j}{2t} for some μ_{1}, μ_{2} ∈ \{±1\}. Suppose that t ⩾ 5 for all quadruples (i, j, μ_{1}, μ_{2}) ∈ {±1}^{4}. Since c | q_{1}, then Lemma 2.2 implies that

α^{\frac{z−1.19}{2}} < \sqrt{P_{z}} < c ⩽ q_{1} < α^{\frac{z+1}{10} −1.16+2 \frac{z−1}{10} +0.2+ \frac{z+1}{10} +0.1} .

When we compare the exponents of α, we arrive at a contradiction.

Assume that t ⩽ 4 and that

\frac{z + μ_{1}i}{2k} = \frac{y + μ_{2}j}{2l}

holds for suitable positive integers k and l such that gcd(k, l) = 1.

Suppose for the moment l > k. Then, we get z = y + 1 since y < z together with y + μ_{2}j > z + μ_{1}i. As both y and z are odd integers, the equation z = y + 1 is impossible.

Now, assume that k = l. Since z + μ_{1}i = y + μ_{2}j, we obtain that z = y + 2. By Lemma 2.1,

P_{z} − 1 = P_{\frac{z+i}{2}} Q_{\frac{z−i}{2}} , P_{z−2} − 1 = P_{\frac{z−2−i}{2}} Q_{\frac{z−2+i}{2}} .

hold for some i ∈ {±1}. If z ≡ 3 (mod 4), then we have the following, by Lemma 2.1

q_{1} = gcd(P_{z} − 1, P_{z−2} − 1) = gcd(P_{\frac{z+1}{2}} Q_{\frac{z−1}{2}} , P_{\frac{z−3}{2}} Q_{\frac{z−1}{2}} )
= P_{gcd}_{(\frac{z+1}{2} , \frac{z−3}{2} )}Q_{\frac{z−1}{2}} = P_{2}Q_{\frac{z−1}{2}} = 2Q_{\frac{z−1}{2}} .

Since q_{1} = 2Q_{\frac{z−1}{2}} , then we have c | Q_{\frac{z−1}{2}} or c | 2Q_{\frac{z−1}{2}} . If c | Q_{\frac{z−1}{2}} , then

Q_{\frac{z−1}{2}} = c_{1}c > c_{1} \sqrt{P_{z}},

and applying Lemma 2.2, we obtain

c_{1}α^{\frac{z−1.19}{2}} < c_{1} \sqrt{P_{z}} < Q_{\frac{z−1}{2}} < α^{\frac{z−1}{2} +0.1} .

So c_{1} < α^{0.2} < 1.2 which gives c_{1} = 1. Then, we obtain c = Q_{\frac{z−1}{2}} . When we put c = Q_{\frac{z−1}{2}} in (3.1), we have a = P_{\frac{z−3}{2}} and b = P_{\frac{z+1}{2}} . From the first equation in (3.1), one can easily see that ab + 1 = P_{\frac{z−3}{2}} P_{\frac{z+1}{2}} + 1 = P 2_{\frac{z−1}{2}} = S_{x}. If x is odd,

then we obtain P_{x} = P 2_{\frac{z−1}{2}} . When we apply the upper and lower bounds for the sequence {P_{n}}, we deduce

P_{x} = P 2_{\frac{z−1}{2}} ⇒ α^{x−1.19} < α^{z−3.32} ⇒ 2.13 < z − x
P_{x} = P 2_{\frac{z−1}{2}} ⇒ α^{z−1−2.38}  < α^{x−1.16} ⇒ z − x < 2.22.

There are no two odd integers x and y such that 2.13 < z − x < 2.22. Similarly, we get 1.35 < z − x < 1.4 when x is even. This is impossible, too.

In the sequel, if 2Q_{\frac{z−1}{2}} = c′_{1}c > c′_{1} \sqrt{P_{z}}, then c′_{1} < 2Q_{\frac{z−1}{2}} /\sqrt{P_{z}} < 2α^{0.2} < 2.4. If c′_{1} = 2, we get c = Q_{\frac{z−1}{2}} which is the same situation as above. If c′_{1} = 1, then we obtain that a = P_{\frac{z−3}{2}} /2 and b = P_{\frac{z+1}{2}} /2. So, P 2_{\frac{z−1}{2}} + 5 = 4S_{x}. If x is an odd integer, then S_{x} = P_{x} which yields that 3.48 < z − x < 3.8. But, this is impossible. Similarly, if x is an even integer, then S_{x} = B\frac{x}{2} . Using the upper and lower bounds for the sequence, we get 2.7 < z − x < 3 which is not possible.

If z ≡ 1 (mod 4), then

q_{1} = gcd(P_{z} − 1, P_{z−2} − 1) = gcd(P_{\frac{z−1}{2}} Q_{\frac{z+1}{2}} , P_{\frac{z−1}{2}} Q_{\frac{z−3}{2}} )
= Q_{gcd}_{(\frac{z+1}{2} , \frac{z−3}{2} )}P_{\frac{z−1}{2}} = Q_{1}P_{\frac{z−1}{2}} = 2P_{\frac{z−1}{2}} .

Assume that q_{1} = 2P_{\frac{z−1}{2}} . Since c | q_{1}, then c | P_{\frac{z−1}{2}} or c | 2P_{\frac{z−1}{2}} . If c | P_{\frac{z−1}{2}} , then P_{\frac{z−1}{2}} = c_{2}c > c_{2} \sqrt{P_{z}}. Since P_{\frac{z−1}{2}} /\sqrt{P_{z}} < 1, we get c_{2} < P_{\frac{z−1}{2}} /\sqrt{P_{z}} < 1 which is impossible. If c | 2P_{\frac{z−1}{2}} , then 2P_{\frac{z−1}{2}} = c′_{2}c > c′_{2} \sqrt{P_{z}}. Since 2P_{\frac{z−1}{2}} /\sqrt{P_{z}} < α^{−0.27} < 0.79, then we obtain c′_{2} < 0.79, which is not possible.

In what follows, assume that k > l with \frac{k}{l} ⩾ 2. Here,

z = \frac{k}{l} (y + μ_{2}i) − μ_{1}j ⩾ 2y − 3.

Together with Lemma 3.1, we deduce that z = 2y − 3 or z = 2y − 2. Since both y and z are odd integers, z = 2y − 2 is impossible. Therefore, there is only one possibility which is z = 2y − 3. Together with Lemma 2.1(4) yields that

q_{1} = gcd(P_{2y−3} − 1, P_{y} − 1) = gcd(P_{y−1} Q_{y−2} , P_{y} − 1).

If y ≡ 1 (mod 4), then

α^{\frac{z−1.19}{2}} < q_{1} = gcd(P_{y−1} Q_{y−2} , P_{y} − 1) = gcd(P_{y−1} Q_{y−2} , P_{\frac{y−1}{2}} Q_{\frac{y+1}{2}} )

⩽ gcd(P_{y−1} , P_{\frac{y−1}{2}} ) gcd(P_{y−1} , Q_{\frac{y+1}{2}} ) gcd(Q_{y−2} , P_{\frac{y−1}{2}} ) gcd(Q_{y−2} , Q_{\frac{y+1}{2}} )

⩽ P_{\frac{y−1}{2}} Q_{2}Q_{1}Q_{3} < α^{\frac{z+1}{4} +5.14}

yields that z < 24, which is not possible.

If y ≡ 3 (mod 4), then

α^{\frac{z−1.19}{2}} < q_{1} = gcd(P_{y−1} Q_{y−2} , P_{y} − 1) = gcd(P_{y−1} Q_{y−2} , P_{\frac{y+1}{2}} Q_{\frac{y−1}{2}} )

⩽ gcd(P_{y−1} , P_{\frac{y+1}{2}} ) gcd(P_{y−1} , Q_{\frac{y−1}{2}} ) gcd(Q_{y−2} , P_{\frac{y+1}{2}} ) gcd(Q_{y−2} , Q_{\frac{y−1}{2}} )

⩽ P_{2}Q_{\frac{y−1}{2}} Q_{3}Q_{1} < α^{\frac{z+1}{4} +5.14} .

But this is also impossible since z > 280. In the sequel, assume that \frac{k}{l} < 2. Note that this condition implies k ⩾ 3. Taking any pair (μ_{1}, μ_{2}) ≠ (μ′_{1}, μ′_{2}), then we have z + μ′_{1}j = \frac{k}{l} (y + μ_{2}i) − μ_{1}j + μ′_{1}j. The main objective is to find an upper bound for q_{1}^{(0)} = gcd( \frac{z+μ′_{1} j}{2} , \frac{y+μ′_{2} i}{2} ). Then

q_{1}^{(0)} = \frac{1}{2} gcd(z + μ′_{1}j, y + μ′_{2}i)

⩽ \frac{1}{2} gcd(k(y + μ_{2}i) − l(μ_{1}j − μ′_{1}j), k(y + μ′_{2}j)) ⩽ |k + l|.

Since 2 ⩽ l < k ⩽ 4 and (k, l) = 1, the cases (k, l) = (4, 3), (3, 2) hold. Therefore,

α^{\frac{z−1.19}{2}} < \sqrt{P_{z}} < gcd(P_{z} − 1, P_{y} − 1) ⩽ α^{\frac{z+1}{4} +3(7+0.1)−1.16} .

Then we get z < 84 which is not possible.

(3) y is even and z is odd

By the definition of the pellans sequence, we get that S_{z} = P_{z} and S_{y} = B_{\frac{y}{2}} since y is an even integer and z is an odd integer. Now, we give a lemma which implies a relation between the integers y and z.

Lemma 3.2. The system

(3.2)  ab + 1 = S_{x}, ac + 1 = B_{\frac{y}{2}} , bc + 1 = P_{z}

satisfies z ⩽ 2y − 3.

Proof. The last two equations yields \sqrt{P_{z}} < c < B_{\frac{y}{2}} . By Lemma 2.2, we have

α^{\frac{z−1.19}{2}} < \sqrt{P_{z}} < c < B_{\frac{y}{2}} < α^{y−1.96} ,

which yields z ⩽ 2y − 3. □

Put q_{2} = gcd(S_{y} − 1, S_{z} − 1) = gcd(B_{\frac{y}{2}} − 1, P_{z} − 1). System (3.2) gives that
\sqrt{P_{z}} < q_{2} = gcd(B_{\frac{y}{2}} − 1, P_{z} − 1).

After applying the properties in Lemma 2.1, we obtain the following for some i ∈ {±1}

q_{2} = gcd(B_{\frac{y}{2}} − 1, P_{z} − 1) ⩽ gcd(B_{\frac{y}{2} −1} B_{\frac{y}{2} +1} , P_{\frac{z−i}{2}} Q_{\frac{z+i}{2}} )

⩽ gcd(B_{\frac{y}{2} −1} , P_{\frac{z−i}{2}} ) gcd(B_{\frac{y}{2} −1} , Q_{\frac{z+i}{2}} ) gcd(B_{\frac{y}{2} +1} , P_{\frac{z−i}{2}} )

× gcd(B_{\frac{y}{2} +1} , Q_{\frac{z+i}{2}} )

⩽ gcd( \frac{1}{2}P_{y−2} , P_{\frac{z−i}{2}} ) gcd( \frac{1}{2}P_{y−2} , Q_{\frac{z+i}{2}} ) gcd( \frac{1}{2}P_{y+2} , P_{\frac{z−i}{2}} )

× gcd( \frac{1}{2}P_{y+2} , Q_{\frac{z+i}{2}} )

⩽ ∏_{j∈\{±2\}} P_{gcd (y+j, \frac{z−i}{2} )} Q_{gcd (y−j, \frac{z+i}{2} )} .

Let gcd(y + η_{1}j, \frac{z+η_{2} i}{2} ) = \frac{z+η_{2} i}{2t} for some η_{1}, η_{2}, i ∈ \{±1\}^{3} and j ∈ \{±2\}. Firstly, suppose that t ⩾ 4. Lemma 2.2 leads to

α^{\frac{z−1.19}{2}} < α^{\frac{z−1}{8} −1.16+ \frac{z+1}{8} −1.16+ \frac{z−1}{8} +0.1+ \frac{z+1}{8} +0.1} .

When we compare the exponents of α, we arrive at a contradiction.

Now, assume that t = 3. Since gcd(y + η_{1}j, \frac{z+η_{2} i}{2} ) = \frac{z+η_{2} i}{6} , then one of the equations \frac{z+η_{2} i}{6} = y + η_{1}j, \frac{z+η_{2} i}{3} = y + η_{1}j, \frac{z+η_{2} i}{3} = \frac{y+η_{1} j}{2} and \frac{z+η_{2} i}{6} = \frac{y+η_{1} j}{5} holds where η_{1}, η_{2}, i ∈ \{±1\}^{3} and j ∈ \{±2\}. If y + η_{1}j = \frac{z+η_{2} i}{6} , then Lemma 3.2 yields that z = 6y + 6η_{1}j − η_{2}i ⩽ 2y − 3. Thus 2z + 6 ⩽ 4y ⩽ η_{2}i − 6η_{1}j − 3 ⩽ 10, contradicting the fact z > 280. If y + η_{1}j = \frac{z+η_{2} i}{3} , then we get z = 3y + 3η_{1}j − η_{2}i. Together with Lemma 3.2, we obtain that y ⩽ η_{2}i − 3η_{1}j − 3 ⩽ 4. Therefore z ⩽ 5, which is impossible.

Assume that \frac{z+η_{2} i}{3} = \frac{y+η_{1} j}{2} = t. The facts z = 3t − η_{2}i and y = 2t − η_{1}j yield that

q_{2} < P_{gcd}_{(2t−η_{1} j−2, \frac{3t−η_{2} i−i}{2} )} Q_{gcd}_{(2t−η_{1} j−2, \frac{3t−η_{2} i+i}{2} )}
× P_{gcd}_{(2t−η_{1} j+2, \frac{3t−η_{2} i−i}{2} )} Q_{gcd}_{(2t−η_{1} j+2, \frac{3t−η_{2} i+i}{2} )}

< α^{2(16−1.16) +16+0.1+ \frac{z+1}{3} +0.1} . (3.3)

As α^{\frac{z−1.19}{2}} < q_{2}, we get z < 280.53 together with (3.3). Similarly, for the case \frac{z+η_{2} i}{6} = \frac{y+η_{1} j}{5} , we find that the upper bound of z is 150. Both cases contradict with the assumption z > 280. In the sequel, suppose that t = 2. The possibilities are y + η_{1}j = \frac{z+η_{2} i}{4} and \frac{y+η_{1} j}{3} = \frac{z+η_{2} i}{4} . If we continue as above, we get the upper bound as 126.9 for the case \frac{y+η_{1} j}{3} = \frac{z+η_{2} i}{4} . This is not possible since z > 280. The case y+η_{1}j = \frac{z+η_{2} i}{4} yields together with Lemma 3.2 that z = 4y+4η_{1}j−η_{2}i ⩽ 2y−3. Consequently, z + 3 ⩽ 2y ⩽ η_{2}i − 4η_{1}j − 3 ⩽ 6, which is not possible.

The case t = 1 leads to z = 2(y + η_{1}j) − η_{2}i. Since 2y − 5 ⩽ z ⩽ 2y − 3 and z is odd integer, then there are two possibilities which are z = 2y − 3 and z = 2y − 5. Assume that z = 2y − 3. When we divide the third equation by the second one in system (3.2), we obtain the following inequality.

α^{y−2.25} < \frac{P_{2y−3}}{P_{y}/2} = \frac{bc + 1}{ac + 1} < \frac{b}{a}.

By multiplying both sides with a^{2}, we get a^{2}α^{y−2.25} < ab < S_{x}. If x is even, then a^{2}α^{y−2.25} < S_{x} = B\frac{x}{2} < α^{x−1.96} . Therefore,

a^{2} < α^{x−y−1.96+2.25} ⩽ α^{−0.71} < 0.6.

But this is not possible since a is positive integer.

Now, assume that x is odd. The inequalities a^{2}α^{y−2.25} < S_{x} = P_{x} < α^{x−1.16} yield

a^{2} < α^{x−y−1.16+2.25} ⩽ α^{0.09} < 1.09.

The only possibility is a = 1. Equation system (3.2) yields that b = P_{x} − 1, c = \frac{P_{y}}{2} − 1 and (P_{x} − 1)( \frac{P_{y}}{2} − 1) = P_{2y−3} − 1. Since

α^{y−2.18} < P_{x} − 1 = \frac{P_{2y−3} − 1}{P_{y}/2 − 1} < α^{y−2.17} ,
α^{x−1.28} < P_{x} − 1 = \frac{P_{2y−3} − 1}{P_{y}/2 − 1} < α^{x−1.16} ,

we get that −1.03 < x − y < −0.8 which gives that y = x + 1. The equation (P_{x} − 1)( \frac{P_{y}}{2} − 1) = P_{2y−3} − 1 yields that (P_{x} − 1)( \frac{P_{x+1}}{2} − 1) = P_{2x−1} − 1. But this is not possible since (P_{x} −1)( \frac{P_{x+1}}{2} −1) < P_{2x−1} −1. Similarly to above, we see that the case z = 2y − 5 is also impossible. In order to avoid unnecessary repetition, we omit this case.

(4) y is odd and z is even Now, we give a lemma.

Lemma 3.3. The system

ab + 1 = S_{x}, ac + 1 = P_{y}, bc + 1 = B\frac{z}{2}

satisfies z ⩽ 2y − 1.

Proof. The previous equation system gives \sqrt{B_{z/2}} < c < P_{y}. By Lemma 2.2, we have α^{(z−1.97)/2} < \sqrt{B_{z/2}} < c < P_{y} < α^{y−1.16} , which yields z ⩽ 2y − 1. □

Put q_{3} = gcd(S_{y} − 1, S_{z} − 1). By the definition of the pellans sequence, we get

q_{3} = gcd(S_{y} − 1, S_{z} − 1) = gcd(P_{y} − 1, B\frac{z}{2} − 1).

The properties in Lemma 2.1 yield that

q_{3} ⩽ gcd(P_{\frac{y−i}{2}} Q_{\frac{y+i}{2}} , B_{\frac{z}{2} −1} B_{\frac{z}{2} +1} )
q_{3} ⩽ gcd(P_{\frac{y−i}{2}} , B_{\frac{z}{2} −1} ) gcd(P_{\frac{y−i}{2}} , B_{\frac{z}{2} +1} ) gcd(Q_{\frac{y+i}{2}} , B_{\frac{z}{2} −1} ) gcd(Q_{\frac{y+i}{2}} , B_{\frac{z}{2} +1} )

⩽ gcd(P_{\frac{y−i}{2}} , P_{z−2} ) gcd(P_{\frac{y−i}{2}} , P_{z+2} ) gcd(Q_{\frac{y+i}{2}} , P_{z−2} ) gcd(Q_{\frac{y+i}{2}} , P_{z+2} )

= ∏ ; j∈\{±2\} gcd(P_{\frac{y−i}{2}} , P_{z+j} ) gcd(Q_{\frac{y+i}{2}} , P_{z−j} )

= ∏_{j∈\{±2\}} P_{gcd}_{(\frac{y−i}{2} ,z+j)} Q_{gcd}_{(\frac{y+i}{2} ,z−j)}

for some i ∈ \{±1\}. Let gcd( \frac{y+ξ_{1} i}{2} , z + ξ_{2}j) = \frac{z+ξ_{2} j}{w} where ξ_{1}, ξ_{2} ∈ \{±1\}. First assume that w ⩾ 8. Then we have

α^{\frac{z−1.97}{2}} < \sqrt{B\frac{z}{2}} < c < q_{3} ⩽ α^{(\frac{z−j}{8} −1.16) +(\frac{z+j}{8} −1.16) +(\frac{z−j}{8} +0.1) +(\frac{z−j}{8} +0.1)} .

The above formula leads to \frac{z}{2} − 0.99 < \frac{z}{2} − 2.12 which is an absurdity.

Now, assume that w ⩽ 7. Further assume that \frac{y+ξ_{1} i}{2k} = \frac{z+ξ_{2} j}{l} holds for a suitable positive integer l coprime to k. If k ⩾ l, then according to y < z,

y + ξ_{2}j < z + ξ_{2}j ⩽ \frac{y + ξ_{1}i}{2}

yields that y ⩽ 5. The inequality z ⩽ 2y − 1 implies that z ⩽ 9 which is impossible. Assume that k < l. First we analyze the case 2 ⩽ \frac{l}{2k}. Then,

z = \frac{l}{2k}(y + ξ_{1}i) − ξ_{2}j ⩾ 2(y − 1) − 2 = 2y − 4,

which together with z ⩽ 2y − 1 implies the following possibilities.

If z = 2y − 4, then by Lemma 2.1 we have the following for some i ∈ {±1}

α^{\frac{z−1.97}{2}} < gcd(P_{y} − 1, B\frac{z}{2} − 1) = gcd(P_{y} − 1, B_{y−2} − 1)

= gcd( P_{\frac{y−i}{2}} Q_{\frac{y+i}{2}} , \frac{1}{2}P_{y−3} Q_{y−1} ) < gcd( P_{\frac{y−i}{2}} Q_{\frac{y+i}{2}} , P_{y−3} Q_{y−1} )

⩽ P_{gcd}_{(\frac{y−i}{2} ,y−3)} Q_{gcd}_{(\frac{y−i}{2} ,y−1)} Q_{gcd}_{(\frac{y+i}{2} ,y−3)} Q_{gcd}_{(\frac{y+i}{2} ,y−1)} .

If i = 1, then
α^{\frac{z−1.97}{2}} = α^{\frac{2y−5.97}{2}}

⩽ P_{gcd}_{(\frac{y−1}{2} ,y−3)} Q_{gcd}_{(\frac{y−1}{2} ,y−1)} Q_{gcd}_{(\frac{y+1}{2} ,y−3)} Q_{gcd}_{(\frac{y+1}{2} ,y−1)}

⩽ P_{2}Q_{\frac{y−1}{2}} Q_{4}Q_{2} < α^{\frac{y}{2} +7.14}

leads to y ⩽ 20. Then z ⩽ 36 which is a contradiction.

If i = −1, then
α^{\frac{z−1.97}{2}} = α^{\frac{2y−5.97}{2}}
⩽ P_{gcd}_{(\frac{y+1}{2} ,y−3)} Q_{gcd}_{(\frac{y+1}{2} ,y−1)} Q_{gcd}_{(\frac{y−1}{2} ,y−3)} Q_{gcd}_{(\frac{y−1}{2} ,y−1)}
⩽ P_{4}Q_{2}Q_{4}Q_{\frac{y−1}{2}} < α^{\frac{y}{2} +7.14} .

When we compare the exponents of α, we arrive at a contradiction. Assume that z = 2y − 2. Then

α^{\frac{z−1.97}{2}} = α^{\frac{2y−3.97}{2}} < ∏_{j∈\{±2\}} P_{gcd}_{(\frac{y−i}{2} ,2y−2+j)} Q_{gcd}_{(\frac{y+i}{2} ,2y−2−j)}
⩽ P_{6}P_{4}Q_{6}Q_{4} < α^{17.88}

yields that z ⩽ 37. But, this is impossible.

Now, assume that \frac{l}{2k} < 2. This implies that l ⩾ 2. If l = 2, then k = 1 as k < l. Together with y < z and \frac{y+ξ_{1} i}{2} = \frac{z+ξ_{2} j}{2} , we get z = y + 1. If y ≡ 1 (mod 4), then z ≡ 1 (mod 4). So,

α^{\frac{z−1.19}{2}} < c < q_{3} = gcd( P_{y} − 1, \frac{P_{z}}{2} − 1)
< gcd(P_{y} − 1, P_{y+1} − 2) = gcd( P_{\frac{y−1}{2}} Q_{\frac{y+1}{2}} , P_{\frac{y+1−2}{2}} Q_{\frac{y+1+2}{2}} )
= P_{\frac{y−1}{2}} Q_{1}.

The inequality \frac{z−1.19}{2} < \frac{y−1}{2} −1.16+0.8 yields that −0.19 < −3.16 since z = y +1. But this is false.

If y ≡ 3 (mod 4), then z ≡ 0 (mod 4). Therefore,

α^{\frac{z−1.19}{2}} < c < q_{3} = gcd( P_{y} − 1, \frac{P_{z}}{2} − 1)
< gcd(P_{y} − 1, P_{y+1} − 2) = gcd( P_{\frac{y+1}{2}} Q_{\frac{y−1}{2}} , P_{\frac{y+1+2}{2}} Q_{\frac{y+1−2}{2}} )
= Q_{\frac{y−1}{2}} P_{1}.

When we compare the exponents of α, we get \frac{z−1.19}{2} < \frac{y−1}{2} +0.1. As z = y +1, we arrive at a contradiction.

In the sequel, assume that l ⩾ 3. Taking any pair (ξ′_{1}, ξ′_{2}) ≠ (ξ_{1}, ξ_{2}), we have

z + ξ′_{2}j = \frac{l}{2k}(y + ξ_{1}i) − ξ_{2}j + ξ′_{2}j.

When we evaluate the upper bound for q_{3}^{(0)} = gcd( \frac{y+ξ′_{1}i}{2} , z + ξ′_{2}j), we have

q_{3}^{(0)} = gcd( \frac{y + ξ′_{1}i}{2} , z + ξ′_{2}j) = gcd( \frac{y + ξ′_{1}i}{2} , \frac{l}{2k}(y + ξ_{1}i) − ξ_{2}j + ξ′_{2}j)
⩽ gcd(l(y + ξ′_{1}i), l(y + ξ_{1}i) − 2k(ξ_{2}j − ξ′_{2}j))
⩽ |l(ξ′_{1}i − ξ_{1}i) − 2k(ξ_{2}j − ξ′_{2}j)|.

The three cases ξ′_{1} ≠ ξ_{1}, ξ′_{2} ≠ ξ_{2} and ξ′_{1} ≠ ξ_{1}, ξ′_{2} = ξ_{2} and ξ′_{1} = ξ_{1}, ξ′_{2} ≠ ξ_{2} imply that q_{3}^{(0)} ⩽ 2(2k + l). Then

α^{\frac{z−1.97}{2}} < q_{3} = gcd(P_{y} − 1, B\frac{z}{2} − 1)

⩽ ∏_{j∈\{±2\}} P_{gcd}_{(\frac{y−i}{2} ,z+j)} Q_{gcd}_{(\frac{y+i}{2} ,z−j)} < α^{\frac{z+2}{l} +2(2k+l) +2·0.1−2·1.16} .

But none of these pairs satisfies the inequality \frac{l}{2k} < 2. The eligible pairs are

(l, k) = P (3, 1), (3, 2), (4, 3), (5, 2), (5, 3), (5, 4),
(6, 5), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6).

The previous argument provides the upper bounds

z < 57.19, 81.19, 77.46, 57.55, 70.89, 84.22,
93.6, 59.23, 70.43, 81.63, 92.83, 104.03,

respectively. But this is impossible since z > 280. Hence, the proof of Theorem 1.1 is completed.

Acknowledgements. The authors express their gratitude to the anonymous referee for the instructive suggestions.

References

[1] M. Alp, N. Irmak, L. Szalay, Balancing Diophantine triples, Acta Univ. Sapientiae 4 (2012), 11–19.

[2] A. Behera, G. K. Panda, On the square roots of triangular numbers, Fibonacci Quarter. 37(2) (1999), 98–105.

[3] R. D. Carmichael, On the numeric factors of the arithmetic forms α^{n} ± β^{n}, Ann. Math. (2) 15(1/4) (1913–1914), 30–48.

[4] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 (2004), 183–214.

[5] C. Fuchs, F. Luca, L. Szalay, Diophantine triples with values in binary recurrences, Ann. Scuola Norm. Sup. Pisa. Cl. Sci. III 5 (2008), 579–608.

[6] V. E. Hoggat, G. E. Bergum, A problem of Fermat and Fibonacci sequence, Fibonacci Quart. 15 (1977), 323–330.

[7] N. Irmak, L. Szalay, Diophantine triples and reduced quadruples with Lucas sequence of recurrence u_{n} = Au_{n−1} − u_{n−2} , Glas. Mat. 49(2) (2014), 303–312.

[8] F. Luca, L. Szalay, Fibonacci Diophantine triples, Glas. Mat. 43(2) (2008), 253–264.

[9] , Lucas Diophantine triples, Integers 9 (2009), 441–457.

[10] G. K. Panda, P. K. Ray, Some links of balancing and cobalancing with Pell and associated Pell numbers, Bull. Inst. Math. Acad. Sin. 6(1) (2011), 41–72.

Department of Mathematics (Received 16 01 2015) Art and Science Faculty (Revised 28 04 2015 and 23 06 2015) Niğde University Niğde Turkey

   nirmak@nigde.edu.tr
   irmaknurettin@gmail.com
   muratalp@nigde.edu.tr