On Pell, Pell-Lucas, and Balancing Numbers Gül Karadeniz Gözeri* - DocsLib

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Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 https://doi.org/10.1186/s13660-017-1599-1

R E S E A R C H Open Access On Pell, Pell-Lucas, and balancing numbers Gül Karadeniz Gözeri*

*Correspondence: [email protected] Abstract Department of Mathematics, Faculty of Science, Istanbul In this paper, we derive some identities on Pell, Pell-Lucas, and balancing numbers University, Vezneciler, Istanbul, and the relationships between them. We also deduce some formulas on the sums, Turkey divisibility properties, perfect squares, Pythagorean triples involving these numbers. Moreover, we obtain the set of positive integer solutions of some speciﬁc Pell equations in terms of the integer sequences mentioned in the text. Keywords: Pell numbers; Pell-Lucas numbers; balancing numbers; perfect squares; divisibility properties; Pell equations

1 Introduction and preliminaries Let p and q be two integers such that d = p2 –4q = 0 (to exclude a degenerate case). We

set two integer sequences Un and Vn by

Un = Un(p, q)=pUn–1 – qUn–2 and Vn = Vn(p, q)=pVn–1 – qVn–2 (1)

for n ≥ 2 with initial values U0 =0,U1 =1andV0 =2,V1 = p. The characteristic equation of (1)isx2 – px + q = 0, and hence its roots are √ √ p + d p – d α = and β = .(2) 2 2

Their Binet formulas are

αn – βn U = and V = αn + βn (3) n α – β n

for n ≥ 0.

Note that, in (1), Un(1, –1) = Fn, Fibonacci numbers (sequence A000045 in OEIS),

Vn(1, –1) = Ln, Lucas numbers (sequence A000032 in OEIS), Un(2, –1) = Pn,Pellnumbers

(sequence A000129 in OEIS), and Vn(2, –1) = Qn, Pell-Lucas numbers (sequence A002203 in OEIS) (for further details, see [1–6]). Balancing numbers have been deﬁned in [7]and[8]. A positive integer n is called a balancing number if the Diophantine equation

1+2+···+(n –1)=(n +1)+(n +2)+···+(n + r)(4)

© The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 2 of 16

holds for some positive integer r,whichiscalledacobalancingnumber.From(4)wehave (n–1)n r(r+1) 2 = rn + 2 ,andso √ √ –(2n +1)+ 8n2 +1 2r +1+ 8r2 +8r +1 r = and n = .(5) 2 2

Let Bn denote the nth balancing number, and let bn denote the nth cobalancing num- 2 ber. Then by (5) Bn is a balancing number if and only if 8Bn +1isaperfectsquare, 2 and bn is a cobalancing number if and only if 8bn +8bn +1isaperfectsquare.So 2 2 Cn = 8Bn +1andcn = 8bn +8bn + 1 are integers, called the nth Lucas-balancing and nth Lucas-cobalancing number, respectively. The Binet formulas for balancing numbers 2n 2n 2n–1 2n–1 2n 2n 2n–1 2n–1 are B = α √–β , b = α √–β – 1 , C = α +β ,andc = α +β ,respectively,where n √ 4 2 n √4 2 2 n 2 n 2 α =1+ 2andβ =1– 2 (for further details, see [9–12]). Later balancing numbers were generalized to the t-balancing numbers (see [13]) for an integer t ≥ 2. A positive integer n is called a t-balancing number if

1+2+···+ n =(n +1+t)+(n +2+t)+···+(n + r + t)(6)

for some positive integer r, which is called a t-cobalancing number. From (6)weobserve that –(2n +2t +1)+ 8n2 +8n(1 + t)+(2t +1)2 r = , 2 √ (7) (2r –1)+ 8r2 +8tr +1 n = . 2

t t Let Bn denote the nth t-balancing number, and let bn denote the nth t-cobalancing t t 2 number. Then from (7)weseethatBn is a t-balancing number if and only if 8(Bn) + t 2 t 8Bn(1 + t)+(2t +1) is a perfect square and that bn is a t-cobalancing number if and only t 2 t if 8(bn) +8tbn +1isaperfectsquare.So t t 2 t 2 t t 2 t Cn = 8 Bn +8Bn(1 + t)+(2t +1) and cn = 8 bn +8tbn +1 (8)

are integers, which are called the nth Lucas t-balancing and the nth Lucas t-cobalancing number (for further details, see [14]). Santana and Diaz-Barrero [15], setting a sequence in Lemma 2 as

an = P2n + P2n+1,(9)

≥ proved that it was a generalized Fibonacci sequence given by an+1√=6an – an–1 for√n 1, αn–βn with initial values a0 =1anda1 =7.SincePn = α–β for α =1+ 2andβ =1– 2, we get

α2n+1 + β2n+1 a = (10) n 2

for n ≥ 0. We can easily see that the sum of the ﬁrst n nonzero terms of an is

n 5a – a –6 a = n n–1 . (11) i 4 i=1 Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 3 of 16

It is proved in [15, Thm. 1] that the sum of the ﬁrst 4n + 1 nonzero terms of Pell numbers is 4n+1 α2n+1 + β2n+1 2 P = . i 2 i=1

We conclude that the sum of the ﬁrst 4n + 1 nonzero terms of Pell numbers is

4n+1 2 Pi = an. i=1

Moreover, Santana and Diaz-Barrero [15] proved that the sum of the ﬁrst 4n+1 nonzero terms of Pell numbers is a perfect square: 4n+1 n 2 2n +1 r Pi = 2 . 2r i=1 r=0

Later, Tekcan and Tayat [16] proved that the sum of the ﬁrst 2n + 1 nonzero terms of Pell numbersisaperfectsquareifn is even or half of a perfect square if n is odd. They proved that the sum of the ﬁrst 2n + 1 nonzero terms of Pell numbers is ⎧ ⎪ n+1 n+1 2n+1 ⎨ α +β 2 ≥ ( 2 ) for even n 2, P = αn+1–βn+1 (12) i ⎩⎪ ( √ )2 i=1 2 ≥ 2 for odd n 1, √ √ where α =1+ 2andβ =1– 2. Considering (12) and setting two integer sequences

αn+1 + βn+1 αn+1 – βn+1 Xn = and Yn = √ (13) 2 2

for n ≥ 0, they proved that the sum of the ﬁrst 4n + 1 nonzero terms of Pell numbers is

4n+1 2 n+1 2 Pi = 2Xn –2XnYn–1 +(–1) i=1

for n ≥ 1.

In this paper, we give several results on the sequences an, Xn, Yn, Bn, bn, Qn, including sums, divisibility properties, perfect squares, and integer solutions of some speciﬁc Pell equations.

2 Results and discussion In this section, we derive our main results.

2.1 Sums and divisibility properties In this subsection, we deal with the sums and divisibility properties of numbers mentioned. First, we reformulate (11) in terms of Pell and Pell-Lucas numbers as follows. Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 4 of 16

Theorem 1 The sum of the ﬁrst n nonzero terms of an is

⎧ n ⎨ 17P2n–1+7P2n–2–3 2 , ai = ⎩ 5Q2n+2Q2n–1–6 i=1 4 .

··· 5an–an–1–6 Proof Since a1 + a2 + + an = 4 by (11) and since an = P2n + P2n+1 by (9), we deduce that

n 5a – a –6 a = n n–1 i 4 i=1 5(P + P )–(P + P )–6 = 2n 2n+1 2n–2 2n–1 4 5P +5(2P + P )–P – P –6 = 2n 2n 2n–1 2n–2 2n–1 4 15(2P + P )+4P – P –6 = 2n–1 2n–2 2n–1 2n–2 4 17P +7P –3 = 2n–1 2n–2 . 2

The second result can be proved similarly.

Theorem 2 For the sequences an, Xn, Yn, Bn, bn, Qn, and Pn, we have:

(1) Xn = Pn+1 + Pn for n ≥ 0, and the sum of the ﬁrst n nonzero terms of Xn is

n Xi = Pn +2Pn+1 –2. i=1

Moreover, Yn = Pn+2 – Pn for n ≥ 0, and the sum of the ﬁrst n nonzero terms of Yn is

n Yi = Pn +3Pn+1 –3. i=1

(2) an = X2n for n ≥ 0 or an = X2n–1 + Y2n–1 for n ≥ 1; Qn+1 =2Xn and Qn+2 – Qn+1 =2Yn ≥ Q2n+1 PnQn ≥ PnQn–1+Pn–1Qn–2 ≥ for n 0; an = 2 and Bn = 2 for n 0, and bn = 4 for n 1.

Proof (1) Let

n n Xn = T1α + T2β (14)

for some T1 and T2.Ifwetaken =0andn = 1, then we have the system of equations α+1 T1 + T2 =1andT1α + T2β = 3. This system of equations has the solution T1 = α–β and Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 5 of 16

–β–1 T2 = α–β .So(14)becomes

n n Xn = T1α + T2β α +1 –β –1 = αn + βn α – β α – β (α +1)αn –(β +1)βn = α – β αn+1 – βn+1 αn – βn = + α – β α – β

= Pn+1 + Pn,

··· Pn+Pn+1–1 as we wanted. Since Xn = Pn + Pn+1 and P1 + P2 + + Pn = 2 ,weeasilydeducethat

n n Xi = (Pi + Pi+1)=Pn +2Pn+1 –2. i=1 i=1

Similarly, it can be shown that Yn = Pn+2 –Pn for n ≥ 0andY1 +Y2 +···+Yn = Pn +3Pn+1 –3. α2n+1+β2n+1 αn+1+βn+1 (2) Since an = 2 and Xn = 2 ,weget

α2n+1 + β2n+1 X = = a . 2n 2 n

Similarly, since Xn = Pn+1 + Pn and Yn = Pn+2 – Pn,wegetXn + Yn = Pn+1 + Pn+2, and hence

X2n–1 + Y2n–1 = P2n + P2n+1 = an.

The remaining cases can be proved similarly.

Theorem 3 Let Pn denote the nth Pell number. (1) If n ≥ 2 is even, then ⎧ 2n–1 ⎨ Qn 2 ( 2 ) , 1+ Pi = ⎩ (2B n +2b n +1)2, i=1 2 2

and if n ≥ 1 is odd, then ⎧ 2n–1 ⎨ Qn 2 ( 2 ) , Pi = ⎩ 2 (b n+3 – B n+1 – B n–1 – b n–1 ) . i=1 2 2 2 2

(2) If n ≥ 2 is even, then

4n+1 2 2 2 Pi = 2Pn+1 –2Pn –1 , i=1 Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 6 of 16

and if n ≥ 1 is odd, then

4n+1 2 2 2 Pi = 2Pn+1 –2Pn +1 . i=1

Proof (1) Let n be even, say n =2k for some positive integer k.SinceP1 + P2 + ···+ Pn = Pn+Pn+1–1 2 ,weeasilyget

2n–1 P + P –1 1+ P =1+ 2n–1 2n i 2 i=1 P + P +1 = 2n–1 2n 2 4k–1 4k–1 4k 4k α –β √ +α –β +1 = 2 2 2 √ α4k(α–1 +1)+β4k(–β–1 –1)+2 2 = √ 4 2 α4k + β4k +2(αβ)2k = 4 α2k + β2k 2 = 2 Q 2 = n . 2

The other cases can be proved similarly.

Theorem 4 For the sequences mentioned, we have

2n+1 2n ai = anan+1, ai =4Bn+1P2n, i=1 i=1 2n+1 2n Bi = anBn+1, Y2i+1 =2anP2n+2, i=1 i=0 ⎧ 2n+1 2n+1 ⎨ 2 (a n ) , neven, 2 Qi =2(P2n+2 –1), Pi = ⎩ i=1 i=1 YnPn+1, nodd, ⎧ 2n 2n ⎨ 2PnPn+1, neven, P2i = anP2n, Pi = ⎩ a n–1 Xn, nodd, i=1 i=1 2 2n 2n X2i =8XnPn+1Bn, Bi = anBn, i=1 i=1 2n 2n Qi =4bn+1, B2i = P2nP2n+1X2nCn, i=1 i=1 2n 2n Y2i–1 =2anP2n, a2i–1 = a2nP2nX2n–1, i=1 i=1 Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 7 of 16

2n 2n P2i+1 = anP2n+1, X2i–1 = P2n+1Y2n–1, i=0 i=1 2n 2n P2i–1 = X2n–1P2n, a2i+1 = anP2n+1X4n+2, i=1 i=0 2n 2n a2i = B2na2n+1, Q2i =2P2n+1Y2n–1. i=1 i=1 2n+1 29a2n–5a2n–1–6 Proof Since i=1 ai = 4 by (11), we get

4n+1 4n+1 4n–1 4n–1 2n+1 29( α +β )–5(α +β )–6 a = 2 2 i 4 i=1 α4n(29α –5α–1)+β4n(29β –5β–1)–12 = 8 √ √ α4n(17 + 12 2) + β4n(17 – 12 2) – 6 = 4 α4n+4 + β4n+4 –6 = 4 α4n+4 + β4n+4 –(αβ)2n+1(β2 + α2) = 4 α2n+1 + β2n+1 α2n+3 + β2n+3 = 2 2

= anan+1.

The other cases can be proved similarly.

From Theorem 4 we have the following result.

Theorem 5 For the divisibility properties, we have 2n+1 2n 2n (1) If n is even, then a n | Pi, Pn| Pi, and Pn+1| Pi, and if n is odd, then 2 i=1 i=1 i=1 2n+1 2n+1 2n 2n Yn| Pi, Pn+1| Pi, a n–1 | Pi, and Xn| Pi. i=1 i=1 2 i=1 i=1 (2) For every n ≥ 1, 2n 2n 2n 2n an a2i+1, P2n+1 a2i+1, X4n+2 a2i+1, a2n a2i–1, i=0 i=0 i=0 i=1 2n+1 2n+1 2n+1 2n+1 an ai, an+1 ai, an Bi, Bn+1 Bi, i=1 i=1 i=1 i=1 2n 2n 2n 2n an Bi, Bn+1 ai, P2n ai, Bn Bi, i=1 i=1 i=1 i=1 2n 2n 2n 2n a2n+1 a2i, bn+1 Qi, an X2i+1, P2n a2i–1, i=1 i=1 i=0 i=1 2n 2n 2n 2n X2n–1 a2i–1, an P2i, P2n P2i, B2n a2i, i=1 i=1 i=1 i=1 Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 8 of 16

2n 2n 2n 2n P2n Q2i–1, an Q2i+1, P2n+1 X2i–1, Pn B2i–1, i=1 i=0 i=1 i=1 2n 2n 2n 2n Y2n–1 X2i–1, an Y2i+1, P2n+2 Y2i+1, an Y2i–1, i=1 i=0 i=0 i=1 2n 2n 2n 2n P2n Y2i–1, P2n+1 B2i+1, an B2i+1, Xn–1 B2i–1, i=1 i=0 i=0 i=1 2n 2n 2n 2n P2n+1 B2i, P2n+1 Q2i, Y2n–1 Q2i, P2n B2i, i=1 i=1 i=1 i=1 2n 2n 2n 2n Xn X2i, Pn+1 X2i, Bn X2i, X2n B2i. i=1 i=1 i=1 i=1

Finally, we give the following result.

Theorem 6 For the sequences an, Xn, Yn, Bn, Qn, and Pn, we have 2n a 2n a 2n B P2 i=0 2i+1 i=1 2i–1 i=0 2i+1 2n+1 2n = X4n+2, 2n = a2n, 2n = , i=0 P2i+1 i=1 P2i–1 i=0 Q2i+1 2 2n Q 2n Y i=1 2i i=1 2i–1 2n =2, 2n =2. i=1 X2i–1 i=1 P2i

2.2 Perfect squares 2 We see in (5)thatBn is a balancing number if and only if 8Bn +1isaperfectsquareand 2 that bn is a cobalancing number if and only if 8bn +8bn + 1 is a perfect square. Similarly, we can give the following result.

Theorem 7 For every n ≥ 1, (1) Q4n+2–2 is a perfect square and Q4n+2–2 = a ; 4 4 n (2) 2P2 –1is a perfect square, and 2P2 –1=a ; 2n–1 2n–1 n–1 (3) 2P2 +1is a perfect square, and 2P2 +1=C ; 2n 2n n (4) P2 + P P is a perfect square, and P2 + P P = X ; 2n+1 2n 2n+2 2n+1 2n 2n+2 2n 2 2 2 2 (5) P2n + P2n–1 + P4n –1is a perfect square, and P2n + P2n–1 + P4n –1=Y2n–1.

Proof (1) Applying the Binet formulas, we deduce that Q –2 α4n+2 + β4n+2 –2 α4n+2 + β4n+2 +2(αβ)2n+1 4n+2 = = 4 4 4 α2n+1 + β2n+1 2 α2n+1 + β2n+1 = = = a , 2 2 n

as claimed. The other cases can be proved similarly.

As in Theorem 7, we can give the following result. Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 9 of 16

Theorem 8 For the sequences an, Xn, Bn, bn, Qn, and Pn, we have

2n 2n B2i+1 = anP2n+1, B2i–1 =2CnXn–1Pn, i=0 i=1 2n 2n Q2i+1 i=0 = a , Q =2P . 2 n 2i–1 2n i=1

2n 2n n n 2n+1 2n+1 Proof Since B = α √–β , P = α √–β ,anda = α +β ,weget n 4 2 n 2 2 n 2

2n B2i+1 = B1 + B3 + ···+ B4n+1 i=0 α2 – β2 α6 – β6 α8n+2 – β8n+2 = √ + √ + ···+ √ 4 2 4 2 4 2 1 α8n+6 – α2 β8n+6 – β2 = √ – 4 2 α4 –1 β4 –1 α8n+4 –1 β8n+4 –1 = + 32 32 α4n+2 + β4n+2 –2(αβ)2n+1 α4n+2 + β4n+2 +2(αβ)2n+1 = 8 4 α2n+1 + β2n+1 2 α2n+1 – β2n+1 2 = √ 2 2 2

= anP2n+1.

The other cases can be proved similarly.

2.3 Continued fraction expansion Theorem 9 The continued fraction expansion of an+1 is an

an+1 = 5; (1, 4)n–1,1,6 an

for n ≥ 1(here (x)k means that there are k successive terms ‘x’).

Proof Let n =1.Then

a2 41 1 = =5+ 1 = [5; 1, 6]. a1 7 1+ 6 Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 10 of 16

an Let us assume that it is satisﬁed for n –1,thatis, =[5;(1,4)n–2,1,6].Thenweget an–1

an+1 1 = 5; (1, 4)n–1,1,6 =5+ 1 an 1+ 4+ 1 1+ ··· 1 1+ 6 1 1 =5+ 1 =5+ 1 1+ 1 1+ an –1+5+ –1+ a 1+ n–1 ··· 1 1+ 6

1 an – an–1 6an – an–1 =5+ an–1 =5+ = . 1+ an an an–an–1

So it is true for all n ≥ 1sincean+1 =6an – an–1.

2.4 Companion matrix 21 The companion matrix for Pell numbers is 10 .Itisknownthat n P P 21 n+1 n = . (15) Pn Pn–1 10

2 n So Pn+1Pn–1 – Pn =(–1) , which known as the Cassini identity, is an immediate conse- quence of the matrix formula [17]. If we take the nth power of the matrix in the left side of (15), then we can give the following theorem, which can be proved by induction on n.

Theorem 10 For the Pell numbers Pn, we have n P P P 2 P 2 n+1 n = n +1 n Pn Pn–1 Pn2 Pn2–1

for n ≥ 1.

2.5 Pythagorean triples

It is known that the Pell numbers Pn have a close connection with square triangular numbers, that is,

2 2 k 2 (P + P ) ((P + P ) –(–1) ) (P + P )P = k–1 k k–1 k . (16) k–1 k k 2

Note that the left side of (16) describes a square number, whereas the right side describes a triangular number, so the result is a square triangular number (see [18]). Notice that if a right triangle has integer side lengths a, b, c (necessarily satisfying the Pythagorean theorem a2 + b2 = c2), then (a, b, c) is known as a Pythagorean triple. As Martin [19]described, Pell numbers can be used to form Pythagorean triples in which a and b are one unit apart, Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 11 of 16

corresponding to right triangles that are nearly isosceles. For instance, 2 2 2 2 2PnPn+1, Pn+1 – Pn, Pn+1 + Pn

is a Pythagorean triple. Now we can give the following theorem related to Pythagorean triples. √ Theorem 11 2B2n, P2n–1 + P2n, and B2n + b2n +1form a Pythagorean triple, that is, √ 2 2 2 ( 2B2n) +(P2n–1 + P2n) =(B2n + b2n +1) , √ 2 2 and (PnQn +2bn +1) –8Bn, 2P2n, and X2n–1 form a Pythagorean triple, that is, √ 2 2 2 2 2 (PnQn +2bn +1) –8Bn +( 2P2n) = X2n–1.

Proof ApplyingtheBinetformulas,wededucethat √ 2 2 ( 2B2n) +(P2n–1 + P2n) √ α4n – β4n 2 α2n–1 – β2n–1 α2n – β2n 2 = 2 √ + √ + √ 4 2 2 2 2 2 α8n –2(αβ)2n + β8n 2α4n +4(αβ)2n +2β4n = + 16 8 (α4n + β4n)2 +4(α4n + β4n)+4 = 16 α4n(1 + α–1)–β4n(1 + β–1) 1 2 = √ + 4 2 2 α4n – β4n α4n–1 – β4n–1 1 2 = √ + √ – +1 4 2 4 2 2 2 =(B2n + b2n +1) .

The other case can be proved similarly.

2.6 The Pell equation Let d be any positive nonsquare integer, and let N be any ﬁxed integer. Then the equation

x2 – dy2 = ±N (17)

is known as a Pell-type equation; x2 – dy2 = N is the positive Pell-type equation, and x2 – dy2 =–N is the negative Pell-type equation. It is named after John Pell (1611-1685), a mathematician who searched for integer solutions to equations of this type in the seven- teenth century. Ironically, Pell was not the ﬁrst to work on this problem, nor did he con- tribute to our knowledge for solving it. Euler (1707-1783), who brought us the ψ-function, accidentally named the equation after Pell, and the name stuck. For N =1,thePellequationx2 – dy2 = ±1 is known as the classical Pell equation. The Pell equation x2 –dy2 = 1 was ﬁrst studied by Brahmagupta (598-670) and Bhaskara (1114- 1185). Its complete theory was worked out by Lagrange (1736-1813), not Pell. It is often Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 12 of 16

said that Euler (1707-1783) mistakenly attributed Brouncker’s (1620-1684) work on this equation to Pell. However, the equation appears in a book by Rahn (1622-1676), which was certainly written with Pell’s help: some say that it is entirely written by Pell. Perhaps Euler knew what he was doing in naming the equation. In 1657, Fermat stated (without giving proof) that the positive Pell equation x2 – dy2 = 1 has an inﬁnite number of solutions. If (m, n)isasolution,thatis,m2 – dn2 =1,then(m2 + dn2,2mn)isalsoasolution since 2 2 m2 + dn2 – d(2mn)2 = m2 – dn2 =1.

So the Pell equation x2 – dy2 = 1 has inﬁnitely many integer solutions. Later, in 1766, La- grange proved that the Pell equation x2 – dy2 = 1 has an inﬁnite number of solutions if

d is positive and nonsquare. The ﬁrst nontrivial solution (x1, y1) =( ±1, 0) of this equation is√ called the fundamental√ solution from which all others are easily computed since n xn + yn d =(x1 + y1 d) for n ≥ 1 can be found using, for example, the cyclic method [20], known in India in the 12th century, or using the slightly less eﬃcient but more reg- ular English method [20] (17th century). There are other methods to compute this so- called fundamental solution, some of which√ are based on a continued fraction expansion of the square root of d given as follows. Let d =[m0; m1, m2,...,ml]denotethecontin-

ued fraction expansion of period length l.SetA–2 =0,A–1 =1,Ak = mkAk–1 + Ak–2 and Ak B–2 =1,B–1 =0,Bk = mkBk–1 + Bk–2 for nonnegative integers k.ThenCk = is the kth √ Bk 2 2 convergent of d, and the fundamental solution of x – dy =1is(x1, y1)=(Al–1, Bl–1)

if l is even or (A2l–1, B2l–1)ifl is odd. Also, if l is odd, then the the fundamental solu- 2 2 tion of x – dy =–1is(x1, y1)=(Al–1, Bl–1) (for further details on Pell equations, see [21– 23]). It is known that there is a connection between integer sequences and Pell equations. For instance, Olajas [9] gave the integer solutions to x2 –5y2 = ±4 as follows.

2 2 Theorem 12 ([9, Thm. 2.17]) The only solutions of the equation x –5y = ±4 are x = ±Lm

and y = ±Fm, where Lm and Fm are the mth terms of the Lucas and Fibonacci sequences, respectively.

2 { }∞ For integers A and B such that A –4B = 0 (to exclude a degenerate case), R = Ri i=0 = R(A, B, R0, R1) is a second-order linear recurrence if the recurrence relation for i ≥ 2

Ri = ARi–1 – BRi–2 (18)

2 2 holds for its terms and R0, R1 are ﬁxed integers. For the Pell equation x –8y =1,Liptai [24] proved the following:

Theorem 13 ([24,Thm.1]) The terms of the second-order linear recurrence R(6, –1, 1, 6) are the solutions of the equation z2 –8y2 =1for some integer z.

Now we can return to our main problem. We consider the integer solutions of the Pell equations

x2 –8y2 =1, x2 –2y2 = ±4, and x2 –8y2 = ±4. Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 13 of 16

Theorem 14 (1) For the positive Pell equation x2 –8y2 =1, we have: Q2n P2n ≥ (a) The integer solutions are (xn, yn)=( 2 , 2 ) for n 1. (b) The integer solutions satisfy the recurrence relation

5(Q + Q )–Q 5(P + P )–P x = 2n–2 2n–4 2n–6 and y = 2n–2 2n–4 2n–6 n 2 n 2

for n ≥ 3. (2) The negative Pell equation x2 –8y2 =–1has no integer solutions. (3) For the positive Pell equation x2 –2y2 =4, we have:

(a) The integer solutions are (xn, yn)=(Q2n, Y2n–1) for n ≥ 1.

(b) The integer solutions (xn, yn) satisfy the recurrence relation

xn =5(Q2n–2 + Q2n–4)–Q2n–6 and yn =5(Y2n–3 + Y2n–5)–Y2n–7

for n ≥ 4. (4) For the negative Pell equation x2 –2y2 =–4, we have:

(a) The integer solutions are (xn, yn)=(Q2n–1, Y2n–2) for n ≥ 1.

(b) The integer solutions (xn, yn) satisfy the recurrence relation

xn =5(Q2n–3 + Q2n–5)–Q2n–7 and yn =5(Y2n–4 + Y2n–6)–Y2n–8

for n ≥ 4. (5) For the positive Pell equation x2 –8y2 =4, we have:

(a) The integer solutions are (xn, yn)=(Q2n, P2n) for n ≥ 1.

(b) The integer solutions (xn, yn) satisfy the recurrence relation

xn =5(Q2n–2 + Q2n–4)–Q2n–6 and yn =5(P2n–2 + P2n–4)–P2n–6

for n ≥ 3. (6) For the negative Pell equation x2 –8y2 =–4, we have:

(a) The integer solutions are (xn, yn)=(Q2n–1, P2n–1) for n ≥ 1.

(b) The integer solutions (xn, yn) satisfy the recurrence relation

xn =5(Q2n–3 + Q2n–5)–Q2n–7 and yn =5(P2n–3 + P2n–5)–P2n–7

for n ≥ 4.

n n Proof (1a) Notice that Q = αn + βn and P = α √–β .So n n 2 2

2n 2n 2 α2n–β2n 2 Q 2 P 2 (α + β ) –8( √ ) x2 –8y2 = 2n –8 2n = 2 2 2 2 4 α4n +2(αβ)2n + β4n –(α4n –2(αβ)2n + β4n) = 4 Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 14 of 16

4(αβ)2n = 4 =1

since αβ =–1. (1b)

5(Q + Q )–Q 5Q +5Q –(2Q + Q )+2Q 2n–2 2n–4 2n–6 = 2n–2 2n–4 2n–5 2n–6 2n–5 2 2 5Q +2(2Q + Q ) = 2n–2 2n–4 2n–5 2 2Q + Q = 2n–1 2n–2 2 Q = 2n 2

= xn.

5(P2n–2+P2n–4)–P2n–6 Similarly, it can be shown that yn = 2 . √ (2) The negative Pell equation x2 –8y2 = –1 has no integer solutions since 8=[2;1, 4], that is, the length of 2, which is even. n+1 n+1 (3a) Notice that Q = αn + βn and Y = α √–β .So n n 2 2n 2n 2 2 α – β x2 –2y2 = Q2 –2Y 2 = α2n + β2n –2 √ 2n 2n–1 2 = α4n +2(αβ)2n + β4n – α4n –2(αβ)2n + β4n

=4(αβ)2n

=4.

(3b) Similarly, we get

5(Q2n–2 + Q2n–4)–Q2n–6 =5Q2n–2 +5Q2n–4 –(2Q2n–5 + Q2n–6)+2Q2n–5

=5Q2n–2 +2(2Q2n–4 + Q2n–5)

=4Q2n–2 + Q2n–2 +2Q2n–3

=2(2Q2n–2 + Q2n–3)+Q2n–2

=2Q2n–1 + Q2n–2

= Q2n

= xn

The other cases can be proved similarly.

It is known that there are a number of applications of Pell and Fibonacci sequences on the theory of numbers. For instance, Sellers proved in [25, Thm. 2.1] that the number of

domino tilings of the graph W4 × Pn–1 equals the product of the nth Fibonacci and Pell Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 15 of 16

numbers for all n ≥ 2. Also, there has been a connection between Diophantine quadruples

and Fibonacci numbers. Recall that a set of m positive integers {a1, a2,...,am} is called a

Diophantine m-tuble if aiaj + 1 is a perfect square for all 1 ≤ i < j ≤ m and is called a D(n)-

m-tuble (or a Diophantine m-tuble with the property D(n)) if aiaj + n is a perfect square for all 1 ≤ i < j ≤ m. n 2 Cassini’s identity for Fibonacci number is FnFn+2 +(–1) = Fn+1 and is the basis for the construction of so-called Hoggatt-Bergum’s quadruple. Hoggatt and Bergum [26]proved that, for any positive integer k,theset

{F2k, F2k+2, F2k+4,4F2k+1F2k+2F2k+3}

is a Diophantine quadruple. Later Morgado [27]andHoradam[28] generalized this result. The identity 2 2 2 Fk–3Fk–2Fk–1Fk+1Fk+2Fk+3 + Lk = Fk 2Fk–1Fk+1 – Fk

is known as the Morgado identity. Using Fibonacci numbers, Dujella [29] deﬁned the elliptic curve (see [30])

2 Ek : y =(F2kx +1)(F2k+2x +1)(F2k+4x +1)

and determined the integer points on it by terms of Fibonacci numbers when the rank of

Ek(Q)is1. It is known that there are several identities on Fibonacci and Lucas numbers. Some of 2 2 2 2 2 them can be given as 4Fk–1Fk+1 + Fk = Lk and 4Fk–1Fk Fk+1 +1=(Fk + Fk–1Fk+1) .Using 2 these, Dujella [31] obtained some quantities on D(Fk )-quadruples 3 2 2 2Fk–1,2Fk+1,2Fk Fk+1Fk+2,2Fk+1Fk+2Fk+3 2Fk+1 – Fk , 2 2 3 Fk–1,4Fk+1, Fk–2Fk–1Fk+1 2Fk – Fk–1 , Fk Fk+2Fk+3 , (19) 3 4Fk–1, Fk+1, Fk–2F2k–2F2k–1, Fk LkLk+1 .

As in (19), the set 2 2 2 2 Fk–3Fk–2Fk+1, Fk–1Fk+2Fk+3, FkLk,4Fk–1FkFk+1 2Fk–1Fk+1 – Fk

2 is a D(Lk)-quadruple. Dujella and Ramasamy [32] considered the Fibonacci numbers and D(4)-quadruple. They proved that the set

{F2k,5F2k,4F2k+2,4L2kF4k+2}

is a D(4)-quadruple. Also, they considered integer solutions of the Pell equations by using a D(4)-quadruple. In the future work, we are planing to study D(n)-quadruples for the sequences mentioned for some n. Karadeniz Gözeri Journal of Inequalities and Applications (2018)2018:3 Page 16 of 16

3Conclusion In this work, we deduced some new results on Pell, Pell-Lucas, and balancing numbers including sums, divisibility properties, perfect squares, and integer solutions of some spe- ciﬁc Pell equations.

Acknowledgements This work was supported by the Scientiﬁc Research Projects Coordination Unit of Istanbul University (No. BEK-2017-25878).

Competing interests The author declares that she has no competing interests.

Authors’ contributions The author read and approved the ﬁnal manuscript.

Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 21 July 2017 Accepted: 9 December 2017

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