PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES. Nurettin Irmak and Murat Alp

PUBLICATIONS DE L INSTITUT MATHÉMATIQUE Nouvelle série, tome 100114 016, 59 69 DOI: 10.98/PIM161459I 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,..., 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 { 1 16, 16, 68 16, 105 16 }. Then, Fermat found the first integer quadruples as {1,, 8, 10}. In [6], the general form of this set was found by Hoggatt and Bergum as follows {F k, F k+, F k+4, 4F k+1 F k+ F k+ }, 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,, }. 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 010 Mathematics Subject Classification: Primary 11D09; Secondary 11B9. Key words and phrases: Diophantine triples, pell numbers, balancing numbers, pellans sequence. Communicated by Žarko Mijajlović. 59

60 IRMAK AND ALP to have finitely many triples. For an integer A, 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 with the initial conditions u 0 = 0 and u 1 = 1. 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 = P n 1 + P n 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 = and Q 1 =. The terms of Balancing sequence {B n } n 0 are defined by B n = 6B n 1 B n 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 n = 1 P n and S n+1 = P n+1. The first few terms of the pellans sequence are 0, 1, 1, 5, 6, 9, 5,.... It is obvious that the terms of pellans sequence {S n } n 0 satisfy S n = 6S n S n 4 with S 0 = 0, S 1 = 1, S = 1 and S = 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.. Preliminaries Now, we present required properties to prove Theorem 1.1. Lemma.1. Let m and n are positive integers. Then 1 gcds n, S m = { S gcdn,m. { P n Q n+1, n is even P n+1 1 = P n+1 Q n, n is odd and P P n+1 Q n 1 n is even n = P n 1 Q n+1 n is odd where {Q n } is the Pell Lucas sequence. B n = P n = P n Q n. 4 B n 1B n + 1 = B n 1 B n+1. 5 B n+1 1 = B n C n+1 where {C n } is associated sequence of {B n }. 6 C n = Q n. 7 If n is odd integer, then B n 1 = 1 P n 1Q n+1. Proof. The third and sixth identities can be found in [10]. For the fourth one, we refer to []. 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

PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES 61 means of []. 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 1 P n 1Q n+1 = αn 1 β n 1 α n+1 + β n+1 α β = 1 α n β n + = κn τ n + 1 = B n + 1 α β κ τ where α and β are the roots of the equation x x 1 = 0 and we use the facts κ = α and τ = β. The remaining identities can be proven by using Binet formulas. Lemma.. For all integers n, 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 n 1.97 < B n < α n 1.96 where α is the dominant root of the equation x x 1 = 0. Proof. The third inequality is from [1]. For n, we have P n αn β α β α n 1 β /α α β α n 1.19, which gives the lower bound for the sequence {P n }. Similarly, P n αn + β α β α n 1 + β /α α β α n 1.16 gives the upper bound. The bounds for the sequence {Q n } can be proven in a similar way.. Proof of Theorem 1.1 Since 0 < a < b < c, then 1 + 1 ab + 1 = S x. We get that x. From now on, the proof is split into two parts. Case 1. Assume that z 80. 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 S x 1S y 1 a =, x < y < z 80. S z 1 Going through all the eligible values for x, y and z, and we found no integer solution. Case. Assume that z > 80.

6 IRMAK AND ALP 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 = and l = 1 in [1]. If we take S x = P x 1 in the case k =, l = 1 in [1], then we can follow the similar way to complete the proof of theorem. Hence, we omit this case. 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.1. The system.1 ab + 1 = S x, ac + 1 = P y, bc + 1 = P z satisfy z y. Proof. The last two equations of 1.1 give P z < c < P y. By Lemma., we have α z 1.19/ < P z < c < P y < α y 1.16 which yields z y. Put q 1 = gcds 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.1, we get q 1 = gcdp y 1, P z 1 = gcd P y i Q y+i, P z j Q z+j, y+µi P gcd y i, z j Q gcd y i, z+j Q gcd y+i, z j Q gcd y+i, z+j where i, j {±1}. Let gcd z+µ 1j = z+µ 1j for some µ 1, µ {±1}. Suppose that t 5 for all quadruples i, j, µ 1, µ {±1} 4. Since c q 1, then Lemma. implies that < P z < c q 1 < α z+1 10 t z 1 z+1 1.16+ 10 +0.+ 10 +0.1. When we compare the exponents of α, we arrive at a contradiction. Assume that t 4 and that z + µ 1 i k = y + µ j l holds for suitable positive integers k and l such that gcdk, l = 1. Suppose for the moment l > k. Then, we get z = y + 1 since y < z together with y + µ 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 j, we obtain that z = y +. By Lemma.1, P z 1 = P z+i Q z i, P z 1 = P z i Q z +i.

PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES 6 hold for some i {±1}. If z mod 4, then we have the following, by Lemma.1 q 1 = gcdp z 1, P z 1 = gcd P z+1 Q z 1, P z Q z 1 = P gcd z+1, z Q z 1 = P Q z 1 Since q 1 = Q z 1, then we have c Q z 1 Q z 1 and applying Lemma., we obtain = Q z 1 or c Q z 1 = c 1 c > c 1 Pz, c 1 < c 1 Pz < Q z 1 < α z 1 +0.1... If c Q z 1, then So c 1 < α 0. < 1. which gives c 1 = 1. Then, we obtain c = Q z 1 c = Q z 1 in.1, we have a = P z.1, one can easily see that ab + 1 = P z then we obtain P x = P z 1 sequence {P n }, we deduce P x = P z 1 P x = P z 1 and b = P z+1. When we put. From the first equation in P z+1 + 1 = P z 1 = S x. If x is odd,. When we apply the upper and lower bounds for the α x 1.19 < α z..1 < z x α z 1.8 < α x 1.16 z x <.. There are no two odd integers x and y such that.1 < z x <.. Similarly, we get 1.5 < z x < 1.4 when x is even. This is impossible, too. In the sequel, if Q z 1 = c 1c > c 1 Pz, then c 1 < Q z 1 / P z < α 0. <.4. If c 1 =, we get c = Q z 1 which is the same situation as above. If c 1 = 1, then we obtain that a = P z / and b = P z+1 /. So, P z 1 + 5 = 4S x. If x is an odd integer, then S x = P x which yields that.48 < z x <.8. But, this is impossible. Similarly, if x is an even integer, then S x = B x. Using the upper and lower bounds for the sequence, we get.7 < z x < which is not possible. If z 1 mod 4, then q 1 = gcdp z 1, P z 1 = gcd P z 1 = Q gcd z+1, z P z 1 = Q 1 P z 1 Q z+1 = P z 1 Assume that q 1 = P z 1. Since c q 1, then c P z 1 = c c > c Pz. Since P z 1 impossible. If c P z 1, then P z 1 0.79, then we obtain c P z 1., P z 1 or c P z 1 / P z < 1, we get c < P z 1 = c c > c Pz. Since P z 1 < 0.79, which is not possible. In what follows, assume that k > l with k l. Here, z = k l y + µ i µ 1 j y. Q z. If c P z 1, then / P z < 1 which is / P z < α 0.7 < Together with Lemma.1, we deduce that z = y or z = y. Since both y and z are odd integers, z = y is impossible. Therefore, there is only one

64 IRMAK AND ALP possibility which is z = y. Together with Lemma.14 yields that q 1 = gcdp y 1, P y 1 = gcdp y 1 Q y, P y 1. If y 1 mod 4, then < q 1 = gcdp y 1 Q y, P y 1 = gcd P y 1 Q y, P y 1 Q y+1 gcd P y 1, P y 1 gcd Py 1, Q y+1 gcd Qy, P y 1 gcd Qy, Q y+1 P y 1 Q Q 1 Q < α z+1 4 +5.14 yields that z < 4, which is not possible. If y mod 4, then < q 1 = gcdp y 1 Q y, P y 1 = gcd P y 1 Q y, P y+1 Q y 1 gcd P y 1, P y+1 gcd Py 1, Q y 1 gcd Qy, P y+1 gcd Qy, Q y 1 P Q y 1 Q Q 1 < α z+1 4 +5.14. But this is also impossible since z > 80. In the sequel, assume that k l <. Note that this condition implies k. Taking any pair µ 1, µ µ 1, µ, then we have z + µ 1 j = k l y + µ i µ 1 j + µ 1j. The main objective is to find an upper bound for q 0 1 = gcd z+µ 1 j, y+µ i. Then q 0 1 = 1 gcd z + µ 1j, y + µ i 1 gcd ky + µ i l µ 1 j µ 1j, k y + µ j k + l. Since l < k 4 and k, l = 1, the cases k, l = 4,,, hold. Therefore, < P z < gcdp z 1, P y 1 α z+1 4 +7+0.1 1.16. Then we get z < 84 which is not possible. 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 y 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.. The system. ab + 1 = S x, ac + 1 = B y, bc + 1 = P z satisfies z y. Proof. The last two equations yields P z < c < B y. By Lemma., we have < P z < c < B y < αy 1.96, which yields z y.

PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES 65 Put q = gcds y 1, S z 1 = gcd B y 1, P z 1. System. gives that Pz < q = gcd B y 1, P z 1. After applying the properties in Lemma.1, we obtain the following for some i {±1} q = gcd B y 1, P z 1 gcd B y 1 B y +1, P z i Q z+i gcd B y 1, P z i gcd B y 1, Q z+i gcd B y +1, P z i gcd B y +1, Q z+i gcd 1 P y, P z i gcd 1 P y+, Q z+i j {±} 1 gcd P y, Q z+i P gcdy+j, z i Q gcdy j, z+i. gcd 1 P y+, P z i Let gcd y + η 1 j, z+ηi = z+η i t for some η 1, η, i {±1} and j {±}. Firstly, suppose that t 4. Lemma. leads to < α z 1 8 1.16+ z+1 8 z 1 z+1 1.16+ 8 +0.1+ 8 +0.1. When we compare the exponents of α, we arrive at a contradiction. Now, assume that t =. Since gcd y + η 1 j, z+ηi = z+η i 6, then one of the equations z+ηi z+η 6 = y + η 1 j, i z+η = y + η 1 j, i = y+η1j and z+ηi 6 = y+η1j 5 holds where η 1, η, i {±1} and j {±}. If y + η 1 j = z+ηi 6, then Lemma. yields that z = 6y + 6η 1 j η i y. Thus z + 6 4y η i 6η 1 j 10, contradicting the fact z > 80. If y + η 1 j = z+ηi, then we get z = y + η 1 j η i. Together with Lemma., we obtain that y η i η 1 j 4. Therefore z 5, which is impossible. Assume that z+ηi = y+η1j = t. The facts z = t η i and y = t η 1 j yield that. q < P gcdt η1j, t η i i Q gcdt η 1j, t η i+i P gcdt η1j+, t η i i Q gcdt η 1j+, t η i+i z+1 16 1.16+16+0.1+ < α +0.1. As < q, we get z < 80.5 together with.. Similarly, for the case z+η i 6 = y+η1j 5, we find that the upper bound of z is 150. Both cases contradict with the assumption z > 80. In the sequel, suppose that t =. The possibilities are y + η 1 j = z+ηi 4 and y+η1j = z+ηi 4. If we continue as above, we get the upper bound as 16.9 for the case y+η1j = z+ηi 4. This is not possible since z > 80. The case y+η 1 j = z+ηi 4 yields together with Lemma. that z = 4y+4η 1 j η i y. Consequently, z + y η i 4η 1 j 6, which is not possible.

66 IRMAK AND ALP The case t = 1 leads to z = y + η 1 j η i. Since y 5 z y and z is odd integer, then there are two possibilities which are z = y and z = y 5. Assume that z = y. When we divide the third equation by the second one in system., we obtain the following inequality. α y.5 < P y P y / = bc + 1 ac + 1 < b a. By multiplying both sides with a, we get a α y.5 < ab < S x. If x is even, then a α y.5 < S x = B x < αx 1.96. Therefore, a < α x y 1.96+.5 α 0.71 < 0.6. But this is not possible since a is positive integer. Now, assume that x is odd. The inequalities a α y.5 < S x = P x < α x 1.16 yield a < α x y 1.16+.5 α 0.09 < 1.09. The only possibility is a = 1. Equation system. yields that b = P x 1, c = Py 1 and P x 1 P y 1 = P y 1. Since α y.18 < P x 1 = P y 1 P y / 1 < αy.17, α x 1.8 < P x 1 = P y 1 P y / 1 < αx 1.16, we get that 1.0 < x y < 0.8 which gives that y = x + 1. The equation P x 1 P y 1 = P y 1 yields that P x 1 P x+1 1 = P x 1 1. But this is not possible since P x 1 P x+1 1 < P x 1 1. Similarly to above, we see that the case z = y 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.. The system satisfies z y 1. ab + 1 = S x, ac + 1 = P y, bc + 1 = B z Proof. The previous equation system gives B z/ < c < P y. By Lemma., we have / < B z/ < c < P y < α y 1.16, which yields z y 1. Put q = gcds y 1, S z 1. By the definition of the pellans sequence, we get q = gcds y 1, S z 1 = gcd P y 1, B z 1.

PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES 67 The properties in Lemma.1 yield that q gcd P y i Q y+i, B z 1 B z +1 q gcd P y i, B z 1 gcd P y i, B z +1 gcd Q y+i, B z 1 gcd Q y+i, B z +1 gcd P y i, P z gcd P y i, P z+ gcd Q y+i, P z gcd Q y+i, P z+ = { } gcd P y i, P z+j gcd Q y+i, P z j j ± = j {±} P gcd y i,z+jq gcd y+i,z j for some i {±1}. Let gcd y+ξ 1i, z + ξ j = z+ξj w where ξ 1, ξ {±1}. First assume that w 8. Then we have < B z < c < q α z j z+j z j z j 8 1.16+ 8 1.16+ 8 +0.1+ +0.1 8. The above formula leads to z 0.99 < z.1 which is an absurdity. Now, assume that w 7. Further assume that y+ξ1i k = z+ξj l holds for a suitable positive integer l coprime to k. If k l, then according to y < z, y + ξ j < z + ξ j y + ξ 1i yields that y 5. The inequality z y 1 implies that z 9 which is impossible. Assume that k < l. First we analyze the case l k. Then, z = l k y + ξ 1i ξ j y 1 = y 4, which together with z y 1 implies the following possibilities. If z = y 4, then by Lemma.1 we have the following for some i {±1} < gcdp y 1, B z 1 = gcdp y 1, B y 1 = gcd If i = 1, then P y i Q y+i, 1 P y Q y 1 < gcd P y i Q y+i, P y Q y 1 P gcd y i,y Q gcd y i,y 1Q gcd y+i,y Q gcd y+i,y 1. = α y 5.97 P gcd y 1,y Q gcd y 1,y 1Q gcd y+1,y Q gcd y+1,y 1 P Q y 1 Q 4 Q < α y +7.14 leads to y 0. Then z 6 which is a contradiction.

68 IRMAK AND ALP If i = 1, then = α y 5.97 P gcd y+1,y Q gcd y+1,y 1Q gcd y 1,y Q gcd y 1,y 1 P 4 Q Q 4 Q y 1 < α y +7.14. When we compare the exponents of α, we arrive at a contradiction. Assume that z = y. Then = α y.97 < j {±} P gcd y i,y +jq gcd y+i,y j P 6 P 4 Q 6 Q 4 < α 17.88 yields that z 7. But, this is impossible. l Now, assume that k <. This implies that l. If l =, then k = 1 as k < l. Together with y < z and y+ξ1i = z+ξj, we get z = y + 1. If y 1 mod 4, then z 1 mod 4. So, < c < q = gcd P y 1, P z 1 < gcdp y 1, P y+1 = gcd = P y 1 Q 1. < y 1 P y 1 Q y+1, P y+1 Q y+1+ The inequality z 1.19 1.16+0.8 yields that 0.19 <.16 since z = y +1. But this is false. If y mod 4, then z 0 mod 4. Therefore, < c < q = gcd P y 1, P z 1 < gcdp y 1, P y+1 = gcd = Q y 1 P 1. P y+1 Q y 1, P y+1+ Q y+1 When we compare the exponents of α, we get z 1.19 +0.1. As z = y +1, we arrive at a contradiction. In the sequel, assume that l. Taking any pair ξ 1, ξ ξ 1, ξ, we have < y 1 z + ξ j = l k y + ξ 1i ξ j + ξ j. When we evaluate the upper bound for q 0 = gcd y+ξ 1 i, z + ξ j, we have y + ξ 1 i y + ξ 1 i q 0 = gcd, z + ξ j = gcd, gcd l y + ξ 1 i, ly + ξ 1 i k ξ j ξ j l ξ 1 i ξ 1 i k ξ j ξ j. l k y + ξ 1i ξ j + ξ j

PELLANS SEQUENCE AND ITS DIOPHANTINE TRIPLES 69 The three cases ξ 1 ξ 1, ξ ξ and ξ 1 ξ 1, ξ = ξ and ξ 1 = ξ 1, ξ ξ imply that q 0 k + l. Then < q = gcd P y 1, B z 1 P gcd y i j {±},z+jq gcd y+i,z j < α z+ l +k+l+ 0.1 1.16. But none of these pairs satisfies the inequality l k <. The eligible pairs are l, k = P, 1,,, 4,, 5,, 5,, 5, 4, 6, 5, 7,, 7,, 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., 9.6, 59., 70.4, 81.6, 9.8, 104.0, respectively. But this is impossible since z > 80. 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 01, 11 19. [] A. Behera, G. K. Panda, On the square roots of triangular numbers, Fibonacci Quarter. 7 1999, 98 105. [] R. D. Carmichael, On the numeric factors of the arithmetic forms α n ± β n, Ann. Math. 151/4 191 1914, 0 48. [4] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 004, 18 14. [5] C. Fuchs, F. Luca, L. Szalay, Diophantine triples with values in binary recurrences, Ann. Scuola Norm. Sup. Pisa. Cl. Sci. III 5 008, 579 608. [6] V. E. Hoggat, G. E. Bergum, A problem of Fermat and Fibonacci sequence, Fibonacci Quart. 15 1977, 0. [7] N. Irmak, L. Szalay, Diophantine triples and reduced quadruples with Lucas sequence of recurrence u n = Au n 1 u n, Glas. Mat. 49 014, 0 1. [8] F. Luca, L. Szalay, Fibonacci Diophantine triples, Glas. Mat. 4 008, 5 64. [9], Lucas Diophantine triples, Integers 9 009, 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. 61 011, 41 7. Department of Mathematics Received 16 01 015 Art and Science Faculty Revised 8 04 015 and 06 015 Niğde University Niğde Turkey nirmak@nigde.edu.tr irmaknurettin@gmail.com muratalp@nigde.edu.tr