Balancing, Pell and Square Triangular Functions

Miskolc Mathematical Notes HU e-ISSN 1787-2413 Vol. 16 (2015), No. 2, pp. 1219–1231 DOI: 10.18514/MMN.2015.1724

BALANCING, PELL AND SQUARE TRIANGULAR FUNCTIONS

AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS Received 22 June, 2015

Abstract. In this work, we derive some functions on balancing, cobalancing, Lucas–balancing, Lucas–cobalancing, Pell, Pell–Lucas and square triangular numbers. At the end of this article we investigated common values of combinatorial numbers and Lucas–balancing numbers.

2010 Mathematics Subject Classiﬁcation: 11B37, 11B39, 11D25 Keywords: balancing numbers, Pell numbers, Pell–Lucas numbers, balancing functions, square triangular numbers

1. INTRODUCTION A positive integer n is called a balancing number (see [1] and [3]) if the Diophant- ine equation 1 2 .n 1/ .n 1/ .n 2/ .n r/ (1.1) C C C D C C C C C C holds for some positive integer r which is called cobalancing number (or balancer). For example 6;35;204;1189 and 6930 are balancing numbers with balancers 2;14;84; 492 and 2870, respectively. If n is a balancing number with balancer r, .n 1/n r.r 1/ then from (1.1) one has rn C ; and so 2 D C 2 .2n 1/ p8n2 1 2r 1 p8r2 8r 1 r C C C and n C C C C : (1.2) D 2 D 2 th th Let Bn denote the n balancing number, and let bn denote the n cobalancing number. Then they satisfy B1 1;B2 6;Bn 1 6Bn Bn 1 and b1 0;b2 D D C D D 2D 2;bn 1 6bn bn 1 2 for n 2. The zeros of the characteristic equation x C D C 6x 1 0 for balancing numbers are ˛1 3 p8 and ˇ1 3 p8. Ray derived someC niceD results on balancing and cobalancingD C numbers in hisD Phd thesis [14, p.19]. Since x is a balancing number if and only if 8x2 1 is a perfect square, he set y2 8x2 1 for some y 0, which is a Pell equation.C The fundamental solution is D ¤ n n .y1;x1/ .3;1/. So yn xnp8 .3 p8/ and yn xnp8 .3 p8/ for n 1. D C D C D This research was (partially) carried out in the framework of the Center of Excellence of Mechat- ronics and Logistics at the University of Miskolc.

c 2015 Miskolc University Press

1220 AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS

.3 p8/n .3 p8/n Thus xn C which is the Binet formula for balancing numbers and D 2p8 is denoted by Bn. Let ˛ 1 p2 and ˇ 1 p2 be the roots of the characteristic equation for Pell D C D and Pell–Lucas numbers defined by P0 0;P1 1;Pn 2Pn 1 Pn 2 and Q0 D D D C 2 D Q1 2;Qn 2Qn 1 Qn 2 for n 2, respectively. Notice that ˛ 3 p8 and 2 D D C ˛D2n ˇC2n ˇ 3 p8. So the Binet formula for balancing numbers is Bn : Thus D D 4p2 P2n there is a connection between balancing and Pell numbers by Bn 2 . Similarly, ˛2n 1 ˇ 2n 1 D 1 the Binet formula for cobalancing numbers is bn , and so bn D 4p2 2 D P2n 1 1 2 . 2 We note that Bn is a balancing number if and only if 8Bn 1 is a perfect square, 2 C and bn is a cobalancing number if and only if 8bn 8bn 1 is a perfect square. Thus (1.2) implies that C C q q 2 2 Cn 8B 1 and cn 8b 8bn 1 (1.3) D n C D n C C are integers called the nth Lucas–balancing and nth Lucas–cobalancing number, re- 2n 2n 2n 1 2n 1 ˛ ˇ ˛ ˇ spectively. Their Binet formulas are Cn C2 and cn C2 , respectively (for further details see [9, 11–13]).D D Balancing numbers and their generalizations have been investigated by several authors, from many aspects. In [5, Theorem 4], Liptai proved that there is no Fibonacci balancing number except 1, and in [6, Theorem 4] he proved that there is no Lucas balancing number. In [15], Szalay considered the same problem and obtained some nice results by a different method. In [4], Kovacs,´ Liptai and Olajos extended the concept of balancing numbers to the .a;b/-balancing numbers defined as follows: Let a > 0 and b 0 be coprime integers. If .a b/ .a.n 1/ b/ .a.n 1/ b/ .a.n r/ b/ C C C C D C C C C C C for some positive integers n and r, then an b is an .a;b/-balancing number. The C .a;b/ sequence of .a;b/-balancing numbers is denoted by Bm for m 1. In [7], Liptai, Luca, Pinter´ and Szalay generalized the notion of balancing numbers to numbers defined as follows: Let y;k;l ZC such that y 4. Then a positive integer x with x y 2 is called a .k;l/-power2 numerical center for y if Ä 1k .x 1/k .x 1/l .y 1/l : C C D C C C They studied the number of solutions of the equation above and proved several ef- fective and ineffective finiteness results for .k;l/-power numerical centers. For positive integers k;x, let ˘ .x/ x.x 1/:::.x k 1/: (1.4) k D C C BALANCING, PELL AND SQUARE TRIANGULAR FUNCTIONS 1221

It was proved in [4, Theorem 3 and Theorem 4] that the equation Bm ˘k.x/ for fixed integer k 2 has only infinitely many solutions and for k 2;3;4 Dall solutions were determined. In [17, Theorem 1] Tengely, considered the2 f case kg 5, that is, D Bm x.x 1/.x 2/.x 3/.x 4/; and proved that this Diophantine equation has no solutionD C for m C0 andCx Z.C Recently in [2], Dash, Ota2 and Dash considered the so-called t-balancing numbers for an integer t 2. A positive integer n is a t-balancing number if 1 2 n .n 1 t/ .n 2 t/ .n r t/ C C C D C C C C C C C C C holds for some positive integer r which is called t-cobalancing number. The t- t t balancing numbers are denoted by Bn and t-cobalancing numbers by bn. In [16], Tekcan, Tayat and Ozbek¨ derived the general terms of t-balancing numbers by solv- ing the Diophantine equation 8x2 y2 8x.1 t/ .2t 1/2 0: Balancing functions are also interesting C in theC theoryC ofC balancingD numbers. There were first considered by Behera, Panda and Ray in [1,14], who derived the following results. Lemma 1 ([1, Theorem 2.1]). For any balancing number x, the functions p p p F .x/ 2x 8x2 1, G.x/ 3x 8x2 1 and H.x/ 17x 6 8x2 1 D C D C C D C C are also balancing numbers. Lemma 2 ([1, Theorem 2.2]). If x any balancing number, then p K.x/ 6x 8x2 1 16x2 1 D C C C is an odd balancing number. Lemma 3 ([1, Theorem 4.1]). If x and y are balancing numbers, then q p f .x;y/ x 8y2 1 y 8x2 1 D C C C is also a balancing number. Lemma 4 ([14, 2.4.12 Theorem]). If x;y and ´ are balancing numbers, then q p p p f .x;y;´/ x 8y2 1 8´2 1 y 8x2 1 8´2 1 D C C C C C p q ´ 8x2 1 8y2 1 8xy´ C C C C is also a balancing number. Similarly to balancing numbers, Panda and Ray [12] defined the following functions for any two cobalancing numbers x and y: p f .x/ 3x 8x2 8x 1 1; D C C C C p g.x/ 17x 6 8x2 8x 1 8; D C C C C 1222 AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS p h.x/ 8x2 8x 1 .2x 1/ 8x2 8x 1; D C C C C C C 1h q t.x;y/ 2.2x 1/.2y 1/ .2x 1/ 8y2 8y 1 D 2 C C C C C C p p q i .2y 1/ 8x2 8x 1 8x2 8x 1 8y2 8y 1 1 : C C C C C C C C C They proved the following two lemmas. Lemma 5 ([12, Theorem 2.1]). For any two cobalancing numbers x and y, f .x/, g.x/, h.x/ and t.x;y/ are all cobalancing numbers. Lemma 6 ([12, Theorem 2.2]). If x any cobalancing number, then the cobalancing number next to x is f .x/ 3x p8x2 8x 1 1, and consequently, the previous D C C C C one is fe .x/ 3x p8x2 8x 1 1. D C C C Here we note, [12] originally contained a type at h.x/, we always use the correct version.

2. MAIN RESULTS In this section, we deal with balancing, Pell, Pell–Lucas and square triangular functions. Before considering balancing functions, we need the following theorem. Theorem 1. For any integers n;k;l 0, we get Bn k l BkCl Cn CkBl Cn CkCl Bn 8BkBl Bn: (2.1) C C D C C C Proof. It is known that Bn k CkBn BkCn and Ck l CkCl 8BkBl for integers n;l;k 0. So we easilyC getD C C D C B C Cn C B Cn C C Bn 8B B Bn k l C k l C k l C k l Cn.B C C B / Bn.C C 8B B / D k l C k l C k l C k l CnBk l BnCk l D C C C Bn k l D C C as we wanted. In Lemmas1–5, Behara, Panda and Ray did not determine precisely value of the functions F .x/, G.x/, H.x/, K.x/, f .x;y/, f .x;y;´/ and f .x/, g.x/, h.x/, t.x;y/. In the following theorem, we are able to do that. Theorem 2. (1) For the balancing functions F .x/;G.x/;H.x/;K.x/;f .x;y/; f .x;y;´/, we have

F .Bn/ B2n; G.Bn/ Bn 1; H.Bn/ Bn 2; K.Bn/ B2n 1; D D C D C D C f .Bn;Bk/ Bn k and f .Bn;Bk;Bl / Bn k l : D C D C C BALANCING, PELL AND SQUARE TRIANGULAR FUNCTIONS 1223

(2) For the cobalancing functions f .x/;g.x/;h.x/;t.x;y/, we have

f .bn/ bn 1; g.bn/ bn 2; h.bn/ b2n and t.bn;bm/ bn m: D C D C D D C p 2 Proof. (1) Since 8B 1 Cn, we easily get n C D Â˛2n ˇ2n ÃÂ˛2n ˇ2n Ã ˛4n ˇ4n F .Bn/ 2 C B2n: D 4p2 2 D 4p2 D Taking k 1 and l 0 in (2.1), one obtains D D q 2 Bn 1 3Bn Cn 3Bn 8Bn 1 G.Bn/: C D C D C C D If k 2 and l 0, then it leads to D D q 2 Bn 2 17Bn 6Cn 17Bn 6 8Bn 1 H.Bn/ C D C D C C D and taking k 1 and l n, we get D D q 2 2 2 B2n 1 6BnCn 16Bn 1 6Bn 8Bn 1 16Bn 1 K.Bn/: C D C C D C C C D Finally, if l 0, we have D q q 2 2 Bn k BnCk BkCn Bn 8Bk 1 Bk 8Bn 1 f .Bn;Bk/: C D C D C C C D The last assertions is obvious from (2.1) since

Bn k l BnCkCl BkCnCl Bl CnCk 8BnBkBl C C D C C C q q q q 2 2 2 2 Bn 8B 1 8B 1 B 8B 1 8B 1 D k C l C C k n C l C q q 2 2 B 8B 1 8B 1 8BnB B C l n C k C C k l f .Bn;B ;B /: D k l (2) It can be proved similarly. Apart from above theorem, we can give the following theorem. Theorem 3. (1) If x is the nth balancing number and y is the nth cobalancing number, then the function p 8xy p8x2 1 8y2 8y 1 1 B1.x;y/ C C C C C D 4 is the nth balancing number, and the function q p 2 2 B2.x;y/ x 2y 8x 1 8y 8y 1 1 D C C C C C C C is the .n 1/st balancing number. C 1224 AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS

th (2)( The generalized type of Lemma2 ) Let Bk denote the k balancing number, tk tk let bk denote the k cobalancing number and let Ck denote the k Lucas– balancing number for an integer k 0. Then for any balancing number x, the function n 2 1 p 2 2 X Bk .x/ Bkx 8x 1 Ckx 1 2b2n k B2i k 1 D C C C C C C C i 0 D th 1 is a balancing number. In fact if x is the n balancing number, then Bk .x/ th 1 is the .2n k/ balancing number. So Bk .x/ is odd balancing number for odd k andC even balancing number for even k. th tk (3) Let Bk denote the k balancing number, let bk denote the k cobalancing th number and let Ck denote the k Lucas–balancing number for an integer k 0. Then for any cobalancing number x, the function p Á b1.x/ C x B 8x2 8x 1 1 b k D k C k C C C C k is a cobalancing number. In fact, if x is the nth cobalancing number, then b .x/ is the .n k/th cobalancing number (Note that b1.x/ f .x/ and k C 1 D b1.x/ g.x/). 2 D (4) If x is the nth balancing number, y is the mth cobalancing number and ´ is the nth Lucas–balancing number, then the function 2x p8x2 1 1 b1.x/ C C D 2 is the .n 1/th cobalancing number and the function C p 2x 8y2 8y 1 2yp8x2 1 ´ 1 b2.x;y;´/ C C C C C D 2 is the .n m/th cobalancing number. C ˛2n ˇ 2n ˛2n 1 ˇ 2n 1 1 ˛2n ˇ 2n Proof. (1) Notice that Bn ;bn ;Cn C and 4p2 4p2 2 2 2n 1 2n 1 D D D ˛ ˇ cn C2 . So for any balancing number x and cobalancing number y, we haveD p 8xy p8x2 1 8y2 8y 1 1 B1.x;y/ C C C C C D 4 ˛2n ˇ 2n Á ˛2n 1 ˇ 2n 1 1 Á ˛2n ˇ 2n Á ˛2n 1 ˇ 2n 1 Á 8 C C 1 4p2 4p2 2 C 2 2 C D 4 ( ˛4n 1 ˛2nˇ 2n 1 ˇ 2n˛2n 1 ˇ 4n 1 ˛2n ˇ 2n ) C C 4 C p2 ˛4n 1 ˛2nˇ 2n 1 ˇ 2n˛2n 1 ˇ 4n 1 C C C 1 C 4 C D 4 BALANCING, PELL AND SQUARE TRIANGULAR FUNCTIONS 1225

.˛ˇ/2n.˛ 1 ˇ 1/ ˛2n ˇ 2n C 1 2 C p2 C D 4 ˛2n ˇ2n D 4p2 Bn D since ˛ˇ 1 and ˛ 1 ˇ 1 2: The others can be proved similarly. D C D As in Theorem3, there are inﬁnitely many Lucas–balancing and Lucas–cobalancing functions given below which can be proved similarly.

th th Theorem 4. Let Bk denote the k balancing number, Ck denote the k Lucas– tk balancing number and let ck denote the k Lucas–cobalancing number. Then for any balancing number x and any cobalancing number y with the same order, say n, (1) the function q p 2 2 Ck.x;y/ Ck 1 8x 1 ck 1 8y 8y 1 cn k 1 D C C C C C C st is the .n k 1/ Lucas–balancing number, that is, Ck.x;y/ Cn k 1 for k 2: C D C (2) the function q 2 p 2 ck.x;y/ Ck 1 8y 8y 1 ck 1 8x 1 Cn k 1 D C C C C C C st is the .n k 1/ Lucas–cobalancing number, that is, ck.x;y/ cn k 1 for n kC 1 1; and the function D C q c .y/ 4B .2y 1/ C 8y2 8y 1 k D k C C k C C th is the .n k/ Lucas–cobalancing number, that is, ck.y/ cn k: C D C For Pell and Pell–Lucas numbers, we can give the following theorem.

th th Theorem 5. Let Bk denote the k balancing number, let bk denote the k co- tk balancing number, let Ck denote the k Lucas–balancing number, let ck denote the tk th k Lucas–cobalancing number, let Pk denote the k Pell number and let Qk denote the kth Pell–Lucas number for an integer k 0. Then for any balancing number x, for any cobalancing number y, for any Lucas–balancing number ´ and for any Lucas–cobalancing number w with the same order, say n, (1) the function p 8B .x y/ C .p8x2 1 8y2 8y 1/ P 1.x;y/ k C C k C C C C P k D 2 C 2k 1226 AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS

th 1 is the .2n 2k/ Pell number, that is, Pk .x;y/ P2n 2k, the function C D C p C p8x2 1 c 8y2 8y 1 P 2.x;y/ k C C k C C k D 2 st 2 is the .2n 2k 1/ Pell number, that is, Pk .x;y/ P2n 2k 1, and the function C D C p ´p8x2 1 w 8y2 8y 1 P1.x;y;´;w/ C C C C D 2 st is the .4n 1/ Pell number, that is, P1.x;y;´;w/ P4n 1: (2) the function D p Q .x/ 32C x2 32B x 8x2 1 Q k D k C k C C 2k th is the .4n 2k/ Pell–Lucas number, that is, Qk.x/ Q4n 2k. (3) We set C D C p q G .x;y/ B 8x2 1 b 8y2 8y 1: k D k C C k C C th (a) If k 1, then the function G1.x;y/ is the n Lucas–balancing number, D that is, G1.x;y/ Cn. D st (b) If k 2, then the function G2.x;y/ is the four times of the .2n 1/ D C Pell number, that is, G2.x;y/ 4P2n 1: D C (c) The function Gk.x;y/ is the sum of Lucas–balancing numbers from n to .n k 1/, that is, C n k 1 CX G .x;y/ Ci k D i n D for every k 1, or is the sum of Pell numbers, that is, 8 k ˆ P2 ˆ 4 P2n 4i 3 for even k 2 < i 1 C Gk.x;y/ D k 3 D 2 ˆ P ˆ P2n P2n 1 4 P2n 4i 3 for odd k 3: : C C i 0 C C D Cn 1 Proof. (1) Notice that Cn cn 2P2n, Bn bn and P2n 2Bn. So C D C D 2 D p 8B .x y/ C .p8x2 1 8y2 8y 1/ P 1.x;y/ k C C k C C C C P k D 2 C 2k 8B .Bn bn/ C .Cn cn/ k C C k C 2B D 2 C k Cn 1 8Bk. / 2CkP2n 2 C 2B D 2 C k BALANCING, PELL AND SQUARE TRIANGULAR FUNCTIONS 1227

2BkCn CkP2n D C ! ! ˛2k ˇ2k Â˛2n ˇ2n Ã ˛2k ˇ2k Â˛2n ˇ2n Ã 2 C C D 4p2 2 C 2 2p2 ˛2.k n/ ˛2kˇ2n ˇ2k˛2n ˇ2.n k/ C C C ˛2.k n/ ˛2kˇ2n ˇ2k˛2n ˇ2.k n/ C C C C D 4p2 ˛2.n k/ ˇ2.n k/ C C D 2p2 P2n 2k: D C The others can be proved similarly. There is a connection between balancing numbers and triangular numbers denoted n.n 1/ by Tn which are the numbers of the form Tn 2C for n 0. There are inﬁnitely many triangular numbers that are also squareD numbers which are called square trian- th gular numbers and is denoted by Sn. For the n square triangular number Sn, we can 2 tn.tn 1/ write Sn sn 2C , where sn and tn are the sides of the corresponding square and triangle.D TheirD Binet formulas are 2n 2n 2n 2n 2n 2n ˛ ˇ 2 ˛ ˇ ˛ ˇ 2 Sn . / ; sn and tn C : D 4p2 D 4p2 D 4 In [10], Ozkoc¸,¨ Tekcan and Gozeri¨ derived some nice results on triangular, square triangular and balancing numbers. For the square triangular numbers, squares and triangles, we can give the following theorem. Theorem 6. For any balancing number x and any cobalancing number y with the same order, say n, (1) the function p 4x2 4y.y 1/ p8x2 1 8y2 8y 1 1 S.x;y/ C C C C C C C D 8 th is the n square triangular number, that is, S.x;y/ Sn: (2) the function D p 6y 2p8x2 1 3 8y2 8y 1 3 s.x;y/ C C C C C D 2 st is the .n 1/ square, that is, s.x;y/ sn 1: (3) the function D p 2.x y 1/ p8x2 1 8y2 8y 1 t.x;y/ C C C C D 2 1228 AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS

th is the n triangle, that is, t.x;y/ tn: D Proof. (1) Since 1 ˛ 2 2˛ 1 1 ˇ 2 2ˇ 1 2 and ˛ˇ 1, we get C C D C C D D p 4x2 4y.y 1/ p8x2 1 8y2 8y 1 1 S.x;y/ C C C C C C C D 8 8 2 9 ˛2n ˇ 2n Á ˛2n 1 ˇ 2n 1 1 Áh ˛2n 1 ˇ 2n 1 1 i < 4 4 1 = 4p2 C 4p2 2 4p2 2 C ˛2n ˇ 2n Á ˛2n 1 ˇ 2n 1 Á : C C 1 ; C 2 2 C D 8 ˛4n.1 ˛ 2 2˛ 1/ ˇ4n.1 ˇ 2 2ˇ 1/ 4 C C C C C D 64 ˛4n 2.˛ˇ/2n ˇ4n C D 32 2 Â˛2n ˇ2n Ã D 4p2 Sn: D The others can be proved similarly.

3. NUMERICAL RESULTS In [8], the authors dealt with the common values of balancing–type sequences. At the end of this paper we investigate this problem between Lucas–balancing and combinatorial numbers. Let us consider the equation Cn f .x/; (3.1) D th where f .x/ is a polynomial with integer coefﬁcients and Cn is the n Lucas– balancing number. x In this paper we study (3.1) when f .x/ is one of the polynomials k , and S .x/ 1k 2k .x 1/k; k D C C C ˘ .x/ x.x 1/:::.x k 1/ k D C C for ﬁxed integer k 2. We have to mention that we can deduce the following property of Lucas–balancing from equation (1.3):

Lemma 7. If x Cn and y Bn, then the equation D ˙ D ˙ x2 8y2 1 D is valid. Our numerical result is the following. BALANCING, PELL AND SQUARE TRIANGULAR FUNCTIONS 1229

Theorem 7. Let 2 k 4 and f .x/ be one of the polynomials x, ˘ .x/, Ä Ä k k Sk 1.x/. Then the solutions .Cn;x/ of equation (3.1) are in the table Cm Bm f .x/ x k x 3 1 k 3 2 3 1 Sk 1.x/ 3 2 x x Proof. Consider the equation (3.1) when f .x/ one of polynomials 2 , 3 and x. Using the multiplications by suitable powers of 2 and transformations X x, 4 D X 2.x 1/2 1, X x2 3x 1 respectively, for the polynomials above. Then byD Lemma7 weC obtain D C .23y/2 2X 4 4X 3 2X 2 8; D C 2 243y X 3 7X 2 15X 297; D C 2 253y 2X 4 4X 2 1150; D where y is a balancing number. These types of equations are solvable by MAGMA (IntegralQuarticPoints and IntegralPoints), so after testing them we get the solutions above. The suitable command at the ﬁrst case is IntegralQuarticPoints([2,-4,2,0,-8],[3,8]), because the point Œ3;8 is on the curve. We get the solutions (2,8),(2,0),(1,0),(3,8) for .X;23y/. Using these results we realize that there is only one solution for Bn, Cn and x, where .Bn;Cn;x/ .1;3;3/. Let us considerD the second equation. In this case there is no solution, since from x the property (1.3) follows that all Lucas–balancing numbers are odd, while 3 is even. We use in the third case the command IntegralQuarticPoints([2,0,-4,0,-1150]) and we found no solution. Remark 1. We mention that we can use IntegralQuarticPoints([]) in the case also when the constant term is not square, but the odd powers are missing from the right hand side. x Now let f .x/ be equal to Sk 1.x/. If k 2, then S1.x/ 2 , and its unique solutions has already been determined. D D Applying the transformations X 2.2x 1/2, X x, respectively, to the equa- D D 2 tion (3.1) when f .x/ S2.x/ and f .x/ S3.x/ we get D D 2 263y X 3 4X 2 4X 4608; D C .22y/2 2X 4 2: D Using the commands IntegralPoints(EllipticCurve([0,-4,4,0,-4608])) and IntegralQuarticPoints([2,0,0,0,-2]), it provides only the solution .1;0/, in the second case, that is there is no solution for Lucas–balancing number. 1230 AHMET TEKCAN, MERVE TAYAT, AND PETER´ OLAJOS

Finally let f .x/ ˘2.x/; ˘3.x/; ˘4.x/ and by using the transformations X 2x 1, X 2.x D1/2 1 and X x2 3x we arrive that D C D C C D C .24y/2 2X 4 4X 2 30; D .23y/2 X 3 7X 2 15X 17; D C y2 2X 4 8X 3 8X 2 2: D C C By MAGMA, there exist no solution in the ﬁrst and the second cases. In the third case we can easily conclude the same as follows. Since X is always even, therefore y is also even and we have y2 0.mod 4/ and 2X 4 8X 3 8X 2 2 2.mod 4/; Á C C Á which is impossible.

ACKNOWLEDGMENTS The authors wish to thank the anonymous referees for their detailed comments and suggestions which have improved the presentation of the paper.

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Authors’ addresses

Ahmet Tekcan Uludag University, Faculty of Science, Department of Mathematics, Bursa-Turkiye E-mail address: [email protected]

Merve Tayat Uludag University, Faculty of Science, Department of Mathematics, Bursa-Turkiye E-mail address: [email protected]

Peter´ Olajos University of Miskolc, Institute of Mathematics, Department of Applied Mathematics, Hungary E-mail address: [email protected]