Equations with Solution in Terms of Fibonacci and Lucas Sequences

Transcription

1 DOI: /auom An. Şt. Univ. Ovidius Constanţa Vol. (3),014, 1 Equations with Solution in Terms of Fibonacci and Lucas Sequences Titu Andreescu and Dorin Andrica Dedicated to the memory of Ion Cucurezeanu ( ) Abstract The main results characterize the equations (.1) and (.10) whose solutions are linear combinations with rational coefficients of at most two terms of classical Fibonacci and Lucas sequences. 1 Special Pell s equations The Diophantine equations u Dv = ±4 (1.1) are called special Pell s equations. In some ways, solutions to (1.1) are more fundamental than solutions to Pell s equations u Dv = ±1. For general aspects concerning the theory of the positive and negative Pell s equation we refer to the authors book []. As with the ±1 equations, all solutions to (1.1) can be generated from the minimal positive solution. If (u 1, v 1 ) is the minimal positive solution to Pell s special equation u Dv = 4, Key Words: special Pell s equation, negative special Pell s equation, Fibonacci and Lucas sequences 010 Mathematics Subject Classification: 11B39, 11D7 Received: 30 June, 013 Revised: 30 September, 013 Accepted: 4 October, 013

2 SEQUENCES 6 then for the n th solution (u n, v n ) we have ( ) n 1 (u u 1 + v 1 D n + v n D) =. (1.) The general solution (1.) is completed with the trivial solution (u 0, v 0 ) = (, 0), obtained for n = 0. Also, it can be extended to negative integers n. We can express all integer solutions to equation (1.1) by the following formula 1 (u n + v n D) = εn ( u 1 + v 1 D ) n, n Z, (1.3) where ε n is 1 or 1. Indeed, for n > 0 and ε n = 1 we get all negative solutions. For n > 0 and ε n = 1 we obtain all solutions (u n, v n ) with u n and v n negative. For n < 0 and ε n = 1 we have (u n, v n ) with u n > 0 and v n < 0, while n < 0 and ε n = 1 gives u n < 0 and v n > 0. The trivial solutions (, 0) and (, 0) are obtained for n = 0. Formula (1.3) captures all symmetries of equation (u, v) ( u, v), (u, v) (u, v), (u, v) ( u, v). Therefore, in D the points (u n, v n ) represents the orbits of the action of the Klein four-group Z Z, i.e. points obtained by the 180 degree rotation, the vertical reflection, and by the horizontal reflection. Now suppose the negative special Pell s equation u Dv = 4 has solutions and let (u 1, v 1 ) be its minimal positive solution. Define the integers u n, v n by (1.3). Then, for n odd, (u n, v n ) is a solution to u Dv = 4, and for n even, (u n, v n ) is a solution to u Dv = 4. All integer solutions to these equations are generated in this way. The main results In this section we find all integers a for which the solutions to the quadratic equations x + axy + ay = ±1, (.1) are linear combinations with rational coefficients of the classical Fibonacci and Lucas numbers: F 0 = 0, F 1 = 1, F n+1 = F n +F n 1, n 1 and L 0 =, L 1 = 1, L n+1 = L n + L n 1, n 1. This problem is related to the so-called Diophantine representation of a sequence of integers, and for some results we refer to the papers of M.J. DeLeon [3], V.E. Hoggatt, M. Bicknell-Johnson [6], J.P. Jones [7], [8], [9], and W.L. McDaniel [11]. Also, it is connected to the Y.V. Matiyasevich and J. Robertson way to solve the Hilbert s Tenth Problem, and it has applications

3 SEQUENCES 7 to the problem of singlefold Diophantine representation of recursively enumerable sets. In the recent paper of R. Keskin, N. Demiturk [10] the equations x kxy + y = 1, x kxy y = 1 are solved in terms of generalized Fibonacci and Lucas numbers. Let us mention that S. Halici [4] defines Hankel matrices involving the Pell, Pell-Lucas and modified Pell sequences, computes their Frobenius norm, and investigate some spectral properties of them. Recall the Binet s formulas for F n and L n : F n = L n = ( ) n ( 1 ) n ], (.) 1 + ) n ( 1 ) n +. (.3) These formulas can be extended to negative integers n in a natural way. We have F n = ( 1) n 1 F n and L n = ( 1) n L n, for all n. Theorem.1. The solutions to the positive equation (.1) are linear combinations with rational coefficients of at most two Fibonacci and Lucas numbers if and only if a = a n = ±L n +, n 1. For each n, all of its integer solutions (x k, y k ) are given by x k = ε k L kn a n F kn F n y k = ± 1 F n F kn, (.4) where k 1, signs + and depend on k and correspond, while ε k = ±1. Proof. The equation x + axy + ay = 1 is equivalent to the positive special Pell s equation From (1.3) it follows that x m + ay m = ε m (x + ay) (a 4a)y = 4 (.) ) m ( ) m ] u 1 + v 1 D u 1 v 1 D + and y m = εm D ) m ( u 1 + v 1 D ) m ] u 1 v 1 D,

4 SEQUENCES 8 where m Z, ε m = ±1, D = a 4a, and (u 1, v 1 ) is the minimal positive solution to u Dv = 4. we have (u 1, v 1 ) = (a, 1), and combining the above relations it follows and y m = x m = ε m ε m a 4a [ ( 1 ) ( a a + a 4a a 4a ( ) ( a a ) m ] a a a 4a ) m (.6) a + ) m ( a 4a a ) m ] a 4a. (.7) Taking into account Binet s formulas, solution (x m, y m ) is representable in terms of F m and L m only if a 4a = s, for some positive integer s. This is equivalent to the special Pell s equation (a ) s = 4, (.8) whose minimal solution is (a 1, s 1 ) = (3, 1). The general integer solution to (.8) is 3 + ) n ( 3 ) n ] a n = ε n + = ε n L n, and s n = ε n 3 + ) n ( 3 ) n ] = ε n F n, where n is an integer and ε n = ±1. The equation (.8) was also used by D. Savin [1] in the study of the equation x 4 6x y + y 4 = 16F n 1 F n+1, and R. Keskin, N. Demiturk [10]. From (x + ay) (a 4a)y = 4 we find (x + a n y) (s n y) = 4, with integer solution (x m, y m ) given by Hence x m + a n y m = ε m L m and s n y m = ±F m. x m = 1 [ ] F m ε m L m a n, y m = ± F m, (.9) F n F n where signs + and correspond, and ε m = ±1.

5 SEQUENCES 9 Taking into account that F n divides F m if and only if n divides m (see [1, pp. 180] and [4, pp. 39]), it is necessary that m = kn, for some positive integer k. Formulas (.9) become (.4). A parity argument shows that in the equation (x + ay) (a 4a)y = 4, x is even, so x k in (.4) is always an integer. The following two tables give the integer solutions to equation (.1) at level k, including the trivial solution obtained for k = 0. n a n = L n + Equation (.1) Solutions 1 x + xy + y = 1 x = ε k L k F k, y = ±F k 9 x + 9xy + 9y = 1 x = ε k L 4k F 6 4k, y = ± 1 F 3 4k 3 0 x + 0xy + 0y = 1 x = ε k L 6k F 4 6k, y = ± 1 F 8 6k 4 49 x + 49xy + 49y = 1 x = ε k L 8k 7 F 6 8k, y = ± 1 F 1 8k 1 x + 1xy + 1y = x + 34xy + 34y = 1 x = ε k L 10k 10k, y = ± 1 F 10k x = ε k L 1k 9 F 8 1k, y = ± k n a n = L n + Equation (.1) Solutions 1 1 x xy y = 1 x = ε k L k ± 1 F k, y = ±F k x xy y = 1 x = ε k L 4k ± F 6 4k, y = ± 1 F 3 4k 3 16 x 16xy 16y = 1 x = ε k L 6k ± F 6k, y = ± 1 F 8 6k 4 4 x 4xy 4y = 1 x = ε k L 8k ± 1 14 F 8k, y = ± 1 1 F 8k 11 x 11xy 11y = 1 x = ε k L 10k F 10k, y = ± 1 F 10k 6 30 x 30xy 30y = 1 x = ε k L 1k 10 9 F 1k, y = ± F 1k Next we will consider the negative equation of the type (.1): Similar negative type equations x + axy + ay = 1 (.10) x kxy y = 1, x kxy y = ±(k + 4) are considered by R. Keskin, N. Demiturk [10]. Unlike the result in Theorem.1, there are only two values of a for which the corresponding property holds. Theorem.. The solutions to the negative equation (.10) are linear combinations with rational coefficients of at most two Fibonacci and Lucas numbers if and only if a = 1 or a =.

6 SEQUENCES 10 If a = 1, all of its integer solutions (x m, y m ) are given by x m = ε m L m+1 ± 1 F m+1, y m = ±F m+1, m 0. (.11) If a =, all integer solutions (x m, y m ) are x m = ε m L m+1 F m+1, y m = ±F m+1, m 0. (.1) The signs + and depend on m and correspond, while ε m = ±1. Proof. As in the proof of Theorem.1 the equation is equivalent to (x + ay) (a 4a)y = 4. Suppose that this negative special Pell s equation is solvable. Its solution (x m, y m ) is representable in terms of Fibonacci and Lucas numbers as a linear combination with rational coefficients only if a 4a = s. As in the proof of Theorem.1 we obtain a n = ±L n + and s n = ±F n, n 1. The equation (x + ay) (a 4a)y = 4 becomes (x + ay) (s n y) = 4, whose integer solutions are x m + ay m = ε m L m+1 and s n y m = ±F m+1. It follows that y m = ± F m+1 F n, m 1. If n, then F n, and since n does not divide m + 1, it follows that F n does not divide F m+1 (see [1], pp. 180 and [], pp. 39), hence y m is not an integer. Thus n = 1 and so a = ±L +, i.e. a = 1 or a =. For a = 1, it follows y m = ±F m+1 and x m y m = ε m L m+1, and we obtain solutions (.11). If a =, then y m = ±F m+1 and x m + y m = ε m L m+1, yielding the solutions (.1). Remark. On the other hand, it is more or less known Zeckendorf s theorem in [13], which states that every positive integer can be represented uniquely as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include two consecutive Fibonacci numbers. Such a sum is called Zeckendorf representation and it is related to the Fibonacci coding of a positive integer. Our results are completely different, because the number of terms is reduced to at most two, and the sum in the representation of solutions is a linear combination with rational coefficients.

7 SEQUENCES 11 References [1] Andreescu, T., Andrica, D., Number Theory. Structures, Examples, and Problems, Birkhäuser, Boston-Basel-Berlin, 009. [] Andreescu, T., Andrica, D., Cucurezeanu, I., An Introduction to Diophantine Equations, Birkhäuser, Boston-Basel-Berlin, 010. [3] DeLeon, M.J., Pell s equation and Pell number triples, The Fibonacci Quarterly 14(1976), no., [4] Halici, S., On some inequalities and Hankel matrices involving Pell, Pell- Lucas numbers, Mathematical Reports, 1(6)(013), no. 1, [] Hoggatt, V.E., Jr., Fibonacci and Lucas Numbers, The Fibonacci Association, Santa Clara, [6] Hoggatt, V.E., Bicknell-Johnson, M., A primer for the Fibonacci XVII: generalized Fibonacci numbers satisfying u n+1 u n 1 u n = ±1, The Fibonacci Quarterly 16(1978), no., [7] Jones, J.P., Diophantine representation of Fibonacci numbers, The Fibonacci Quarterly 13(197), no. 1, [8] Jones, J.P., Diophantine representation of the Lucas numbers, The Fibonacci Quarterly 14(1976), no., 134. [9] Jones, J.P., Representation of solutions of Pell equations using Lucas sequences, Acta Academiae Pedagogicae Agriensis, Sectio Matematicae 30(003), [10] Keskin, R., Demiturk, N., Solutions to Some Diophantine Equations Using Generalized Fibonacci and Lucas Sequences, Ars Combinatoria, Vol.CXI, July, 013, [11] McDaniel, W.L., Diophantine representation of Lucas sequences, The Fibonacci Quarterly 33(199), no.1, [1] Savin, D., About a Diophantine Equation, Analele Sţ. Univ. Ovidius Constanţa, 17(3)(009), [13] Zeckendorf, E., Représentation des nombres naturels par une somme de nombres de Fibonacci ou des nombres de Lucas, Bull. Soc. R. Sci. Liège, 41, (197).

8 SEQUENCES 1 Titu ANDREESCU University of Texas at Dallas School of Natural Sciences and Mathematics Richardson, TX titu.andreescu@utdallas.edu Dorin ANDRICA Babeş-Bolyai University Faculty of Mathematics and Computer Science Cluj-Napoca, Romania dandrica@math.ubbcluj.ro