On the intersections of Fibonacci, Pell, and Lucas numbers

We describe how to compute the intersection of two Lucas sequences of the forms {Un(P,±1)}n=0 or {Vn(P,±1)} ∞ n=0 with P ∈ Z that includes sequences of Fibonacci, Pell, Lucas, and Lucas-Pell numbers. We prove that such an intersection is finite except for the case Un(1,−1) and Un(3, 1) and the case of two V-sequences when the product of their discriminants is a perfect square. Moreover, the intersection in these cases also forms a Lucas sequence. Our approach relies on solving homogeneous quadratic Diophantine equations and Thue equations. In particular, we prove that 0, 1, 2, and 5 are the only numbers that are both Fibonacci and Pell, and list similar results for many other pairs of Lucas sequences. We further extend our results to Lucas sequenceswith arbitrary initial terms. In 1962 Stein [11] using elementary arguments proved that if two Fibonacci sequences (with different initial terms) share three terms then they are identical from some terms on. This result was later generalized by Revuz [9] to arbitrary sequences satisfying the same second order linear recurrence. A similar problem is to determine the intersection of two sequences satisfying distinct linear recurrences. Mignotte, 1978 [6] proved that the intersection of two sequences is finite unless the roots of their characteristic polynomials are multiplicatively dependent. Mátyás, 1981 [5] gave a criterion for determining whether two second order linear sequences have nonempty intersection. Posing the problem as a system of Pellian equations, Pinch, 1988 [7] and Tzanakis, 2002 [15] proposed computational methods for solving such systems. In the current paper we focus on the computational matters of finding the intersection and provide some explicit results for a class of Lucas sequences. While the general structure of such intersections is well known [6, 5, 7, 15], our approach (outlined in the proof of Theorem 6) is different from previous studies (see [15] for a review of various computational methods). In short, we reduce the problem to a finite number of Thue equations, each of which has a finite number of solutions. In practice the intersection of particular Lucas sequences can be computed with PARI/GP computer algebra system [13] ∗Department of Computer Science and Engineering, University of South Carolina. Email: [email protected] providing a functionality for solving Thue equations, based on Bilu and Hanrot’s improvement [1] of Tzanakis and de Weger’s method [16]. We also characterize the cases when the intersection is infinite and show that it also forms a Lucas sequence. While the current paper was under review, our attention was drawn to the paper of Szalay, 2007 [12], which employs a similar approach for solving systems of Pellian equations. We therefore feel obliged to underline the differences between [12] and our paper. In Theorem 6 we consider a general system of two homogeneous quadratic equations in two indeterminants and formulate conditions under which such a system has a finite number of integer solutions. The systems of Pellian equations considered in [12] represent a particular case of the system from our Theorem 6, while the algorithm and the corresponding proof of solutions’ finiteness in [12] are flawed. In [12] it is used the result from [4] that the denominator in the general solution to a Diophantine equation Ax + By + Cz = 0, derived from a particular solution (x0, y0, z0) with z0 , 0, divides 2ABCz 0 . In Theorem 5 we prove a stronger result that this denominator actually divides 2 lcm(A,B)Cz20, leading to a smaller number of Thue equations to solve. This result also corrects an error in Corollary 6.3.8 of [2]. We further focus on multiple issues specific to the intersections of Lucas sequences which are out of scope of [12]. Said that, we refer to [12] for many examples of systems of Pellian equations and a review of their applications and related literature. The paper is organized as follows. In Section 1 we give definition and basic properties of Lucas sequences. We develop an algorithmic approach to solving systems of quadratic Diophantine equations in Section 2 and apply it to the problem of finding intersections of Lucas sequences in Section 3. Some numerical results are given and a number of open questions are posed in Section 4. Finally, we discuss generalizations of our approach to Lucas sequences with arbitrary initial terms in Section 5. 1 Basic properties of Lucas sequences The pair of Lucas sequencesUn(P,Q) andVn(P,Q) are defined by the same linear recurrent relation with the coefficient P,Q ∈ Z but different initial terms: U0(P,Q) = 0, U1(P,Q) = 1, Un+1(P,Q) = P ·Un(P,Q) −Q ·Un−1(P,Q), n ≥ 1; V0(P,Q) = 2, V1(P,Q) = P, Vn+1(P,Q) = P · Vn(P,Q) −Q · Vn−1(P,Q), n ≥ 1. Some Lucas sequences have their own names: Sequence Name Initial terms Un(1,−1) Fibonacci numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, . . . Vn(1,−1) Lucas numbers 2, 1, 3, 4, 7, 11, 18, 29, . . . Un(2,−1) Pell numbers 0, 1, 2, 5, 12, 29, 70, 169, . . . Vn(2,−1) Pell-Lucas numbers 2, 2, 6, 14, 34, 82, 198, . . . In [12] it is incorrectly claimed that the constructed quartic equation (2.5) represents a Thue equation. We found a counterexample with (a1, b1, c1) = (5,−1,−1) and (a2, b2, c2) = (20,−1, 1) in (2.1) − (2.2) which yields (a, b, c) = (25,−1,−1) in (2.4). From a basic solution (X0,Y0,Z0) = (1, 0, 5) it further leads to the reducible quartic polynomial T(s, r) = 3125s − 2250sr + 5r = 5(25s − 20sr − r)(25s + 20sr − r) in the l.h.s. of (2.5). Other examples include Jacobsthal numbers Un(1,−2), Mersenne numbers Un(3, 2) etc. In the current paper we focus on the case of Q = 1 or Q = −1. We also notice that Un(−P,Q) = (−1)Un(P,Q) and Vn(−P,Q) = (−1)Vn(P,Q) and restrict our attention to Lucas sequences with P ≥ 0. Similarly, we exclude from consideration terms with negative indices, noticing that they may deviate only in signs: U−n(P, 1) = −Un(P, 1), V−n(P, 1) = Vn(P, 1), U−n(P,−1) = (−1)Un(P,−1), V−n(P,−1) = (−1)Vn(P, 1). From further consideration we also exclude the following degenerate cases: (P,Q) Un(P,Q) Vn(P,Q) (0, 1) U2m = 0 V2m = 2 · (−1)m U2m+1 = (−1)m V2m+1 = 0 (0,−1) U2m = 0 V2m = 2 U2m+1 = (−1)m V2m+1 = 0 (1, 1) U3m = 0 V3m = 2 U3m+1 = (−1)m V3m+1 = (−1)m U3m+2 = (−1)m V3m+2 = (−1)m+1 (2, 1) Um = m Vm = 2 It is easy to see that under the described restrictions bothUn(P,Q) andVn(P,Q) are positive for n > 0. The characteristic polynomial of Lucas sequences {Un(P,Q)} and {Vn(P,Q)} isλ2−Pλ+Q with the discriminant D = P − 4Q. For non-degenerate sequences, the discriminant D is a positive non-square integer. Let α = P+ √ D 2 and β = P− √ D 2 be the roots of the characteristic polynomial, then the following explicit (Binet-type) formulas take place Q = αβ D = (α − β) Un(P,Q) = αn−βn α−β Vn(P,Q) = α n + β In particular, these formulas imply that Vn(P,Q) 2 −D ·Un(P,Q) = 4Q. (1) For |Q| = 1, it means that the pairs (Vn(P,Q),Un(P,Q)) form solutions to the equation: x −Dy = ±4. (2) The converse statement can be used to prove that given positive integers belong to {Vn(P,Q)} or {Un(P,Q)} respectively (stated without a proof for Fibonacci numbers in [8]): Here and everywhere below ± in the r.h.s. of an equation means that we accept both signs as solutions. Theorem 1. Let P, Q be integers such that P > 0, |Q| = 1, (P,Q) , (3, 1), and D = P − 4Q > 0. If positive integers u and v are such that v −Du = ±4, then u ∈ {Un(P,Q)} and v ∈ {Vn(P,Q)}. Proof. We notice thatD is congruent to 0 or 1modulo 4, implying that the general solution to equation (2) has the form (x, y) = (xk, yk), where xk + yk √ D 2 = ± 