In memoriam Péter Kiss

Transcription

1 Kálmán Liptai Eszterházy Károly Collage, Leányka út 4, 3300 Eger, Hungary (Submitted March 2002-Final Revision October 2003) In memoriam Péter Kiss ABSTRACT A positive integer n is called a balancing number if (n 1) = (n + 1) + (n + 2) + + (n + r) for some natural number r. We prove that there is no Fibonacci balancing number except INTRODUCTION The sequence {R n } n=0 = R(A, B, R 0, R 1 ) is called a second order linear recurrence if the recurrence relation R n = AR n 1 + BR n 2 (n > 1) holds for its terms, where A, B 0, R 0 and R 1 are fixed rational integers and R 0 + R 1 > 0. The polynomial x 2 Ax B is called the companion polynomial of the second order linear recurrence sequence R = R(A, B, R 0, R 1 ). The zeros of the companion polynomial will be denoted by α and β. Using this notation, as it is well known, we get where a = R 1 R 0 β and b = R 1 R 0 α (see [6]). A positive integer n is called a balancing number [3] if R n = aαn bβ n, (1) α β (n 1) = (n + 1) + (n + 2) + + (n + r) for some r Z +. Here r is called the balancer corresponding to the balancing number n. For example 6 and 3 are balancing numbers with balancers 2 and 14. In a joint paper A. Behera and G. K. Panda [3] proved that the balancing numbers fulfil the following recurrence relation B n+1 = 6B n B n 1 (n > 1) (2) where B 0 = 1 and B 1 = 6. We call a balancing number a Fibonacci balancing number if it is a Fibonacci number, too. In the next section we prove that there are no Fibonacci balancing numbers. 2. The equation x 2 Dy 2 = N with given integers D and N and variables x and y, is called Pell s equation. First we prove that the balancing numbers are solutions of a Pell s equation. 330

2 Theorem 1: The terms of the second order linear recurrence B(6, 1, 1, 6) are the solutions of the equation z 2 y 2 = 1 (3) for some integer z. Proof: Let B(6, 1, 1, 6) be a second order linear recurrence and denote by α and β the zeros of their companion polynomials and D the discriminant of the companion polynomial. Using (1) and the definition of α and β we get therefore, with y = B n, we have B n = ( )α n (3 2 2)β n 4 2 and αβ = 1, 1 + y 2 = 1 + Bn 2 = 1 + ( ( ) 2 α 2n 32 2( )(3 2 2)α n β n + (3 2 2) 2 β 2n) ( (3 + 2 ) 2 ( 2)α n = (3 2 2)β n 2 = ( ( )α n 2 ) 2 + (3 2 2)β n = z 2. 2 ) 2 Using that α = , β = and the binomial formula it can be proved that z is a rational integer. To prove our main result we need the following theorem of P. E. Ferguson [4]. Theorem 2: The only solutions of the equation x 2 y 2 = ±4 (4) are x = ±L n, y = ±F n (n = 0, 1, 2,... ), where L n and F n are the n th terms of the Lucas and Fibonacci sequences, respectively. Using the method of A. Baker and H. Davenport we prove that there are finitely many common solutions of the Pell s equations (3) and (4). We remark that this result follows from a theorem of P. Kiss [], too. In the process we show that there are no Fibonacci balancing numbers. In the proof we use the following theorem of A. Baker and H. Wüstholz [2]. Theorem 3: Let α 1,..., α n be algebraic numbers not 0 or 1, and let Λ = b 1 log α b n log α n, where b 1,..., b n are rational integers not all zeros. 331

3 We suppose that B = max( b 1,..., b k, e) and A i = max{(h(α i ), e} (i = 1, 2,... n). Assume that the field K generated by α 1, α 2,..., α n over the rationals has degree at most d. If Λ 0 then log Λ > (16nd) 2(n+2) log A 1 log A 2... log A n log B. (H(α) is equal to the maximum of absolute values of the coefficients of the minimal defining polynomial of α.) The following theorem is the main result of this paper. Theorem 4: There is no Fibonacci balancing number except 1. Proof: First we show that there are finitely many common solutions of the equations (, (6) and ( ), (6 ) and x = y 2 ( x 2 4 = y 2 ( ) x = z 2 (6) x = z 2 (6 ) The equations ( and ( ) can be written as (y + x (y x = 4 (7) (y + x (y x = 4. () y + x = (y 0 + x 0 (9 + 4 m where m 0, it is easily verified (by combining this equation with its conjugate) that y 0 is always positive but x 0 is negative if m is large. Hence we can choose m so that x 0 > 0; but if x 1 is defined by y 0 + x 0 = (y1 + x 1 (9 + 4 then x 1 0. Since y 0 + x 0 = (9y1 + 20x 1 ) + (9x 1 + 4y 1 ) we have y 0 = 9y x 1 and x 0 = 9x 1 + 4y 1. From the previous equations we have x 1 = 9x 0 4y 0 and x 0 4y 0 9. Using equation ( we have y = x y2 0. Hence y 0 = 3, 7, 1 and x 0 = 1, 3,, respectively. Thus the general solution of equation ( is given by y + x = (3 + (9 + 4 m (9) y + x = (7 + 3 (9 + 4 m (10) where m = 0, 1, 2... y + x = (1 + (9 + 4 m (11) 332

4 Using the same method as before with ( ), we find that y 1 = 9y 0 20x 0 0 (in this case x 0 is always positive), whence y0 2 = x x2 0, so that x 0 = 1, 2, and y 0 = 1, 4, 11, respectively. Thus the general solution of equation ( ) is given by y + x = (1 + (9 + 4 m (12) y + x = (4 + 2 (9 + 4 m (13) y + x = (11 + (9 + 4 m (14) where m = 0, 1, 2,... The general solution of equations (6) and (6 ) is given by z + x = (3 + ) n (1 where n = 0, 1, 2,... We are looking for the common solutions of the equation (9), (10), (11), (12), (13), (14) with the equations (1. Using (9), (1 and their conjugates we have 2x = (3 + ) n (3 ) n = 3 + (9 + 4 m 3 (9 4 m and so 1 (3 + ) n (3 + ) n = + 3 ( m + (9 + 4 m. (16) Putting Q = 1 (3 + ) n, P = + 3 (9 + 4 m, in equation (16) we obtain Q 1 Q 1 = P 4 P 1. (17) Since Q P = 1 Q 1 4 P 1 < 4 (Q 1 P 1 ) = 4 P Q QP and plainly P > 1 and Q > 1, we have Q < P. Also P Q = 4 P 1 1 Q 1 < 4 P 1 and P >

5 0 < log P ( Q = log 1 P Q ) < 4 ( ) 2 4 P P 2 + P 2 = 4 P P 4 < 0.1P 2 < 0.1 ( Using the previous inequality and the definitions of P and Q, we get 0 < m log(9 + 4 n log(3 + ) + log (3 + < We apply Theorem 3 with n = 3 and α 1 = α 2 = 3 + α 3 = ( ( (1) We use that 0.1(( ) m < e.77m. The equations satisfied by α 1, α 2, α 3 are α 2 1 1α = 0 α 2 2 6α = 0 2α α = 0. Hence A 1 = 1, A 2 = 6, A 3 = 1120 and d = 4. Using Theorem 3 and the previous inequality we have.77 (16 3 4)10 log 1 log 6 log 1120 log 0 24 log m. Thus we have Using the same method we investigate the equations (10) and (1. We have 2x = (9 + 4 m 7 3 (9 4 m = (3 + ) n (3 ) n that is (9 + 4 m 7 3 (9 + 4 m = (3 + ) n (3 + ) n. (19) P 1 = (9 + 4 m, Q = 1 (3 + ) n (20) 334

6 using (19) and (20) we have Using the previous method we have Q 1 Q 1 = P 1 4 P < log P 1 Q < 0.1P 2 1 < ( Substituting from (20), we obtain 0 < m log(9 + 4 n log(3 + ) + log (7 + 3 < We apply Theorem 3 with n = 3 and α 1 = α 2 = 3 + α 3 = ( ( (21) The equation 2α α = 0 is satisfied by α 3, that is A 3 = 720. Using Theorem 3 as above we have.77 (16 3 4)10 log 1 log 6 log 720 log 0 24 log m From the equations (11) and (1 we have 2x = 1 + (9 + 4 m 1 (9 4 m = (3 + ) n (3 ) n that is 1 + (9 + 4 m 1 (9 + 4 m = (3 + ) n (3 + ) n. (22) P 2 = 1 + (9 + 4 m, Q = (3 + ) n. (23) 33

7 Using (22) and (23) we have Q 1 Q 1 = P 2 4 P 1 2. As before we obtain Substituting from (23) we have 0 < log P ( 2 Q = log 1 P ) 2 Q < P ( < m log(9 + 4 n log(3 + ) + log (1 + < We apply Theorem 3 as above with n = 3 and α 1 = α 2 = 3 + α 3 = ( ( (24) The equation 2α 4 120α = 0 is satisfied by α 3 and so A 3 = (16 3 4)10 log 1 log 6 log 120 log 0 24 log m From the equations (12) and (1 we have 2x = 1 + (9 + 4 m 1 (9 4 m = (3 + ) n (3 ) n that is 1 + (9 + 4 m 1 (9 + 4 m = (3 + ) n (3 + ) n. (2 P 3 = 1 + (9 + 4 m, Q = (3 + ) n. (26) 336

8 Using (2 and (26) we have From the previous equation we have Q 1 Q 1 = P P 1 3. that is Q > P 3 and Q P 3 = 4 P Q 1 > 0 Q P 3 < 4 P P 1 3 = P < log Q ( = log 1 + Q P ) 3 < 37 P 3 P 3 40 P 3 2 and so ( ) P 3 2 < 0.926P3 2 1 < ( < m log(9 + 4 n log(3 + ) + log Using the previous method and that the equation is satisfied by α 3 = 1024α 4 40α = 0 (1+ and so A 3 = (1 + < ( (27).77 (16 3 4)10 log 1 log 6 log 1024 log 0 24 log m From the equations (13) and (1 we have (9 + 4 m 4 2 (9 4 m = (3 + ) n (3 ) n. 337

9 we get similar inequalities as before. We have P 4 = (9 + 4 m, Q = (3 + ) n and 0 < log Q < 0.926P < 0.06 P 4 ( < m log(9 + 4 n log(3 + ) + log (4 + 2 < ( (2) The equation 62464α 4 20α 2 + = 0 is satisfied by α 3 = (4+2 and so A 3 = We use the Theorem of A. Baker and G. Wüstholz again and we have.77 (16 3 4)10 log 1 log 6 log log 0 24 log m Finally let s consider the equations (14) and (1. We have 11 + (9 + 4 m 11 (9 4 m = (3 + ) n (3 ) n. and we use the previous steps we have P = 11 + (9 + 4 m, Q = (3 + ) n and 0 < log Q < 0.926P 2 1 < P ( < m log(9 + 4 n log(3 + ) + log (11 + < ( (29) The equation 1024α α = 0 33

10 is satisfied by α 3 = (11+ and so A 3 = We apply again Theorem 3 and we have.77 (16 3 4)10 log 1 log 6 log 1960 log 0 24 log m We get that there are finitely many common solutions of simultaneous equations, that is, there are finitely many Fibonacci balancing numbers. Since the bounds for m are too high (0 26 ) we can t investigate all of them. In order to get a lower bound we use the following lemma of A. Baker and H. Davenport [1]. Lemma: Suppose that K > 6. For any positive integer M, let p and q be integers satisfying 1 q KM, θq p < 2(KM) 1. Then, if qβ 3K 1, there is no solution of the equation in the range mθ n + β < C m log K 2 M log C < m < M. (It is supposed that θ, β are real numbers and C > 1. z denotes the distance of a real number z from the nearest integer.) We divide the inequality (1) by log(3 + ) and we show the steps of reduction. In the other cases, (21), (24), (27), (2) and (29), the method is similar. Using the lemma we have and C = ( = , θ = log(9 + 4 log(3 + ) β = log (3 + (log(3 + )) 1. In our case we take M = 10 26, K = 100. Let θ 0 be the value of θ correct to 6 decimal places, so that θ θ 0 < Let p q be the last convergent to the continued fraction for θ 0 which satisfies q < 10 2 ; then qθ 0 p < We therefore have qθ p q θ θ 0 + qθ 0 p < In this case the first and the second inequalities of the Lemma are satisfied. The values of θ, β and q computed by Maple are given in Appendix. We have that qβ = It follows from the lemma, since qβ 0.03, there is no solution of (1) in the range log log 321, 997 <

11 That is m less than 12, so we can calculate by hand that there is no Fibonacci balancing number in this case. We get the same result in the other cases, too. ACKNOWLEDGMENTS I would like to thank Professor L. Hajdú and Á. Reszegi for their valuable and useful advices. Research supported by the Hungarian OTKA Foundation No. T 0329 and the György Békésy Foudation. APPENDIX θ = q = β = qβ = β 1 = qβ 1 = β 2 = qβ 2 = β 3 = qβ 3 = β 4 = qβ 4 = β = qβ = REFERENCES [1] A. Baker and H. Davenport. The Equations 3x 2 2 = y 2 and x 2 7 = z 2. Quart. J. Math. Oxford 20.2 (1969): [2] A. Baker and G. Wüstholz. Logarithmic Forms and Group Varieties. J. Reine Angew. Math. 442 (1993): [3] A. Behera and G. K. Panda. On the Square Roots of Triangular Numbers. The Fibonacci Quarterly 37.2 (1999): [4] D. E. Ferguson. Letter to the editor. The Fibonacci Quarterly (1970): 9. [] P. Kiss. On Common Terms of Linear Recurrences. Acta Math. Acad. Sci. Hungar 40 (192): [6] T. N. Shorey and R. Tijdeman. Exponential Diophantine Equations. Cambridge University Press (196). AMS Classification Numbers: 11B39, 11D4 340