ELEMENTARY PROBLEMS AND SOLUTIONS

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1 EDITED BY RUSS EULER AND JAWAD SADEK Please submit all new problem proposals their solutions to the Problems Editor, DR. RUSS EULER, Department of Mathematics Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 64468, or by at All solutions to others proposals must be submitted to the Solutions Editor, DR. JAWAD SADEK, Department of Mathematics Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO If you wish to have receipt of your submission acknowledged, please include a self-addressed, stamped envelope. Each problem solution should be typed on separate sheets. Solutions to problems in this issue must be received by February 5, 04. If a problem is not original, the proposer should inform the Problem Editor of the history of the problem. A problem should not be submitted elsewhere while it is under consideration for publication in this Journal. Solvers are asked to include references rather than quoting well-known results. The content of the problem sections of The Fibonacci Quarterly are all available on the web free of charge at BASIC FORMULAS The Fibonacci numbers F n the Lucas numbers L n satisfy F n+ F n+ +F n, F 0 0, F ; L n+ L n+ +L n, L 0, L. Also, α (+ 5/, β ( 5/, F n (α n β n / 5, L n α n +β n. PROBLEMS PROPOSED IN THIS ISSUE B-3 Proposed by Hideyuki Ohtsuka, Saitama, Japan For integers a b, prove that (Fa +Fa+ +Fa+(F a F b +F a+ F b+ +F a+ F b+ (FaF 3 b +Fa+F 3 b+ +Fa+F 3 b+ ; ( (Fa +F a+ +F a+ (F afb 3 +F a+fb+ 3 +F a+fb+ 3 (Fb +F b+ +F b+ (F3 a F b +Fa+ 3 F b+ +Fa+ 3 F b+. ( 74 VOLUME 5, NUMBER 3

2 B-3 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade General School, Buzău, Romania Prove that F n +F n+ +5F n+ > 4 6 F n F n+ F n+ for any positive integer n; ( L n +L n+ +5L n+ > 4 6 L n L n+ L n+ for any positive integer n. ( B-33 Proposed by Mohammad K. Azarian, University of Evansville, Indiana Determine the value of the following infinite series S B-34 Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Francesc Gispert Sánchez, CFIS, Barcelona Tech, Barcelona, Spain Let n be a positive integer. Prove that [ ( F n F n F n+ n k k + n k F k ] ( n k F ( /n k. B-35 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade General School, Buzău, Romania Prove that Fn+ Fn+ Fn(F 3 n F n+ +Fn+ + Fn+ 3 (F n+f n+ +Fn + Fn Fn+ 3 (F nf n+ +Fn+ > 3 ; ( F n F n+ F n+ L n+ L n+ L 3 n(l n L n+ +L n+ + L 3 n+ (L n+l n+ +L n + for any positive integer n. L n L 3 n+ (L nl n+ +L n+ > 3 L n L n+ L n+, ( AUGUST 03 75

3 THE FIBONACCI QUARTERLY SOLUTIONS Fibonacci Numbers Divided by B- Proposed by Mircea Merca, University of Craiova, Romania (Vol. 50.3, August 0 Let n be a positive integer. Prove that Fn+ + 3n 5 k. Solution by Marielle Silvio Kasey Zemba (jointly students at California University of Pennsylvania (CALURMA, California, PA. It is enough to prove that k To show this, we prove that k < F n+ + 3n 5 k. F n+ + 3n 5 +l n ( for some rational number l n such that 0 l n <. We start with some preliminaries. We denote by (n mod m the remainder of n divided by m for n,m N. Thus, the division algorithm can be written in the form n n m m see [, Id. 3., p. 8]. It is also known that n n+m m m (n mod m, ( m F k F n+, (3 see [, Ex., p. 96] [, Th. 5., p. 69], respectively. Using ( (3, we can easily see that +0 k k (F k +0 ((F k +0 mod k k F n+ + 0n ((F k +0 mod. (4 76 VOLUME 5, NUMBER 3 k k

4 Since the Pisano period of {F k } k N modulo is 0, see [3], the period of {F k +0} k N modulo is also 0. Therefore, n 0 n mod 0 ((F k +0 mod ((F k +0 mod + ((F k +0 mod 0 k k k n mod n ((F k +0 mod 0 k ( n mod n (n mod ((F k +0 mod, (5 0 0 k k where we used ( in (5. A straightforward computation from (4 (5 shows that F n+ + 3n 5 +l n, where l n 7 n mod 0 (n mod m ((F k +0 mod. 55 Since (n mod 0 {0,...9}, one can easily see that l n {0, 7 This proves (. Thus, the proposed equality follows. k References 55, 34 55, 46 55, 53 55, 0, 3 55, 4 5, 6 55, 3 5 }. [] R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, Addison-Wesley, New York, 990. [] T. Koshy, Fibonacci Lucas Numbers with Applications, John Wiley, New York, 00. [3] OEIS Foundation Inc. (0, The On-Line Encyclopedia of Integer Sequences, Also solved by Paul S. Bruckman, Dmitry Fleishman, Russell Jay Hendel, David Stone John Hawkins (jointly, the proposer. The Fifth Seven Powers of Fibonacci Numbers B- Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain (Vol. 50.3, August 0 Let n be a nonnegative integer. Show that ( F 7 n+ Fn+ 7 F7 n is a fifth perfect power. F 5 n +F5 n F n+ F nf n+ Solution by Robinson Higuita (student, Universidad de Antioquia, Columbia. From [] we know that Fn+ 5 Fn 5 Fn+ 5 5F n+ F n+ F n [Fn+ +( n+ ] ( Fn+ 7 F7 n F7 n+ 7F n+f n+ F n [Fn+ +( n+ ]. ( AUGUST 03 77

5 THE FIBONACCI QUARTERLY Dividing ( by ( we have Fn+ 5 F5 n F5 n+ Fn+ 7 F7 n F7 n+ 5 7[F n+ +( n+ ]. So, Fn+ 5 Fn 5 +Fn+ 5 5(Fn+ 7 + F7 n F7 n+ 7[Fn+ +(F n+ +( n+ ]. (3 Using the Catalan Identity (see [, p. 83] we have This (3 imply that Note that F n+ F n F n+ +( n+. Fn+ 5 F5 n +F5 n+ + 5(F7 n+ F7 n Fn+ 7 7[Fn+ +F. (4 n+f n ] Fn+ +F n+f n Fn+ +F n+(f n+ F n+ Fn+ +F n+(f n+ F n+ Fn+ F nf n+. ( This (4 imply that Fn 5 +F5 n+ + 5 F 7 n+ Fn+ 7 F7 n 7 Fn+ 5. F n+ FnF n+ References [] L. Carlitz, Problem H-, The Fibonacci Quarterly, 5. (967, 7. [] T. Koshy, Fibonacci Lucas Numbers with Applications, John Wiley, New York, 00. Also solved by Brian D. Beasley, Paul S. Bruckman, Kenneth B. Davenport, Dmitry Fleischman, Russell Jay Hendel, Courtney Killian Tricia Stoner (jointly, Harris Kwong, Kathleen Lewis, Carl Libis, Ángel Plaza, the proposer. Nesbitt Type Inequality with Fibonacci Numbers B-3 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Bucharest Neculai Stanciu, George Emil Palade Secondary School, Buzău, Romania (Vol. 50.3, August 0 Prove that Fm (F q F n +F q+ F p + Fn (F q F p +F q+ F m + Fp (F q F m +F q+ F n 3 Fq+, for any positive integers m, n, p, q. Solution by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain. Since F q+ F q+ +F q, the proposed inequality is a particular case of the following more general inequality involving positive numbers: for any positive integers a, b, x, y, z. x (ay +bz + y (az +bx + z (ax+by 3 (a+b, ( 78 VOLUME 5, NUMBER 3

6 The last general inequality is a consequence of Chebyshev s sum inequality of the following Nesbitt type inequality []. If a, b, x, y, z are positive real numbers then x ay +bz + y az +bx + z ax+by 3 a+b. Then, by Chebyshev s sum inequality, the LHS of equation ( satisfies x ay+bz LHS + y az+bx + z ( ax+by x 3 ay +bz + y az +bx + ( 3 a+b 3 3 (a+b. References z ax+by [] S. Wu, O. Furdui, J. Seibert, P. Trojovský, A note on a conjectured Nesbitt type inequality, Taiwanese J. Math., 5. (0, Also solved by Paul S. Bruckman, Dmitry Fleischman, the proposer. A Simple Inequality B-4 (Correction Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Bucharest Neculai Stanciu, George Emil Palade Secondary School, Buzău, Romania (Vol. 50.3, August 0 If f : R + R + such that f(x > x for all x R +, prove that f((+ > 4F n F n+ for any positive integers n. k A composite solution by Amos G. Gera, Ashdod, Israel, Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain (independently. We have f((+ > k k n+ 4 (+F k + k k. k F 4 k AUGUST 03 79

7 THE FIBONACCI QUARTERLY Now, since F nf n+, the conclusion follows. k Also solved by Paul S. Bruckman, Charles K. Cook, Dmitry Fleischman, the proposer. A Product Involving a Series With Inverse Fibonacci Numbers B-5 Proposed by Hideyuki Ohtsuka, Saitama, Japan (Vol. 50.3, August 0 Given a positive integer m, prove that ( ( lim k F m k F m kn+ k { L m 5Fm (if m is even, (if m is odd. Solution by the proposer. We have m lim F m k n Fn m lim F m kn+ k lim k lim k lim ( α k β k m α n β n ( α k n β k α n α (k nm k n lim j0 α jm α m, lim kn+ lim kn+ lim kn+ β n α n j0 ( α n β n α k β k α jm m ( β n α n m α k n β k α n α (k nm α m α m α m. j m α jm 80 VOLUME 5, NUMBER 3

8 Therefore, lim ( k F m k The desired identity follows. ( F m kn+ k lim ( F m k F m k n α m α m ( F m n F m kn+ k α m +α m α m +( β m. Also solved by Paul S. Bruckman Dmitry Fleischman. We would like to belatedly acknowledge Gera Amos for solving problem B-08. AUGUST 03 8