ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS EDITED BY HARRIS KWONG Please submit solutios ad problem proposals to Dr. Harris Kwog Departmet of Mathematical Scieces SUNY Fredoia Fredoia NY or by at If you wish to have receipt of your submissio acowledged by mail please iclude a selfaddressed stamped evelope. Each problem or solutio should be typed o separate sheets. Solutios to problems i this issue must be received by November If a problem is ot origial the proposer should iform the Problem Editor of the history of the problem. A problem should ot be submitted elsewhere while it is uder cosideratio for publicatio i this Joural. Solvers are ased to iclude refereces rather tha quotig well-ow results. The cotet of the problem sectios of The Fiboacci Quarterly are all available o the web free of charge at BASIC FORMULAS The Fiboacci umbers F ad the Lucas umbers L satisfy F +2 = F +1 +F F 0 = 0 F 1 ; L +2 = L +1 +L L 0 = 2 L 1. Also α = (1+ /2 β = (1 /2 F = (α β / ad L = α +β. PROBLEMS PROPOSED IN THIS ISSUE B-1206 Proposed by José Luis Díaz-Barrero Barceloa Tech Barceloa Spai. Let 2 be a iteger ( Fi F j+1 2 F i+1 F j 2 F i F j 1 i<j i which the subscripts are tae modulo. 1 F +1 F B-1207 Proposed by D. M. Bătieţu-Giurgiu Matei Basarab Natioal College Bucharest Romaia ad Neculai Staciu George Emil Palade School Buzău Romaia. for ay iteger > 1. F 4 +F4 1 F 2 +F F 4 +F4 +1 F 2+1 > F F VOLUME NUMBER 2

2 ELEMENTARY PROBLEMS AND SOLUTIONS B-1208 Proposed by Iva V. Feda Vasyl Stefay Precarpathia Natioal Uiversity Ivao-Fraivs Uraie. For every positive iteger fid all real solutios of the followig liear system of equatios: F 1 x 1 + x 2 = F 3 F 2 x 1 + F 1 x 2 + x 3 = F 4 F 3 x 1 + F 2 x 2 + F 1 x 3 + = F F 1 x 1 + F 2 x 2 + F 3 x x = F +1 F x 1 + F 1 x 2 + F 2 x F 1 x + x +1 = F +2 F +1 x 1 + F x 2 + F 1 x F 2 x + F 1 x +1 = F B-1209 Proposed by Hideyui Ohtsua Saitama Japa. The Triboacci umbers T satisfy T 0 = 0 T 1 = T 2 ad for ay iteger 1. T = T 1 +T 2 +T 3 for 3. ( 2 T 2 T 2 1 = T 2 1 B-1210 Proposed by Taras Goy Vasyl Stefay Precarpathia Natioal Uiversity Ivao-Fraivs Uraie. t 1 +2t 2 + +t = where s = t 1 +t 2 + +t. ( 1 s s! t 1!t 2! t! Ft 1 1 Ft 2 2 Ft ( 1 2 SOLUTIONS A Telescopig Lucas Sum B-1186 Proposed by Hideyui Ohtsua Saitama Japa. =0 ( 1 L 2 L = 0. Solutio by Ágel Plaza Uiversidad de Las Palmas de Gra Caaria Spai. Sice L 2 +2( 1 = L 2 for ay eve umber m we have L 2m = L 2 m 2. This applies to every deomiator i the give expressio for > 0 ad hece ( 1 L 2 L = ( 1 L 2 L = ( 1 (L (L 2 1(L 2 +1 = ( 1 L ( 1 L MAY

3 THE FIBONACCI QUARTERLY Therefore the give sum telescopes from the secod term; thus N ( 1 L 2 L = ( 1N L 2 N+1 +1 from which the result follows. =0 Also solved by Jeremiah Bartz Bria Bradie Key B. Daveport Steve Edwards I. V. Feda Dmitry G. Fleischma G. C. Greubel Sai Gopal Rayaguru Jaroslav Seibert David Terr Da Weier ad the proposer. The Same Fiboacci Number B-1187 Proposed by José Luis Díaz-Barrero Barceloa Tech Barceloa Spai. Let 1 be a positive iteger. Fid all real solutios of the followig system of equatios: x 3 +L x+y = F (1+L +F x 2 F y 3 +F 2 y +z = F (1+F 2 +F 2 y 2 L z 3 +L 2 z +x = F (1+L 2 +F 2 z 2. Solutio by Bria D. Beasley Presbyteria College Clito SC. We use the idetity F 2 = F L to rewrite the system as (x 2 +L (x F = F y F (y 2 +L (y F = F z L (z 2 +L (z F = F x. The substitutio yields P (x F = F x where P = L (z 2 +L F (y 2 +L (x 2 +L. Sice 1 ad x y ad z are real we have P > 0 ad hece x = F. Thus y = F ad z = F as well. Note that the result also holds whe = 0 as the uique solutio of the system i that case is x = y = z = 0 = F 0. Also solved by I. V. Feda Dmitry Fleischma G. C. Greubel ad the proposer. A Telescopig Fiboacci Sum B-1188 Proposed by Key B. Daveport Dallas PA. Fid a closed form expressio for F F 3. Solutio by Bria Bradie Christopher Newport Uiversity Newport News VA. 180 VOLUME NUMBER 2

4 Worig from the Biet formula for F ELEMENTARY PROBLEMS AND SOLUTIONS F = α β we fid Therefore ad F 3 = α3 β 3 3(αβ (α β F = F 3 3( 1 F. F 3 3 = F ( 1 3 F 3 = F F 3 F 3 = F 3 +1 F 3 = F 3 +1 F 1 = F Editor s Remar: A similar result for the Lucas umbers L L 3 = L 3 +1 L 1 = L appeared as Elemetary Problem B-1176 i Volume 3.4 (November 201. Also solved by Itzal De Urioste (studet Steve Edwards I. V. Feda Dmitry Fleischma G. C. Greubel Harris Kwog Hideyui Ohtsua Ágel Plaza Sai Gopal Rayaguru Jaroslav Seibert David Terr Da Weier ad the proposer. Our Old Fried Biomial Theorem B-1189 Proposed by Ágel Plaza Uiversidad de Las Palmas de Gra Caaria Spai. Fid a closed form for 2 ( 2 L 2 2. Solutio by Harris Kwog SUNY Fredoia Fredoia NY. We shall derive a geeral result. Tae ote that for ay oegative iteger m α m 2 = α m (α 1 = (α+α 1 m = (α β m. I a similar maer we also fid β m 2 = (β α m. Therefore L m 2 = ( { m 0 if m is odd (α m 2 +β m 2 = 2 m/2 if m is eve. MAY

5 THE FIBONACCI QUARTERLY Editor s Remar: G. C. Greubel oted that 2 2 ( F2 2 = 0. Usig the same argumet show above we see that for ay oegative iteger m α m 2 β m 2 { 2 (m 1/2 if m is odd F m 2 = = 0 if m is eve. Also solved by Bria Bradie Key B. Daveport Itzal De Urioste (studet Steve Edwards I. V. Feda Dmitry Fleischma G. C. Greubel Ralph P. Grimaldi Russell Jay Hedel Hideyui Ohtsua Sai Gopal Rayaguru Jaroslav Seibert Jaso L. Smith David Terr ad the proposer. A Cyclic Sum B-1190 Proposed by José Luis Díaz-Barrero Barceloa Tech Barceloa Spai. Let 1 be a positive iteger. Compute ( F +2 F +F+1 F +2 F F +1 F 1 +F+1 1 +F F +3 F +1 F F +F +1 F +2 F ( F +1 +F +2 F F 1 +F+1 1 +F 1 +2 ( F +2 +F F +1 F 1 +F+1 1 +F 1 +2 Solutio 1 by Bria Bradie Christopher Newport Uiversity Newport News VA.. Because it follows that F +2 F F +1 = F +1 +F F F +1 F + 1 F +1 F +3 F +1 F +2 = F +2 +F +1 F +1 F +2 F F +2 2F +F +1 F +1 F +2 = F +F +2 F +2 F F F F +2 F F +1 (F +F +1 F +2 = F 1 +F F (F +1 F F +1 (F F +2 F +3 F +1 F +2 (F +1 +F +2 F = F+1 1 +F (F+2 F F + 1 (F+1 F +1 F +2 2F +F +1 F +2 F (F +2 +F F +1 = F+2 1 +F (F F F (F+2 +2 F F VOLUME NUMBER 2

6 ELEMENTARY PROBLEMS AND SOLUTIONS Therefore ( ( F +2 F +F+1 F +2 F F +1 F 1 +F F +3 F +1 +F+2 F +F 1 F F +2 F 1 ( + 2F +F +1 F +2 +F F +1 = 2(F 1 F +2 F F 1 F 1 +F+1 1 +F F+1 1 +F F+1 1 +F F+1 1 +F 1 +2 Solutio 2 by Ágel Plaza Uiversidad de Las Palmas de Gra Caaria Spai. The result is 2 for all itegers 1. Let us cosider the followig more geeral expressio 1 a+b E = a 1 +b 1 +c 1 (a +b c. ab Notice that a+b ab (a +b c abc from which we fid E = 2. abc c(a+b(a +b c = 2. [c(a +1 +b +1 (a+bc +1 +abc(a 1 +b 1 ] = 2abc(a 1 +b 1 +c 1 abc = 2(a 1 +b 1 +c 1 Also solved by Key B. Daveport Itzal De Urioste (studet Steve Edwards I. V. Feda Dmitry G. Fleischma G. C. Greubel Marcus Harbol ad Lue Tiscareo (studets George A. Hisert Wei-Kai Lai ad Joh Risher (studet (joitly Hideyi Ohtsua Sai Gopal Rayaguru Jaroslav Seibert Jaso L. Smith ad the proposer. Belated Acowledgmet: The editor would lie to belatedly acowledge the solutios to Problems B-117 B-1182 ad B-1183 by Key B. Daveport ad the solutio to Problem B-118 by Dmitry G. Fleischma. MAY