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 acnowledged, 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 August 1, 01. 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 ased to include references rather than quoting well-nown results. The content of the problem sections of The ibonacci Quarterly are all available on the web free of charge at BASIC ORMULAS The ibonacci numbers n the Lucas numbers L n satisfy n+ = n+1 + n, 0 = 0, 1 = 1; L n+ = L n+1 +L n, L 0 =, L 1 = 1. Also, α = (1+ )/, β = (1 )/, n = (α n β n )/, L n = α n +β n. PROBLEMS PROPOSED IN THIS ISSUE B-1114 (Correction) Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade Secondary If f : R + R + such that f(x) > x for all x R +, prove that n f ( (1+ )) > 4 n n+1 for all positive integers n. 84 VOLUME 1, NUMBER 1

2 B-111 Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade General for any positive integer n. n+4+4 n n+1 > 4 n+ (1) n+4+4l n L n+1 > 4L n+ () B-11 Proposed by Harris Kwong, SUNY redonia, redonia, NY, given any integer r 4, if gcd(r 1, r r 1) = 1, then n+φ(r r 1) n (mod r r 1) for all nonnegative integers n. Here, φ denotes Euler s phi-function. B-11 Proposed by José Luis Díaz-Barrero, BARCELONA TECH, Barcelona, Spain Mihály Bencze, Braşov, Romania Let n be a positive integer. 1 n n n n n n 1 n+ 1. B-114 Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade General for any positive integer n. n ( ) +1 + > n n ( ) L L +1 + > n L + L +L +1 (1) () B-11 Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade General (L 1 +1)(L +1) L 1 L +1 for any positive integer n. + (L +1)(L +1) L L (L n 1 +1)(L n +1) + (L n +1)(L 1 +1) L n+ 6, L n 1 L n +1 L n L 1 +1 EBRUARY 01 8

3 THE IBONACCI QUARTERLY SOLUTIONS The Ubiquitous AM-GM Inequality B-1101 Proposed by D. M. Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania Neculai Stanciu, George Emil Palade Secondary arctan n +n+1 L n +L n+1 +arctan arctan n+ +arctan L n+. Almost all solvers provided a version of the following solution. ( n n+1 ) 0 n +n+1 n n+1 (n +n+1) ( n + n+1 ) = n+ ( n +n+1 ) n+ n +n+1 n+ 4 since the arctan function is increasing, it follows that n arctan + n+1 arctan n+. An identical argument wors for the Lucas inequality. Thus, adding the two inequalities yields the desired result. Also solved by Paul S. Brucman, Charles K. Coo, Kenneth B. Davenport, Robinson Higuita Alexer Ramirez (jointly), Zbigniew Jaubczy (student), Harris Kwong, ONU-Solve Problem Group, Ángel Plaza, Jaroslav Seibert, the proposer. rom the Weighted AM-GM Inequality B-110 Proposed by Diana Alexrescu, University of Bucharest, Bucharest, Romania José Luis Díaz-Barrero, BARCELONA TECH, Barcelona, Spain Let n be a positive integer. n + L n + n+ n+1 L n+ L n+1 < 4. Solution by Zbigniew Jaubczy (student), Warsaw, Pol. 86 VOLUME 1, NUMBER 1

4 Since the function f(x) = x is concave for all x 0, Jensen s inequality yields (x+y) x > + y, 4 for x 0, y 0, x y. Under the same conditions, this inequality can be equivalently written as x + y < (x+y). or x = n y = n+1, we obtain n + n+1 < ( n + n+1 ) = n+. This implies n + n+ Similarly, letting x = L n y = L n+1 yields L n + L n+ n+1 L n+1 Multiplying inequalities (1) () yields the desired result. <. (1) <. () Also solved by Paul S. Brucman, Charles K. Coo, Kenneth B. Davenport, Robinson Higuita (student), ONU-Solve Problem Group, Ángel Plaza, Jaroslav Seibert, David Stone John Hawins (jointly), the proposer. An Extension to Negative Subscripts B-110 Proposed by Hideyui Ohtsua, Saitama, Japan If a+b+c = 0 abc 0, find the value of ( a L a + b L b + c L c L a L b L c a b c Solution by ONU-Solve Problem Group, Ohio Northern University We will show that if a+b+c = 0 abc 0, then ( L a L b L c a + b + ) c =. (1) a b c L a L b L c After a straightfoward algebraic manipulation, (1) is equivalent to ). L a L b c +L a b L c + a L b L c + a b c = 0. () By using c = (a+b) together with the formulas for the natural extensions of the ibonacci Lucas numbers to negative subscripts n = ( 1) n 1 n L n = ( 1) n L n we can rewrite () in the following form L a b L a+b + a L b L a+b = L a L b a+b + a b a+b. () EBRUARY 01 87

5 THE IBONACCI QUARTERLY By using the explicit formulas n = αn β n L n = α n + β n, where α = (1 + )/ β = (1 )/, we have L a b + a L b = a+b, (4) Indeed, L a L b + a b = L a+b () L a b + a L b = 1 (α a +β a )(α b β b )+ 1 (α a β a )(α b +β b ) = (α a+b β a+b ) = a+b L a L b + a b = (α a +β a )(α b +β b )+(α a β a )(α b β b ) = (α a+b +β a+b ) = L a+b. We now use the identities (4) () in order to prove (). Indeed, from (4), the left-h side of () becomes (L a b + a L b )L a+b = a+b L a+b. On the other h, from (), the righth side of () becomes (L a L b + a b ) a+b = L a+b a+b. This concludes the proof of the desired identity. Also solved by Paul S. Brucman, Charles K. Coo, Robinson Higuita (student), Zbigniew Jaubczy (student), abian Maple, Ángel Plaza, Jaroslav Seibert, the proposer. A Symmetrical Identity B-1104 Proposed by Javier Sebastián Cortés (student), Universidad Distrital rancisco José de Caldas, Bogotá, Colombia n+(+1) L n+(+1)i = L n+(+1) n+(+1)i. Solution by Robinson Higuita (student), Universidad de Antioquia, Colombia. We claim that Indeed, (+1)( i) = = (+1)( i) + (+1)i + (+1)( i) = 0. (1.1) (+1)( i) = i= ( 1) (+1)i+1 (+1)i = (+1)(i) + (+1)( i) (+1)i (+1)i = 0. rom [1, page 9] we now that m L n = n+m +( 1) n m n. This implies that n+(+1) L n+i(+1) = n+(+1)(i+) +( 1) n (+1)( i), (1.) 88 VOLUME 1, NUMBER 1

6 L n+(+1) n+i(+1) = n+(+1)(i+) +( 1) n (+1)( i), Thus, from (1.1), (1.), (1.) we obtain n+(+1) This (1.1) imply that n+(+1) L n+(+1)i = L n+(+1)i = = n+(+1)(i+) +( 1) n+1 (+1)( i). (1.) n+(+1)(i+) +( 1) n (+1)( i). n+(+1)(i+) +( 1) n+1 = L n+(+1) n+(+1)i. References (+1)( i) [1] T. Koshy, ibonacci Lucas Numbers with Applications, John Wiley, New Yor, 001. Also solved by Paul S. Brucman, Kenneth B. Davenport, Harris Kwong, Ángel Plaza, Jaroslav Seibert, the proposer. ibonomial Coefficients B-110 Proposed by Paul S. Brucman, Nanaimo, BC, Canada Let where [ m+1] m+1 G m (x) = =0 ( 1) (+1)/ [ m+1 is the ibonomial coefficient 1... m+1 [ ] [ ] m+1 m+1 = = 1. 0 m+1 ] x m+1,m = 0,1,,..., ( 1... )( 1... m+1 ) Let G m (1) = U m. Prove the following, for n = 0,1,,...: (a) U 4n = 0; (b) U 4n+ = ( 1) n+1 {L 1 L L...L n+1 } ; (c) U n+1 = ( 1) (n+1)(n+)/ {L 1 L L...L n+1 }. Solution by E. Kilic I. Aus (jointly). rom [1], we have m+1 [ ] m+1 G m (x) = ( 1) (+1)/ x m+1 = =0, 1 m; also define m (1 α j β m j ), EBRUARY 01 89

7 THE IBONACCI QUARTERLY where [ n] sts for the usual ibonomial coefficients, α = (1 + )/, β = (1 )/, α+β = 1, αβ = 1. (a) Notice that G 4n (1) = U 4n. Since 4n+1 [ ] 4n+1 4n U 4n = ( 1) (+1)/ = (1 α j β 4n j ) =0 includes the factor (1 (αβ) n ) = 0, U 4n = 0. (b) Using the Binet formula for the Lucas numbers, we obtain 4n+ [ ] 4n+ 4n+ U 4n+ = ( 1) (+1)/ = (1 α i β 4n+ i ) (c) or m = n+1, U n+1 = =0 n+1 = (1+( 1) i α i )(1+( 1) i β i ) n+1 = ( 1) i (α i +( 1) i +β i ) n+1 = ( 1) n+1 (α i +β i ) n+1 = ( 1) n+1 = n+ =0 L i. [ ] n+ ( 1) (+1)/ = (1 α j β n+1 j ) n+1 n ( 1) j+1 (α n+1 j +β n+1 j ) = ( 1) (n+ ) n L j+1. References [1] L. Carlitz, The characteristic polynomial of a certain matrix of binomial coefficients, The ibonacci Quarterly,. (196), All solvers gave, more or less, a similar proof. Also solved by Harris Kwong, Ángel Plaza Sergio alcón (jointly), the proposer. We wish to belatedly acnowledge the solution to problem B-1099 by Amos Gera. 90 VOLUME 1, NUMBER 1