ELEMENTARY PROBLEMS AND SOLUTIONS

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1 EDITED BY RUSS EULER AND JAWAD SADEK Please submit all new problem proposals and their solutions to the Problems Editor, DR. RUSS EULER, Department of Mathematics and 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 and 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 and solution should be typed on separate sheets. Solutions to problems in this issue must be received by May 15, 014. 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 and 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+ 5)/, β = 1 5)/, n = α n β n )/ 5, and L n = α n +β n. PROBLEMS PROPOSED IN THIS ISSUE B-1136 Proposed by Hideyui Ohtsua, Saitama, Japan. +1 ) 3 = +1). =1 =1 NOVEMBER

2 THE IBONACCI QUARTERLY B-1137 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, and n 1) =0 =0 1) n ) L m +L m+1 L m+ ) m + m+1 m+ = Ln m +Ln m+1 L n m+ = n m +n m+1 n m+ for any positive integer n; 1) for any positive integer n. ) B-1138 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, if m > 0 and p > 0, then ) m+1 L m for any positive integer n. =1 =1 ) p+1 L p n+ 1) m+p+ L n 3) m+p, B-1139 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, 1+ +L ) > 4 n n+1 +L n L n+1 ) 8, =1 for any positive integer n. B-1140 Proposed by José Luis Díaz-Barrero, BARCELONA TECH, Barcelona, Spain. Let n be a positive integer. Show that 1 n 3 n+1 n+1 n ) + n+1 3 n n n+1 ) + n+ 3 ) n+1 n+ n+1 ) is an integer and determine its value. 368 VOLUME 51, NUMBER 4

3 SOLUTIONS In recognition of his invaluable contributions to the ibonacci Quarterly, we are dedicating this issue to the late Paul S. Brucman. Paul had solved all of the problems appearing in this section during our tenure as co-editors, and we are featuring his solutions to all the problems in this issue. He will be missed. Another Division by 5 B-1116 Proposed by M. N. Deshpande, Nagpur, India. Vol. 50.4, November 01) Let n be a nonnegative integer and let T n be the nth triangular number. each of the following is divisible by 5: Solution by Paul S. Brucman, Nanaimo, BC, Canada. nl n+1 + n 1) T n L n+1 +n+1) n ) The sequence {L n mod 5)} n=1 is periodic with period 4; the repeating) period is {1,3, 4,}. Thisisthesameasthesequence{L n+1 mod 5)} n=0. Thesequence{nL n+1 mod 5)} n=0 is readily seen to be periodic with period 0. We find that its repeating period is as follows: {0,3,3,1,4,0,4,4,3,,0,,,4,1,0,1,1,,3}. Also, the sequence { n mod 5)} n=0 is periodic with period 0. Its repeating period is seen to be as follows: {0,,,4,1,0,1,1,,3,0,3,3, 1,4,0,4,4,3,}. It then readily follows that the sequence {nl n+1 + n ) mod 5)} n=0 is periodic with a period of 1, and its repeating period is {0}. That is, the first given sequence is divisible by 5 for all n 0. If u n = nl n+1 + n and v n = T n L n+1 +n+1) n, we have shown that 5 u n for all n 0. Since T n = nn+1)/, we see that v n = n+1) u n. If n is odd, this immediately implies that 5 v n for all odd n 1. However, v n is clearly an integer. It must therefore be true that 5 v n for all even n 0, which proves that 5 v n for all n 0. Also solved by St H. Brown, Michael R. Bacon and Charles C. Coo jointly), Edwardo H. M. Brietze, Kenneth B. Davenport, Dmitry leischman, Amos E. Gera, Ralph P. Grinaldi, Russell Jay Hendel, Robinson Higuita, Zbigniew Jacubczyc, Parviz Khalili, Harris Kwong, Kathleen E. Lewis, Zachary McCaslin, Ángel Plaza and Sergio alcón jointly), David Stone and John Hawins jointly), Rattanapol Wasutharat, and the proposer. Two Sums with Square Roots B-1117 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Vol. 50.4, November 01) NOVEMBER

4 THE IBONACCI QUARTERLY : ) 4 +1 < n n+1 1) L 4 L +1+ L 1 ) L 4 +1 < L n L n+1 ) =1 =1 for any positive integer n. Solution by Paul S. Brucman, Nanaimo, BC, Canada. Let S n = = ) 4 +1,T n = We also use the following nown identities: = n n+1 ; =1 =1 L 4 L +1+ L 1 ) L 4 +1,n = 1,,... 1) L = L nl n+1. ) =1 It suffices to prove the following, valid for all x 1: x x+1+ x 1 x x. 3) +1 or if 3) is true, then setting x = or x = L in 3) and using ) implies S n n n+1, T n L n L n+1. This is a slightly weaer result than the desired original, since we find that equality holds for n = 1 i.e. x = 1). Proof of 3). Given x 1, let Gx) = x x x+1 x 1 x +1 = x3 +1 x+1 x x+1 = x3 +1 x+1 which is manifestly 0 for all x 1. This proves 3). x 3 +1 x+1 Also solved by Charles C. Coo, Kenneth B. Davenport, Dmitry leischman, Amos E. Gera, Russell Jay Hendel, Robinson Higuita, Zbigniew Jacubczyc, Parviz Khalili, Harris Kwong, Kathleen E. Lewis, Ángel Plaza, Marielle Silvio and Kasey Zemba jointly), and the proposer. Quadratic Identities B-1118 Proposed by Gordon Clare, Brisbane, Australia. Vol. 50.4, November 01) If n is a nonnegative integer, prove that: n +n+1 ) n+ +n+3 ) = n ) n +n+ ) n+4 +n+6 ) = n+6 + n+3 ±5). ) Solution by Paul S. Brucman, Nanaimo, BC, Canada. 370 VOLUME 51, NUMBER 4

5 Part 1). We use the following nown identity: n + n+1 = n+1. Then also, n+1 + n+3 = n+5. Therefore, the left side of 1) equals n+1 n+5. We now use the following nown identity: m m+4 m+ = 1)m 1. Therefore, setting m = n+1 in the left side of 1) yields n+1 n+5 = n Part ). We use the following nown identities: n + n+ = 3 n+1 1) n and m + m+4 = 7 m+ + 1) m. Then also n+4 + n+6 = 3 n+5 1)n. The left side of ) then becomes {3 n+1 1)n }{3 n+5 1)n } = 9 n+1 n+5 ) 6 1) n { n+1 n+5 }+4; from Part 1), with m = n+1, n+1 n+5 = n+3 + 1)n. Also, with m = n+1, we obtain n+1 + n+5 = 7 n+3 1)n. Then the left side of ) equals The left side of ) then becomes 9{ n+3 + 1)n } 6 1) n { n+1 + n+5 }+4. 9{ n+3 + 1) n } 6 1) n {7 n+3 1) n }+4 = 9 4 n ) 1) n n ) = 9 4 n+3 4 1) n n+3 +5 = 5 4 n+3 4 1)n n+3 +{ n+3 5 1)n } = n+3 {5 n+3 4 1)n }+{ n+3 5 1)n } = n+3 L n+3 +{ n+3 5 1)n } = n+6 +{ n+3 5 1) n }. George A. Heisert used a nontraditional approach to solve the problem. The technique called dual polynomial method) is explained in detail in his article A different method for Deriving ibonacci Power Identities, JP Journal of Algebra, Number Theory and Applications, 4 01), 1 6. Also solved by Charles C. Coo, Kenneth B. Davenport, Dmitry G. leishman, Amos E. Gera, George A. Heisert, Russell Jay Hendel, Zbigniew Jaubczy, Harris Kwong, Carl Libis, Ángel Plaza and Sergio alcón jointly), and the proposer. A Trig and ibonacci Amalgam B-1119 Proposed by D. M. Bătineţu Giurgiu, Matei Basarab National College, Vol. 50.4, November 01) NOVEMBER

6 THE IBONACCI QUARTERLY tan x =1 x n+1 x ) n n, for any x 0, π ). n n+1 4 Solution by Paul S. Brucman, Nanaimo, BC, Canada. In the following manipulations, all trigonometric functions are well-defined and positive in the open interval x 0,π/4). We first observe that the proposed inequality, as presently stated, is false. To see this, we let n = 1, x = π/6. The left member of the proposed inequality is 1 tan π 1) = ; the right memberis{π/1) 4π/3)} /4 = )/ Thus,theproposedinequality is clearly false. We believe that the proposer intended to show the following inequality: tan x =1 { x n ) n x)} / n n n+1,n = 1,,..., for all x 0,π/4). 1) We first mae the following definitions for = 1,,...,n: u = x ; a = tanu, b = ; therefore, a b = tanu. Let tan u S = Sx,n) = = a. =1 We also recall the following well-nown identity: b = = n n+1. =1 =1 We now invoe the Cauchy-Schwarz inequality: { } { a b =1 Therefore, { n } tanu =1 n n+1 S or { S =1 =1 tanu a }{ =1 =1 b }. } / n n+1. ) We now employ the following trigonometric identity, valid for z 0, π/) z z = tanz. 3) We may easily verify 3) from the definitions of the indicated functions. Setting z = u, we then have tanu = u u = x ) x 1 ) ; then tanu = x/ ) x/ 1 ) 1. rom this, it follows by telescoping that tanu = u n n x. 4) =1 37 VOLUME 51, NUMBER 4

7 Proof of 1). By ) and 4), S { x n ) n x) } / n n+1, or S { x ) / n x)} n n n n+1. 5) Equality occurs if and only if n = 1; in this case both sides are equal to ) tanx/) ) x/) x) =. or n = 1 and x = π/6, each side equals Higuita too a different route and corrected the proposed inequality by starting the sum at = 0. Also solved by Kenneth B. Davenport, Dmitry leischman, Robinson Higuita, and the proposer. Periodic Sequences B-110 Proposed by the Problem Editor. Vol. 50.4, November 01) Prove or disprove: n n3 n mod 5) for all nonnegative integers n. Solution by Paul S. Brucman, Nanaimo, BC, Canada. The sequence { n mod 5)} n=0 is periodic with period 0. Its repeating period is found to be as follows: {0,1,1,,3,0,3,3,1,4,0,4,4,3,,0,,,4,1}. On the other hand, the sequence {n mod 5)} n=0 is periodic with period 5; its repeating period is {0,,4,1,3}. Also the sequence {3 n mod 5)} n=0 is periodic with period 4; its repeating period is {1,3,4,}. Therefore, the sequence {n3 n mod 5)} n=0 is periodic with period 0 equal to GCM4,5). We find that its repeating period is the same as that of { n mod 5)} n=0. Therefore, n n3 n mod 5) for all n 0. The conjecture is true. Also solved by Edwardo H. M. Brietze, Michael J. Bucmarter, Charles C. Coo, Kenneth B. Davenport, Dmitry leischman, Amos E. Gera, Ralph P. Grimaldi, Russell J. Hendel, Robinson Higuita, Harris Kwong, Kathleen E. Lewis, Carl Libis, Marielle Silvio and Kasey Zemba jointly), David Stone and John Hawins jointly), Matt Zinle, and the proposer. Solution to Problem 1114 was published in the August, 013 issue. We list here the name of additional solvers: Brian Beasley, Paul S. Brucman, Russell Jay Hendel, Gurdial Aroroa and Sindhu Unnithan jointly). We would lie to belatedly acnowledge Charles Coo for solving problem B-111. NOVEMBER