ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS 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 acknowledged, 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 November 1, 011 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 wwwfqmathca/ BASIC FORMULAS The Fibonacci numbers F n and the Lucas numbers L n satisfy F n+ = F n+1 +F n, F 0 = 0, F 1 = 1; L n+ = L n+1 +L n, L 0 =, L 1 = 1 Also, α = 1+ /, β = 1 /, F n = α n β n /, and L n = α n +β n PROBLEMS PROPOSED IN THIS ISSUE B-1086 Proposed by José Luis Díaz-Barrero, Polytechnical University of Catalonia, Barcelona, Spain Let n be a positive integer Prove that F n+1 +F n F n+1 +1 F n+ + F i F j is an integer and determine its value 1 i<j n 180 VOLUME 49, NUMBER

2 ELEMENTARY PROBLEMS AND SOLUTIONS B-1087 Proposed by Br J Mahon, Kensington, Australia Evaluate i=3 F i 1 +F i 1 F i 1 F i 1 B-1088 Proposed by José Luis Díaz-Barrero, Polytechnical University of Catalonia, Barcelona, Spain Let {a n } n 1 be a sequence of real numbers defined by a 1 = 3, a = and for all n 3, a n+1 = 1 a n +1 Prove that L k 1+ < L nl n+1 1+ak B-1089 Proposed by Mohammad K Azarian, University of Evansville, Indiana Let the sequence a 1,a,a 3,,a n, be defined by the recurrence relation a n+ = a n+1,n 0,a 0 = 1,a 1 = a n Also, let p 1,p,p 3,,p n, represent a permutation of a 1,a,a 3,,a n Show that n! a i p i is divisible by n!+f n!+ 1 for n 0 j=1 i=1 a j B-1090 Proposed by N Gauthier, Kingston, ON, Canada a Let x be an arbitrary variable and k, m, n, r nonnegative integers, with 0 k n and 0 r m Also let {S r m : 0 r m} be the set of Stirling numbers of the second kind, which satisfy the following recurrence, with initial value S 0 0 = 1 and boundary conditions S m 1 = 0 and Sm m+1 = 0: S m+1 r = rs m r +S m r 1 Finally, let n r = nn 1 n r + 1 for r 0, with n 0 = 1, and adopt the convention that k 0 = 1 for all values of k, including zero Prove that: n m k m x k = n r S r m x r 1+x n r k k=0 b Use a to show that n k 4 F k = anf n 3 +bnf n 4 k k=0 r=0 and determine the polynomial coefficients, an and bn MAY 0181

3 THE FIBONACCI QUARTERLY SOLUTIONS A Sequence of Nested Radicals B-1066 Proposed by Hideyuki Ohtsuka, Saitama, Japan Vol 48, May 010 Determine the value of 1+F 1+F4 1+F 6 1+F n Solution by Kenneth B Davenport, Dallas, PA A solution to this intriguing relation depends on the identity F n 1 = F n F n+ 1 This relation can be established via the known result appearing as I 13 in Fibonacci and Lucas Numbers, by F n 1 F n+1 F n = 1n, n 1 Substituting n for n in yields F n 1 F n+1 Fn = 1 3 Now in the RHS of 1 we substitute F n F n 1 for F n, and F n +F n+1 for F n+ to get or which is F n F n 1 F n +F n+1 F n F nf n 1 +F n F n+1 F n 1 F n+1 F n +F n F n+1 F n 1 F n 1 F n+1 = F n +F n F n 1 F n+1 Using and 3, the above simplifies to Fn 1 and this proves relation 1 This implies F 4 = 1+F F 6 and F 6 = 1+F 4 F 8 This substitution yields F 4 = 1+F 1+F4 F 8 Continuing in this fashion by substituting 1+F 6 F 10 for F 8, it is straightforward to derive an endless number of nested squareroots This, then, establishes Ohtsuka s relation, ie 1+F4 F 4 = 3 = 1+F 1+F 6 1+F n Also solved by Paul S Bruckman, Russell J Hendel, and the proposer 18 VOLUME 49, NUMBER

4 ELEMENTARY PROBLEMS AND SOLUTIONS Close The Sum! B-1067 Proposed by N Gauthier, Royal Military College of Canada, Kingston, ON, Canada Vol 48, May 010 Let n be a positive integer Find a closed-form expression for kfk 3 Solution by Charles K Cook, Sumter, SC 910 Differentiating n xk = x1 xn 1 x and multiplying by x yields kx k = x1 x[1 n+1xn ]+x 1 x n 1 x 1 Next α Fk 3 k = β k 3 = α3k β 3k 3 1 k α k β k Using αβ = 1, 1 α 3 = 4α, 1 β 3 = 4β, 1+α = α, and 1+β = β as needed, it follows from 1 that A = kα 3k = α [1 n+1α 3n ]+α 4 1 α 3n 4 B = kβ 3k = β [1 n+1β 3n ]+β 4 1 β 3n 4 C = k α k = α+n+1 1n α n n α n α and Subtracting yields D = k β k = β +n+1 1n β n n β n β A B = α β +n+1α 3n+ β 3n+ +α 4 β 4 α 3n+4 β 3n+4 4 = 4 [1+n+1F 3n+ F 3n+4 ] and C D = α β α β +n+1 1 n α n 1 β n 1 1 n α n β n = [ n+1 1 n F n n F n ] MAY 0183

5 THE FIBONACCI QUARTERLY Substituting these expressions into and simplifying yields kfk 3 = +nf 3n+ F 3n n nf n 1 F n 3 0 Also solved by Paul S Bruckman, Kenneth B Davenport, G C Greubel, Russell J Hendel, Jaroslav Seibert, David Terr, and the proposer Limit of a Sequence B-1068 Proposed by Mohammad K Azarian, University of Evansville, Indiana Vol 48, May 010 If the sequence s 0,s 1,s,s 3, is defined by the difference equation ks k+1 +1 ks k s k 1 = 0,k 1,s 0 = F n,s 1 = F n, then write lim k s k in terms of Lucas numbers Solution by Francisco Perdomo and Ángel Plaza jointly, Universidad de Las Palmas de Gran Canaria, Las Palmas G C, Spain The difference equation defining s k may be written for k > 0, as follows s k+1 s k = 1 k s k s k 1 It follows that s k+1 s k = 1 k 1 k! s 1 s 0 and so lim s k = s 0 +s 1 s 0 +s s 1 +s 3 s + k 1 k 1 = s 0 +s 1 s 0 k! k=0 = s 0 +s 1 s 0 e 1/ = F n + F n F n e = L n 1 +L n+1 1+ L n 1 e Also solved by Paul S Bruckman and the proposer A Fibonacci Dis Array! B-1069 Proposed by Pantelimon George Popescu, Politechnica University, Bucharest, Romania and José Luis Díaz-Barrero, Polytechnical University of Catalonia, Barcelona, Spain Vol 48, May 010 Let n,a,b,c,d be positive integers and A be a square matrix for which A n Fa+n 1 L = b+n 1 L c+n 1 F d+n VOLUME 49, NUMBER

6 Show that b = c and A = 1 0 or A = ELEMENTARY PROBLEMS AND SOLUTIONS 0 1 Solution by Steve Edwards, Southern Polytechnic State University, Marietta, GA Correction: The matrix should consist of all Fibonacci numbers Also, a and d should be non-negative rather than positive Fa F Since A = b, A F c F Fa+1 F = b+1, but also, d F c+1 F d+1 A F = AA = a +F b F c F a F b +F b F d F a F c +F c F d F b F c +Fd we have F b+1 = F b F a +F d and F c+1 = F c F a +F d, from which it follows that b = c But since F b+1 F b = F a +F d, F b+1 F b is an integer, which is only possible for b = 1 or For b =, F a +F d =, but this implies that F a and F d have values 0, 1, or, all of which lead to contradictions For b = 1, F b+1 F b = F F 1 = 1 = F a +F d, so either F a = 1 and F d = 0 or vice versa Both F 1 and F equal 1, but it is easily seen that we need a = and d = 0 or a = 0 and d = in order F F 1 F0 F or A = 1 Finally, F 1 F 0 F 1 F for the equation for A n to be satisfied, ie A = note that for A = 1 0 A k+1 Fk+1 F = A k = F k F k The case A = is similar, the equation with A n is satisfied since, Fk+1 +F k F k +F k 1 Fk+ F = k+1 F k F k 1 F k+1 F k A corrected version was also solved by Paul S Bruckman and Russell J Hendel A Trigonometric Identity B-1070 Proposed by Roman Witula and Damian Slota, Silesian University of Technology, Poland Vol 48, May 010 Prove that x+ tan 1 arctan α tanx β tanx α, for all x k 1 π, x kπ +arctanβ, x kπ +arctanα, x kπ 1 arctan and k Z Solution by Paul S Bruckman, 38 Front St, Unit #30, Nanaimo, BC, V9R 0B8, Canada MAY 018

7 THE FIBONACCI QUARTERLY For the moment, we assume that x is such that the expression in the statement of the problem is well-defined Now tan 1 = cos, so 1 tan 1 = tan = tan 1 = tan 1 = tan 1 +1 α Therefore, tan x+ 1 arctan = tanx+ 1 α 1 tanx α = αtanx+1 α tanx Then, tan x+ 1 arctan α tanx β tanx = αtanx+1 β tanx = αtanx β β tanx = α; therefore, tan x+ 1 arctan α tanx β tanx = α We must exclude the possible values of x that might be singularities First of all, tanx must bewell-defined, which implies thatx k 1 π whenk Z Secondly, α tanx andβ tanx must not vanish; that is, x kπ+arctanα and x kπ+arctanβ Finally, tanx+ 1 arctan must be well-defined, so x k 1 pi 1 arctan, or x k 1pi +arctanβ We may combine two of these conditions as follows: x k pi +arctanβ = kpi 1 arctan Also solved by Charles K Cook, Kenneth B Davenport, Steve Edwards, G C Greubel, Russell J Hendel, Zbigniew Jakubczyk, Jaroslav Seibert, David Terr, and the proposers 186 VOLUME 49, NUMBER