ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. H1LLMAN Uiversity of New Mexico, Albuquerque, New Mex. Sed all commuicatios regardig Elemetary Problems ad Solutios to Professor A* P. Hillma, Departmet of Mathematics ad Statistics, Uiversity of New Mexico, Albuquerque, New Mexico Each problem or solutio should be submitted I legible form, preferably typed i double spacig, o a separate sheet or sheets, i the format used below. Solutios should be received withi three moths of the publicatio date, B-148 Proposed by David Eglud, Rockford C o l l e g e, Rockford, I l l i o i s, ad M a l c o l m Tollma, Brookly^ N e w Y o r k. Let F ad L deote the Fiboacci ad Lucas umbers ad show that 11 F, t v = F L L 0 L L /o f-i \ (2%) 2 4 (2 1 TI) B-149 Proposed by V. E. Hoggatt, J r., Sa Jose State C o l l e g e, Sa Jose, C a l i f. Show that L M L L + 4(-l) + 1 = 5F F ± i. +i B-150 Proposed by V. E. Hoggatt, J r., Sa Jose State C o l l e g e, Sa Jose, C a l i f. Show that L 2 - F 2 = 4F F ^. -i +i B Proposed by Hal Leoard, Sa Jose State C o l l e g e, Sa Jose, C a l i f. Let Let m = Lj + L2 + + L be the sum of the first Lucas umbers. P (x) = ( l + x 1 ) = a 0 + ajx a m xm. i=i Let q be the umber of itegers k such that both 0 < k < m ad a, = 0. Fid a recurrece relatio for the q. 400

2 Dec ELEMENTARY PROBLEMS AND SOLUTIONS 401 B-152 Proposed by Phil Maa, Uiversity of New Mexico, Albuquerque, N. Mex. Prove that m+ m+i +i m-i -i B Proposed by KIaus-Guther Recke, G o t t i g e, Germay. P r o v e that F{F% + F2F3 + F3F9 + + F F3 = FF +if 2 +i «SOLUTIONS GOLDEN RATIO AGAIN? B a Proposed by Sidey K r a v i t z, D o v e r, N. Jersey. A eterprisig etrepreeur i a amusemet part challeges the public to play the followig game. The player is give five equal circular discs which he must drop from a height of oe ich oto a larger circle i such a way that the five smaller discs completely cover the larger oe. What is the maximum ratio of the diameter of the larger circle to that of the smaller oes so that the player has the possibility of wiig? Partial Solutio by the Proposer. With the ceters of the smaller circles placed at the vertices of a regular petago, the smaller circles cover the larger oe with a ratio of diameters equal to the golde ratio (1 + \ / 5 ) / 2. There may exist aother arragemet of the five circles which results i a smaller ratio. EVEN AND ODD SEQUENCES B-131a Proposed by Charles R. Wall, Uiversity of Teessee, Koxville, Te. Let JH I be a geeralized Fiboacci sequece, i. e., H 0 = q* Hi = p, H +2 = H +1 + H * Exted, by the recursio formula, the defiitio to iclude egative subscripts,, Show that if SH I = IH I for all, the JH is a costat multiple of either the Fiboacci or the Lucas sequece.

3 402 ELEMENTARY PROBLEMS AND SOLUTIONS [Dec. Solutio by David Zeitli / Mieapolis, Miesota. ad sice F = (-1) F, we have IH I = (-l) (qf - (p - q)f )j = IqF 4 - (p - q)f I. J - J ' H F H/ F -i I p - i If! H t = H.i, the (a) p - q = p or (b) p - q = -p. If (a) holds, the q = 0 1 ad H ~ pf ; if (b) holds, the q = 2p, ad H = 2pF. - p F = pl. K +i ^ ^ Remark. Let U ad V be solutios of m W, = aw, + bw +2 +i f where U 0 = 0, Ui = 1 ad V 0 = 2, V t = a (if a = b = 1, the U = F ad V ' = L ). If 7 \ huw -\ = W for all, the {W } is a costat multiple of either {u } or {v } Also solved by Herta T. Freitag, Joh I v i e, D. V. Jaiswal ( I d i a ), Bruce W. K i g, C. B. A. Peck, A. C. Shao (Australia), ad the proposer. EXPONENT PROBLEM B-132 Proposed by Charles R. W a l l, U i v e r s i t y of Teessee, K o x v i l l e, Te. Let u ad v be relatively prime itegers. We say that u belogs to the expoet d modulo v if d is the smallest positive iteger such that u = 1 (mod v). For > 3 show that the expoet to which F belogs modulo F, is 2 if is odd ad 4 if is eve, +i

4 1968] ELEMENTARY PROBLEMS AND SOLUTIONS 403 Solutio by the proposer. From we have F A F 4 - F 2 = (-l) +i -i F^ E (-l) + 1 (mod F + 1 ), Now F 1 (mod F ') as 1 / F < F, for > 3. If is odd the +i +i F 2 = 1 (mod F, ). If is eve the F 2 = -1 (mod F, ). Now ~r"i "+i F 3 = - F = F (mod F, ) -i +i ad F / 1 as ^ 4 (sice is eve). But the F 4 = (~1) 2 = 1 (mod F _,_ ). +i Also solved by D. V. Jaiswal (Idia) ad A. C. Shao (Australia). AN OLD P R O B L E M IN FIBONACCI CLOTHES B-133 Proposed by Douglas Lid / Uiversity of Virgiia, Charlottesville, V a. s r L e t r = FJOOO ac * s = F iooi ^ ^ e ^wo umbers r ad s, which is the larger? Solutio by Phil Maa, Uiversity of New Mexico, Albuquerque, N. Mexico. Sice (I x)/x is mootoically decreasig for x > e, (I r ) / r > (I s)/s or I r*/ r > I s 1 / 8.

5 404 ELEMENTARY PROBLEMS AND SOLUTIONS p e c. Sice I x is mootoically icreasig for x > 0, this implies that v i ' v> s^s. Hece r S > s r. Also solved by William D. Jackso, George F. Lowerre, Arthur Marshall, C.B.A. Peck, D. Z e i t l i, ad the proposer. A TELESCOPING SUM B-134 Proposed by Douglas Lid, Uiversity of Virgiia, Charlottesville, Va. Defie the sequece j a } by a A = a 2 = 1, a 2 k+i = a 2 k + a 2 k-i» ad a 2k = \ for k > 1. Show that 2_s\ k=i = a 2+i - h 2^ a 2 k ~ 1 = a " a 2+i k=i Solutio by M. N. S. Swamy # Nova Scotia Techical College, Halifax, Caada. ]T}a k = ^ a 2 k k=i k=i = (a 3 - a t ) + (a 5 - a 3 ) + + (a 2 +i - a 2 _i) = a 2+ i - at = a 2 +t - 1. The, h 2 a2k -i = S a 2k-i + Z a 2k " X a 2k = a k - ]Ta k k=i k=i k=i k=-i k=i k=i = ( a 4+i ~!) - (a 2+ j - 1) = a^+j - a 2 +i.

6 1968] ELEMENTARY PROBLEMS AND SOLUTIONS 405 Also solved by L. Carlitz, Herta T. Freitag, Joh Ivie, D. V. Jaiswal {Idia}, Bruce W. Kig, George F. Lowerre, C. B. A. Peck, A. C. Shao (Austrqlia), C. R. Wall, Howard L. Walto, David ad the proposer. GENERALIZED SUMS B-135 Proposed by L. Carlitz, Duke Uiversity, Durham, No. Carolia. Put -1-1 Show that, for 1, F,' = 2 - F L' = L + ' :. +2 ' 2 Solutio by Charles R. Wall, Uiversity of Koxville; Te,essee. Let {H } be a geeralized Fiboacci sequece, ad defie -1 H' "" H 2 - k - 1 LJ k k=o The we claim that (A) H' for all 1. Idetity (A_ ca be verified for small ; assume that (A) holds for. The 'sice (H -H )+H -(H -H) we have

7 406 ELEMENTARY PROBLEMS AND SOLUTIONS Dec H + i = E H ^ - H + 2H; = 2 +i H 2-2H +2 + H» 2«+\ - H ^. k=o Thus (A) holds for all > 1. To obtai the idetities give by Carlitz, we ote that F 2 = 1, L 2 = 3. Also solved by Herta T. Freitag, D 9 V. Jaiswal (Idia), Bruce W. Kig, C.B A. Peck, A. C. Shao (Australia), David Zeitli, ad the proposer. * * ERRATA Please make the followig correctio i the October Elemetary Problems ad Solutios: I the third equatio from the bottom, o p. 292, delete F 2k F 2 k F2k*i F 2 k-i ad add, istead, F 2k+2 ^2kH K F 2 k < " T ^ ~ F 2 k F 2 k+2 F 2 k+! F 2 k-i F 2k+2 X Fifcf? K F 2 k + 2 K "FiF * * * * * [Cotiued from p ] Hece, by (13), p \ DJ I each case we have foud a reduced arithmetic progressio o prime member of which is a factor of a certai D. Hece, by Lemma 1, II), there is a ifiitude of composite DJ +j. REFERENCES 1. R. D. Carmiehael, "O the Numerical Factors of the Arithmetic Forms a ± / 3, " Aals of Mathematics, 15 ( ), pp , 2. W. J. LeVeque, Topics i Number Theory, I (1958). * * *