ELEMENTARY PROBLEMS AND SOLUTIONS. Edited by Stanley Rabinowitz
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1 Edited by Staley Rabiowitz Please submit all ew problem proposals ad correspodig solutios to the Problems Editor, DR. RUSS EULER, Departmet of Mathematics ad Statistics, Missouri State Uiversity, 800 Uiversity Drive, Maryville, MO All solutios to others 9 proposals must be submitted to the Solutios Editor, DR. JAWAD SADEK, Departmet of Mathematics ad Statistics, Missouri State Uiversity, 800 Uiversity Drive, Maryville, MO Proposers of problems should ormally iclude solutios. Although this Elemetary Problem sectio does ot isist o origial problems, we do ask that proposers iform us of the history of the problem, if it is ot origial. A problem should ot be submitted elsewhere while it is uder cosideratio for publicatio i this colum. Each solutio should be o a separate sheet (or sheets) ad must be received withi six moths of publicatio of the problem. Solutios typed i the format used below will be give preferece. B A S I C F O R M U L A S The Fiboacci umbers F ad the Lucas umbers L satisfy F *+2 = F +1 + F > F 0 =0, F x = 1; Ai+2 ~ Ai+1 + A M) = 2, L x = 1. Also, a = (l + V5)/2, fi = (l-j5)/2, F = (a - fi ) / fil md L = a"+fl. P R O B L E M S P R O P O S E D IN T H I S ISSUE B-895 Proposed by Idulis Strazdis, Riga Tech Uiv., Latvia Fid a recurrece for F 2. B-896 Proposed by Adrew Cusumao, Great Neck, NY Fid a iteger k such that the expressio F + 2F*F +l + kf^f^x - 2F F^+l + F* +l is a costat idepedet of. B-897 Proposed by Bria D. Beasley, Presbyteria College, Clito, SC Defie (a ) by a +3 = 2a +2 +2a +l -a for > 0 with iitial coditios a 0-4, a x = 2 9 ad a Express a i terms of Fiboacci ad/or Lucas umbers. B-898 Proposed by Alexadra Lupa, Sibiu, Romaia Evaluate k=o v / B-899 Pi oposed by David M. Bloom, Brookly College of CUNY, Brookly, NY I a.sequece of coi tosses, a sigle is a term (H or T) that is ot the same as ay adjacet term. For example, i the sequece HHTHHHTH, the sigles are the terms i positios 3, 7, ad 8. Let S(, r) be the umber of sequeces of coi tosses that cotai exactly r sigles. If > 0 adp is a prime, fid the value modulop of ~S( + p-l,p-l). 2000] 181
2 SOLUTIONS A Recurrece for F B-879 Proposed by Mario DeNobili, Vaduz, Lichtestei Let (c ) be defied by the recurrece c w+4 = 2c +3 + c +2-2c +l - c with iitial coditios c 0 = 0, c x - 1, Cj = 2, ad c$ = 6. Express c i terms of Fiboacci ad/or Lucas umbers. Solutio by H.-J. Seiffert, Berli, Germay Let (c ) satisfy the give recurrece, but with iitial coditios c 0 = 0, c x =a, c 2 = 2b, ad c 3 = 3(a + b), where a ad ft are ay fixed umbers. We shall prove that c =H, where the sequece (H ) satisfies the recurrece H +2 = H +l + H with iitial values H x =a ad H 2 -b. Direct computatio shows that the equatio c - H holds for w = 0,1,2, ad 3. Suppose that it is true for all k e {0,1,..., + 3}, where > 0. We the have C +4 = 2c +3 + C +2-2c +l - C = 2{ + 3)H +3 + ( + 2)H +2-2{ + l)h +l - H = { + 4)(# w+3 + H +2 ) + K^w+ 3" 2# w+1 - # ) + 2(# w+3 - H +2 - H +l ) = ( + 4)H +4 ; ote that H +3 - H +2 +H +l = 2JF/ W+1 + #. This completes the iductio proof. It ca be show that this equatio holds for egative as well. To solve the preset proposal, take a-b-\. The c - F. With a - 1 ad b = 3, we have (H ) ~{L ), ad therefore, c = L. Solutios also received by Paul S. Bruckma, Charles K. Cook, Leoard A. G. Dresel, N. Gauthier, Joe Howard, Harris Kwog, James A. Sellers, Idulis Strazdis, ad the proposer. A Sum for F 2m+2 B-880 Proposed by A. J. Stam, Wisum, The Netherlads Express 2/<wA i terms of Fiboacci ad/or Lucas umbers. Solutio by Harris Kwog, SUNY College at Fredoia, NY Deote the give sum by s m. The geeratig fuctio of the sequece (s m ) is m=q wi=0 m=0 22i<»A / < m v " ^ ' fc=0 k=0 2/<fc+A 2i<k+i' ' Sice 1-3x 2 + x 4 = (1 - x 2 ) 2 - x 2, we fid ' 1 / _ IV ik-i k+i = I if*1(-1)'(3^2' = Z(3z-z 2 )* = 1 + z [MAY
3 Therefore, ^,=F 2lltf2. m=0 Redmod showed that Y2m+2 _ X 2 _ 1 [ X X 1 ( l - X 22 ) 2 ) 2 - X 2 " 2 l l - x - X X - X 2 / = i l l F» x "+1 ^ ( - * ) 4 = I ^* 2-2/</w ' where U m (x) is the /w r/l Chebyshev polyomial of the secod kid. Cook oted that the problem is a slight variatio of a problem posed by Mrs. William Squire i [1], solved bym. N. S. Swamy i [2], ad further explored by H. W. Gould i [3J. Refereces 1. Mrs. W. Squire. "Problem H-83." The Fiboacci Quarterly 4.1 (1966):57'. 2. M. N. S. Swamy. "Solutio of H-83." The Fiboacci Quarterly 6.1 (1968): H. W. Gould. "A Fiboacci Formula of Lucas ad Its Subsequet Maifestatios ad Rediscoveries." The Fiboacci Quarterly 15.1 (1977): Seiffert reported that the idetities ad x - h{ m 7 i )(- x yy( x+ yt~ 2i = *"+y, "i>\, 2i<m m l^ J are due to E. Lucas (Theorie des Nombres [Paris: Blachard, 1961], Ch. 18). Solutios also received by Paul S. Bruckma, Charles K. Cook, Leoard A. G Dresel, Do Redmod, H.-J. Seiffert, Idulis Strazdis, ad the proposer. Diophatie Pair B-881 Proposed by Mohammad K. Azaria, Uiversity of Evasville, IN Cosider the two equatios Z A ^ = ^ + 3 ad I^^-^i- Show that the umber of positive iteger solutios of the first equatio is equal to the umber of oegative iteger solutios of the secod equatio. Solutio by L. A. G. Dresel, Readig, Eglad We have (L l + L 2 + -~ + L ) = E( /+2~^/+i) = Ai+2~^2- Subtractig this from the first of the give equatios ad usig the idetity L +2 = F +3 -\-F +l, we obtai HL j (x j -1) = L 2 -F +l. Puttig y t = Xj - 1, it follows that y t is a oegative iteger wheever x t is a positive iteger. Similarly, startig with the secod of the give equatios, we ca obtai the first give equatio. 2000] 183
4 Thus, the two give equatios are equivalet ad have the same umber of solutios of the specified kids. Geeralizatio by Harris Kwog, SUNY College at Fredoia 9 NY We give a geeralizatio. Let (u ) be a sequece that satisfies the recurrece u - u _ x + u _ 2. It ca be proved, for istace, by iductio, that Ef =1 u t = u +2 - u 2. Let (v ) be aother sequece. Cosider the equatios Jl U i X i= V + 3 ^ d E ^ = V «+ 3 - ^ % - /=! ;=1 Every positive iteger solutio (a h a 2,...,a ) of the first equatio yields a oegative solutio (a { - l,a 2-1,...,a w -1) of the secod equatio. Coversely, ay oegative solutio (b x, b 2,..., b ) of the secod equatio leads to a positive solutio (b x + \b 2 + \,...,b + \) of the first equatio. Therefore, the positive solutios of the first equatio ad the oegative solutios of the secod equatio are i oe-to-oe correspodece. I particular, whe u -L ad v -F ^ the two equatios reduce to the oes i the problem statemet, because L +2 - F +3 + F +l. Solutios also received by Paul S. Bruckma, H.-J. Seiffert, Idulis Strazdis, ad the proposer. A Multiple of F +1 B-882 Proposed by A. J. Stai, Wisum,-The Netherlads Suppose the sequece (A ) satisfies the recurrece A = A _ l + A _ 2. Let k=q Prove that B = A$F +l for all oegative itegers, Solutio by H.-J. Seiffert, Berli, Germay Let x be ay complex umber ad suppose that the sequece (A (x)) satisfies the recurrece A (x) = xa _ x {x) + A _ 2 {x). Defie * (*) = I H ) * 4,-2*00- k=0 We shall prove that B (x) = A 0 (x)f ri+l (x) for all oegative itegers, where (F (x)) deotes the sequece of Fiboacci polyomials which satisfies the same recurrece as (A (x)) y but with give iitial coditios F 0 (x) = 0 ad F X (x) = 1. If >2, the k=0 k=0 = xg(-l)*a-i-2t(*) + Z(-l)*^2-2t(*) k=0 k=q + {-ixa x {x) + A 2 {x) - A_ {x)) 184 [MAY
5 or B (x) = xb _ l (x) + B _ 2 (x). Sice B 0 (x) = A Q (x) = A^F^x) ad ^(x) = A x (x)- A_ x (x) = xa 0 (x) = A 0 (x)f 2 (x), the desired equatio ow follows by a simple iductio argumet. The proposal's result is obtaied whe takig x = l. Solutios also received by Paul S. Bruckma, Charles K Cook, Leoard A, G Dresel, K Gauthier, Petti Haukkae, Joe Howard, Harris Kwog, Do Redmod, James A. Sellers, Idulis Strazdis, ad the proposer. Property of a Periodic Sequece B-8S3 Proposed by L. A, G Dresel, Readig, Eglad (VoL 37, o. 3, August 1999) Let {f ) be the Fiboacci sequece F modulop, wherep is a prime, so that we have f = F (mod/?) ad 0 < f <p for all > 0. The sequece (f ) is kow to be periodic. Prove that, for a give iteger c i the rage 0 < c < p, there ca be at most four values of for which f = c withi ay oe cycle of this period. Solutio by the proposer From the idetities F +l + F _ x = L ad F +l -F _ l = F, we obtai 2F +l - L + F, ad we also have ]} -5(F ) 2 +4(-l) w. We shall assume first that p*2, ad that F = c (mod p) for some eve value of. The it follows that L = ±^j(5c 2 + 4) ad 2F +l = c±^(sc 2 + 4) (modp); this gives two possible values for f +l, say s l ad ^. It Is possible that we also have f = c occurrig for some odd value of w, so that we have f +l = c±^j(5c 2-4) (mod p\ givig two further possible values.% ad s 4, say. These values may ot all be distict, but clearly there are at most four differet values of s which ca follow c i the sequece </>. But if a give cosecutive pair of values c, s were to occur a secod time, the sequece (f) would repeat itself because of the recurrece relatio. Hece, the value c ca occur at most four times i (J) withi oe cycle of the period, amely, at most twice for a eve value of ad at most twice for a odd value of. For the special case p = 2, we see that a complete cycle is (/> = 0,1,1 (mod 2). Corollary: Sice there are oly p values of c i the rage 0 < c < p-1, it follows that the period K(p) of the sequece F modulo/? satisfies K(p) <4p. I the special case p = 5, we do i fact obtai #(5) = ] 185
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