ADVANCED PROBLEMS AND SOLUTIONS. Edited by Florian Luca

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1 Edited by Floria Luca Please sed all commuicatios cocerig ADVANCED PROBLEMS AND SOLU- TIONS to FLORIAN LUCA, IMATE, UNAM, AP. POSTAL 6-3 (XANGARI), CP , MORELIA, MICHOACAN, MEXICO, or by at as files of the type tex, dvi, ps, doc, html, pdf, etc. This departmet especially welcomes problems believed to be ew or extedig old results. Proposers should submit solutios or other iformatio that will assist the editor. To facilitate their cosideratio, all solutios set by regular mail should be submitted o separate siged sheets withi two moths after publicatio of the problems. PROBLEMS PROPOSED IN THIS ISSUE H-635 Proposed by Jayatibhai M. Patel, Ahmedabad, Idia For ay positive iteger, prove that F F + F +2 F +2 F F + F + F +2 F = 2(F 3 + F+), 3 ad that the same holds with the Fiboacci umbers replaced by the correspodig Lucas umbers. H-636 Proposed by Charles K. Coo, Sumter, SC Evaluate the determiat of the matrix F 2 + L 2 P 2 R 2 2(L P F R ) 2(F P + L R ) 2(F R + L P ) F 2 L 2 + P 2 R 2 2(P R F L ) 2(L R F P ) 2(F L + P R ) F 2 L 2 P 2 + R 2 where F, L, P ad R are the Fiboacci, Lucas, Pell ad Pell-Lucas umbers, respectively. H-637 Proposed by Ovidiu Furdui, Kalamazoo, MI Prove that 2 + F 2 2 F 2 2 +, 2 + ad 2 F F are the sides of a triagle whose circumradius is /2 for all 0. 9

2 H-638 Proposed by José Luis Díaz-Barrero, Barceloa, Spai Let be a positive iteger. Prove that F + log ( + F + /F ) < F + + 3F +2. SOLUTIONS A idetity for Lucas polyomials H-62 Proposed by Mario Catalai, Torio, Italy (Vol. 43, o., February 2005) Let L (x, y) be the bivariate Lucas polyomials, defied by L (x, y) = xl (x, y) + yl 2 (x, y), L 0 (x, y) = 2, L (x, y) = x. Assume x 2 + 4y 0. Prove the followig idetity ( ) + ( y) x (+) L + (x, y) = x. Solutio by the proposer Cosider the polyomials L (, y). The roots of the characteristic equatio, α α(, y) ad β β(, y), are such that α + β =, αβ = y. Idetity.78 i [2] says that ( ) + [( x) + x + x + ( x) ] =. I the left had side, write x = α to get ( ) + (β + α + α + β ) = ( ) + ( y) L + (, y). () I [], it is proved that L (x, y) = (iγ) L (iγx, γ 2 y), where i 2 = ad γ is a complex umber. Tae γ = ix. The L (x, y) = x L (, y x 2 ), 92

3 that is L (x, y) = x L (, y ) x 2. (2) I the right had side of formula () replace y by y/x 2 ad the use formula (2) to get ( ) + ( y ) x (+ ) x 2 L + (x, y) =, which reduces easily to the desired idetity. [] M. Catalai. Geeralized Bivariate Fiboacci Polyomials, [2] H. W. Gould. Combiatorial Idetities, Morgatow, W. Va., 972. Also solved by Paul S. Brucma ad G. C. Greubel. Lucas Sequeces ad Sophie Germai Primes H-622 Proposed by Lawrece Somer, The Catholic Uiversity of America, Washigto, DC (Vol. 43, o. 2, May 2005) Let u(a, b) be the Lucas sequece defied by u 0 = 0, u = ad u +2 = au + bu ad havig discrimiat D(a, b) = a 2 4b, where a ad b are itegers. Let ω() deote the umber of distict prime divisors of. Let c > be a fixed positive iteger. Show that there exist 2 ω(c) distict Lucas sequeces u(a, c 2 ) such that 2p + u p (a, c 2 ) for every Sophie Germai prime p such that 2p + D(a, c 2 ), where (a, c) = ad ad(a, c 2 ) 0. Recall that p is a Sophie Germai prime if both p ad 2p + are primes. Combied solutio by the proposer ad the editor It follows, by the Biet formula, that u ( a, c 2 ) = ( ) + u (a, c 2 ). Thus, we ca assume without loss of geerality that a > 0. It suffices to fid 2 ω(c) positive itegers a coprime to c such that 2p + u p (a, c 2 ) for all Sophie Germai primes p such that 2p + cd(a, c 2 ). By a theorem of D. H. Lehmer (see [, pg. 44]), if u(a, b) is a Lucas sequece ad q is a odd prime such that q D ad (b/q) =, the q u (q (D/q))/2, where (b/q) ad (D/q) are Legedre symbols. Thus, if D(a, c 2 ) = a 2 4c 2 is equal to a ozero square m 2, the 2p + u p (a, c 2 ) for all Sophie Germai primes p such that 2p + cd(a, c 2 ). Cosequetly, it suffices to show that there exist 2 ω(c) positive itegers a coprime to c such that for some positive iteger m. a 2 4c 2 = a 2 (2c) 2 = m 2 () 93

4 Rewritig () as (a m)(a + m) = (2c) 2 = 4c 2, ad recallig that m ad c are coprime, we see that if we write α for the exact order at which 2 divides 4c 2, the there exist two coprime odd divisors d ad d 2 of c 2 whose product is c 2 /2 α 2 such that a m = 2 α d ad a + m = 2d 2. Note that α is eve ad that c = 2 α/2 c, where c is odd. It is clear that there are precisely 2 ω(c ) possibilities for a odd divisor d of c 2 such that c 2 /d = d 2 is coprime to d. Further, for each such divisor d, we ca put a m = mi{2 α d, 2d 2 } ad a + m = max{2 α d, d 2 }, whose solutio is a = 2 α 2 d + d 2 ad m = 2 α 2 d d 2. Note that we do have m > 0 sice m = 0 implies α = 2, d = d 2 =, therefore c =, which is ot allowed. We ow ote that half of the values of a obtaied i this way are distict for distict divisors d of c if c is odd, ad all of them are distict if c is eve. Ideed, if two such a s are equal, the by () ad the fact that m > 0, we get that the correspodig m s are also equal. Thus, the pair of umbers a m ad a + m are the same. Whe c is odd the d is oe of (a m)/2 ad (a + m)/2. Whe c is eve, the exactly oe of a m ad a + m is a multiple of 2 α ad d is the correspodig cofactor. Sice ω(c ) = ω(c) if c is eve ad ω(c ) = ω(c) if c > is odd, the required coclusio ow follows. [] D. H. Lehmer. A Exteded Theory of Lucas Fuctios, A. of Math. 3 (930) : Also solved by Paul S. Brucma. A Large Product H-623 Proposed by José Luis Díaz-Barrero, Barceloa, Spai (Vol. 43, o. 2, May 2005) Let be a positive iteger. Prove that 2 (2 ) F 2 ( F2 F 2 ) F2 F2. Solutio by H.-J. Seiffert, Berli, Germay It is ow (see equatio (3.32) i []), that F F + F F +2 = ( ) +. Also, we have that F F +2 = (F + F )(F + + F ) = F+ 2 F 2. Now, it is easily verified that F 2 = (( + )F F + )F + (F F )F + ( ). Summig over =, 2,..., 2, ad otig the telescopig o the right had side gives 2 F 2 = (2F 2 F 2 )F 2. () 94

5 From equatio (I 3 ) i Hoggatt s list, we ow that S := 2 F 2 = F 2 F 2. (2) Subtractig () from the 2-times (2) yields T := (2 )F 2 = F2. 2 (3) 2 The atural logarithm l is cocave o the iterval (0, + ). Hece, by Jese s iequality, 2 F 2 l(2 ) S l(t /S ). Usig (2) ad (3), after simplifyig ad upo expoetiatio, oe obtais the desired iequality. [] A. F. Horadam & Bro. J. M. Maho. Pell ad Pell-Lucas Polyomials, The Fiboacci Quarterly 23. (985) : Also solved by Said Amghibech, Paul S. Brucma, Ovidiu Furdui ad the proposer. Limits of Itegrals H-624 Proposed by Ovidiu Furdui, Wester Michiga Uiversity, Kalamazoo, MI (Vol. 43, o. 2, May 2005) Prove that + t F + t F lim 0 + t L dt = ad lim + t L dt = 0. Solutio by Said Amghibech, St. Foy, Caada For t [0, ], we have + tf + t L + tf ; 95

6 thus, 0 + t F + t L dt 0 ( + t F )dt = + + F, so + t F lim 0 + t L dt =. For t [, ], we have 0 + tf + t L t L + t L F. Thus, 0 + t F + t L dt ( t L + t L F ) dt = L + L F, which gives + t F lim + t L dt = 0 Also solved by Paul S. Brucma, H.-J. Seiffert ad the proposer. Correctio: I H-663 (volume 43.4) the A + B = C should have bee A = B + C. 96