ADVANCED PROBLEMS AND SOLUTIONS
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1 ADVANCED PROBLEMS AND SOLUTIONS EDITED BY VERNER E, HOGGATT 9 JR 0, SAN JOSE STATE COLLEGE Sed all commuicatios cocerig Advaced P r o b l e m s ad Solutios to Verer E. Hoggatt, J r., Mathematics Departmet, Sa Jose State College, Sa J o s e, Califoria. be ew or extedig old r e s u l t s. This departmet especially welcomes problems believed to P r o p o s e r s should submit solutios or other iformatio that will a s s i s t the editor. To facilitate t h e i r cosideratio, s o l u- tios should be submitted o separate siged sheets withi two moths after publicatio of the problems. H- 24 Proposed by the late Morga Ward, Califoria Istitute of Techology, Pasadea, Califoria. Let cp (x) = x + x 2 / x /, ad let k(x) = k (x) = ( x p _ 1 - l ) / p, where p is a odd p r i m e g r e a t e r tha 5. (The fuctio k(x) is called the "quotiet of F e r m a t " i the literature. ) Let P = P be the rak of apparitio of p i the sequece 0, 1, 1, 2, 3, 5, - - \, F, (so P 1 3 = 7, P 7 = 8 ad so o). The Fp = 0 mod p 2 if ad oly if 0 ( p 1 ) / 2 ( 5 / 9 ) = 2k (3/2) mod p. H-25 Proposed by Joseph Erbacker ad Joh A. Fuchs. Uiversity of Sata Clara, adf.d* Parker, Suy, Buffalo, N.Y. P r o v e : D = I a.. I = 36, for all, I ij I where a.. = F 3... (i,j = 1,2,3) x J j IJ +i+j-2 H-26 Proposed by Leoard Carlitz, Duke Uiversity, Durham, N,C that L e t R k = ^ w h e r e b r s = ( k s)> * e s h 0 W R k = ( a r s > S U c h 47
2 48 ADVANCED PROBLEMS AND SOLUTIONS [Dec. a = V f r 7 M( k + J " r ) F k ^ r ~ s + J F r + s J F J rs V J A s j ) H-27 Proposed by Harla L. Umasky, Emerso High School, Uio City,N*J. Show that F k = ^(-^Vsk-sj + ^ k ^. k " 4 H-28 Proposed by H.W. Gould, West Virgiia Uiversity, Morgatow, W/ e Va, Let C.(r,) be the umber of umbers, to the base r (r 2) with at most digits f ad the sum of the digits equal to j. Sum the series: OO ^ y rs C. i (r, ) \ a ji^r--i J J b j=0 J SOLUTIONS TRINOMIAL COEFFICIENTS H-9 Proposed by Olga Taussky, Califoria Istitute of Techology, Pasadea, Calif, Fid the umbers a, where > 0 ad r are itegers, for which the relatios a + a -, + a 0 = a, _.,r, r - l $ r-2 +l,r ad fo r 4= 0 a =6 =1 o.r o.r 1 A? * V. 1 r = 0 hold. Solutio by the proposer* It ca be show that a is the coefficiet of x I the expasio of (l + x + x 2 ) s This is certaily true for = 0 ad It follows for > 0 by usig the geeratig fuctios v r 2^ a x r f r For 9 multiplyig this sum by (1 + x + x 2 ) ad usig the recurrece relatios it follows that
3 1963] A D V A N C E D P R O B L E M S A ND S O L U T I O N S 49 This proves the assertio. (1 + x + x 2 ) 2 a x r = 2 a, x r. r,r r + l s r SOME FIBONACCI SUMS H Proposed by R,L. Graham, Bell Telephoe Laboratories, Murray Hill, New Jersey Show that oo oo +1 2J = = p -p p- =l =l Solutio by Leoard Carlitz? Duke Uiversity, Durham, N. C.., F o o F. F -. ~ F 2 oo / x v J L _ y = y (-1) s o F : F J M t F, F F, - ^ F - F F, oo F so F - F, = 2 ^±l zl = = = 2 2 F - l F + l 2 F - l F + l 2 \ F - 1 F + 1 / F l F 2 oo oo y _ O. y _,! L - ± i _ F ~ ^ ^ T? F F I Also solved by Zvi Dreser, FIBONACCI AND FOURIER H-ll Proposed by Joh L.Brow,Jr,.Ordace Re search Laboratory, The Pesylvaia State Uiversity, Uiversity Park, Pea, Fid the fuctio whose formal F o u r i e r s e r i e s Is co F si x f(x) - 2 ~, =l! where F is the th Fiboacci umber. Solutio by Lucile Morto, Sata Clara, Calif... v co.. oo z(cos x+i si x) Vl (cos x + i s m x l z ^ cos x e \ / _ 2 - = 2 j z =0! =0 * Therefore OO,. si x j z
4 50 ADVANCED PROBLEMS AND SOLUTIONS [Dec. Recallig the z cos x.,. N e s m (z s m x) = _ 2 si x ; z =0 F = (a - /3 ), where a = (1 + \fe)/2 ad js = (1 - \fe)/2, r /. I r a c o s x... v 8 cos x.,.. x-\ f(x) = -T3 <! e s m (a s m x) - e^ s m (/3 s m x)> ad.. a cos x...., B cos x. /. v s m x T g(x) = e s m (a s m x) + e^ s m (J3 s m x) = i - L Also solved by the proposer. A CURIOUS SEQUENCE H Proposed by D.E, Thoro, Sa Jose State College,Sa Jose,Calif. Fid a formula for the th t e r m i the sequece: 1, 3, 4, 6, 8, 9, 1 1, 1 2, 1 4, 1 6, 1 7, 1 9, 2 1, 2 2, 2 4, 2 5, -. Solutio by Malcolm Tallma, Brookly, N. Y. N 1, 3, 4, 6, 8, 9,11,12 1 4, 1 6, 1 7, 1 9, 2 1, 2 2, 2 4, 2 5 M <; 2 7, 2 9, 3 0, 3 2, 3 4, 3 5, 3 7, , 4 2, 4 3, 4 5, 4 7, 4 8, 5 0, 5 1 Let M = 8m + 1,2, 3*, 4, 5, 6, 7, 8 N M = 13m + 1,3,4*, 6, 8, 9,11,12 What is the 19th t e r m? M = 19 = 8 x *, thus N 19 = 13 x 2 + 4* = 30. Also solved by Maxey Brooke ad the proposer. Editorial Commet: If T t = 1, T 2 = 3, T 3 = 4, T 4 = 6, T 5 = 8, T 6 = 9, T 7 = 11, T 8 = 12, the T g m + k = 13m + T k, k > 0, m = 1, 2, 3, -. A MATRIX DERIVED IDENTITY H Proposed by H.W.Gould, West Virgiia Uiversity,Morgatow, W.Va, ad Verer E.Hoggatt,Jr,,Sa Jose State College,Sa Jose,Calif, Show that = I / r x r - j j. V j / k-1 k +i-rk See p. 65 of "A P r i m e r for the Fiboacci Numbers Part III, rt Oct., 1963, Fiboacci Quarterly. Also solved by Leoard Carlitz ad Merritt E l m o r e.
5 1963 J ADVANCED PROBLEMS AND SOLUTIONS 51 IDENTITY FOR FIBONACCI CUBES xi-14 Proposed by David Zeitli f Mieapolis f Miesota, ad F.D.Parker, Uiversity of Alaska, College, Alaska. Prove the Fiboacci idetity F 3 _ 3 F 3 _ 6 F F 3 x 1 + F 3 = Solutio by Maxey Brooke, Sweey, Texas. F r o m "Fiboacci F o r m u l a s, " page 60, April, 1963, Fiboacci Quarterly, obtais, from paragraph 3, oe (1) F 3 ' + F 3 - F 3, = F 0 v ' ad the c o r r e c t e d versio of Jekuthiel Gisburg's idetity t h e r e i s (2) F 3-3 F 3 + F 3 0 = F v ; Multiplyig equatio (1) through by 3 ad equatig the ew left side of (1) to the left side of (2) ad simplifyig yields F F 3, - 6 F F 3, + F 3 0 = Also solved by J. A. H. Huter, Zvi D r e s e r ad the p r o p o s e r s. SOME CHOICE IDENTITIES H-16 Proposed by H.W.Gould, West Virgiia Uiversity,Morgatow,W.Va. Show that x ~x Defie the ordiary Hermite polyomials by H = (-1) e D (e ). oo (i) 7 2 H (x/2) 7 ^~ = 1!, ' =0 (ii) 2 H (x/2) - F = 0 =0!
6 52 ADVANCED PROBLEMS AND SOLUTIONS [Dec. 1963] x 11 2 X (III) 7 X H (x/2) ^ L = 2 e ' 9 '! =0 where F ad L a r e the th Fiboacci ad th Lucas u m b e r s, respectively. Solutio by Zvi D r e s e r, Tel-Aviv, Israel. (i) x 0 X \ X, S H i ~77 I 7 = e^ =0 [ 2 l! I =0 x=x 0 /2 x 0 x 0 y" y The sum i b r a c e s o the right is the expasio of e about the poit + with x = - x 0 / 2. Hece 2 H x 0 \ x 0 =0 2 e + x 0 / 4 ( e - x. 0 * / 4 \ = X 2 J (ii) I the s a m e way (a = 1 + \I5 1 - N/5 ad F = - ^ ( «- / 3 ) ), x \ x 1 x^/4 J D e ~ x ' 2 H F = - i - e X / 4 / 2 ^ JL e *o ^5 x=x 0 /2 -x 0 af - (-x 0 /3) )) /4{ e -(^)! e -(^#)! } = 0. (iii) Similarly (L = a + /3 ) x 0 \ x 0 2 H ( L \ x 2 / =0 ' S/4 ~x^ D e =0 : x=x 0 /2 (-Xoa)"+ (-x 0 is: = e x S / 4 ( 2 e - 5 x o 2 / 4 ) = 2 e ^. Also solved by L. Carlitz ad the proposer. Correctio to Problem H-20 i the October issue H ( C o r r e c t e d ) P r o p o s e d by Ve mei~j E.Hoggatt,Jr.ad Charles H.Kig, Sa Jose State Co I lege, Sa Jose,Califoria. If Q show D ( e Qlj where D(A) is the determiat of matrix A ad L is the th Lucas umber. AWWyvywyVWy^^
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