ADVANCED PROBLEMS AND SOLUTIONS
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1 ADVANCED PROBLEMS AND SOLUTIONS Edited by RAYMOND E. WHITNEY Lock Have State College, Lock Have, Pesylvaia Sed all couicatios cocerig Advaced Probles ad Solutios to Rayod E. Whitey, Matheatics Departet, Lock Have State College, Lock Have, Pesylvaia This departet especially welcoes probles believed to be ew or extedig old results. Proposers should subit solutios or other iforatio that will assist the editor. To facilitate their cosideratio, solutios should be subitted o separate siged sheets withi two oths after publicatio of the probles. H-136 Proposed by V. E. Hoggatt, J r., Sa Jose State C o l l e g e, Sa Jose, C a l i f o r i a, ad D. A. Lid, Uiversity of V i r g i i a, C h a r l o t t e s v i l l e, V i r g i i a. Let J H j be defied by Hi = p, H 2 = q, H +2 = H H ( I> 1), where p ad q are o-egative itegers. Show there are itegers N ad k such that F,, < H ^ F,,,, for all > N. Does the coclusio hold +k +k+l if p ad q are allowed to be o-egative reals istead of itegers? H-137 Proposed by J. L Brow, J r., Ordace Research Laboratory, State C o l l e g e, Pa. GENERALIZED FORM OF H-70: Cosider the set S cosistig of the first N positive itegers ad choose a fixed iteger k satisfyig 0 < k ^ N. How ay differet subsets A of S (icludig the epty subset) ca be fored with the property that a f - a M f k for ay two eleets a f, a TT of A: that is, the itegers i ad i + k do ot both appear i A for ay i = 1,2,, N - k. H-138 Proposed by George E. Adrews, Pesylvaia State Uiversity, Uiversity Park, Pa. If F deotes the sequece of polyoials Fj = F 2 = 1, F = F + x ~ F, prove that 1 + x + x x p divides F for ay prie p = ±2 (od 5). 250
2 Oct ADVANCED PROBLEMS AND SOLUTIONS 251 H-139 Proposed by L. Carlitz, Duke Uiversity, Durha, North Carolia. Put A = F +k-i +i F " +k-i F +k-2 F F +i +2 +k +(-i)k M = +(-i)k \i+(-2)k +k +2k Evaluate det M. F o r = k = 2 the proble reduces to H-117 (Fiboacci Vol. 5, No. 2 (1967), p. 162). Quarterly, H-140 Proposed by Douglas Lid, Uiversity of Virgiia, Charlottesville, Virgiia F o r a positive iteger, let a = a() be the least positive iteger such that F = 0 (od ). Show that the highest power of a p r i e p dividig F i F 2 F is L k=i (p ) w h e r e [ x ] deotes the g r e a t e s t iteger cotaied i x. Usig this, show that the Fiboacci bioial coefficiets F F e ' e - i FiF 2^ F - r + i (r > 0) a r e itegers.
3 252 ADVANCED PROBLEMS AND SOLUTIONS H-141 Proposed by H. T. Leoard, Jr., ad V. E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria. Show that [Oct (a) '3 + 2 F [-i] l J^J \( 2 k + l) ) L 2(-(2k+i)) 2k+i (b) L 2 - L l "*1 L^ ( 2 k + v L 2k+i (e) L 2 + L W L2k H-142 Proposed by H. W. Gould, West Virgiia Uiversity, Morgatow, W. Va. With the usual otatio for Fiboacci ubers, F 0 = 0, Fj = 1, F "+i = F + F. show that -l v / 1 +VITA / 1 + V 5., =0 \ k - k where (t) = x(x - l)(x - 2)- (x - j + l)/j! is the usual bioial coefficiet sybol.
4 1968 ADVANCED PROBLEMS AND SOLUTIONS 253 SOLUTIONS ORIGINAL COMPOSITION H-88 Proposed by Verer E. Hoggatt, J r,, Sa Jose State College, Sa Jose, Califoria (Corrected). Prove that / \ Yi / A F 4k []A = L 2 F 2I1 2 Solutio by M. N. S, Sway, N o v a Scotia Techical C o l l e g e, H a l i f a x, Caada. L e t S -2>-*(l) ~»-= k ^ 0 k=o w k=o where = - 4 [(H-p 4 ) - (l + q^) ] p + q = 1, pq = - 1 ; or (pq) 2 l = 1 Hece, s = - L ri( p q)2 + p4^ _ ( p q ) 2 i + q i J = 1 I"p2i ( p 2i + q2hi) _ q2( p 2 + q2) l V5"L ^ = ( p 2 + qzj W ~ q ) L 2 F 2
5 254 ADVANCED PROBLEMS AND SOLUTIONS [Oct. Therefore, i r ^ / \ / J F 4k ( k ) k=o = L 2F 2 Also solved by Joh Wesser, L. C a r l i t z, ad F. D. Parker. FINE BREEDING H-96 Proposed by Maxey Brooke, Sweey, Texas, ad V. E. Hoggatt, J r., Sa Jose State C o l l e g e, Sa Jose, Califoria (Corrected). Suppose a feale rabbit produces F (L ) feale rabbits at the " 1 tie poit ad her feale offsprig follow the sae birth sequece, the show that the ew arrivals, C, (D ) at the tie poit satisfies ad C, = 2C + C ; Ci = 1, Co = 2 1 A +2 +i. D +l. =3D +'(-l) + 1 Solutio by Douglas L i d, Uiversity of V i r g i i a. Hoggatt ad Lid ["The Dyig Rabbit Proble, ff to appear, Fiboacci Quarterly] have proved the followig result: Let a feale rabbit produce B th feale rabbits at the tie poit, her offsprig do likewise, ad put B(x) QO =i The the uber R of ew arrivals has the gee ratig fuctio R(x) = 2^xI1 = r-rw =0
6 1968] ADVANCED PROBLEMS AND SOLUTIONS 255 where we use the covetio that R 0 = 1 (the origial feale beig bor at the 0 tie poit)» We apply this result to the cases (i) B = F, ad (ii) B L. (i) If B = F, the B(x) = > F x = - ^ _ =l 1 - x - x 2 SO R(x) = x - x J It is clear fro the geeratig fuctio that here the R = C obey the recurrece relatio C. = 2C. + C alog with Ci = 1, C? = 2 9 thus estab+2 +i * & l lishig the desired result. (ii) The recurrece relatio proposed is icorrect., the proper oe beig show below* If B = L, the 00 B( X ) = > L x = ^LlJxL,4-,,- =i l - x - x 2 so that Now D / \ x + 2x 2 R(x) =, 0 7" = 1 +! 2Lt2xi_ L2X-3X* 1 - X - X 2 A 1 x + 2x 2 _ 2, 12 ^ 4 - _ - + -f _ _ 1-2x - 3x 2 1-3x 1 + x
7 256 ADVANCED PROBLEMS AND SOLUTIONS [Oct. so that D = R = (5/12) (3 ) + (1/4) (-l) ( > 1). +l It follows that DH 1 = 1 9 ad that D, = 3D + (-1), the correct relatio. +i BINOMIAL, ANYONE? H-97 Proposed by L. Carlitz, Duke Uiversity, Durha, North Carolia. Show / v2 Solutio by David Zeitll, Mieapolis, Miesota. If P W. >»v * k ad QW - >. r " ) r J S)( k ) ( K - i ) " k k=o the P(x) = Q(x) is a kow idetity (see eleetary proble E799, Aerica
8 1968] ADVANCED PROBLEMS AND SOLUTIONS 257 Math. Mothly, 1948, p 30). If a ad /3 are roots of x 2 - x - 1 = 0, the L = a 1 + /5, F = (a - /3 )/V5, ad thus (a) T?(a) + P(/3) = Q(a) + Q(p), sice (a - l ) " k + (fi - l ) " k = (apf- k + «Sa2) - k (b) = ( - l ) " \ _ k, (P(or) - P(/3))/V5 = (Q(a) - Q(/3))/V5 sice <* - D - k (^ - D - k = (-D -V~ k - ^ k ) = _ ( - l ) - k ( ^ ) F _ k PRODUCTIVE SUMS H-99 Proposed by Charles R. Wall, Harker Heights, Texas. Usig the otatio of H-63 (April 1965 FQJ, p. 116), show that if a ( l + \ / S ) / 2, VSF a a^ = (-l) ( - 1)/2 F(,)a- ( +1 ) =l =l A T - ^ i, 1x(+l)/2 _,. -(+l) j j L ^ a = 1 + L K) F(,)«v ' 5 =l =l where F F F _, v F(,) = 1 2 Solutio by Douglas Lid, Uiversity of Virgiia* We use the failiar idetity -i W "TT'd-A) = E (-l) q ( - l)/2 pjx, =o =o
9 258 where ADVANCED PROBLEMS AND SOLUTIONS [Oct. [l /- w - -ix M -+i (1 - q )(1 - q ) ««(1 - q ( l - q ) ( l - q 2 ) -.. ( l - q ) If p = (1 - \/E)/2, the A/5F a = 1 - (p/a). Puttig q = p/a, the *-. = a F (, ), ad puttig x = q i (*) gives o X/EV. T* 1 o /i \ V / i x w \ (+l)/2 2 - li V 5 F a = U (1 - q ) = 2-» (-D F(,)q " or =i =i =o = E v(-i)/2_ /. -(+i) ( - l ) U l - W / ' F ( i i f i ) a - =o where we have used op = - 1. Siilarly, La = 1 - (P/a), so puttig q = p/a ad x = -q i (*) gives =i L a ~ /i. iiv V (l + q > = 2L / i\-r,/ \ (-i)/2 2 -/, (-1) F(,)q a (-q) =i =o =o (+i)/2 2 - F( 5 )q " a = Z ( - l ) ( + l ) / 2 F ( > ) a - ( + 1 ) =o Also soived by M. N. S. Sway.
10 1968] ADVANCED PROBLEMS AND SOLUTIONS 259 PYTHAGOREANS AND ALL THAT STUFF H-101 Proposed by Harla Uasky, Cliffside Park, N. J., ad Malcol Tolla, Brookly, N. Y. Let a,b 9 c 9 d be ay four cosecutive geeralized Fiboacci ubers (say H 1 = p ad H 2 = q ad H + 2 = H H ^ > 1), the show (cd - ab) 2 = (ad) 2 + (2bc) 2 Let A = L k L k + 3? B = 2 1 ^ 1 ^, ad C - L k L ^. Te show A 2 + B 2 = C 2. Solutio by M. N. S. Sway, N o v a Scotia Techical C o l l e g e, H a l i f a x, Caada* Now (ed-ab) 2 = [c(b + c) - b(e - b)] 2 = (c 2 + b 2 ) 2 = ( c 2 - b 2 ) 2 + (2bc) 2 = (c + b) 2 (c - b) 2 + (2bc) 2 = d 2 a 2 + (2bc) 2 Hece (1) (cd - ab) 2 = (ad) 2 + (2bc) 2 with Sice L,, the Lucas uber, is also a geeralized Fiboacci sequece Li = p = 1, L 2 = q = 3 5 we have that for the four cosecutive Lucas ubers L k> L k + 1, L k+2? L^+3 ' (2) (L. M L T MO - L. L. _,, ) 2 = (L. L. M ) 2 + (2L.,,L.,_ ) 2 = A 2 + B 2 ' k+2 k+3 k k+i k k+3 k+i k+2 Now
11 260 ADVANCED PROBLEMS AND SOLUTIONS Oct ( L k+2 L k+ 3 " L k L k + i ) = L k+2 (L k+2 + L k + l ) _ L k+l ( L k+2 " L k + i ) = L + k L + 2 k + 1 = ( F k F k + i ) 2 + ( F + k F + 2 k ) 2 = ( F k F k + i ) 2 + ( 2 F k F k + i ) 2 (3) = ^ U ^ k i * = 5F 2k + 3 = 2F + (F - F ) + (F + F ) + F ^ 2k+3 U 2k+5 * 2k+4' U 2k+2 2k+i' 2k+3 2k+2 2k+4 Thus, fro (2) ad (3) we have, A 2 + B 2 = C 2, Also solved by J. A. H. Huter ad A. G. Shao. [Cotiued fro p. 285] RECURRING SEQUENCES - LESSON 1 ANSWERS TO PROBLEMS 1. a = ( + 1 ) ; T ^ = 3T - 3T + T i 2. a = 3-2; T M = 2T ^ - T +2 +i 3. a = 3 ; T, i = 4T, - 6T ^ + 4T, - T i 4 - Tg+k = 1, 3, 3, 1, 1 / 3, 1 / 3, for k = 1,2, 3,4, 5, 6, respectively 5. T ^ = V l + T 2 +i 6. T ^ = 4T ^ - 6T ^ + 4T ^ - T i 7. T, 4 = at 9. T +i 2 _! = a, T 2 = l / a 8. ± = 3 T ± - 3 T, + T 10. T +3 _ Lj = 1/( T ) +i +l * *
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