ELEMENTARY PROBLEMS AND SOLUTIONS

ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. H1LLMASM Uiversity of Mew Mexico, Albuquerque, New Mexico 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, 8716. 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. Cotributors (i the Uited States) who desire ackowledgemet of receipt of their cotributios are asked to eclose self-addressed stamped postcards. B-l 78 Proposed by James E. Desmod, Florida State Uiversity, Tallahassee, Florida. For all positive itegers show that ad F = Y 2 ~ i F + 2 *2+2.A; *2i-l 1=1 1=1 Geeralize. B-l 79 Based o Douglas hid's Problem B-l65. + + Let Z cosist of the positive itegers ad let the fuctio b from Z to Z + be defied by b(l) = b(2) = l f b(2k) = b(k), ad b(2k + 1) = b(k + 1) + b(k) f o r k = 1,2,. Show that every positive iteger m is a value of b() ad that b( + 1) <> b() for all positive itegers. 15

16 ELEMENTARY PROBLEMS AND SOLUTIONS [Feb, B-18 Proposed by Reube C. Drake, North Carolia A T Uiversity, Greesboro, North Carolia. Eumerate the paths i the Cartesia plae from (,) to (,) that cosist of directed lie segmets of the four followig types: Type 1 I i IV Iitial Poit (k,) (k,) (k,l) ( k, l ) T e r m i a l Poit (k,d (k + 1,) ( k + 1, 1 ) (k + 1,) B-181 Proposed by J. B. Roberts, Reed College, Portlad, Orego. Let m be afixed iteger ad let G - = -, G c = 1, G = G., + G -1 o ' -1-2 for > 1. Show that G Q, G, G 2 f G~ * e " is the sequece of upper left pricipal miors of the ifiite matrix 1 1 r m-2 G m - 2 + G m 1 (~D m G m - 2 + G m 1 (-D m G -2 m- + G m (- l ) m B-182 Proposed by James E. Desmod, Florida State Uiversity, Tallahassee, Florida. Show that for ay prime p ad ay iteger, F = F F (mod p) ad p p N ^ L = L L = L (mod p) p p N ^' B-183 Proposed by Gustavus J Simmos, Sadia Corporatio, Albuquerque, New Mexico. A positive iteger is a palidrome if its digits read the same forward or backward. The least positive iteger such that 2 is ot is 26. Let S be the set of such that 2 ot. Is S empty, fiite, or ifiite? is a palidrome but is a palidrome but is

197] ELEMENTARY PROBLEMS AND SOLUTIONS 17 SOLUTIONS FIBONACCI PYTHAGOREAN TRIPLES B-16 Proposed by Robert H. Agli, Da River Mills, Daville, Virgiia. Show that if x = F F + 3, y = 2 F + 1 F + 2, ad z = F ^, the X 2 + y 2 = z 2. Solutio by Michael Yoder, Studet, Albuquerque Academy, Albuquerque, New Mexico. Let' tr = F, ad v = F, -. +2 +1 The u 2 - v 2 _ ( u + v)(u - v) = F + 3 F = x, 2uv = y, u 2 + v 2 = z, ad hece x 2 + y 2 = z 2, Also solved by Herta T. Freitag, Bruce W. Kig, Douglas Lid, Joh W. Milsom, A. G. Shao (Boroko, T. P. N. GJ, Gregory Wulcyz, ad the Proposer. P E L L NUMBER IDENTITIES B-161 Proposed by Joh Ivie, Studet, Uiversity of Califoria, Berkeley, Califoria. Give the Pell u m b e r s defied by? + 2 = 2 P + 1 + V^ P = S P t = 1 9 show that for k > : a) p k = [<k-l)/2] t-d/2j,/ k \ r= Ur \ + l) 2 / W) P m 2k = i i i ; j 2 r p r Solutio by Douglas Lid, Cambridge Uiversity, Cambridge, Eglad. Let a = 1 + \ / 2, b = 1 - ViT

18 ELEMENTARY PROBLEMS AND SOLUTIONS [Feb, be the roots of the characteristic polyomial x 2-2x - 1. It follows from the theory of differece equatios that there are costats A ad B such that P = Aa 11 + Bb. Solvig the system of simultaeous equatios resultig by settig =, 1, we fid A = 1/2V ", B = -1/2V?. Hece P = _L ( a k - b k ) = -L= E ("P /2 K - <-D J 2 j/2 l 2VT 2V5 j= V J A -I Cioc-i)] [!(k-d] 2 r Also, sice a ad b satisfy x 2 = 2x + 1, we have,= -J_ (a 2k Po,,--= - ^ ( a ^ - b 2 k ) = - ^ ( f e a + l ) k - (2b + l ) b ) 2 k 2V2 ' 2V2~ UH'bf) - a * r 4/so so/ve<i fry i/erfa T. Freitag, Bruce W. Kig, Gregory Wulcyz, Michael Yoder, ad the Proposer. A REPRESENTATION THEOREM B-162 Proposed by V. E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria. Let r be a fixed positive iteger ad let the sequece uj, u 2,* * satisfy u = u -,+u + + u for > r ad have iitial coditios u. = 2 J ~ -1-2 -r 3 for j = 1,2,«,r. Show that every represetatio of U as a sum of distict u. must be of the form u itself or cotai explicitly the terms u -, u 9 1 -* x &, u - ad some represetatio of u

197] ELEMENTARY PROBLEMS AND SOLUTIONS 19 Solutio by Michael Yoder, Studet, Albuquerque Academy, Albuquerque, New Mexico. For r = 1, the theorem is trivial; we therefore assume r ^ 2. First we show by iductio that E u. < u l. l +2 i=l For = l, 2, ' ", r this is obvious; ad if J2 u. < u + 2 * where < r, i=l +1 S 1=1 u i < \ + 2 + V l ^ Vf-3' Now let k< c(k) = or 1 for all k 9 be a represetatio of u some j with - r + 1<. j <; - 1* The ad assume c(j) = for -1 c(k)u^< k< < k=l J ( - r - 1-1 \ E \\Z \)-Vr + l k=l k=-r / / - r - 1 \ 1 j ^. k -r+11

11 ELEMENTARY PROBLEMS AND SOLUTIONS [Feb. which is a cotradictio. Thus we must have u = u - + + u,- + S, - 1 - r + 1 where S i s some represetatio of u _. See "Geeralized Fiboacci Numbers ad the Polygoal N u m b e r s, " Joural of Recreatioal Mathematics, July* 1968, pp. 144-15. Also solved by the Proposer. A VARIANT OF THE EULER-BINET FORMULA B-163 Proposed by Phil Maa, Uiversity of New Mexico, Albuquerque, New Mexico. Let be a positive iteger. Clearly (1 + V 5 ) = a + b V 5 with a ad b itegers. Show that 2 i s a divisor of a ad of b. Solutio by David Zeitli, Mieapolis, Miesota. Let a = (1 + V ) / 2 ad ]8 = (1 - V )/2. Elimiatio of ^ from L = a + fi 1 ad V 5 F = a - /3 11 gives 2c/ 1 = L + V 5 >. T h u s, where a = 2 ~ L ad b = 2 "~ F. (1 + V 5 ) = 2 ~" 1 L + V5"(2 "" 1 F ), Also solved by Juliette Daveport, Herta T. Freitag, Joh E. Homer, Jr., Joh Ivie, Bruce W. Kig, Douglas Lid, Peter A. Lidstrom, A. G. Shao (Boroko, T. P. N GJ, Michael Yoder, ad the proposer.

197] ELEMENTARY PROBLEMS AND SOLUTIONS 111 A GENERALIZED SEQUENCE WITH CHARACTERISTIC 1 1, 1 H B-164 Proposed by J. A. H. Huter, Toroto, Caada. A Fiboacci-type sequece i s defied by: = G. G, - + G, +2 +1 with Gi = a ad G 2 = b. Fid the miimum positive values of itegers a ad b, subject to a beig odd, to satisfy: G, G ^ - G 2 = - l l l l l ( - l ) for > 1. - 1 +1 Solutio by Michael Yoder, Studet, Albuquerque Academy, Albuquerque, New Mexico. If the give equatio is true for ay oe value of, it is true for all values of ; hece takig = 2, we get a (a + b) - b 2 = -11111, 4a 2 + 4ab - 4b 2 = -44444, (2a + b) 2 = 5b 2-44444. Now 5b 2-44444 > b 2 leads to b > 15; tryig b = 16, 17, i s u c c e s - sio, oe fids the smallest value of b to make 5b 2-44444 a square b = 111. However, this gives 2a + b = 131, a = 1, ad a i s supposed to be odd. Cotiuig with b = 112, 113, 9, we fid 166 2 = 5(12) 2-44444, which gives a = 23, b = 12 as the smallest solutio,, Also solved by Christie Aderso, Herta T. Freitag, Joh E. Homer, Jr., Gregory Wulczy, ad the Proposer. A MONOTONIC SURJECTION FROM Z + TO z" B-165 Proposed by Douglas Lid, Uiversity of Virgiia, Charlottesville, Virgiia. Defie the sequece b()} by b(l) = b(2) = 1, b(2k) = b(k), ad

112 ELEMENTARY PROBLEMS AND SOLUTIONS Feb. 197 b(2k + 1) = b(k + 1) + b(k) for k > 1. For > 1, show the followig: (a) b([2 + 1 + (-l) /3) = F + 1. (b) bqi-^+i-ifysi = L. Solutio by Michael Yoder, Studet, Albuquerque Academy, Albuquerque, New Mexico. (a) For =,1 the formula is easily verified. - 2 ad - 1 with >, 2; the if is eve, Assume it is true for Similarly, if is odd, b[(2 + 1 + l)/3] = b[(2 - l)/3 + b (2 + 2)/3] = F + b [(2 _ 1 + l)/3] = F + F - = F ^. -1 +1 b[(2 + 1-1/3] = F +1 ' (b) For = 1,2 the theorem is true; ad by exactly the same argumet as i (a), it follows by iductio for all positive itegers. Also solved by Herta T. Freitag ad the Proposer. * * * (Cotiued from page 11.) SOLUTIONS TO PROBLEMS 1.. 5 3-4 2 + 3-8. 2. 3 2-8 + 4 ad the Fiboacci sequece: 1,4,5,9,14,'. 3. 7 3 + 3 2-5 + 2 + 3x2. 4. 4 + 3 + 3(-l). 5. 2 3-3 2 - + 5 ad the Fiboacci sequece 4L *