ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited By A, P. HtLLMAN Uiversity of New Mexico, ASbuquerque, New Mexico Sed all commuicatios regardig Elemetary Problems ad Solutios to Professor A. P. Hillma; 709 Solao Dr., S. E.; Albuquerque, New Mexico Each solutio or problem should be o a separate sheet (or sheets). Preferece will be give to those typed with double spacig i the format used below. Solutios should be received withi four moths of the publicatio date. DEFECTIONS The Fiboacci umbers F ad the Lucas umbers L satisfy F+2 = F+1 + F, F 0 = 0, F 1 = 7 ad L +2 = L +1 + L, L 0 = 0, L 1 = I Also a ad b desigate the roots of (1 + V5 )/2 ad (1 - V5 )/2, respectively, of x 2 - x - 1 = 0. PROBLEMS PROPOSED IN THSS ISSUE B-328 Proposed by Walter Haseli, Mill Valley, Califoria, ad V. Hoggatt, Jr., Sa Jose, Califoria. Show that is always a sum 6( ) m 2 +(m 2 + 1) + (m 2 + 2) + -+(m 2 +r) of cosecutive itegers, of which the first is a perfect square. B-329 Proposed by He rta T. Pre/'tag, Roaoke, Virgiia. Fid r, s, ad fas liear fuctios of/? such that 2P 2 r- F s F t is a itegral divisor of L +2 + L for/7 = 1,2,». B-330 Proposed by George Berzseyi, Lamar Uiversity, Beaumot, Texas. Let G = F+29F + 4 +F+8 - Fid the greatest commo divisor of the ifiite set of itegers j G 0, G u G 2, - t - B-331 Proposed by George Berzseyi, Lamar Uiversity, Beaumot, Texas. F % +1 s ; ( mod 24 >- B-332 Proposed by Phil Ma a, Albuquerque, New Mexico. Let a() be the umber of ordered pairs of itegers (r,s) with both 0 < r<s ad 2r + s =. Fid the geeratig fuctio A(x) = a(0) +xa(1) +x 2 a(2) +... B-333 Proposed by Phil Maa, Albuquerque, New Mexico. Let S be the set of ordered pairs of itegers (a,b) with both 0\<a <b ad a + b <. Let T be the set of ordered pairs of itegers (c,d) with both 0 <c <d < ad c + d >. For > 3, establish at least oe bijectio (i.e., 1-to-l correspodece) betwee S ad T +f. SOLUTIONS SO BEE ST B-304 Proposed by Sidey Kravitz, Dover, New Jersey. Accordig to W. Hope-Joes'The Bee ad the Petago/' The Mathematical Gazette, Vol. X, No. 150, 1921 (Reprited Vol. LV, No. 392, March 1971, Page 220) the female bee has two parets but the male bee has a mother 188

2 April 1876 ELEMENTARY PROBLEMS AND SOLUTIONS 180 oly. if we go back/7 geeratios for a female bee she will have F male acestors i that geeratio ad F +] female acestors, makig a total of F +2 acestors. Solutio by Sister Mario Beiter, Rosary Hill College, Buffalo, New York. The proof is by iductio, let P() be the statemet of the problem. P/W holds for/7 = /. Oe geeratio back a female bee will have F i = 1 male acestor ad Fj. + 1 = 1 female acestor, a total of F l+2 = 2. UP() holds for = k, it holds for = k+ 1: If we go back k geeratios for a female bee she will have F^ male acestors i that geeratio ad F^+i female acestors, makig a total of F^-^2 acestor s. The if we go back k + 1 geeratios, she will have F^+i male acestors (from the F^+f females i the k th geeratio), ad Fk+2 female acestors (from the total F^+2 acestors i the k th geeratio). This makes a total of F^+j + F k+2 = F k+3 Hece, P() holds for all atural umbers. Also solved by George Berzseyi, Paul S. Bruckma, Herta T. Freitag, Graham Lord, A. G. Shao, ad the Proposer. A TELESCOPING SUM B-305 Proposed by Frak Higgis, North Cetral College, Napervil/e, Illiois. F 8 = L 2 Yl L 2+4k-2 Solutio by Graham Lord, Uiversite Laval, Quebec, Caada. The followig steps use stadard idetities: L2 5 3 L 2+4k-2 = L2 ( 5 3 \ F 2+4k-2+2~ = L2(F6~ F 2) = L2' F 2'U = F4 'L4 = Fs «F 2+4k-2-2 Also solved by George Berzseyi, Wray G. Brady, Paul S. Bruckma, Herta T. Freitag, Ralph Garfield, C. B. A. Peck, Bob Prielipp, Jeffrey Shallit, A. G. Shao, ad the Proposer. SOMETHING SPECIAL B-306 Proposed by Frak Higgis, North Cetral College, Naperville, Illiois. F 8+f~ 1 = L 2 /2, ^2+4k-1 Solutio by George Berzseyi, Lamar Uiversity, Beaumot, Texas. For geeralized Fiboacci umbers defied by lettig //#ad / / ; be arbitrary itegers ad H = / / _ ; + > 2, it is kow that 5 3 H 4k-1 = F 2^2+1 (See, for example, Idetity (9) i Iyer's article, FQ, 7 (1969), pp ;) More geerally, / M ^2+4k-1 = F 2 H4+1 \ J H ^2^

3 190 ELEMENTARY PROBLEMS AND SOLUTIONS [April Specializig this idetity to Lucas umbers ad usig (l ) ad (/ 24 ) of Hoqqatt's Fiboacci ad Lucas Numbers, oe obtais L L 2+4k-1 = ^2 F 2U+J = F4 L 4+1 = F 8+1 ~ 1> Also solved be Wray G Brady, Paul S. B rue km a, Herta T. Freitag, Ralph Garfield, Graham Lord, C. B. A. Peck, Bob Prielipp, Jeffrey Shallit, A. G Shao, ad the Proposer. MODULARLY MOVING MAVERICK B-307 Proposed by Verer E. Hoggatt, Jr., Califoria State Uiversity, Sa Jose, Califoria. Let (where, of course, a tk = 0 for k > 2). Also let (1+x+x 2 ) = a f o + a g1 x+a t 2X 2 +-, OO OO OO OO A = ^ L, a,4j, B = 2_^ a f 4j+i, C =zl a,4j+2, D = 2 j a,4j+3 1=0 j=0 j=0 j=0 Fid ad prove the relatioship of A,B, C ad D to each other. I particular, show the relatioship amog these four sums for = 333. Solutio by George Berzseyi, Lamar Uiversity, Beaumot, Texas. Oe may easily prove by iductio o/7 that the triomial coefficiets^^ satisfy the recursio formula a,k - a- 1,k-2 + a - 1,k-1 + a - 1,k for/7 > 0 with iitial values f 1 if k = 0 a O,k = I 0, otherwise. By lettigx = 1 i the defiig equatio oe may also deduce that This last fact ad two applicatios of the recurrece relatioship readily yield the followig idetities for/? >2: A = C - 2, C = A. 2. B = 2-3"- 2 + D - 2. D = 2-3 ~ 2 + B. 2. Iteratio o, upo summatio of the resultig geometric series, yields the followig formula for each Xe{A,B,C,D\, m e \o, 1,2,3,], = 0,1,2,-: X 4+m = y43 4+m -3 m )+X m. Less compactly, but more i the spirit of compariso oe fids B4 = C4 = D4 = A4-1, B4 +1 = C4 +i = A4 +1 = D , R4+2 = D4+2 = ^4+2 = C4+2~ *, ^4+3 = ^4+3 = &4+3 = B * > for each = 0, 1, 2,, I particular, ^333 = *... = ^333 * D^ + 1 = %(3 + 1>. Also solved by Paul S. Bruckma, Herta T. Freitag, Graham Lord, David Zeitli, ad the Proposer.

4 1976] ELEMENTARY PROBLEMS AMD SOLUTIONS 101 A GARBLED HINT B-308 Proposed by Phil Maa, Albuquerque, New Mexico. (b) Let r be a real umber such that cos (m) = p/q, with p ad q relatively prime positive itegers ad q ot i J1, 2,4, 8,. r is ot ratioal. [The (a) part has bee deleted due to a error i i t ] Solutio by Paul S. Bruckma, Uiversity of Illiois, Chicago Circle, III. (b) We first recall the multiple-agle formula from trigoometry: i/2] (1) cos/?0 = # YJ ( ~ 1)k ITTk [ ~k k ] tfcosw"- 2 *, = 1,2,3,-. We also recall, or we may easily show, that this is a polyomial with iteger coefficiets, i cos 6. Suppose ow that r = u/v is ratioal (with u ad v relatively prime atural umbers), ad satisfies: (2) cos (rir) = p/q, wherep ad q are relatively prime atural umbers ad (q,2) = 1 (i.e.,q is odd), except qf\. Mote that this restricts q more tha i the origial problem, but we will deal with the remaiig values of q later. Lettig 6 = m ad = v i (1), we get: lv/2] v-2k <y\t-1 V "i <-" - -"- *? '-"* A (v) (*p - ^ v ~ 3 v " - 2 v-2 k=0 q q *=. = 2 v - 1 p v + c 2 M where M is some iteger. Sice (2,q) = (p,q) = 1, it follows that (2 v v-1~v "'p v + J. q*m, ~2 j, q V\ v ) = _ 7; -f.u..+ but +i, the q v caot + divide A\*.\A (2 V ~'p /iv-1~v_ v + q*m), L 2 l ad their ratio caot be (-1) u = ±1. This cotradictio shows that r caot be ratioal, whe q is as stated above. Suppose ow, as before, that (2) holds for some ratioal r, where q = 2 s t, (2,p) = (p,t) = (2,t) = 1,s ad t are atural umbers, t > 3. As before, (-l)u = (2 v - 1 p v + q 2 M)/q v = (2 y ~ 1 p v + 2 2s t 2 M)/2 sv t v. Sice (2,t) = (p,t) = 1, the idicated ratio caot be a iteger, ad we have agai reached a cotradictio. Hece, we have proved that the oly possible values of q satisfyig (2) are q = 1 ad q = 2; this, i tur, implies that cos (rir) = 0, ± 1 /2, ±1 are the oly possible values, correspodig to r = + Y2, ± 1/3, ad, respectively, where is a arbitrary iteger. This is a stroger result that origially sought. Also solved by the Proposer. The error i Part (a) was poited out by Paul S. Bruckma ad Herta T. Freitag. B-309 Corrected versio ofb-284. AN ANALOGUE OF a =af + P ^ Letz 2 = xz + y ad let k, m, ad be oegative itegers. (a) z = p (x,y)z + Q (x,y), where p ad Q are polyomials ix ad y with iteger coefficiets adp has degree \ i xfor> 0. (b) There are polyomials r, s, ad t, ot all idetically zero ad with iteger coefficiets, such that z k r(x,y)+z m s(x,y)+z t(x,y) = 0. Composite of solutios by David Zeitli, Mieaspolis, Miesota, ad the Proposer. (a) Let UQ = 0, Uj = 1, ad U +2 = xu +] z =zu + U -i usig z ^ 7 = xz k + yz k ~ 1. + yu for = O t 1, -. The oe easily proves by iductio that

5 192 ELEMENTARY PROBLEMS AND SOLUTIONS April 1976 (b) z satisfies a quadratic equatio over the field F = Q(x,y) of polyomials i * ad y with ratioal coefficiets. Hece F[z] is a vector space of degree 2 over F. Thus ay three powers of z are liearly depedet over F. Clearig deomiators, gives the desired result. Also solved by Paul S. B rue km a ad He rta T. Freitag. *******