ADVANCED PROBLEMS AND SOLUTIONS

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1 ADVANCED PROBLEMS AND SOLUTIONS Edited by RAYIVIOWDE. WHITNEY Lock Have State College, Lock Have, Pesylvaia Sed all commuicatios cocerig Advaced Problems ad Solutios to Eaymod E D Whitey, Mathematics Departmet, Lock Have State College 9 Lock Have* Pesylvaia 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 e To facilitate their cosideratio, solutios should be submitted o separate siged sheets withi two moths after publicatio of the problems. H-192 Proposed by Roald Alter, Uiversity of Ketucky, Lexigto, Ketucky. If 3+l c = 3=o {%: i ) M i, i prove that c = N, (N odd 9 > 0) H-193 Proposed by Edgar Karst, Uiversity of Arizoa, Tucso, Arizoa. Prove or disprove: If x + y + z = ad x 3 + y 3 + z 3 = , the ad 2-1 are primes. H-194 Proposed by H. V. Krisha, Maipal Egieerig College, Maipal, Idia. Solve the Diophatie equatios* 283

2 284 ADVANCED PROBLEMS AND SOLUTIONS [Apr* (i) x 2 + y 2 ± 5 = 3xy (il) x 2 + y 2 ± e = 3xy 9 where e = p 2 - pq - q 2 p,q positive itegers. SOLUTIONS BINET GAINS IDENTITY H-180 Proposed by L Carlitz, Duke Uiversity, Durham, North Carolia. Show that \^ ( \ 3 _ \*^ ( + k)i F Y ^ / \ 3 = «( + k)i T S * ' ^ " ^ W ( a - 2k)! (2 " 3k) ' th where F, ad L, deote the k Fiboacci ad Lucas umbers s respectively. Solutio by David Zeitli, Mieapolis, Miesota. A more geeral result is that «Ef^Yb " k av = V ^( + k)l - b k a k W 2-3k ' k=0 2k< where W l 0 = aw,- + bw, = 0, 1, * *. For a = b = 1, we obtai the desired results with W, = F. ad W, = L,.

3 1972 ADVANCED PROBLEMS AND SOLUTIONS 285 Proof. From a well-kow resultt we ote that E (IP - E (2) V ( t I"** " V '" * "" x k <X + I)"" 2 " k=0 ' 2k^ ^ ( Q ~ 2k)I 7 k=q 2k=r- Set x = (ay)/b i (2) to obtai: (3) E / \ 3 u - k k k \ ^ ( + k)i, k k k,,, v I k ) b a y = b a y ( a y + b) z^ r^r f k=0 L _ A \ x '/ 2k< o i ^ (k!) 3 ( - 2k)l Let a 9 fi be the roots of y 2 = ay + b* Notig that W = CjQf + C 2 j8, we obtai (1) by additio of (3) for y = a ad y = 8 e Remarks,, If a = 2x b = - 1, the with W. = T. (x), the Chebyshev polyomial of the first kid^ we obtai from (1) k=0 X 7 2k< lks) l " m For a = 2 g b = 1, oe may choose W. = P., the Pell sequece. Let V 0 = 2, Vi = a, ad V k + 2 = av k bv fe. The, from (1), we obtai the geeral result k=0 ^! < - i > m t l ^ i " k w m k + p s S " ^ ^ J ( ( - 1 ) m + l b m V m ) W m(2-3k) + p for m 5 p = 0 g 1, It should be oted that (1) is valid for equal roots* i. e*, a = /3. *J S Riorda, Combiatorial Idetities, p* 41.

4 286 ADVANCED PROBLEMS AND SOLUTIONS [Apr, Also solved by F. D. Parker, A. G. Shao, ad the Proposer. SUM-ER TIME H-181 Proposed by L Carlitz, Duke Uiversity, Durham, North Carolia. Prove the idetity E m? =0 where m - (am + e) (bm + d) r-r- = u w^, v., 9 m! l (1 - ax) (1 - dy) - boxy u = x e - ^ - ^ y ), v = ye- ( c x 4 d y> Solutio by the Proposer. / J ( a m + c) m (bm + d) m ^ ~ m 9 =0 m = (am + c) m (bm + d) i j X - e-(am-*)x-(bm-hi)y m 9 = o o u- TT «" % 4 / m 4 - r > r -i ^ "H I r ( K m - l - r l m l = (am + c) m (bm + d) ^ (-1) J ^ ^ (-l) k ( b m^, d l l ), But m 9 =0 j=0 k=0 m E E E E w» i+lt (T)(k) ««>-» <*.-«m * -» *» " m 9 =0 j=0 k=0 00 m E l & E E w) m+ - j - k (T)(^)w + «m w +»" m 9 =0 j=0 k=0

5 1972] ADVANCED PKOBLEMS AND SOLUTIONS 287 m j=0 i=0 k=0 lr= \ / \ / S m m m r =0=0 X 7 j=0 ^ ' ^) " k (^) k=0 X ',d r + s Sice m j=0 m) m) we eed oly cosider those terms i (*) such that [m + - r - s ^ m 1 r + s ^ 5 that is s r + s = * We therefore get mi(m,).^ / m \ / \ m-r,-r / u \T J» AAA = f AX 8 "! ^ V ( rj( rja d (be) f m g Z ^ r=0 so that

6 288 ADVANCED PROBLEMS AND SOLUTIONS [Apr. oo V (am + c ) m ( b m + d ) - V r - m 9 =0 mi(m,) E X m y \ «% f m \ / \ m - r j i - r,, vi.l \ v){v) a d (bc) m,=g r=0 oo oo = < bc^r z ( m ; r )( ; r )(-) m (dy) r=0 m, = r co = (bcxy) r (l - a x ) - r _ 1 ( l - d y ) r=0 _1 fi - rfvrmi _ ^cxy = (1 - ax) * (! - dy)~ = {(1 - ax)(l - dy) - bcxy) ( r (1 - ax)(l - dy) X ARRAY OF HOPE H-183 Proposed by Vemer Hoggatt, Jr., &/? Jose State College, Sa Jose, Califoria. Cosider the display idicated below Pascal Rule of Formatio Except for P r e s c r i b e d Left Edge. (i) Fid a expressio for the row s u m s. (ii) Fid a geeratig fuctio for the row s u m s.

7 1972 ADVANCED PROBLEMS AND SOLUTIONS 289 (ill) Fid a geeratig fuctio for the risig diagoal sums. Solutio by the Proposer. i) A ispectio of the array reveals that the row sums are F - ( = 0, 1, 2, ) 9 e e ii) If the colums are multiplied by 1, 2 $ 3 9 sequetially to the right 9 the the row sums have the geeratig fuctio, (1 - x) (1 - x) (1-3x + x 2 ) (1-2x) Thus the row sums are the covolutio of the two sequeces: a) Ai = 1, A = F 2 + i ( - D ad b) Bi = 1, B = 2-l ( > 1). iii) The risig diagoal sums, E, are the covolutio of the two sequeces: c) C = F x ad d) D =\ 2a _* ( = 0 s l, 2,. -. ). Hece (1 - x) 3 LmJ (1 - x - x 2 ) ( l - 3x + x 2 ) Q = 0 FIBO-CYCLE H-184 Proposed by Raymod Whitey, Lock Have State College, Lock Have, Pesylvaia. Defie the cycle a ( = 1, 2, ) as follows^ i) a = (1, 2, 3, 4, -, F ), II where F deotes the Fiboacci umber. Now costruct a sequece of permutatios 00 «* [. ( = 1, 2, )

8 290 ADVANCED PROBLEMS AND SOLUTIONS [Apr F F F (ii) a i + 2 = a i x i + 1 (i - 1). N Fially 9 defie a sequece 1 =l as follows: u is the period of OO Ft. i. e., u is the smallest positive iteger such that (iii) a 1+u = a 1 (i > N). x v a) Fid a closed form expressio for u. b) If possiblej show N = 1 is the miimum positive iteger for which iii) holds for all = e e e. Solutio by the Proposer. Sice a is of order F, it follows that the expoets of a may be replaced by residues mod F ad u is thus the period of the Fiboacci s e- quece mod F. Therefore u A = u 2 = l s u 3 = 3* Cosider the First r e s - idue classes of the Fiboacci seuqece 9 mod F ( ^ 4); 1, 1, 2, 3, oa 9 st F 1 9(K The ( +1) residue class is F - = 1 + (F - - 1) ad (2-1) class is However, F - + F - (F - - 1) = F 2, ad F ^ = F + F ( > 2) N +1-1

9 1972] ADVANCED PROBLEMS AND SOLUTIONS 291 Implies F, F ^ - F 2 = (-l) ( > 2) F 2 - = (»l) (mod F ). -1 ir If is eve ( ^ 4), we have F 2 - = 1 (mod F ) ad u = 2* If is ' -1 odd ( > 4), F 2 - = -1 (mod F ) ad u = From the above, it is obvious that N = 1 is the smallest positive Iteger for which (III) holds for all = 1, 2, * 9. It Is iterestig to ote that (u ( = 1, 2, } fl ( F j = 1, 2, } = {F l f F 4? F 6, F B, F 12, }. [Cotiued from page 282. ] by NOTE ON SOME SUMMATION FORMULAS S 0 + S 1 + S 2 ^ TT (k + St + 2s 2 + 3s I) i=l s 0 is 1 Is 2 t... REFERENCE 1, L 9 CarlitZg "Some Summatio Formulas, M Fiboacci Quarterly, Vol. 9 (1971), pp

<aj + ck) m (bj + dk) n. E E (-D m+n - j - k h)(i) j=0 k=0 \ / \ /

$<aj + ck) m (bj + dk) n. E E (-D m+n - j - k h)(i) j=0 k=0 \ / \ /$ ADVANCED PROBLEMS AND SOLUTIONS Edited by RAYMOND j=. WHITNEY Loc Have State College Loc Have # Pesylvaia Sed all commuicatios cocerig Advaced Problems ad Solutios to Raymod E. Whitey, Mathematics Departmet,