ADVANCED PROBLEMS AND SOLUTIONS

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1 Edited by Raymond E. Whitney Please send all communications concerning ADVANCED PROBLEMS AND SOLUTIONS to RAYMOND E. WHITNEY, MATHEMATICS DEPARTMENT, LOCK HAVEN UNIVERSITY, LOCK HA VEN, PA This department especially welcomes problems believed to be new or extending old results. Proposers should submit solutions or other information that will assist the editor. To facilitate their consideration, all solutions should be submitted on separate signed sheets within two months after publication of the problems. PROBLEMS PROPOSED IN THIS ISSUE B-572 Proposed by Paul S* Bruckman, Berkeley\ CA Prove the following, where <p - a~ l : {<p 5n+l l (5/i +1) + <p 5n+3 / (5n + 2) - tp 5n+4 1 (5n + 3) - <p 5n+4 / (5/i + 4)} = (x/25)(50-l0j5) y2. H-573 Proposed by N. Gauthier, Royal Military College of Canada "By definition, a magic matrix is a square matrix whose lines, columns, and two main diagonals all add up to the same sum. Consider a 3 x 3 magic matrix O whose elements are the following combinations of the rfi 1 and (w + l)* Fibonacci numbers: O ll = 3F n+l +F n ; % 2 = F n+h l3 = 2F n+l +2F ; * 2 1 = Fn+1 + 2F n> ^ 2 2 = 2F W F ; d> 23 = 3F n+h *3i=2/Wi; ^32 = 3F n+l + 2F n ; <D 33 - F n+l+ F n. Find a closed-form expression for <P W, where m is a positive Integer, and determine all the values of m for which it too is a magic matrix." SOLUTIONS Geometric? H-561 Proposed by K Gauthier, Dept of Physics, Royal Military College of Canada (Vol 38, no. 2, May 2000) Let n be an integer and set s n+l = a n + a n ~ l fi+ + a$ n ~ l +fi n y where a + $ = a, a$ = b, with a & 0, b * o two arbitrary parameters. Then prove that: (b) ffa = (-l){ 'M^W+pr+n, ( C ) S ip+q S qr+n = 2^\ r P S p+q S {2p+q)t-pr+m where r > 0, w, p(* 0), and q(* 0,±p) are arbitrary integers. 2001] 187

2 Solution by Paul S. Bruckman^ Berkeley 9 CA Let the sums indicated in the right-hand members of parts (a), (b), and (c) be denoted by $& = <9&(p,q,r,ri), < 3b = ( 3b(p,q,r,n), %-%{p,q,r,n), respectively, We distinguish two pos bilities: Case 1, where a 2 ^4b, i.e., a^jb; Case 2, where a 2 = 4b, i.e., a-(i. For Case 1, the Binet formula holds: s n - (a n - j3 n )/(a-/j); for Case 2, we have: s n = na n ~ l P in either case for all integers n. We first deal with Case 1; we also make the substitution 0-a-p. The sums may be evaluated by application of the appropriate formula for s n and the Binomial theorem. We will also employ some auxiliary results, given in the form of lemmas. Lemma 1: h q s p _ q + a p s q = a q s p ; b q s p _ q +/?% = /?%. also, Proof: b q $ p _ q +a p s q = {{ap) q {a p - q ~ p p ~ q ) + a p {a q - fi*)}/0=a p (a q -fi q )/0^a q s p ; b q s p _ q +@ p $ q = {(a/3) q (a p - q -p p - q )+@ p (a q - Lemma 2: s p+q - a q s p p s q ; s p+q -p q s p = a p s q. Proof: In Lemma 1, replace/? by p + q, then divide by a q (or J3 q ). D Lemma 3: 0 p+ % + a p s p+q = a p+q s p +/J p s p+q = s,^. Proof: Replace q by p + q in Lemma 2. We may now proceed to the proof of the problem, at least for Case 1. (a) si = rc k b^- k \s q ) k (s p _ q y k (a pk+ "-j3 pk+ ")/0 k=0 = (a"/0w Sp _ q + ap Sq Y-(P"/0)(b%_ q+^s q y = (a* /6)(a q s p ) r - {fi" 16)(fi q s p ) r (using Lemma 1) = (s p na" r+ "-p^")l0 = (s p Ys qr+n. D (t) % = (-i) fc rc k ( Sp ) k ( Sp+q y- k (a gk+pr+ "-^k+ p r+n )/e = (a pr+ "/0)(s p+q - a%) r - (fi pr+ "/0)(s p+q - P"s p y = a pr+ " 10)p pr {s q ) r -(J3 pr+n 10)a pr (s q ) r (usinglemma2) = b pr (s q y{a"-p n )i0 = b pr { Sq y Sn. (c) ^tc^^'^'^^^/tafw^"-/;^)^")/^ k=q = (a~ pr+ "/0)(b p+q s p + a 2p+q s p+q y-(p- pr+ "/0)(b p+ ' 1 s p + p 2p+q Sp+qy = ( a - pr+n+pr+qr 10)(fi p + q s p + a p s p+q ) r - (fi-pr^pr^r / ff)^^ + P p s p+q ) r ; hence, using Lemma 3, ^ = (a qr+ "/0)(s 2p+q y-(fi qr+ "/0)(s 2p+q y =(s 2p+q Ys qr+. D 188 [MAY

3 It remains to prove (a), (b), and (c) for Case 2. We first show that Lemmas 1, 2, and 3 are valid for this case as well; the modified versions of these lemmas are denoted by a "prime" punctuation. Lemma 1 f : b q s p _ q + a p s q = a q s p. Proof: b q s p _ q + a p s q = (p- q)a 2q (a p - q ~ l ) + qa p (a q ~ l ) = pa p+q ~ l = pa p ~ l a q = a q s p. D Lemma 2 f : s p+q - a q s p = a p s q. Proof: Replace p by p + q in Lemma V and divide by a q. D Lemma 3 f : a p+q s p + a p s p+q = & lp q. Proof: Replace qby p + q'mlemma T. = P r Z r-ic k b^- x - k \s q r\s p _J- x - k a^^"- x +nt rc k b^- k \s q f(s p _ q y- k a pk+ "- 1 k=0 k=0 - pr$ q a p+n ~ l {b q $ p _ q + a p s q Y~ l +na n ~ l {h q s p _ q + a p s q } r = pra p+n - l+q{r - l) s(sj- 1 +na n - l+qr q\»p, """ (sj \ p/ (usinglemma 1') = a^"- 2+ i r (sy- l (pqr + np) = pa p ~\qr + n)a* r +»-\s p T l = (sj Sqr+. D (*>)' =T(-i) k rc k ( Sp ) k (s p+q y- k (qk + pr + n)a<> k+ r r+ "- 1 /?/ qr+n' + 0'/-+/i)a^-, X(-l) k rc t (^)*(* p +,r t a» t = -qrs p a" +pr+ "-\s p+q - a*s p T x + {pr + n)a pr+ "-\s p+q - a«s p J = -qrs p a q+pr+ "-\a p s q y- 1 + {pr+ ri)a pr+ "-\a p s q ) r (using Lemma 2') = { - r a ^ a ^ " + (pr + n)s q a pr+n - 1+p }(a p s q y- 1 = {-ra p a 2pr+n - ; ' + (pr + «)a 2pr+ "-! }(s? ) r = {-pra 2^"-' + (pr + n)a 2pr+ "~ l }(s q y = na 2pr+ "-\s q y = a 2pr (s q ys = S ^ ) ' *,,. (c)': % = 2 r Q ^ + ^ - t ) ( 5 ;, ) r - * ( V g ) i {(2p +?)* - pr+n) k=0 CPP+IV-P""-* = (2/? +? )m 2^-^+"- V, Z r-iq A ^ ^ ' - ^ ^ r 1 - * ^ ) * ^ 2 ^ ^ + ifc=0 2001] 189

4 + {-pr + n)a' pr+n ~ l Y, rc k b (p+^r-k \s p J- k (s p+q ) k a i2p+q)k k=0 = (2p + q)ra 2p+q - pr+ "'\ +q (b p+ % + a 2p+ % +q y~ l + (-pr + n)a- pr+h - 1 (b p **s p + a 2p+q s p+q ) r = (2p + q)ra 2p+q - pr+n - l a (p^r- X) s p+q (a p+q s p + a p s p+q ) r (-pr + n)a- pr+ "- l a (p+ ' l)r (a p+ ' } s p +a p s p+q y = (2p + qyao-^s^i^j- 1 + (-pr+n)a^^\s 2pHi y = r a ^ ^ - ^ ^ s ^ i ^ J + (-pr+nw^i^j = r(p + q)a- p+qr+ "- q+p+q -\s 2p+q Y + (-pr+^a^'x^j = r(p + q)a qr+ '-\ S2p+q y+(-pr +r,)a qr+ "- l (s 2p+q y = (qr + n)a^"- l (s 2p+q y = (^Js qr+n. This completes the proof of the problem. Also solved by H. Kwong, H.-J. Seiffert, and the proposer. Greatest ProbSem H-562 Proposed by H.-J. Seiffert, Berlin, Germany (Vol 38, no. 2, May 2000)-corrected Show that, for all nonnegative integers n, where [ ] denotes the greatest integer function. Solution by the proposer Define the Fibonacci polynomials by F 0 (x) - 0, F x (x) - 1, and F +2 (x) = xf n+l (x) + F n (x), for n GN 0 (natural numbers). Let A n : = i n ~ l F n (ia) 9 nen 0, where i = ^/(-l). Writing n = 5m + r, where men Q and re {0,1,2,3,4}, a simple induction proof on m yields A, 0 ifn = 0(mod5), 1 if w = 1 (mod 5), -a ifn = 2 (mod 5), (1) a ifn = 3 (mod 5), [-1 if«= 4(mod5). From H-518 [identity (8)], we know that, for all complex numbers x and all nonnegative integers n, tc-^w-^ 190 [MAY

5 With x = / a, this gives ;? 0 U 2 - J ^ cr (2) a x Since 1 / (4 - a 2 ) = a IJ5,. we have or, by a r = (L r + j5f r )/2, r ez, n _ 2n 4 " - a 1 A-a 2 S (4"a-a2n+1 ). 4 " _ a2n /c i 4 - a 2 10 On the other hand, from (1), it follows that j=o \ - - = (4" - ^ i ) 4 ( 4 " ~ ^ i ) - (3) [ ] [?], J y 4=0 v y fc=o v Usinga 2 = (3 + V5)/2, we then obtain ±U?,)4- J=0 [ fv In \W( In L{n-5k-2)+^o{n-5k- + *,, where q n is a rational number. Since V5 is irrational, (2), (3), and the latter equation imply that This proves the desired identity, because XL-5^-2) + XL-5i-3)"5 ( 4 "~^+i> fc=0 V J k=q V y J w-2' {n-5k-2y{n~5k-3) = {n-%-2j> - k ~ where we set ( 2 /) = 0 for j < G. ^4/so solved by H. Kwong and P. Bruckmait A Stirling Problem H-563 Proposed by N. Gauthier, Dept of Physics, Royal Military College of Canada (Vol 38, mo. 2, May 2000) Let /w>0, n>0, p&q, q*-p,0, and s be integers and, for \<k<n, let {ri) k :=n(n~l)... (n - k +1) and S^ be a Stirling number of the second kind. Prove the following identity for Fibonacci numbers: I(-l) r ; ^[F/F p+ J r F 9r+^(-ir[^/F p+ J"X(-l)^fc («) fc^>[f p /FJ fe V^ np+sk=\ 2001] 191

6 Solution hy the proposer For x an arbitrary variable, consider the binomial expansion identity z(;v=o-x)-, r = 0 V / and apply the operator D := x~ to it m times (m > 0) to get ^r m ( n f\x r =D m (l + x) n. By a well-known result, where {8^ : 1 < k < m; \<m} are the Stirling numbers of the second kind. Consequently, Next, for integers p (^ 0), q(^0 9 -p), solve the following for w and w: One readily finds Inserting x = i/a g, 1 + x = wa~ p n 1 + tfa^vc"^; l + ufi q =w/t p. «= -lf p /F p+q ]; w = (-\y[f q IF p, q l (**) in (*) and multipying the resulting equation by a s then gives m f \ X r U )u r a qr+s = S%\n) k ita#+^kcfi>- n p. Finally, replace a and /? in this result, subtract from the above, and divide by V5 to get n m f \ I ' " r «r / V" = l^\n) k u k w"-% +q)k _ np+s. V / lc=l Inserting the values for u and w from (**) then establishes the result claimed in the problem statement. The case m = 0 is readily dealt with, and one gets F Q r+s K A y ' \ F -pn+s- \ J p+gj VP+QJ 192