ELEMENTARY PROBLEMS AND SOLUTIONS. Edited by A. P. HILLMAN University of New Mexico, Albuquerque, NM 87131

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. HILLMAN University of New Mexico, Albuquerque, NM 873 Send all communications regarding ELEMENTARY PROBLEMS AND SOLUTIONS to PROFESSOR A. P. HILLMAN, 79 Solano Dr., S.E., Albuquerque, New Mexico 878* Each solution or problem should be on a separate sheet (or sheets). Preference will be given to those typed with double spacing in the format used below. Solutions should be received within 4 months of the publication date. DEFINITIONS The Fibonacci numbers F n and Lucas numbers L n satisfy F n + 2 =F n + i + F n 9 F =, F = and L = n + 2 L n + i + ^ J ^ O = 2 5 L = I. Also a and b designate the roots ( + /5)/2 and ( - /5)/2, respectively, of x 2 - x - =. PROBLEMS PROPOSED IN THIS ISSUE B-43 Proposed by M. Wachtel, H. Klauser, and E. Schmutz, Zurich^ Switz. For every positive integer a s prove that (a 2 + a - I) (a 2 4-3a + ) + is a product m(m + ) of two consecutive integers,, B-43 Proposed by V. E. Hoggatt, Jr., San Jose State University, San Jose, CA. For which fixed ordered pairs (h s for all integers n? k) of integers does F n( L n+h " F n + h) ~ F n+i,^ln + k " F n + k) B-^32 Proposed by V. E. Hoggatt, Jr., San Jose State University, San Jose, CA. Let G = F F 2 Q - F 3. Prove that the terms of the sequence alternate in sign, G Q9 G I9 G 2,... B-433 Proposed by J. F. Peters and R. P etcher, St. John s University, Collegeville, MN. Let {T(n)} For each positive integer n 9 let q n and r n be the integers with be defined by n = 3q n + v n and <_ v n < 3. T() =, T(l) = 3 9 T(2) = 4 9 and T(n) = hq n + T(r n ) for n > 3. Show that there exist integers a 9 b 9 c such that m/ N fan + 5P< " > - L 5 J * where [x] denotes the greatest integer in x. 273

2 27k ELEMENTARY PROBLEMS AND SOLUTIONS [Oct. B-**3*t Proposed by Herta T. Freitag, Roanoke, VA. For which positive integers n 9 if any, is a perfect square? L 3n -(-l) n L n B-^35 Proposed by M. Wachtel, H. Klauser, and E. Schmutz, Zurich, Switz. For every positive integer a, prove that no integral divisor of is congruent to 3 or 7, modulo. a 2 + a - SOLUTIONS F!rst Term as GCD B-46 Proposed by Wray G. Brady, Slippery Rock State College, PA. Let x n = 4L 3n - L\ and find the greatest common divisor of the terms of the sequence x x, x 2, # 3,.... Solution by Paul S. Bruckman, Concord, CA. or *n = L n^l 3n /L n - L 2 ) = L n \ka 2n - 4<a6) n + kb 2n - a 2n - 2(ab) n - b 2n ] = 3L n [a 2n - 2(ab) n + & 2n ]= 3L n (a n - b n ) 2 = 5L n F 2, x n = 5F n F 2n, n =, 2, 3,.... It fol- Note that a?! = 5F F 2 = 5. Hence, a^l^n, n =, 2, 3 lows that the greatest common divisor of {x n } is x-^ = 5. Also solved by Herta T. Freitag, John W. Milsom, Bob Prielipp, E. Schmutz, A.G. Shannon, Sahib Singh, Lawrence Somer, M.Wachtel, Gregory Wulczyn, and the proposer. Generator of Pascal Triangle B-47 Proposed by Robert M. Giuli, University of California, Santa Cruz, CA. Given that - x - xy E JL a nk xn y l n = k = is a double ordinary generating function for a nk ; determine a nk. Solution by Paul S. Bruckman, Concord, CA. ( - x -xy)-± = ( - x(l + z/))-i = ]T* n (l + 2/> n = L xn ib(k) n= n= k=^ ' oo n i \ -EE(J)*" The binomial coefficient ( r, ) is defined to be zero for k > n. we may extend the second sum above over all nonnegative k, i.e., 2/* Hence, [ -x-xy)-i = I ) E(fe)x"y n= & = o X '

3 98] ELEMENTARY PROBLEMS AND SOLUTIONS 275 T l m S ' (n\ a<nk = I l» (n* k = *» 2, ' ). Also solved 2?y W. O. J. Moser, A. G. Shannon, Sahib Singh, and the proposer. B-48 No solutions received. Proposal Tabled Exact Divisor B-49 Proposed by Gregory Wulczyn, Bucknell University, Lewisburg, PA. Let P n = F n F n + a. Must P n + Br - P n be an integral multiple of V - ~P J- n + hr J ~n + 2r for all nonnegative integers a and p? Solution by Sahib Singh, Clarion State College, Clarion, PA. Yes. Using we see that divisibility of P - P bv P - P n + 6r n -^ n+hr n + 2r is equivalent to divisibility of L 2n+2r + a ^2n+a ^ y ^ 2 n + 8 r + a 2n + t +r + a'' The result follows immediatly by seeing that ^2n+2r+a ~ ^2n + a = ^2h + 8r + a " ^2n+>r + a^^m-r + *) Also solved by Paul S. Bruckman, Bob Prielipp, and the proposer. Golden Limit B-4 Proposed by M. Wachtel, Zurich, Switz. Some of the solutions of in positive integers x and y are: 5(x 2 + x) + 2 = y 2 + y (x s y) = (, ), (, 3 ), (,- 23), (27 s 6). Find a recurrence formula for the x n and y n of a sequence of solutions (i n, 2/n) Also find l±m(x n + /x n ) and lim(# 27# n )- as n -> in terms of a = ( i + ; / 5 ) / 2. Solution by Paul S. Bruckman, Concord, CA. Multiplying the given Diophantine equation throughout by 4 9 completing the square 9 and simplifying yields: () Y 2-5X 2 = 4 5 where (2) X = 2x +, J = 2z/ +. The solutions of () in positive integers are known to be

4 276 ELEMENTARY PROBLEMS AND SOLUTIONS [Oct. However, due to (2), X and Y must also be odd. By inspection of the first few values (mod 3) of the Fibonacci and Lucas sequences, it is apparent that these values are even if and only if their subscripts are multiples of 3. Hence, we must have m ~ ± (mod 3) in (3), Distinguishing between these cases, we obtain two distinct sets of solutions; (5) a 2 ). *J 2> > = <%+*. i>*,+x.o-' In terms of the original problem, this yields the following distinct solution sequences: (6) (atf>, y^) -{W 6n + 2 -».Mhn + 2 ~ D};. 5 (7) (*< 2). J/i 2) ) - { ^ e» + * " D. ^ e +, - D } :. O - It is apparent from the fact that the successive indices of the Fibonacci and Lucas sequences in (6) and (7) "increase by sixes," that we are interested in the second-order equation for a 6, which must be the same for b s. Since a 6 = 8a+5 and a 2 = 44a+89 (special cases of a v = af r + F r _ )* it is evident that a and b satisfy the common equation: (8) s 2-8s 6 + =. Let (9) D n «-3 n * n + + s n, where z n ='*<*> or z/<*>, fc» or 2. We see from (6) and (7) that D n = %(-l ) [using (8)], or () D n = 8, n =,, 2,.... This is a recursion for the x and t/, as required. A homogeneous recursion may be obtained by noting simply that This is equivalent to the following third-order recursion: (ID s n + 3-9s n a B. + - a B -, n =.,, 2,.... It is evident from (6) and (7) that and (3) lim x z lx = lira y$jy = a 2 <fc = or 2). n-»-oo Also solved by the proposer. n-)-oo Trldlagonal Determinants B 4 Proposed by Bart Rice, Crofton, MD. Tridiagonal n by n matrices A n = (a^-) n of the form (a real) for 3 = i for 3 = i ±, otherwise occur in numerical analysis. Let d n - det A n. (i) Show that {d } satisfies a second-order homogeneous linear recur-

5 98] ELEMENTARY PROBLEMS AND SOLUTIONS 277 (ii) Find closed-form and asymptotic expressions for d n. (iii) Derive the combinatorial idantity [<n-l)72] -D./2] i v Solution by Paul S. Bruckman, Concord, CA ( i ) We see that ' Taking determinants along the first row 9 we find that d n = d n _ - det B n, where #..-. ^ B - i? I ^n»2 ^ ^ //(«-l) x(n-l) Taking the determinant of 5 n along its first columns we see easily that det B = d n _ 2 * Hence, we have the following recursions (3) d n d n+ + d n = 9 n =, Note also the initial values of the recursions (4) d x = 9 d 2 = 4a 2 -. The characteristic polynomial of (3) is (b) p(x) = x 2 - x +, which has the zeros (6) u = a + /a 2 -, y = a - /a 2 -. It follows that d n is of the form pu n + at? n 9 for some constants pand q which are determined from the initial conditions. Note that We then find uv =, u + v -. (7) d - ~ ~ ^ ^ «= The behavior of d n as n > depends on the magnitude of a s and we distinguish several cases. Co6e J: <. a <. Let a = cos 9* Then M = e %Q, i? = e~* e, and so (8) <f n = s i n ( n +' l ) 6 / s i n 9.

6 278 ELEMENTARY PROBLEMS AND SOLUTIONS [Oct. In this case, the sequence (d n ) is dense in (-esc 6, esc ) = (-( - a 2 Y h 9 ( - a 2 )'*) and oscillates within this interval without lending itself to approximation by an asymptotic expression. Co6e II: a = -. Then u = V = -. Since n thus k = k = Clearly, d n oscillates between these two values only, in this case. C(Ud llli a =. Here u = V =. Hence, n. d» = X w n "V = = n +. k=o fe=o Therefore, for this case, d n ^ n as n ->. Co6e!(/: a < -l. Then y < - < u <, which implies u n > as n ->. Hence,7 a ^ + Co6 I/: a > l. n + l Then < t; < < u 5 which implies y n + and d n ^ ~ u - v To prove the given combinatorial identity, note that n v as n V«>. h(u - y ) ^., = h(u» - v n ) = %E(^)a n - f e (a 2 - )***(- (-)*) t^(tt-l)] = L 2; +,K a " ( a 2 - i ) l t H k = X ' th(n-l)] / \ fc In (9), let x = a 2 -, supposing Case I above, so that we can have x >. Then, since p = t a n " ^ is in (, %TT), Also, sin 3? = ( - a )" 2. v = t a n ' M d - a 2 ) V2 / a } = cos^m a }. Using the notation of Case I, v = if a > and

7 98] ELEMENTARY PROBLEMS AND SOLUTIONS 279 v = IT - if a <. In either case, sin 9 = sin r. Thus, using (8), Also, a 2 ( sin nr/sin p, if a > ;? _! = sin n/sin = < ( (-)"" sin np/sin r 9 if a <. = ( + tf)", which implies that In either case, it follows from (9) that ( + a:)"^"" ', a > ; "(-ly^a + a?)-^'-", a < o. ih(n-i)} ( + ao-ho-d ^ ( 2^n+ ij ( ~ x ) f e = s i n ^ / s i n ^> which is equivalent to the desired identity. Also solved by the proposer.