Đang tải... (xem toàn văn)
Tạp chí toán học AMM của Mỹ
PROBLEMS AND SOLUTIONS Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West with the collaboration of Paul T. Bateman, Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall Dougherty, Tam ´ as Erd ´ elyi, Zachary Franco, Chris- tian Friesen, Ira M. Gessel, Jerrold Grossman, Frederick W. Luttmann, Vania Mas- cioni, Frank B. Miles, Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, John Henry Steelman, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ull- man, Charles Vanden Eynden, and Fuzhen Zhang. Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before August 31, 2008. Additional information, such as general- izations and references, is welcome. The problem number and the solver’s name and address should appear on each solution. An asterisk (*) after the number of a problem or a part of a problem indicates that no solution is currently available. PROBLEMS 11355. Proposed by Jeffrey C. Lagarias, University of Michigan, Ann Arbor, MI. De- termine for which integers a the Diophantine equation 1 x + 1 y + 1 z = a xyz has infinitely many integer solutions (x, y, z) such that gcd(a, xyz) = 1. 11356. Proposed by Michael Poghosyan, Yerevan State University, Yerevan, Armenia. Prove that for any positive integer n, n k=0 n k 2 (2k + 1) 2n 2k = 2 4n (n!) 4 (2n)!(2n + 1)! . 11357. Proposed by Mehmet S¸ahin, Ankara, Turkey. Let I a , I b , I c and r a , r b , r c be respectively the excenters and exradii of the triangle ABC.Ifρ a , ρ b , ρ c are the inradii of triangles I a BC, I b CA,andI c AB, show that ρ a r a + ρ b r b + ρ c r c = 1. 11358. Proposed by Marian Tetiva, National College “Gheorghe Ros¸ca Codreanu”, B ˆ arlad, Romania. Let d be a square-free positive integer greater than 1. Show that there are infinitely many positive integers n such that dn 2 + 1dividesn!. 11359. Proposed by J. Monterde, University of Valencia, Valencia, Spain. Find an ex- plicit parametric formula for the geometric envelope on the interval (0, 1) of the fam- ily of Bernstein polynomials B n s (t),definedbyB n s (t) = n s t s (1 − t) n−s for s ∈[0, n], where n s = n! (s+1)(n−s+1) . 11360. Proposed by Cezar Lupu, student, University of Bucharest, Bucharest, and Tu- dorel Lupu, Decebal High School, Constanta, Romania. Let f and g be continuous April 2008] PROBLEMS AND SOLUTIONS 365 real-valued functions on [0, 1]satisfying the condition 1 0 f (x)g(x) dx = 0. Show that 1 0 f 2 1 0 g 2 ≥ 4 1 0 f 1 0 g 2 and 1 0 f 2 1 0 g 2 + 1 0 g 2 1 0 f 2 ≥ 4 1 0 f 1 0 g 2 . 11361. Proposed by Finbarr Holland, University College Cork, Cork, Ireland. The Lemoine point of a triangle is the unique point L inside the triangle such that the distances from L to the sides are proportional to the corresponding side lengths. Given a circle G and distinct fixed points B, C on G,letK be the locus of the Lemoine point of ABC as A traverses the circle. Show that K is an ellipse. SOLUTIONS A Vanishing Alternating Sum 11212/11220 [2006, 268/367]. Proposed by David Beckwith, Sag Harbor, NY. Show that when n is a positive integer, n r=0 (−1) r n r 2n − 2r n − 1 = 0. Solution I by Harris Kwong, SUNY, Fredonia, NY. From (1 − t) −n = ∞ k=0 k+n−1 n−1 t k and (1 − t 2 ) n = n r=0 (−1) r n r t 2r , we obtain (1 + t) n = (1 − t 2 ) n (1 − t) n = ∞ k=0 n r=0 (−1) r n r k + n − 1 n − 1 t k+2r . Comparing coefficients of t n+1 yields the desired identity. Solution II by Rob Pratt, Raleigh, NC. More generally, if k < n,then n r=0 (−1) r n r 2n − 2r k = 0. Since k < n,nok-person committees can be formed from n couples using at least one member of each couple. To count these committees using inclusion-exclusion, let A i be the set of k-person committees using neither member of couple i.Now i∈S A i = 2n−2|S| k . Hence the sum on the left is the inclusion-exclusion sum to count the empty set. Solution III by Nicholas Singer, Annandale, VA. Indeed, n r=0 (−1) r n r P(r) = 0for any polynomial P of degree less than n.Thenth forward difference of the sequence (P(0), P(1), )has first term (−1) n n r=0 (−1) r n r P(r).SinceP is a polynomial of degree less than n, its nth forward difference is identically zero. Also solved by 62 other readers and the proposer. Sometimes an Integer 11213 [2006, 268]. Proposed by Stanley Rabinowitz, Chelmsford, MA. For positive integers n and m with n odd and greater than 1, let S(n, m) = (n−1)/2 k=1 sec 2m ( kπ n+1 ). (a) Show that if n is one less than a power of 2, then S(n, m) is a positive integer. (b ∗ ) Show that if n does not have the form of part (a), then S(n, m) is not an integer. 366 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 Solution to (a) by NSA Problems Group, Fort Meade, MD. Let n + 1 = 2 s with s ≥ 2. We use induction on s. The base case s = 2 is easy: sec 2m (π/4) = 2 m . Note that S(2 s+1 − 1, m) = 2 s −1 k=1 sec 2m kπ 2 s+1 = S(2 s − 1, m) + 2 s−1 j=1 sec 2m (2 j − 1)π 2 s+1 . By the induction hypothesis, it suffices to show that the last sum is integral. The Chebyshev polynomial T r is a polynomial of degree r defined by T r (x) = cos(r arccos(x)). In particular, T 2 s is a polynomial of degree 2 s with T 2 s (x) = 0when x = cos((2 j − 1)π/2 s+1 ) for 1 ≤ j ≤ 2 s . Furthermore, since T 2 (x) = 2x 2 − 1and T 2 s (x) = 2cos 2 (2 s−1 arccos(x)) − 1 = 2T 2 s−1 (x) 2 − 1, it follows that T 2 s (x) ∈ Z[x] and, apart from the constant term, all of its coeffi- cients are even. Since T 2 s (0) = cos(2 s π/2) = 1fors ≥ 2, the reciprocal polynomial x 2 s T 2 s (1/x) is a monic integral polynomial that has a zero at each of the 2 S numbers x k := sec((2 j − 1)π/2 s+1 ),1≤ j ≤ 2 s . Furthermore, every elementary symmetric function in 2 S variables evaluates to an even integer at (x 1 , ,x 2 S ).Now 2 s−1 j=1 sec 2m (2 j − 1)π 2 s+1 = 1 2 2 s j=1 sec 2m (2 j − 1)π 2 s+1 = 1 2 2 S j=1 x 2m j , and from the foregoing observations, this is an integer, as desired. Editorial comment. Unfortunately, due to an error by the editors, part (b ∗ ) of this prob- lem was incorrectly stated. It should have read, “Show that if n does not have the form of part (a), then there exists a positive integer m such that S(n, m) is not an integer.” Several solvers noticed that part (b ∗ ) as stated is false, with a simple counterexample being S(9, 1) = 16. The intended statement of part (b ∗ ) remains untreated. Also solved by D. R. Bridges, R. Chapman (U. K.), P. P. Dalyay (Hungary), Y. Dumont (France), J. W. From- meyer, J. Grivaux (France), G. Keselman, O. P. Lossers (The Netherlands), M. A. Prasad (India), A. Stadler (Switzerland), A. Stenger, R. Stong, M. Tetiva (Romania), A. Tissier (France), BSI Problems Group (Ger- many), GCHQ Problem Solving Group (U. K.), and the proposer. The Number Between 1 and n That is Least Prime 11218 [2006, 366]. Proposed by Gary Gordon, Lafayette College. Consider the follow- ing algorithm, which takes as input a positive integer n and proceeds by rounds, listing in each round certain positive integers between 1 and n inclusive, ultimately producing as output a positive integer f (n), the last number to be listed. In the 0th round, list 1. In the first round, list, in increasing order, all primes less than n. In the second round, list in increasing order all numbers that have not yet been listed and are of the form 2p,wherep is prime. Continue in this fashion, listing numbers of the form 3 p,4p, and so on until all numbers between 1 and n have been listed. Thus f (10) = 8 because the list eventually reaches the state (1, 2, 3, 5, 7, 4, 6, 10, 9, 8), while f (20) = 16 and f (30) = 27. (a)Find f (2006). (b) Describe the range of f . (c) Find lim n→∞ f (n)/n and lim n→∞ f (n)/n. Solution by Bruce S. Burdick, Roger Williams University, Bristol, RI. In describing the first round of the process, “less than n” should be “at most n”. April 2008] PROBLEMS AND SOLUTIONS 367 (b)LetR be the range of f ,andletS ={2 i 3 j : i is a nonnegative integer and j = 0 or 3 ·2 l−1 ≤ 3 j < 2 l+1 for some l ∈ N}. We show that R = S. Let g(1) = 0. For k ≥ 2, let g(k) = k/ p,where p is the largest prime divisor of k. When k ≤ n, the number k appears in round g(k). Thus f (n) is the largest element of [n] that maximize g. Let T ={k ∈ N : g(k) ≥ g(i) for 1 ≤ i < k}.Ifk ∈ T ,then f (k) = k; hence T ⊆ R.Furthermore, f (n) = k requires k ∈ T ,soR = T . To compare T and S,we show first that a number in T has no prime divisor outside {2, 3}.Ifg(k) = k/ p with p > 4, then there is a positive integer m such that 2k/ p < 2 m < k.Nowg(2 m ) = 2 m−1 > k/ p = g(k) and k /∈ T . Now consider the exponents. Note that T contains all powers of 2, since g(2 m ) = 2 m−1 > g(k) for k < 2 m .If2 i 3 j ∈ T with j = 0, then 3 j−1 ≥ 2 l−1 whenever 2 l < 3 j ; otherwise 2 i+l < 2 i 3 j and g(2 i 3 j )<g(2 i+l ). Letting 2 l be the largest power of 2 less than 3 j , we obtain 3 · 2 l−1 ≤ 3 j < 2 l+1 . Hence T ⊆ S. Finally, if 2 i 3 j ∈ S,then2 i 3 j is listed after all smaller numbers divisible by 3 and after all smaller powers of 2. Hence S ⊆ R, and we have R = S. The condition for 2 i 3 j ∈ S when j = 0 is that the second bit from the left in the binary expansion of 3 j is 1. It is also equivalent to the fractional part of j log 2 3being at least log 2 3. The first few values of j with this property are 1, 3, 5, 8, 10, 13, 15, 18, and 20. (a) f (2006) = 1944. To find f (2006) using (b), we create a list of candidates by taking powers of 2 and numbers of the form 2 i 3 j where j ∈{1, 3, 5} and 2 i is the largest power of 2 bounded by 2006/3 j . The candidates are 1024, 1536, 1728, and 1944. The largest, 1944, is f (2006). (c) lim n→∞ f (n)/n = 1. When n ∈ R,thevalueof f (n)/n is 1. There are infinitely many such n,and f (n) ≤ n for all n,so lim n→∞ f (n)/n = 1. lim n→∞ f (n)/n = 2/3. When n = 3 ·2 i −1, the value of f (n) is 2 i+1 ,so f (n)/n = 2 i+1 /(3 · 2 i − 1) = 2/(3 − 2 −i ). These numbers tend to 2/3. If 2 i+1 ≤ n ≤ 3 · 2 i − 1, then f (n) = 2 i+1 ,and f (n)/n > 2/3. Similarly, if 3 · 2 i ≤ n ≤ 2 i+2 − 1, then f (n) ≥ 3 · 2 i ,and f (n)/n > 3/4. Hence lim n→∞ f (n)/n = 2/3. Solved also by M. R. Avidon, D. Beckwith, P. Corn, D. Cranston, D. Fleischman, J. W. Frommeyer, J P. Gri- vaux (France), C. C. Heckman, E. A. Herman, M. Hildebrand, M. Huibregtse, G. Keselman, J. H. Lindsey II, O. P. Lossers (The Netherlands), D. Lovit, F. B. Miles, D. Opitz, W. Y. Pong, M. A. Prasad (India), A. Stadler (Switzerland), R. Staum, J. H. Steelman, R. Stong, M. Tetiva (Romania), L. Zhou, BSI Problem Solving Group (Germany), Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hungary), GCHQ Problem Solv- ing Group (U. K.), Manchester Problem Solving Group (U. K.), Microsoft Research Problems Group, and Northwestern University Problem Solving Group. A Romanian Olympiad Problem Generalized 11248 [2006, 760]. Proposed by P ´ al P ´ eter D ´ alyay, De ´ ak Ferenc High School, Szeged, Hungary. Let n be a positive integer, and let f be a continuous real-valued func- tion on [0, 1] with the property that 1 0 x k f (x) dx = 1for0≤ k ≤ n − 1. Prove that 1 0 ( f (x)) 2 dx ≥ n 2 . (The case n = 2, due to Ioan Ras¸a, appeared on the 55th Roma- nian Mathematical Olympiad, and the solution was published in Gazeta Matematic ˘ a (CIX) 5-6 (2004), page 227.) Solution by Jaime Vinuesa, University of Cantabria, Santander, Spain. The clas- sical Legendre polynomial P k is a polynomial of degree k given by 2 k k!P k (x) = 368 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 (d k /dx k )[(x 2 − 1) k ]. These polynomials satisfy P k (1) = 1and 1 −1 P j (x)P k (x) dx = ⎧ ⎪ ⎨ ⎪ ⎩ 0ifj = k, 2 2k + 1 if j = k. Let Q k (x) = √ 2k + 1 P k (2x − 1).Now{Q 0 , Q 1 , ,Q n−1 } is an orthonormal set in L 2 [0, 1]. Moreover, f, Q k = 1 0 f (x)Q k (x) dx = Q k (1) = √ 2k + 1, so Bessel’s inequality yields 1 0 ( f (x)) 2 dx =f 2 ≥ n−1 k=0 | f, Q k | 2 = n−1 k=0 (2k + 1) = n 2 . Editorial comment. As some contributors noted, the inequality is sharp, and it holds for all f ∈ L 2 [0, 1]. Some solvers observed that it suffices to construct a polynomial p of degree at most n − 1 satisfying 1 0 ( p(x)) 2 dx = 1and p(1) = n. The desired result then follows from the Schwarz inequality using p, f =p(1). Various devices—e.g., orthogonal polynomials, Lagrange interpolation, Hilbert matrices—were used in con- structing such a polynomial. Also solved by U. Abel (Germany), O. J. L. Alfonso (Colombia), S. Amghibech (Canada), K. F. Andersen (Canada), A. Bandeira & E. Dias (Portugal), M. Benito & ´ O. Benito & E. Fern ´ andez (Spain), M. W. Botsko, P. Bracken, D. R. Bridges, M. A. Carlton, R. Chapman (U. K.), A. Chaudhuri & R. Selukar, P. R. Chernoff, K. Dale (Norway), Y. Dumont (France), B. Dunn III, J. Fabrykowski & T. Smotzer, S. Foucart, J P. Grivaux (France), J. Groah, E. A. Herman, F. Holland (Ireland), E. J. Ionascu (Romania), M. E. H. Ismail, D. Jespersen, J. Kalis, G. Keselman, O. Kouba (Syria), A. V. Kumchev, K W. Lau (China), J. H. Lindsey II, O. P. Lossers (The Netherlands), R. Martin (U. K.), W. Matysiak (Poland), T. Nowak (Austria), M. A. Prasad (India), H. Ricardo, J. Rooin & M. Hassani (Iran), H J. Seiffert (Germany), J. G. Simmonds, N. C. Singer, A. Stadler (Switzerland), V. Stakhovsky, A. Stenger, R. Stong, J. Sun, M. Tetiva (Romania), S. Vagi, Z. V ¨ or ¨ os (Hun- gary), R. Whitley, L. Zhou, BSI Problems Group (Germany), Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hungary), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, and the proposer. A Symmetric Inequality 11255 [2006, 848]. Proposed by Slavko Simic, Mathematical Institute SANU, Bel- grade, Serbia. Let n be a positive integer, x 1 , ,x n be real numbers, and let p 1 , , p n be real numbers summing to 1. For k ≥ 1, let S k = n i=1 p i x k i − n i=1 p i x i k . Show that for m ≥ 1, S 2m S 2m+2 ≥ 1 − 1 m(2m + 1) S 2 2m+1 . Solution by O. P. Lossers, Eindhoven University of Technology, Eindhoven, The Netherlands. We assume that p 1 , , p n are nonnegative real numbers summing to 1, because if p = (1, −3, 3) and x = (1, 2, 3),thenS 2 S 4 <(2/3)S 2 3 and the in- equality fails. Define f : R 2 → R by f (x, t) = t 2 x 2m+2 (2m + 2)(2m + 1) + 2tx 2m+1 2m(2m + 1) + x 2m 2m(2m − 1) . Now ∂ 2 ∂x 2 f (x, t) = (tx m + x m−1 ) 2 ≥ 0, so f (x, t) is a convex function of x for all t. April 2008] PROBLEMS AND SOLUTIONS 369 From Jensen’s inequality it then follows that, for all t, n i=1 p i f (x i , t) − f n i=1 p i x i , t ≥ 0, or, equivalently, t 2 S 2m+2 (2m + 2)(2m + 1) + 2tS 2m+1 2m(2m + 1) + S 2m 2m(2m − 1) ≥ 0. The expression on the left, considered as a polynomial in t, must therefore have a nonpositive discriminant: 2S 2m+1 2m(2m + 1) 2 − 4 S 2m+2 (2m + 2)(2m + 1) · S 2m 2m(2m − 1) ≤ 0, and this is equivalent to the desired inequality. Editorial comment. The requirement that p i ≥ 0 was given by the proposer but lost by the editors. A more subtle oversight is the unstated requirement that m should be an integer. Byron Schmuland provided a counterexample with m = 3/2, x 1 = x 2 = 1, x 3 =−1, and p 1 = p 2 = p 3 = 1/3, in which case S 3 S 5 <(5/6)S 2 4 . Also solved by B. Schmuland (Canada), Microsoft Research Problems Group, and the proposer. Perimeter by Integral 11256 [2006, 848]. Proposed by Finbarr Holland, University College Cork, Cork, Ire- land. For complex a, b,andc,let f (x) = max{Re(ae ix ), Re(be ix ), Re(ce ix )}.Find 2π 0 f (x) dx. Solution by Michel Bataille, Rouen, France. Let I = 2π 0 f (x) dx. We show that I = |a − b|+|b − c|+|c − a|; this is the perimeter of the triangle formed by the three complex numbers. Let g(x) = min{Re(ae ix ), Re(be ix ), Re(ce ix )}.Now 2π 0 g(x) dx =− 2π 0 max{−Re(ae ix ), −Re(be ix ), −Re(ce ix )}dx =− 2π 0 max{Re(ae i(x+π) ), Re(be i(x+π) ), Re(ce i(x+π) )}dx =− 3π π max{Re(ae it ), Re(be it ), Re(ce it )}dt =−I, since f is 2π-periodic. Now max{u,v}= 1 2 (u + v +|u − v|) with u,v ∈ R and 2π 0 e ix dx = 0, so with the notation J a,b = 2π 0 max{Re(ae ix ), Re(be ix )}dx it follows that J a,b = 1 2 Re 2π 0 (ae ix + be ix ) dx + 1 2 2π 0 |Re(ae ix ) − Re(be ix )|dx = 1 2 2π 0 |Re(ae ix ) − Re(be ix )|dx. 370 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 Now let a = α +iα and b = β + iβ where α, α , β,andβ are real numbers. Then Re(ae ix ) − Re(be ix ) = (α − β)cos x − (α − β ) sin x = A sin(x + φ), where A = ((α − β) 2 + (α − β ) 2 ) 1/2 =|a − b|,sinφ = (α − β)/A,andcosφ = −(α − β )/A. Thus, 2π 0 [Re(ae ix ) − Re(be ix )|dx =|a − b| 2π 0 |sin(x + φ)|dx =|a − b| 2π 0 |sin x|dx = 4|a − b|, and J a,b = 2|a − b|. Similarly, with K a,b = 2π 0 min{Re(ae ix ), Re(be ix )}dx,wehave K a,b =−2|a − b|. Finally, for real numbers u, v,andw,wehave max{u,v,w}−min{u,v,w}= 1 2 (max{u,v}+max{v, w}+max{w, u} − min{u,v}−min{v, w}−min{w, u}), so 2I = 2π 0 ( f (x) − g(x)) dx = 1 2 (J a,b + J b,c + J c,a − K a,b − K b,c − K c,a ) = 1 2 2|a − b|+2|b − c|+2|c − a|+2|a − b|+2|b − c|+2|c − a| = 2|a − b|+2|b − c|+2|c −a|. This is the stated result. Also solved by S. Amghibech (Canada), M. R. Avidon, D. R. Bridges, B. S. Burdick, R. Chapman (U. K.), A. Demis (Greece), J P. Grivaux (France), E. A. Herman, F. Hjouj, G. Keselman, J. H. Lindsey II, M. D. Meyerson, J. G. Simmonds, V. Stakhovsky, R. Stong, R. Tauraso (Italy), E. I. Verriest, J. Vinuesa (Spain), H. Widmer (Switzerland), J. H. Zacharias, L. Zhou, Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hun- gary), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, and the proposer. A Choice Sum 11263 [2006, 940]. Proposed by Gregory Keselman, Oak Park, MI, and formerly of Lvov Polytechnic Institute, Ukraine. Show that when n is a positive integer and a is real, n/2 k=0 (−1) k a k n − k k = ⎧ ⎪ ⎨ ⎪ ⎩ (1+ √ 1−4a) n+1 −(1− √ 1−4a) n+1 2 n+1 √ 1−4a , if a < 1/4; (n + 1)/2 n , if a = 1/4; a n/2 cos(nβ) +sin(nβ)/ √ 4a − 1 , if a > 1/4. Here, β denotes arcsin √ 1 − 1/(4a). Solution by E. Leonard, University of Alberta, Edmonton, AL, Canada. For each non- negative integer n,define b n = n/2 k=0 (−1) k a k n − k k = n k=0 (−1) k a k n − k k . April 2008] PROBLEMS AND SOLUTIONS 371 Using m+1 r = m r + m r−1 and m r = 0ifr > m, we see that the sequence b n satisfies b 0 = 1, b 1 = 1, b n+1 = b n − ab n−1 , n ≥ 1. This linear recurrence has characteristic equation λ 2 − λ + a = 0, which has roots: λ = 1 2 ± 1 2 √ 1 −4a if a < 1 4 , λ = 1 2 if a = 1 4 , λ = 1 2 ± i 1 2 √ 4a − 1ifa > 1 4 . (i) For a < 1/4 there are two distinct roots, so the general solution of the recurrence is b n = A 1 + √ 1 − 4a 2 n + B 1 − √ 1 − 4a 2 n . We now compute A and B from the initial conditions b 0 = b 1 = 1toarriveat b n = 1 + √ 1 − 4a n+1 − 1 − √ 1 − 4a n+1 2 n+1 √ 1 −4a . (ii) For a = 1/4 there is a double root, so the general solution of the recurrence is b n = A 2 n + Bn 2 n . Again using the initial conditions to determine A and B, we obtain b n = n + 1 2 n . (iii) For a > 1/4, there are two distinct complex roots, which may be written in polar form as √ a cos β +i sin β = √ ae iβ , and √ a cos β −i sin β = √ ae −iβ , where β = arcsin √ 1 − 1/(4a). The general solution of the recurrence is b n = Aa n/2 e inβ + Ba n/2 e −inβ . Applying the initial conditions to evaluate A and B, and using sin β = √ 4a − 1 2 √ a , cos β = 1 2 √ a , we get b n = a n/2 cos nβ + a n/2 √ 4a − 1 sin nβ. Also solved by U. Abel (Germany), S. Amghibech (Canada), M. R. Avidon, M. Bataille (France), D. Beckwith, J. Borwein (Canada), P. Bracken, R. Chapman (U.K.), P. P. D ´ alyay (Hungary), P. De (Ireland), G.C. Greubel, J P. Grivaux (France), H. Kwong, G. Lamb, O. P. Lossers (The Netherlands), J. Minkus, C. R. Pranesachar (India), T. L. & V. R ˆ adulescu (Romania), H. Roelants (Belgium), J. N. Senadheera, A. Stadler (Switzerland), R. Stong, M. Tetiva (Romania), H. Widmer (Switzerland), C. Zhou (Canada), BSI Problems Group (Germany), GCHQ Problem Solving Group (U.K.), Microsoft Research Problems Group, and the proposer. 372 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 . should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before August 31, 2008. Additional information, such as. University of Cantabria, Santander, Spain. The clas- sical Legendre polynomial P k is a polynomial of degree k given by 2 k k!P k (x) = 368 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 (d k /dx k )[(x 2 −. form of part (a), then S(n, m) is not an integer. 366 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 Solution to (a) by NSA Problems Group, Fort Meade, MD. Let n + 1 = 2 s with s ≥ 2. We