Solutions to the 71st William Lowell Putnam Mathematical Competition Saturday, December 4, 2010

Transcription

1 Solutions to the 71st William Lowell Putnam Mathematical Competition Saturday, December 4, 21 Kiran Kedlaya and Lenny Ng A1 The largest such k is n+1 2 = n 2. For n even, this value is achieved by the partition {1,n},{2,n 1},...; fornodd, it is achieved by the partition {n},{1,n 1},{2,n 2},... One way to see that this is optimal is to note that the common sum can never be less than n, since n itself belongs to one of the boxes. This implies thatk 1+ + n)/n = n + 1)/2. Another argument is that if k > n+1)/2, then there would have to be two boxes with one number each by the pigeonhole principle), but such boxes could not have the same sum. Remark. A much subtler question would be to find the smallest k as a function of n) for which no such arrangement exists. A2 The only such functions are those of the form = cx + d for some real numbers c, d for which the property is obviously satisfied). To see this, suppose that f has the desired property. Then for any x R, 2f x) = fx+2) = fx+2) fx+1))+fx+1) ) = f x+1)+f x). Consequently,f x+1) = f x). Define the functiong : R R by gx) = fx+1), and put c = g), d = f). For all x R, g x) = f x+1) f x) =, sogx) = c identically, and f x) = fx+1) = gx) = c, so = cx + d identically as desired. A3 If a = b =, then the desired result holds trivially, so we assume that at least one of a,b is nonzero. Pick any point a,b ) R 2, and let L be the line given by the parametric equation Lt) = a,b ) + a,b)t for t R. By the chain rule and the given equation, we have d dt h L) = h L. If we writef = h L : R R, then f t) = ft) for all t. It follows that ft) = Ce t for some constant C. Since ft) M for all t, we must have C =. It follows that ha,b ) = ; since a,b ) was an arbitrary point, h is identically over all ofr 2. A4 Put N = 1 11n +1 1n +1 n 1. Write n = 2 m k with m a nonnegative integer and k a positive odd integer. For any nonnegative integer j, 1 2mj 1) j mod 1 2m +1). Since 1 n n 2 m m+1, 1 n is divisible by 2 n and hence by 2 m+1, and similarly 1 1n is divisible by 2 1n and hence by2 m+1. It follows that N )+ 1) mod 1 2m +1). Since N 1 1n > 1 n m + 1, it follows that N is composite. A5 We start with three lemmas. Lemma 1. If x, y G are nonzero orthogonal vectors, then x x is parallel to y. Proof. Put z = x y, so that x,y, and z = x y are nonzero and mutually orthogonal. Then w = x z, so w = x zis nonzero and orthogonal to x andz. However, if x x) y, thenw = x x y) = x x) y = x x) y is also orthogonal toy, a Lemma 2. If x G is nonzero, and there exists y G nonzero and orthogonal to x, then x x =. Proof. Lemma 1 implies that x x is parallel to both y and x y, so it must be zero. Lemma 3. If x,y G commute, thenx y =. Proof. Ifx y, thenx y = x y = y x = y x, sox y y x. We proceed now to the proof. Assume by way of contradiction that there exista,b G witha b. Put c = a b = a b, so that a,b,c are nonzero and linearly independent. Let e be the identity element of G. Since e commutes with a,b,c, by Lemma 3 we have e a = e b = e c =. Since a,b,c span R 3, e x = for all x R 3, soe =. Sinceb,c, andb c = b c are nonzero and mutually orthogonal, Lemma 2 implies b b = c c = b c) b c) = = e. Hence b c = c b, contradicting Lemma 3 because b c. The desired result follows.

2 2 A6 First solution. Note that the hypotheses on f imply that > for all x [, + ), so the integrand is a continuous function of f and the integral makes sense. Rewrite the integral as 1 fx+1) ), and suppose by way of contradiction that it converges to a finite limit L. For n, define the Lebesgue measurable set I n = {x [,1] : 1 fx+n+1) fx+n) 1/2}. Then L n= µi n)), so the latter sum converges. In particular, there exists a nonnegative integer N for which n=n 1 µi n)) < 1; the intersection I = n=n I n = [,1] [,1] I n ) n=n then has positive Lebesgue measure. By Taylor s theorem with remainder, for t [, 1/2], { } log1 t) t+t 2 1 sup t [,1/2] 1 t) 2 = t+ 4 3 t2 5 3 t. For each nonnegative integer n N, we then have n L 1 fx+1) ) N n 1 = 1 fx+i+1) ) i=n fx+i) n 1 1 fx+i+1) ) i=n I fx+i) 3 n 1 fx+i) log 5 i=n I fx+i+1) = 3 n 1 ) fx+i) log 5 I fx+i+1) i=n = 3 log fx+n) 5 fx+n). I For each x I, logfx + N)/fx + n) is a strictly increasing unbounded function of n. By the monotone convergence theorem, the integral I logfx + N)/fx + n)) grows without bound as n +, a Thus the original integral diverges, as desired. Remark. This solution is motivated by the commonlyused fact that an infinite product 1 + x 1 )1 + x 2 ) converges absolutely if and only if the sumx 1 +x 2 + b a converges absolutely. The additional measure-theoretic argument at the beginning is needed because one cannot bound log1 t) by a fixed multiple of t uniformly for all t [,1). Second solution. Communicated by Paul Allen.) Let b > a be nonnegative integers. Then fx+1) = = = b 1 b 1 b 1 fx+k) fx+k +1) fx+k) fx+k) fx+k +1) fx+k) fx+k) fx+k +1) fx+a) fx+a) fx+b). fx+a) Now since, given a, we can choose an integer la) > a for which fla)) < fa + 1)/2; then fx+a) fx+la)) fx+a) 1 fla)) fa+1) > 1/2 for allx [,1]. Thus if we define a sequence of integersa n by a =, a n+1 = la n ), then fx+1) = > an+1 n= a n n= and the final sum clearly diverges. fx+1) 1/2), Third solution. By Joshua Rosenberg, communicated by Catalin Zara.) If the original integral converges, then on one hand the integrand fx+1))/ = 1 fx+1)/ cannot tend to 1 asx. On the other hand, for anya, < fa+1) fa) a+1 < 1 fa) a = 1 fx+1)) fa) a fx+1), a and the last expression tends to asa. Hence by the squeeze theorem,fa+1)/fa) asa, a B1 First solution. No such sequence exists. If it did, then the Cauchy-Schwartz inequality would imply 8 = a 2 1 +a )a 4 1 +a ) a 3 1 +a3 2 + )2 = 9,

3 3 Second solution. Communicated by Catalin Zara.) Suppose that such a sequence exists. If a 2 k [,1] for allk, thena 4 k a2 k for all k, and so 4 = a 4 1 +a a 2 1 +a = 2, There thus exists a positive integerk for which a 2 k 1. However, in this case, for m large, a 2m k > 2m and so a 2m 1 +a 2m 2 + = 2m. Third solution. We generalize the second solution to show that for any positive integerk, it is impossible for a sequence a 1,a 2,... of complex numbers to satisfy the given conditions in case the seriesa k 1 +ak 2 + converges absolutely. This includes the original problem by taking k = 2, in which case the series a a2 2 + consists of nonnegative real numbers and so converges absolutely if it converges at all. Since the sum a i k converges by hypothesis, we can find a positive integer n such that i=n+1 a i k < 1. For each positive integerd, we then have n kd a kd i a i kd < 1. i=n+1 We thus cannot have a 1,..., a n 1, or else the sum n akd i would be bounded in absolute value by n independently of d. But if we put r = max{ a 1,..., a n } > 1, we obtain another contradiction because for any ǫ >, n lim supr ǫ) kd a kd i d >. For instance, this follows from applying the root test to the rational function n 1 1 a k i z = n ) a kd i z d, d= which has a pole within the circle z r 1/k. An elementary proof is also possible.) Fourth solution. Communicated by Noam Elkies.) Since k a2 k = 2, for each positive integer k we have a 2 k 2 and so a4 k 2a2 k, with equality only for a 2 k {,2}. Thus to have k a4 k = 4, there must be a single index k for which a 2 k = 2, and the other a k must all equal. But then k a2m k = 2 m 2m for any positive integerm > 2. Remark. Manjul Bhargava points out it is easy to construct sequences of complex numbers with the desired property if we drop the condition of absolute convergence. Here is an inductive construction of which several variants are possible). Forn = 1,2,... andz C, define the finite sequence ) 1 s n,z = z e2πij/n : j =,...,n 1. This sequence has the property that for any positive integer j, the sum of the j-th powers of the terms of s n,z equals1/z j ifj is divisible byn and otherwise. Moreover, any partial sum of j-th powers is bounded in absolute value byn/ z j. The desired sequence will be constructed as follows. Suppose that we have a finite sequence which has the correct sum of j-th powers for j = 1,...,m. For instance, for m = 1, we may start with the singleton sequence 1.) We may then extend it to a new sequence which has the correct sum of j-th powers for j = 1,...,m+1, by appendingk copies ofs m+1,z for suitable choices of a positive integer k and a complex numberz with z < m 2. This last restriction ensures that the resulting infinite sequence a 1,a 2,... is such that for each positive integerm, the seriesa m 1 +a m 2 + is convergent though not absolutely convergent). Its partial sums include a subsequence equal to the constant valuem, so the sum of the series must equalm as desired. B2 The smallest distance is 3, achieved bya =,),B = 3,), C =,4). To check this, it suffices to check thatab cannot equal 1 or 2. It cannot equal because if two of the points were to coincide, the three points would be collinear.) The triangle inequality implies that AC BC AB, with equality if and only if A,B,C are collinear. If AB = 1, we may assume without loss of generality that A =,), B = 1,). To avoid collinearity, we must have AC = BC, but this forces C = 1/2, y) for some y R, a One can also treat this case by scaling by a factor of 2 to reduce to the case AB = 2, treated in the next paragraph.) IfAB = 2, then we may assume without loss of generality that A =,),B = 2,). The triangle inequality implies AC BC {,1}. Also, forc = x,y), AC 2 = x 2 +y 2 andbc 2 = 2 x) 2 +y 2 have the same parity; it follows that AC = BC. Hence c = 1, y) for some y R, so y 2 and y 2 +1 = BC 2 are consecutive perfect squares. This can only happen for y =, but then A, B, C are collinear, a contradiction again. Remark. Manjul Bhargava points out that more generally, a Heronian triangle a triangle with integer sides and rational area) cannot have a side of length 1 or 2 and again it is enough to treat the case of length 2). The original problem follows from this because a triangle whose vertices have integer coordinates has area equal to half an integer by Pick s formula or the explicit formula for the area as a determinant). B3 It is possible if and only if n 15. Since = = , for n 14, we can start with an initial distribution in which each boxb i starts with at mosti 1 balls so

4 4 in particularb 1 is empty). From such a distribution, no moves are possible, so we cannot reach the desired final distribution. Suppose now that n 15. By the pigeonhole principle, at any time, there exists at least one index i for which the box B i contains at least i balls. We will describe any such index as being eligible. The following sequence of operations then has the desired effect. a) Find the largest eligible indexi. Ifi = 1, proceed to b). Otherwise, move i balls from B i to B 1, then repeat a). b) At this point, only the index i = 1 can be eligible so it must be). Find the largest index j for which B j is nonempty. If j = 1, proceed to c). Otherwise, move 1 ball from B 1 to B j ; in case this makesj eligible, movej balls fromb j tob 1. Then repeat b). c) At this point, all of the balls are in B 1. For i = 2,...,21, move one ball fromb 1 tob i n times. After these operations, we have the desired distribution. B4 First solution. The pairs p, q) satisfying the given equation are those of the form px) = ax+b,qx) = cx+d fora,b,c,d R such thatbc ad = 1. We will see later that these indeed give solutions. Suppose p and q satisfy the given equation; note that neither p nor q can be identically zero. By subtracting the equations px)qx+1) px+1)qx) = 1 px 1)qx) px)qx 1) = 1, we obtain the equation px)qx+1)+qx 1)) = qx)px+1)+px 1)). The original equation implies that px) and qx) have no common nonconstant factor, so px) divides px + 1)+px 1). Since each ofpx+1) andpx 1) has the same degree and leading coefficient as p, we must have px+1)+px 1) = 2px). If we define the polynomialsrx) = px+1) px), sx) = qx+1) qx), we haverx+1) = rx), and similarlysx+1) = sx). Put a = r),b = p),c = s),d = q). Then rx) = a,sx) = c for all x Z, and hence identically; consequently, px) = ax+b, qx) = cx+d for all x Z, and hence identically. For p andq of this form, px)qx+1) px+1)qx) = bc ad, so we get a solution if and only if bc ad = 1, as claimed. Second solution. Communicated by Catalin Zara.) Again, note thatpandq must be nonzero. Write px) = p +p 1 x+ +p m x m qx) = q +q 1 x+ +q n x n with p m,q n, so that m = degp),n = degq). It is enough to derive a contradiction assuming that max{m,n} > 1, the remaining cases being treated as in the first solution. Put Rx) = px)qx+1) px+1)qx). Since m+n 2 by assumption, the coefficient of x m+n 1 in Rx) must vanish. By easy algebra, this coefficient equals m n)p m q n, so we must havem = n > 1. Fork = 1,...,2m 2, the coefficient ofx k inrx) is ) )) j i p i q j p j q i ) k i k j i+j>k,j>i and must vanish. For k = 2m 2, the only summand is fori,j) = m 1,m), so p m 1 q m = p m q m 1. Suppose now that h 1 and that p i q j = p j q i is known to vanish whenever j > i h. By the previous paragraph, we initially have this forh = m 1.) Take k = m + h 2 and note that the conditions i+j > h,j m forcei h 1. Using the hypothesis, we see that the only possible nonzero contribution to the coefficient of x k in Rx) is from i,j) = h 1,m). Hence p h 1 q m = p m q h 1 ; since p m,q m, this impliesp h 1 q j = p j q h 1 wheneverj > h 1. By descending induction, we deduce that p i q j = p j q i whenever j > i. Consequently, px) and qx) are scalar multiples of each other, forcing Rx) =, a Third solution. Communicated by David Feldman.) As in the second solution, we note that there are no solutions where m = degp),n = degq) are distinct and m + n 2. Suppose p,q form a solution with m = n 2. The desired identity asserts that the matrix ) px) px+1) qx) qx+1) has determinant 1. This condition is preserved by replacingqx) withqx) tpx) for any real numbert. In particular, we can choosetso thatdegqx) tpx)) < m; we then obtain a B5 First solution. The answer is no. Suppose otherwise. For the condition to make sense, f must be differentiable. Since f is strictly increasing, we must have f x) for all x. Also, the function f x) is strictly increasing: if y > x then f y) = ffy)) > f) = f x). In particular, f y) > for all y R.

5 5 For any x, if fx ) = b and f x ) = a >, then f x) > a for x > x and thus ax x )+b for x x. Then either b < x or a = f x ) = ffx )) = fb) ab x ) + b. In the latter case, b ax +1)/a+1) x +1. We conclude in either case thatfx ) x +1 for all x 1. It must then be the case that f) = f x) 1 for all x, since otherwise > x + 1 for large x. Now by the above reasoning, iff) = b andf ) = a >, then > a x + b for x >. Thus for x > max{, b /a }, we have > and f) > a x + b. But then f) > 1 for sufficiently large x, a Second solution. Communicated by Catalin Zara.) Suppose such a function exists. Since f is strictly increasing and differentiable, so is f f = f. In particular, f is twice differentiable; also, f x) = f )f x) is the product of two strictly increasing nonnegative functions, so it is also strictly increasing and nonnegative. In particular, we can choose α > andm R such thatf x) > 4α for allx M. Then for all x M, fm)+f M)x M)+2αx M) 2. In particular, for some M > M, we have αx 2 for all x M. Pick T > so that αt 2 > M. Then for x T, > M and so f x) = f) α 2. Now 1 ft) 1 2T f2t) = T f t) 2T ft) 2 dt αdt; T however, as T, the left side of this inequality tends to while the right side tends to +, a Third solution. Communicated by Noam Elkies.) Since f is strictly increasing, for some y, we can define the inverse function gy) of f for y y. Then x = g), and we may differentiate to find that 1 = g )f x) = g )f). It follows that g y) = 1/fy) for y y ; since g takes arbitrarily large values, the integral y dy/fy) must diverge. One then gets a contradiction from any reasonable lower bound on fy) for y large, e.g., the bound αx 2 from the second solution. One can also start with a linear lower bound βx, then use the integral expression for g to deduce that gx) γ log x, which in turn forces to grow exponentially.) B6 For any polynomial px), let [px)]a denote the n n matrix obtained by replacing each entry A ij of A by pa ij ); thus A [k] = [x k ]A. Let Px) = x n + a n 1 x n a denote the characteristic polynomial ofa. By the Cayley-Hamilton theorem, = A PA) = A n+1 +a n 1 A n + +a A = A [n+1] +a n 1 A [n] + +a A [1] = [xpx)]a. Thus each entry ofais a root of the polynomialxpx). Now supposem n+1. Then = [x m+1 n px)]a = A [m+1] +a n 1 A [m] + +a A [m+1 n] since each entry of A is a root of x m+1 n px). On the other hand, = A m+1 n PA) = A m+1 +a n 1 A m + +a A m+1 n. Therefore if A k = A [k] for m + 1 n k m, then A m+1 = A [m+1]. The desired result follows by induction onm. Remark. David Feldman points out that the result is best possible in the following sense: there exist examples of n n matrices A for which A k = A [k] for k = 1,...,n but A n+1 A [n+1].