The 66th William Lowell Putnam Mathematical Competition Saturday, December 3, 2005

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1 The 66th William Lowell Putnam Mathematical Competition Saturday, December 3, 5 A Show that every positive integer is a sum of one or more numbers of the form r 3 s, where r and s are nonnegative integers and no summand divides another. (For example, A Let S {(a, b a,,..., n, b,, 3}. A rook tour of S is a polygonal path made up of line segments connecting points p, p,..., p 3n in sequence such that (i p i S, (ii p i and p i+ are a unit distance apart, for i < 3n, (iii for each p S there is a unique i such that p i p. How many rook tours are there that begin at (, and end at (n,? (An example of such a rook tour for n 5 was depicted in the original. A3 Let p(z be a polynomial of degree n, all of whose zeros have absolute value in the complex plane. Put g(z p(z/z n/. Show that all zeros of g (z have absolute value. A4 Let H be an n n matrix all of whose entries are ± and whose rows are mutually orthogonal. Suppose H has an a b submatrix whose entries are all. Show that ab n. A5 Evaluate ln(x + x + dx. A6 Let n be given, n 4, and suppose that P, P,..., P n are n randomly, independently and uniformly, chosen points on a circle. Consider the convex n-gon whose vertices are P i. What is the probability that at least one of the vertex angles of this polygon is acute? B Find a nonzero polynomial P (x, y such that P ( a, a for all real numbers a. (Note: ν is the greatest integer less than or equal to ν. B Find all positive integers n, k,..., k n such that k + + k n 5n 4 and + +. k k n B3 Find all differentiable functions f : (, (, for which there is a positive real number a such that ( f a x x f(x for all x >. B4 For positive integers m and n, let f(m, n denote the number of n-tuples (x, x,..., x n of integers such that x + x + + x n m. Show that f(m, n f(n, m. B5 Let P (x,..., x n denote a polynomial with real coefficients in the variables x,..., x n, and suppose that ( x + + x P (x,..., x n n and that x + + x n divides P (x,..., x n. Show that P identically. (identically B6 Let S n denote the set of all permutations of the numbers,,..., n. For π S n, let σ(π if π is an even permutation and σ(π if π is an odd permutation. Also, let ν(π denote the number of fixed points of π. Show that σ(π ν(π + n ( n+ n +.

2 Solutions to the 66th William Lowell Putnam Mathematical Competition Saturday, December 3, 5 Manjul Bhargava, Kiran Kedlaya, and Lenny Ng A We proceed by induction, with base case 3. Suppose all integers less than n can be represented. If n is even, then we can take a representation of n/ and multiply each term by to obtain a representation of n. If n is odd, put m log 3 n, so that 3 m n < 3 m+. If 3 m n, we are done. Otherwise, choose a representation (n 3 m / s + +s k in the desired form. Then n 3 m + s + + s k, and clearly none of the s i divide each other or 3 m. Moreover, since s i n 3 m < 3 m+ 3 m, we have s i < 3 m, so 3 m cannot divide s i either. Thus n has a representation of the desired form in all cases, completing the induction. Remarks: This problem is originally due to Paul Erdős. Note that the representations need not be unique: for instance, A We will assume n hereafter, since the answer is for n. First solution: We show that the set of rook tours from (, to (n, is in bijection with the set of subsets of {,,..., n} that include n and contain an even number of elements in total. Since the latter set evidently contains n elements, so does the former. We now construct the bijection. Given a rook tour P from (, to (n,, let S S(P denote the set of all i {,,..., n} for which there is either a directed edge from (i, to (i, or from (i, 3 to (i,. It is clear that this set S includes n and must contain an even number of elements. Conversely, given a subset S {a, a,..., a r n} {,,..., n} of this type with a < a < < a r, we notice that there is a unique path P containing (a i, + ( i, (a, for i,,..., r. This establishes the desired bijection. Second solution: Let A n denote the set of rook tours beginning at (, and ending at (n,, and let B n denote the set of rook tours beginning at (, and ending at (n, 3. For n, we construct a bijection between A n and A n B n. Any path P in A n contains either the line segment P between (n, and (n,, or the line segment P between (n, and (n,. In the former case, P must also contain the subpath P which joins (n, 3, (n, 3, (n,, and (n, consecutively; then deleting P and P from P and adding the line segment joining (n, 3 to (n, results in a path in A n. (This construction is reversible, lengthening any path in A n to a path in A n. In the latter case, P contains the subpath P which joins (n, 3, (n, 3, (n,, (n, consecutively; deleting P results in a path in B n, and this construction is also reversible. The desired bijection follows. Similarly, there is a bijection between B n and A n B n for n. It follows by induction that for n, A n B n n ( A + B. But A and B, and hence the desired answer is A n n. Remarks: Other bijective arguments are possible: for instance, Noam Elkies points out that each element of A n B n contains a different one of the possible sets of segments of the form (i,, (i +, for i,..., n. Richard Stanley provides the reference: K.L. Collins and L.B. Krompart, The number of Hamiltonian paths in a rectangular grid, Discrete Math. 69 (997, This problem is Theorem of that paper; the cases of 4 n and 5 n grids are also treated. The paper can also be found online at the URL kcollins.web.wesleyan.edu/vita.htm. A3 Note that it is implicit in the problem that p is nonconstant, one may take any branch of the square root, and that z should be ignored. First solution: Write p(z c n j (z r j, so that g (z g(z z n j Now if z r j for all j,then z + r j z r j. z + r j (z + r j(z r j z r j z r j z + Im(zr j z r j, and so Re zg (z g(z z j z r j. Since the quantity in parentheses is positive, g (z/g(z can be only if z. If on the other hand z r j for some j, then z anyway. Second solution: Write p(z c n j (z r j, so that g (z g(z n j (. z r j z

3 We first check that g (z whenever z is real and z >. In this case, for r j e iθj, we have z r j (z cos(θ j + sin(θ j i, so the real part of z r j z is z cos(θ j z z cos(θ j + z z z(z z cos(θ j + >. Hence g (z/g(z has positive real part, so g (z/g(z and hence g(z are nonzero. Applying the same argument after replacing p(z by p(e iθ z, we deduce that g cannot have any roots outside the unit circle. Applying the same argument after replacing p(z by z n p(/z, we also deduce that g cannot have any roots inside the unit circle. Hence all roots of g have absolute value, as desired. Third solution: Write p(z c n j (z r j and put r j e iθj. Note that g(e iθ is equal to a nonzero constant times h(θ n e i(θ+θj e i(θ+θj j i n sin(θ + θ j. j Since h has at least n roots (counting multiplicity in the interval [, π, h does also by repeated application of Rolle s theorem. Since g (e iθ ie iθ h (θ, g (z has at least n roots on the unit circle. Since g (z is equal to z n times a polynomial of degree n, g (z has all roots on the unit circle, as then does g (z. Remarks: The second solution imitates the proof of the Gauss-Lucas theorem: the roots of the derivative of a complex polynomial lie in the convex hull of the roots of the original polynomial. The second solution is close to problem B3 from the Putnam. A hybrid between the first and third solutions is to check that on the unit circle, Re(zg (z/g(z while between any two roots of p, Im(zg (z/g(z runs from + to and so must have a zero crossing. (This only works when p has distinct roots, but the general case follows by the continuity of the roots of a polynomial as functions of the coefficients. One can also construct a solution using Rouché s theorem. A4 First solution: Choose a set of a rows r,..., r a containing an a b submatrix whose entries are all. Then for i, j {,..., a}, we have r i r j n if i j and otherwise. Hence a r i r j an. i,j On the other hand, the term on the left is the dot product of r + +r a with itself, i.e., its squared length. Since this vector has a in each of its first b coordinates, the dot product is at least a b. Hence an a b, whence n ab as desired. Second solution: (by Richard Stanley Suppose without loss of generality that the a b submatrix occupies the first a rows and the first b columns. Let M be the submatrix occupying the first a rows and the last n b columns. Then the hypothesis implies that the matrix MM T has n b s on the main diagonal and b s elsewhere. Hence the column vector v of length a consisting of all s satisfies MM T v (n abv, so n ab is an eigenvalue of MM T. But MM T is semidefinite, so its eigenvalues are all nonnegative real numbers. Hence n ab. Remarks: A matrix as in the problem is called a Hadamard matrix, because it meets the equality condition of Hadamard s inequality: any n n matrix with ± entries has absolute determinant at most n n/, with equality if and only if the rows are mutually orthogonal (from the interpretation of the determinant as the volume of a paralellepiped whose edges are parallel to the row vectors. Note that this implies that the columns are also mutually orthogonal. A generalization of this problem, with a similar proof, is known as Lindsey s lemma: the sum of the entries in any a b submatrix of a Hadamard matrix is at most abn. Stanley notes that Ryser (98 asked for the smallest size of a Hadamard matrix containing an r s submatrix of all s, and refers to the URL www3.interscience.wiley.com/cgi-bin/ abstract/5586/abstract for more information. A5 First solution: We make the substitution x tan θ, rewriting the desired integral as Write π/4 log(tan(θ + dθ. log(tan(θ + log(sin(θ + cos(θ log(cos(θ and then note that sin(θ + cos(θ cos(π/4 θ. We may thus rewrite the integrand as log( + log(cos(π/4 θ log(cos(θ. But over the interval [, π/4], the integrals of log(cos(θ and log(cos(π/4 θ are equal, so their contributions cancel out. The desired integral is then just the integral of log( over the interval [, π/4], which is π log(/8. Second solution: (by Roger Nelsen Let I denote the desired integral. We make the substitution x (

4 3 u/( + u to obtain I yielding log( ( + u log(/( + u ( + u log( log( + u + u du du + u I, du ( + u I log( du + u π log(. 8 Third solution: (attributed to Steven Sivek Define the function f(t log(xt + x + so that f( and the desired integral is f(. Then by differentiation under the integral, f (t By partial fractions, we obtain dx x (xt + (x + dx. f (t t arctan(x log(tx + + log(x + (t + f(t whence πt + log( 4 log(t + 4(t, + log( arctan(t and hence f( π log( 4 x x + π log(t + t log(t + 8 t dt + log(t + t dt. + But the integral on the right is again the desired integral f(, so we may move it to the left to obtain f( π log( 4 and hence f( π log(/8 as desired. Fourth solution: (by David Rusin We have ( log(x + ( n x n x dx + n(x dx. + n We next justify moving the sum through the integral sign. Note that n ( n x n dx n(x + is an alternating series whose terms strictly decrease to zero, so it converges. Moreover, its partial sums alternately bound the previous integral above and below, so the sum of the series coincides with the integral. Put J n x n dx x + ; then J arctan( π 4 and J log(. Moreover, Write then J n + J n+ A m B m m i x n dx n +. ( i i m ( i ; i i J n ( n (J A n J n+ ( n (J B n. Now the N-th partial sum of our series equals N n J n n J n n N n ( n n (J B n ( n n (J A n A N (J B N + B N (J A N + A N B N. As N, A N J and B N J, so the sum tends to J J π log(/8. Remarks: The first two solutions are related by the fact that if x tan(θ, then x/( + x tan(π/4 θ. The strategy of the third solution (introducing a parameter then differentiating it was a favorite of physics Nobelist (and Putnam Fellow Richard Feynman. Noam Elkies notes that this integral is number.49#8 in Gradshteyn and Ryzhik, Table of integrals, series, and products. The Mathematica computer algebra system (version 5. successfully computes this integral, but we do not know how. A6 First solution: The angle at a vertex P is acute if and only if all of the other points lie on an open semicircle. We first deduce from this that if there are any two acute angles at all, they must occur consecutively. Suppose the contrary; label the vertices Q,..., Q n in counterclockwise order (starting anywhere, and suppose that the angles at Q and Q i are acute for some i with 3 i n. Then the open semicircle starting

5 4 at Q and proceeding counterclockwise must contain all of Q 3,..., Q n, while the open semicircle starting at Q i and proceeding counterclockwise must contain Q i+,..., Q n, Q,..., Q i. Thus two open semicircles cover the entire circle, contradiction. It follows that if the polygon has at least one acute angle, then it has either one acute angle or two acute angles occurring consecutively. In particular, there is a unique pair of consecutive vertices Q, Q in counterclockwise order for which Q is acute and Q is not acute. Then the remaining points all lie in the arc from the antipode of Q to Q, but Q cannot lie in the arc, and the remaining points cannot all lie in the arc from the antipode of Q to the antipode of Q. Given the choice of Q, Q, let x be the measure of the counterclockwise arc from Q to Q ; then the probability that the other points fall into position is n+ x n if x / and otherwise. Hence the probability that the polygon has at least one acute angle with a given choice of which two points will act as Q and Q is / ( n+ x n dx n n n+. Since there are n(n choices for which two points act as Q and Q, the probability of at least one acute angle is n(n n+. Second solution: (by Calvin Lin As in the first solution, we may compute the probability that for a particular one of the points Q, the angle at Q is not acute but the following angle is, and then multiply by n. Imagine picking the points by first choosing Q, then picking n pairs of antipodal points and then picking one member of each pair. Let R,..., R n be the points of the pairs which lie in the semicircle, taken in order away from Q, and let S,..., S n be the antipodes of these. Then to get the desired situation, we must choose from the pairs to end up with all but one of the S i, and we cannot take R n and the other S i or else Q will be acute. That gives us (n good choices out of n ; since we could have chosen Q to be any of the n points, the probability is again n(n n+. B Take P (x, y (y x(y x. To see that this works, first note that if m a, then m is an integer less than or equal to a, so m a. On the other hand, m + is an integer strictly greater than a, so m + is an integer strictly greater than a, so a m +. B By the arithmetic-harmonic mean inequality or the Cauchy-Schwarz inequality, ( (k + + k n + + n. k k n We must thus have 5n 4 n, so n 4. Without loss of generality, we may suppose that k k n. If n, we must have k, which works. Note that hereafter we cannot have k. If n, we have (k, k {(, 4, (3, 3}, neither of which work. If n 3, we have k + k + k 3, so k 3. Hence (k, k, k 3 {(,, 7, (, 3, 6, (, 4, 5, (3, 3, 5, (3, 4, 4}, and only (, 3, 6 works. If n 4, we must have equality in the AM-HM inequality, which only happens when k k k 3 k 4 4. Hence the solutions are n and k, n 3 and (k, k, k 3 is a permutation of (, 3, 6, and n 4 and (k, k, k 3, k 4 (4, 4, 4, 4. Remark: In the cases n, 3, Greg Kuperberg suggests the alternate approach of enumerating the solutions of /k + +/k n with k k n. This is easily done by proceeding in lexicographic order: one obtains (, for n, and (, 3, 6, (, 4, 4, (3, 3, 3 for n 3, and only (, 3, 6 contributes to the final answer. B3 First solution: The functions are precisely f(x cx d for c, d > arbitrary except that we must take c in case d. To see that these work, note that f (a/x dc(a/x d and x/f(x /(cx d, so the given equation holds if and only if dc a d. If d, we may solve for a no matter what c is; if d, we must have c. (Thanks to Brad Rodgers for pointing out the d restriction. To check that these are all solutions, put b log(a and y log(a/x; rewrite the given equation as Put f(e b y f (e y e b y. g(y log f(e y ; then the given equation rewrites as or g(b y + log g (y + g(y y b y, log g (y b g(y g(b y. By the symmetry of the right side, we have g (b y g (y. Hence the function g(y + g(b y has zero derivative and so is constant, as then is g (y. From this we deduce that f(x cx d for some c, d, both necessarily positive since f (x > for all x. Second solution: (suggested by several people Substitute a/x for x in the given equation: Differentiate: f (x a xf(a/x. f a (x x f(a/x + a f (a/x x 3 f(a/x.

6 5 Now substitute to eliminate evaluations at a/x: f (x f (x x Clear denominators: + f (x f(x. xf(xf (x + f(xf (x xf (x. Divide through by f(x and rearrange: f (x f(x + xf (x xf (x f(x f(x. The right side is the derivative of xf (x/f(x, so that quantity is constant. That is, for some d, f (x f(x d x. Integrating yields f(x cx d, as desired. B4 First solution: Define f(m, n, k as the number of n- tuples (x, x,..., x n of integers such that x + + x n m and exactly k of x,..., x n are nonzero. To choose such a tuple, we may choose the k nonzero positions, the signs of those k numbers, and then an ordered ( k-tuple of positive integers with sum m. There are n k options for the first choice, and k for the second. As for the third, we have ( m k options by a stars and bars argument: depict the k-tuple by drawing a number of stars for each term, separated by bars, and adding stars at the end to get a total of m stars. Then each tuple corresponds to placing k bars, each in a different position behind one of the m fixed stars. We conclude that f(m, n, k k ( m k ( n f(n, m, k; k summing over k gives f(m, n f(n, m. (One may also extract easily a bijective interpretation of the equality. Second solution: (by Greg Kuperberg It will be convenient to extend the definition of f(m, n to m, n, in which case we have f(, m f(n,. Let S m,n be the set of n-tuples (x,..., x n of integers such that x + + x n m. Then elements of S m,n can be classified into three types. Tuples with x + + x n < m also belong to S m,n. Tuples with x + + x n m and x n correspond to elements of S m,n by dropping x n. Tuples with x + + x n m and x n < correspond to elements of S m,n by dropping x n. It follows that f(m, n f(m, n + f(m, n + f(m, n, so f satisfies a symmetric recurrence with symmetric boundary conditions f(, m f(n,. Hence f is symmetric. Third solution: (by Greg Martin As in the second solution, it is convenient to allow f(m, f(, n. Define the generating function G(x, y m n f(m, nx m y n. As equalities of formal power series (or convergent series on, say, the region x, y < 3, we have G(x, y x m y n m n n n y n y n x x k,..., k n Z k + + k n m k,..., k n Z m k + + k n k,..., k n Z y n( n k Z ( + x y n x n x k + + kn x x k n n x y( + x/( x x y xy. x m Since G(x, y G(y, x, it follows that f(m, n f(n, m for all m, n. B5 First solution: Put Q x + + x n. Since Q is homogeneous, P is divisible by Q if and only if each of the homogeneous components of P is divisible by Q. It is thus sufficient to solve the problem in case P itself is homogeneous, say of degree d. Suppose that we have a factorization P Q m R for some m >, where R is homogeneous of degree d and not divisible by Q; note that the homogeneity implies that n i x i R x i dr. Write as shorthand for + + x x ; then n P mnq m R + Q m R + n i Q m R + (mn + 4mdQ m R. m R mx i Q x i Since m >, this forces R to be divisible by Q, contradiction. Second solution: (by Noam Elkies Retain notation as in the first solution. Let P d be the set of homogeneous

7 6 polynomials of degree d, and let H d be the subset of P d of polynomials killed by, which has dimension dim(p d dim(p d ; the given problem amounts to showing that this inequality is actually an equality. Consider the operator Q (i.e., apply then multiply by Q on P d ; its zero eigenspace is precisely H d. By the calculation from the first solution, if R P d, then (QR Q R (n + 4dR. Consequently, Q j H d j is contained in the eigenspace of Q on P d of eigenvalue (n + 4(d j + + (n + 4(d. In particular, the Q j H d j lie in distinct eigenspaces, so are linearly independent within P d. But by dimension counting, their total dimension is at least that of P d. Hence they exhaust P d, and the zero eigenspace cannot have dimension greater than dim(p d dim(p d, as desired. Third solution: (by Richard Stanley Write x (x,..., x n and ( x,..., x n. Suppose that P (x Q(x(x + + x n. Then P ( P (x Q( ( P (x. On the other hand, if P (x α c αx α (where α (α,..., α n and x α x α xαn n, then the constant term of P ( P (x is seen to be α c α. Hence c α for all α. Remarks: The first two solutions apply directly over any field of characteristic zero. (The result fails in characteristic p > because we may take P (x + + x n p x p + + xp n. The third solution can be extended to complex coefficients by replacing P ( by its complex conjugate, and again the result may be deduced for any field of characteristic zero. Stanley also suggests Section 5 of the arxiv e-print math.co/5363 for some algebraic background for this problem. B6 First solution: Let I be the identity matrix, and let J x be the matrix with x s on the diagonal and s elsewhere. Note that J x (x I, being the all s matrix, has rank and trace n, so has n eigenvalues equal to and one equal to n. Hence J x has n eigenvalues equal to x and one equal to x + n, implying det J x (x + n (x n. On the other hand, we may expand the determinant as a sum indexed by permutations, in which case we get det J x sgn(πx ν(π. Integrating both sides from to (and substituting y x yields sgn(π ν(π + as desired. (x + n (x n dx ( n+ (n yy n dy ( n+ n n +, Second solution: We start by recalling a form of the principle of inclusion-exclusion: if f is a function on the power set of {,..., n}, then f(s T S ( T S U T f(u. In this case we take f(s to be the sum of σ(π over all permutations π whose fixed points are exactly S. Then U T f(u if T n and otherwise (since a permutation group on or more symbols has as many even and odd permutations, so f(s ( n S ( n + S. The desired sum can thus be written, by grouping over fixed point sets, as n ( n ( n i n + i i i + i n ( n n ( ( n i ( n i n n i i + i i i n ( ( n i n n + n + i + i ( n+ n n +. Third solution: (by Richard Stanley The cycle indicator of the symmetric group S n is defined by Z n (x,..., x n x c(π n, x cn(π where c i (π is the number of cycles of π of length i. Put F n σ(πx ν(π Z n (x,,,,,... and f(n σ(π ν(π + F n (x dx.

8 7 A standard argument in enumerative combinatorics (the Exponential Formula gives Z n (x,..., x n tn n! exp t k x k k, n yielding n f(n tn n! k exp (xt t + t33 dx e (x t+log(+t dx ( + te (x t dx t ( e t ( + t. Expanding the right side as a Taylor series and comparing coefficients yields the desired result. Fourth solution (sketch: (by David Savitt We prove the identity of rational functions σ(π ν(π + x ( n+ n!(x + n x(x + (x + n by induction on n, which for x implies the desired result. (This can also be deduced as in the other solutions, but in this argument it is necessary to formulate the strong induction hypothesis. Let R(n, x be the right hand side of the above equation. It is easy to verify that R(x, n R(x +, n + (n! ( n+ x n l (n! + ( R(x, n l, (n l! l since the sum telescopes. To prove the desired equality, it suffices to show that the left hand side satisfies the same recurrence. This follows because we can classify each π S n as either fixing n, being an n-cycle, or having n in an l-cycle for one of l,..., n ; writing the sum over these classes gives the desired recurrence.