The Sixty-Sixth William Lowell Putnam Mathematical CompetitionAuthor(s): Leonard F. Klosinski, Gerald L. Alexanderson, Loren C. LarsonSource: The American Mathematical Monthly, Vol. 113, No. 8 (Oct., 2006), pp. 733-743Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/27642034 .Accessed: 22/01/2011 04:24

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The Sixty-Sixth William Lowell Putnam Mathematical Competition

Leonard F. Klosinski, Gerald L. Alexanderson, and Loren C. Larson

The results of the Sixty-Sixth William Lowell Putnam Mathematical Competition, held December 3, 2005, follow. They have been determined in accordance with the

regulations governing the Competition. The contest is supported by the William Lowell

Putnam Prize Fund for the Promotion of Scholarship, an endowment established by Mrs. Putnam in memory of her husband. The annual Competition is held under the

auspices of the Mathematical Association of America.

The first prize, $25,000, was awarded to the Department of Mathematics of Harvard

University. The members of the winning team were Tiankai Liu, Alison B. Miller, and

Tong Zhang; each was awarded a prize of $1,000. The second prize, $20,000, was awarded to the Department of Mathematics of

Princeton University. The members of the winning team were Ana Caraiani, Andrei

Negut, and Aaron C. Pixton; each was awarded a prize of $800. The third prize, $15,000, was awarded to the Department of Mathematics of Duke

University. The members of the winning team were Nikifor C. Bliznashki, Jason Fer

guson, and Lingren Zhang; each was awarded a prize of $600. The fourth prize, $10,000, was awarded to the Department of Mathematics of the

Massachusetts Institute of Technology. The members of the winning team were Tim

othy G. Abbott, Vladimir Barzov, and Daniel M. Kane; each was awarded a prize of

$400. The fifth prize, $5,000, was awarded to the Department of Mathematics of the Uni

versity of Waterloo. The members of the winning team were Olena Bormashenko,

Ralph Furmaniak, and Xiannan Li; each was awarded a prize of $200.

The six highest ranking individual contestants, the Putnam Fellows, in alphabetical order, were Oleg I. Golberg, Massachusetts Institute of Technology; Mathew M. Ince,

Massachusetts Institute of Technology; Daniel M. Kane, Massachusetts Institute of

Technology; Ricky I. Liu, Harvard University; Tiankai Liu, Harvard University; and

Aaron C. Pixton, Princeton University. Each receives an award of $2,500. The next ten highest ranking contestants, in alphabetical order, were: Robert M.

Barrington Leigh, University of Toronto; Thomas D. Belulovich, Massachusetts Insti

tute of Technology; Richard V. Biggs, Carnegie Mellon University; Steven J. Byrnes, Harvard University; Alexander R. Fink, University of Calgary; Po-Ru Loh, California

Institute of Technology; Alison B. Miller, Harvard University; Thanasin Nampaisarn, Massachusetts Institute of Technology; Eric C. Price, Massachusetts Institute of Tech

nology; and Kuat T Yessenov, Massachusetts Institute of Technology. Each receives

an award of $1,000. The next eight highest ranking contestants, in alphabetical order, were: Ralph Fur

maniak, University of Waterloo; Hyun Soo Kim, Massachusetts Institute of Technol

ogy; Zhiwei Calvin Lin, University of Chicago; Roger Mong, University of Toronto; Andrei Negut, Princeton University; Dimiter V. Ostrev, Yale University; Steven W.

Sivek, Massachusetts Institute of Technology; and Lingren Zhang, Duke University. Each receives an award of $250.

October 2006] 66TH putnam mathematical competition 733

The following teams, named in alphabetical order, received honorable mention: Cal

ifornia Institute of Technology, with team members Justin Blanchard, Po-Ruh Loh, and Rumen Zarev; Carnegie Mellon University, with team members Richard V. Biggs, Paul Zagieboylo, and Chunhua Zhang; Stanford University, with team members John

W. Hegeman, Serin Hong, and Robert D. Hough; University of Toronto, with team

members Robert Barrington Leigh, David Tianyi Han, and Jacob Tsimerman; and

Yale University, with team members Joshua D. Batson, Lazar Krstic, and Dimiter V.

Ostrev.

Honorable mention was achieved by the following fifty-one individuals in alphabet ical order: Timothy G. Abbott, Massachusetts Institute of Technology; Boris Alexeev,

Massachusetts Institute of Technology; Jae M. Bae, Harvard University; Jongmin Baek, Massachusetts Institute of Technology; Joshua D. Batson, Yale University; Natth Bejraburnin, Stanford University; Kshipra U. Bhawalkar, Duke University; Justin E. Blanchard, California Institute of Technology; Nikifor C. Bliznashki, Duke

University; Olena Bormashenko, University of Waterloo; Ana Caraiani, Princeton

University; Po-Ning Chen, Massachusetts Institute of Technology; Andrew J. Critch, Memorial University of Newfoundland; Fernando A. Delgado, University of Michi

gan, Ann Arbor; Anand R. Deopurkar, Massachusetts Institute of Technology; Gabriel

E. Gauthier-Shalom, McGill University; Elyot J. L. Grant, University of Waterloo; Mathieu Guay-Paquet, McGill University; John W. Hegeman, Stanford University; Robert D. Hough, Stanford University; Nathaniel J. Ince, Massachusetts Institute of

Technology; Lozan M. Ivanov, California Institute of Technology; Junehyuk Jung,

University of Chicago; Nima Kamoosi, University of British Columbia; Dmytro Karabash, Columbia University; Pramook Khungurn, Massachusetts Institute of Tech

nology; Sungyoon Kim, Massachusetts Institute of Technology; Aaron D. Kleinman, Princeton University; Roman L. Kogan, State University of New York, Stony Brook; Gabriel E. Kreindler, Princeton University; Joel B. Lewis, Harvard University; Joshua

J. Lim, Massachusetts Institute of Technology; Yuncheng Lin, Massachusetts Institute

of Technology; Aleksandar D. Lishkov, Princeton University; Po-Ling Loh, California

Institute of Technology; Thomas J. Mildorf, Massachusetts Institute of Technology; Samuel A. Miner, Pomona College; Paul D. Nelson, Princeton University; Virgil C. P?trea, Massachusetts Institute of Technology; Natee Pitiwan, Williams College;

Wei Quan Julius Poh, Cornell University; Vedran Sohinger, University of California,

Berkeley; Kiat Chuan Tan, Stanford University; Matthew J. Thibault, Massachusetts

Institute of Technology; David W. Vincent, Massachusetts Institute of Technology; Nathaniel G. Watson, Washington University, St. Louis; Ryan B. Williams, Stan

ford University; Yeoil Yoon, California Institute of Technology; Chunhua Zhang,

Carnegie Mellon University; Tong Zhang, Harvard University; and Yan Zhang, Har

vard University. The other individuals who achieved ranks among the top 99 students, in alpha

betical order of their schools, were: California Institute of Technology, Yingkun Li,

Oleg O. Rudenko; Carnegie Mellon University, Yinmeng N. Zhang; Duke University, Aaron J. Pollack; Harvard University, Luke A. Gustafson, Anatoly Preygel; Kansas

State University, Jeffrey M. Amos; Massachusetts Institute of Technology, Nikolay V Andreev, Vladimir V. Barzov, Daniel R. Gulotta, Anand B. Rajagopalan, Gary L.

Sivek, Iliya T. Tsekov; Princeton University, Anton S. Malyshev; Queen's University, Michael T. LeBlanc; Rice University, Max I. Glick; Simon Fraser University, Denis K.

Dmitriev; University of British Columbia, Wei-Lung D. Tseng; University of Califor

nia, Berkeley, Paul R. Monasterio; University of Mississippi, Sam S. Watson; Univer

sity of Nebraska, Lincoln, Robert M. Brase; University of Toronto, Tianyi Han, Jacob

Tsimerman; and University of Waterloo, Tor G. J. Myklebust.

734 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113

The Elizabeth Lowell Putnam Prize, named for the wife of William Lowell Putnam

and awarded "to a woman whose performance on the Competition has been deemed

particularly meritorious,"is awarded this year to Alison B. Miller, Harvard University. The winner is awarded a prize of $1,000.

There were 3545 individual contestants from 500 colleges and universities in

Canada and the United States who participated in the competition of December 3, 2005. There were teams entered by 395 institutions. The Questions Committee for

the sixty-sixth competition consisted of Hugh L. Montgomery (Chair), University of Michigan, Ann Arbor; Titu Andreescu, University of Texas, Dallas; and Steven

Tschantz, Vanderbilt University; they composed the problems and were most promi nent among those suggesting solutions. Alternative solutions have been published in

Mathematics Magazine 79 (2006) 76-81.

PROBLEMS.

A-l.

A-2.

Show that every positive integer is a sum of one or more numbers of the form

2r35, where r and s are nonnegative integers and no summand divides another.

(For example, 23 = 9 + 8 + 6.)

Let S ? {(a,b) : a = 1,2, ... ,n, b ?

1,2,3}.A rook tour of S is a polygonal

path made up of line segments connecting points px,p2, ... , p3n in sequence

such that (i) pi e S, (ii) p? and pi+\ are a unit distance apart for 1 < / < 3n,

(iii) for each p e S there is a unique / such that pt = p. How many rook tours

are there that begin at (1, 1) and end at (n, 1)? (An example of such a rook tour

for n = 5 is depicted.)

A-3. Let p(z) be a polynomial of degree n all of whose zeros have absolute value 1

in the complex plane. Put g(z) =

p(z)/zn/2. Show that all zeros of g'(z) have

absolute value 1.

A-4. Let H be an n x n matrix all of whose entries are ? 1 and whose rows are mu

tually orthogonal. Suppose H has an a x b submatrix whose entries are all 1.

Show thatafr < n. 11

ln(jc + l) A-5. Evaluate

/ Jo

dx.

/o x1 + 1 A-6. Let n be given,

n > 4, and suppose that Px, P2, ... , Pn are n randomly, inde

pendently and uniformly, chosen points on a circle. Consider the convex n-gon whose vertices are the P?. What is the probability that at least one of the vertex

angles of this polygon is acute?

B-l. Find a nonzero polynomial P(x, y) such that P(laj, \_2aJ) = 0 for all real num

bers a. (Note: \y\ is the greatest integer less than or equal to v.)

B-2. Find all positive integers n,kx, ...

,k? such that k\ + + kn = 5n

? 4 and

1 1

October 2006] 66TH putnam mathematical competition 735

B-3. Find all differentiate functions / : (0, oo) ?> (0, oo) for which there is a pos itive real number a such that

rQ fix)

0) ( -^ + + "^ ) p(*i' '*?)= ? (identically)

for all x > 0.

B-4. For positive integers m and n, let f(m,n) denote the number of ^-tuples

(xx,x2, ... ,xn) of integers such that \xx| + \x2\ + + \xn\ < m. Show that

f(m,n) =

f(n,m).

B-5. Let P(x\, ... ,xn) denote a polynomial with real coefficients in the variables

xx, ..., xn, and suppose that

^+'"+' dxl) and that

(b) x2 -\-Y x2n divides P(xx, ..., xn).

Show that P = 0 identically.

B-6. Let Sn denote the set of all permutations of the numbers 1,2,... , n. For n e

Sn, let a(7t) = 1 if t? is an even permutation and a (it) = ?1 if n is an odd

permutation. Also, let v(n) denote the number of fixed points of n. Show that

y> or(TT) = w+1 n

?fvOO + l n + 1'

SOLUTIONS. In the 12-tuples (nxo, n9, n8, n7, n6, n5, nA, n3, n2, nx, nQ, n_x) follow

ing each problem number, n?, for 10 > / > 0, is the number of students among the top 196 contestants achieving / points for the problem and n^x is the number of those not

submitting solutions.

A-l. (132, 17, 6, 0, 0, 0, 0, 0, 10, 4, 16, 11)

Solution. We argue by induction. Clearly 1 = 2?3?, so the number 1 is represented. If

n is even, then by the induction hypothesis n/2 has such a representation, and it suffices

to multiply each summand by 2. Suppose n is odd, and let k be the largest integer such

that 3k < n. If n = 3k, then we are done. If 3k < n < 3k+x, then m = (n ?

3?)/2 is

a positive integer. Since m < n, it follows from the inductive hypothesis that we can

write m = J2i 2n 3Si. Then n = 3k -\- ̂ 2r,+13Si. It remains to show that no summand

divides another. Using the inductive hypothesis, 2r/+13i/ divides 2rJ+l3sJ only when

/ = j. Also, 2r,+13*' does not divide 3k because 2 is a factor of 2r/+13?/ but not a

factor of 3*. Finally, m = (n -

3^/2 < (3k+x -

3*)/2 = 3*. Thus, 2r'+135' < 3*, and

therefore 3^ does not divide 2r,+135/.

A-2. (85, 44, 13, 0, 0, 0, 0, 0, 24, 5, 12, 13)

Solution. Let T(n) be the number of rook tours of Sn = S that begin at (1, 1) and

end at (n, 1), and let U(n) be the number of rook tours of Sn that begin at (1, 1) and

end at (n,3). Let j (0 < j < n ? 2) denote the number of moves to the right before

the first vertical move is made. Once the first move up has been made, the only way to visit all squares in the first j + 1 columns without getting trapped is to move left j

times, move up, and then move right j + 1 times. This leaves a player at (j + 2, 3). It

736 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113

remains to traverse a3x(n-l-j) board, starting at (j + 2, 3) and ending at (n, 1). By definition, this can be done in U(n

? I ? j) ways. Thus

n-2

r(Ai) = ?i/(n-l-./).

Similarly,

n-\

U(n) = ^T(n-l-j),

j=o

with the convention that T(0) = 1. Clearly, 7(1) = 0 and t/(l) = 1. We claim that

T(n) = U(n) = 2n~2 when n > 2. This follows directly by induction from the pre

ceding identities.

A-3. (14, 2, 5, 0, 0, 0, 0, 0, 6, 4, 86, 79)

Solution. Write p(z) = c(z

? Z\)(z

? z2) (z

? zn)- Thus

g(z) f^i\z-zk 2z)'

Hence

g(z) frf z-Zk

Fix k, and consider the triangle with vertices z9 Zk, and ? Zk- Since the segment from

?Zk to Zk is a diameter of the unit circle, the angle at z is acute if \z\ > 1. Thus in this case, |arg((z + Zk)/(z

- Zk))\ < tt/2, which is to say Rt((z + zk)l(z

- zk)) > 0.

Since this holds for each k, the right-hand side of (1) has positive real part when \z\ >

1. Thus g'(z) = 0 has no solution when |z| > 1. Similarly, if \z\ < 1, the angle at z is

obtuse, 7T/2 < |arg((z + zk)/(z -

zk))\ < x, and Re((z + zk)/(z -

zk)) < 0 for all k. In this case the real part of the right-hand side of (1) is negative, so again g'(z)

= 0 is

impossible.

A-4. (20, 0, 7, 0, 0, 0, 0, 0, 3, 0, 55, 111)

Solution. Write H = [hij], and let / and J be intervals chosen so that the submatrix in

question is obtained by restricting i to / and j to J. Let the rows of H be i*i, ... , rn, and put v =

Yliei Vi ^n tne one nand> since the rows are orthogonal, it follows that

"vn2 = Yl mr/i

* r/2= YlYi Vi = an

i\el i2eJ i el

On the other hand, the jth coordinate of v is ̂ /e/ hi}-, whence

IMI2 = ?(?A<;)2 7 = 1 \ iel '

Here all the summands are nonnegative, and for j in / the summand is a2. Hence

||u||2 > a2b. It follows that a2b < an, which gives the stated result.

October 2006] 66TH putnam mathematical competition 737

A-5. (10, 10, 0, 0, 0, 0, 0, 0, 2, 2, 40, 132)

Solution I. Let / denote the integral in question. Because x = tan(arctanx) for all x

in [0, 1], with the substitution arc tan x =iwe have

I = / ln(l +tant)dt .

Let u ? it ?4 ? t. Then

I = f\ (l + tan

(^-u)) (-du) = fin (l

+ \

nn/4 / 2 \ r71/4

? I In I - ) du ? I ln2du ? I.

Jo \ 1 + tan u ) Jo

tanw . . du

+ tanw

It follows that 2/ = (tt/4) In 2; that is, / =

(tt/8) In 2.

Solution 2. Let / denote the integral in question. The substitution x = (1 ?

u)/(l + u) leads directly to / =

(ln2)7r/4 - /.

A-6. (5, 0, 2, 0, 0, 0, 0, 0, 3, 0, 65, 121)

Solution. Let A, B, and C denote three consecutive points on the circle. To say that

?ABC is acute is equivalent to saying that the arc on the circle from A through B to C measures more than tz. Thus there is not enough room on the circle to have two acute

angles if they are not adjacent. In the event under consideration, there is a unique point where the angle is obtuse followed (counterclockwise) by an acute angle at the next

point. Suppose that 0 < 6 < 1 and that 0 < 0 < 1/2. The probability that exactly one

of the n points lies on the arc (2jt6, 2tt0 + 2ttA0) and exactly one point lies on the

arc (2tt(0 + 0), 2tt(0 + 0) + 2ttA(?)) is approximately n(n ?

1)A# A0. This estimate

would be exact were it not for the possibility that two or more points might lie on

one of these arcs or that the two arcs might overlap. These possibilities are negligible when A0 and A0 are small. Let A denote the point in the first arc and B denote the

point in the second arc. In order for the angle at B to be acute all remaining points must lie in the semicircle from A to its antipode. The probability of this is l/2n_1. Let A' denote the antipode of A, and B' the antipode of B. We don't want all of these

remaining points to lie between A! and B '. The probability of this latter event is at

least (0 ?

Ar9);?~2 but not more than (0 + A0)n~2. Thus the probability of the event

in question is

Jo Jo n(n-l)l [ (^-4>"~2)d4>d6

n(n ?

1) 1 1 \ n(n-2)

K2n~{ 2n~l(n-l)J 2n~{

B-l. (155, 9, 22, 0, 0, 0, 0, 0, 7, 0, 3, 0)

Solution. One such polynomial is P(x, y) ?

(2x ?

y)(2x ?

y + 1). To see this, let a

be a real number and n ? [a\. If a e [n,n + ^)

then \_2a J ?

2[a\, so P([a\, \_2a\) =

0. If a e [n + \, n + 1) then \_2a\ = 2[a\ + 1, so once again P(la], [2a\) = 0.

738 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113

B-2. (148, 11, 4, O, O, O, O, O, 9, 1, 13, 10)

Solution. The possible solutions are:

n = 1, ki = 1;

n = 3, {ki,k2,k3} = {2,3,6};

n ? \,ki

= k2

= fc3

= fc4

= 4.

From the arithmetic mean-geometric mean (or Cauchy-Schwarz) inequality, we find

that

(ki + --- + kn)(^-

+ -.. + pj>n2,

(1)

with equality if and only if k\ = = kn. Hence 5n ? 4 > n2, which is equivalent to n(n

? 4) < 0. Thus n e {1, 2, 3, 4}. For n = 1 or n = 4, we have equality in (1),

so k\ = =

kn. We obtain the solutions n = 1, k\

= 1, and n =

4, ki =

k2 =

k3 ?

k4 ?

4. We are left with the cases ft = 2 and n ? 3. For n = 2, the system of equations

/cj + ?2 = 6, 1/fei + l/k2 = 1 is not solvable in positive integers. For n = 3, we seek

the triples (k{, k2, fc3) such that fci + k2 + /c3 = 11 and \/k\ + l//c2 + 1/^3 = 1- Let

fci^2 + ^2^3 + ^3^1 = kik2k3 = ^. Then A:l9 fc2, and /c3 are positive integral solutions

to the equation x3 ? \\x2 + qx

? q = 0. It follows that a solution x is an integer

different from 1 and 11, and that

-x3 + 11jc2 7 10 q

=- = -x2 + IOjc + 10 +-. x

? 1 x ?

1

Because q is a positive integer, x ? 1 is a positive divisor of 10 and different from

10. Then x ? I e {1,2,5}, so x e {2,3,6}. A simple case analysis shows that

{*,,fc2,*3} = {2,3,6}.

B-3. (23, 34, 8, 0, 0, 0, 0, 0, 2, 8, 75, 46)

Solution. Define g(x) =

f(x)f(a/x) for x in (0, oo). We claim that g is a constant

function. Indeed, substituting a/x for x in the given condition yields f(a/x)f'(x) =

a/x when x > 0, so

gf(x) =

f(x)f(a/x) + f(x)f(a/x)(-a/x2) =

a/x -

a/x = 0,

ensuring that g is some positive constant b.

From the original equation we can write

?_iW_/w/(i)_/w (?._?_). which gives

fix) = a

f{x) bx'

October 2006] 66TH putnam mathematical competition 739

Integrating each side we obtain In f(x) =

(a/b) lnx + lnc, where c > 0. It follows

that f(x) = cxa/b for x > 0. Substituting back into the original equation yields

a aa/b~x x c '

which is equivalent to

b xalb~x cxa<b'

~2na/b _

By eliminating c we find the family of solutions

a/b (\ a/D

7?) (b > 0)'

B-4. (97, 15, 12, 0, 0, 0, 0, 0, 13, 12, 27, 20)

Solution 1. It suffices to show that

minim, h i / \ / \

since this is symmetric in n and m. First we group the lattice points according to the

number of nonzero coordinates. Call this number k. There are Q) ways of choosing

which coordinates are the nonzero ones. For each nonzero coordinate we must choose

the sign of the coordinate. There are 2k ways of choosing these signs. It remains to

count the number of /c-tuples (ax, ... , a?) of positive integers such that Ylai ? m

Put b[ ?

al ?

1. Then (bx, ...

,bk) is a /c-tuple of nonnegative integers whose sum

is at most m ? k. Let this sum be r. It is well known that the number of ^-tuples of

nonnegative integers with sum r is exactly ^J^). Hence the number of/:-tuples is

m~k /Jk + r-l\ (m

E

This leads to the stated formula for f(m,n).

Solution 2. Extend the definition to include m = 0 or n = 0 but not both. For the case

m = 0 the only /i-tuple that works is (0, 0, ..., 0), so f(0, n) = 0 for n > 0. For the case n ? 0 the only 0-tuple is 0, so f(m, 0)

? 1. Thus /(0, n) ?

f(m, 0) = 1.

By considering the possible values of xn among ?2-tuples we see that if n > 0, then

m

f(m,n) = ]T f(m

- \i\, n)

m =

f(m,n- l) + ^2/(m

-i,n- 1), (1) /=i

for the last expression counts the n-tuples with xn ? i (?m < i < m).

Applying this to m ? 1 for positive m, we obtain

m-\

f(m -l,n) = f(m -l,n- l) +

^2/((m -l)-i,n- I). (2) i = \

740 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113

Subtracting (2) from (1) leads to the recursion formula

f(m, n) = f(m

? 1, n) + f(m, n ?

1) + f(m ?

1, n ? 1). (3)

The result now follows by strong induction on the sum m + n by using (3) twice,

namely,

f(m, n -h 1) = /(m

? 1, ft + 1) + /(m, n) + /(m

? 1, n)

= f(n + 1, m ?

1) + /(ft, m) + /(ft, m ? 1)

= f(n + 1, m).

Solution 3. Let ^(m, ft) denote the set of ft-tuples counted by f(m, n), and let T(m, n) be the set of sequences

(aubi,ei), (a2,b2,e2), ... , (ak,bk,ek)

such that the following conditions hold:

k > 0,

0 < ai < a2 < - < cik < m,

0 < bi < b2 < < bk < ft

ef- = ?1 for 1 < i < k.

Clearly the correspondence (a?, bt, e?) \-> (b?, at, et) induces a bijection from T(m,n) to T(n, m). It therefore suffices to describe a bijection between T(m,n) and S(m,n).

Define (p : T(rn,n) ?> S(m, n) as follows:

(?)((ai,bi,ei), (a2,b2,e2), ... , (ak,bk,ek))

= (*i,*2? ,^w),

where

I^_/(?_/ ?bj-\) if / = <2; (Note: ?>0 is taken to be 0),

0 if / ̂ a,- for any j with 1 < j < k.

Define ^ S(m, n) ?> T(m, n) by

if(xux2, ...

,xn) =

((aubi,ei), (a2,b2,e2), ... , (ak,bk,ek)),

where k = | {jc,- : x? ^ 0} | and

a? ? the index of the /th nonzero element in (x\, x2, ..., xn),

b<= J2 1^1

??z =

sign(xfl|.)

for / = 1,2,... , /: It is easy to show that <j> o i/r and ^ ? 0 are identity functions, so

we have a bijection.

B-5. (4, 0, 1, 0, 0, 0, 0, 0, 0, 0, 54, 137)

October 2006] 66TH putnam mathematical competition 741

Solution. Write P = Y^k Pk, where Pk is homogeneous of degree k. If P has the re

quired properties, then each of the Pk does. Thus we may assume that P is homo

geneous, say of degree d. Put S = x2 + + x2. The ring R[xx, ..., xn] is a unique

factorization domain, and S is irreducible (for otherwise S would factor into linear fac

tors, and then it would have nontrivial zeros in Rn, which it doesn't). Thus if P ^ 0, then there are a unique polynomial Q in R[xx, ..., xn] and a unique m such that

P = SmQ. By hypothesis (b) in the problem, m > 0. We now compute the second

partial derivatives of P. First

dP , 35 ao ? =mSm-x

? Q + Sm ?.

dXi dx? dXi

Taking the partial derivative with respect to xt again, we find that

But dS/dXi ?

2x? and d2S/dx2 = 2, so this reduces to

4m(m -

l)Sm-2x2Q + 2mSm-xQ+4mSm-xXi? + Smd-^. (1)

Now Q is homogeneous of degree d ? 2m. From an identity of Euler it follows that

y^Xiir- = (d-2m)Q.

V dxi

Thus when we sum the expression (1) over / we discover that

0 = 4m(m -

l)Sm~x Q + 2mnSm~x Q + 4m(d -

2m)Sm~x Q + Sm V ?|. i dxt

On cancelling Sm~x from both sides, we deduce that

v-92? -2m(2m

- 2 + n + 2d - 4m)Q = S > ??-.

V ^2 Here the constant on the left side is positive, so it follows that S divides Q, contrary to

the assumption that m is the highest power of S in P.

B-6. (18, 0, 1, 0, 0, 0, 0, 0, 2, 1, 41, 133)

Solution. We sum over the number v(n):

y^^ =

y_]_ y a(n)

= ?7TT ? CH ; = 1 ^ X

* $?_;,v(tt)=0 v/

/ = l v 7 7re5?_/,u(^)=0

742 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113

Now we claim that Dn = J^nes? v(tt)=o cr(7t) *s given recursively by

Dn =

-(n-\)(Dn_i+Dn_2).

To see this, consider ix in Sn with v(7i) = 0. Either n2(n) ? n or 7r2(ft) 7^ ft. If

7i2(ft) = ft, then (ft, 71 (ft)) o 7T leaves ft and ti (ft) fixed, so can be identified with a

derangement of Sn_2. Otherwise, (n,n(n)) on leaves n fixed, and can be identified

with a derangement of Sn-\. There are ft ? 1 choices for n (n) and a ((ft, n(n)) on)) =

?o(tt), which establishes the recurrence relation.

It is clear that D2 = -1 and D3 = 2, and we claim that Dn = (-l)n~{(n

- 1). This

is true for n = 2 and n = 3 and therefore by the recursion formula and the inductive

hypothesis

Dn = -(ft -

1) ((-iy-2(n -

2) + (-iy-3(ft -

3))

= (-\)n~\n

- \)((n

- 2)

- (ft

- 3)) = (-1)"" V

- 1).

Then, noting that D0 = 1 and Dx = 0 (which also fit the formula), we obtain

?b?' + 'W =

<-D"2JT--(;i<-l)i+,n ?=0

/=0

"?-'^te(::.>?

1+1 ft + 1

0

= (-ir+i? ft +1

LFKand GLA: Santa Clara University, Santa Clara, CA 95053-0374

LCL: St. Olaf College, Northfield, MN 55057-1098

October 2006] 66TH putnam mathematical competition 743