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\begin{document}
\title{Solutions to the 72nd William Lowell Putnam Mathematical Competition \\
Saturday, December 3, 2011}
\author{Kiran Kedlaya and Lenny Ng}
\noaffiliation
\maketitle
\begin{itemize}
\item[A1] We claim that the set of points with $0\leq x\leq 2011$ and $0\leq
y\leq 2011$ that cannot be the last point of a growing spiral are as follows:
$(0,y)$ for $0\leq y\leq 2011$; $(x,0)$ and $(x,1)$ for $1\leq x\leq 2011$;
$(x,2)$ for $2\leq x\leq 2011$; and $(x,3)$ for $3\leq x\leq 2011$. This
gives a total of
\[
2012 + 2011 + 2011 + 2010 + 2009 = 10053
\]
excluded points.
The complement of this set is the set of $(x,y)$ with $0 0$, while if $n$ is even, we can write this as
$(x_1-x_2) + \cdots + (x_{n-1}-x_n) < 0$. Thus no point beyond $P_0$ can have
$x$-coordinate $0$, and we have ruled out $(0,y)$ for $0\leq y\leq 2011$.
Next we claim that any point beyond $P_3$ must have $y$-coordinate either
negative or $\geq 4$. Indeed, each such point has $y$-coordinate of the
form $y_1-y_2+\cdots+(-1)^{n-1} y_n$ for $n \geq 2$, which we can write
as $(y_1-y_2) + \cdots + (y_{n-1}-y_n) < 0$ if $n$ is even, and
\[
y_1 +(-y_2+y_3) + \cdots + (-y_{n-1}+y_n) \geq y_1 + 2 \geq 4
\]
if $n \geq 3$
is odd. Thus to rule out the rest of the forbidden points, it suffices to
check that they cannot be $P_2$ or $P_3$ for any growing spiral. But none
of them can be $P_3 = (x_1-x_2,y_1)$ since $x_1-x_2<0$, and none of them
can be $P_2 = (x_1,y_1)$ since they all have $y$-coordinate at most equal
to their $x$-coordinate.
\item[A2]
For $m\geq 1$, write
\[
S_m = \frac{3}{2}\left(1 - \frac{b_1\cdots b_m}{(b_1+2)\cdots(b_m+2)}\right).
\]
Then $S_1 = 1 = 1/a_1$ and a quick calculation yields
\[
S_m-S_{m-1} = \frac{b_1\cdots b_{m-1}}{(b_2+2)\cdots(b_m+2)} = \frac{1}{a_1\cdots a_m}
\]
for $m\geq 2$, since $a_j = (b_j+2)/b_{j-1}$ for $j \geq 2$. It follows
that $S_m = \sum_{n=1}^m 1/(a_1\cdots a_n)$.
Now if $(b_j)$ is bounded above by $B$, then $\frac{b_j}{b_j+2}
\leq \frac{B}{B+2}$ for all $j$, and so $3/2 > S_m \geq
3/2(1-(\frac{B}{B+2})^m)$. Since $\frac{B}{B+2} < 1$, it follows that the
sequence $(S_m)$ converges to $S = 3/2$.
\item[A3]
We claim that $(c,L) = (-1,2/\pi)$ works.
Write $f(r) = \int_0^{\pi/2} x^r\sin x\,dx$. Then
\[
f(r) < \int_0^{\pi/2} x^r\,dx = \frac{(\pi/2)^{r+1}}{r+1}
\]
while since $\sin x \geq 2x/\pi$ for $x \leq \pi/2$,
\[
f(r) > \int_0^{\pi/2} \frac{2x^{r+1}}{\pi} \,dx = \frac{(\pi/2)^{r+1}}{r+2}.
\]
It follows that
\[
\lim_{r\to\infty} r \left(\frac{2}{\pi}\right)^{r+1} f(r) = 1,
\]
whence
\[
\lim_{r\to\infty} \frac{f(r)}{f(r+1)} = \lim_{r\to\infty}
\frac{r(2/\pi)^{r+1}f(r)}{(r+1)(2/\pi)^{r+2}f(r+1)} \cdot
\frac{2(r+1)}{\pi r} = \frac{2}{\pi}.
\]
Now by integration by parts, we have
\[
\int_0^{\pi/2} x^r\cos x\,dx = \frac{1}{r+1} \int_0^{\pi/2} x^{r+1} \sin x\,dx
= \frac{f(r+1)}{r+1}.
\]
Thus setting $c = -1$ in the given limit yields
\[
\lim_{r\to\infty} \frac{(r+1)f(r)}{r f(r+1)} =
\frac{2}{\pi},
\]
as desired.
\item[A4] The answer is $n$ odd. Let $I$ denote the $n\times n$ identity
matrix, and let $A$ denote the $n\times n$ matrix all of whose entries
are $1$. If $n$ is odd, then the matrix $A-I$ satisfies the conditions of
the problem: the dot product of any row with itself is $n-1$, and the dot
product of any two distinct rows is $n-2$.
Conversely, suppose $n$ is even, and suppose that the matrix $M$ satisfied
the conditions of the problem. Consider all matrices and vectors mod
$2$. Since the dot product of a row with itself is equal mod $2$ to the
sum of the entries of the row, we have $M v = 0$ where $v$ is the vector
$(1,1,\ldots,1)$, and so $M$ is singular. On the other hand, $M M^T =
A-I$; since
\[
(A-I)^2 = A^2-2A+I = (n-2)A+I = I,
\]
we have $(\det M)^2 = \det(A-I) = 1$ and $\det M = 1$, contradicting the
fact that $M$ is singular.
\item[A5] (by Abhinav Kumar) Define $G : \RR \to \RR$ by $G(x) = \int_0^x
g(t)\,dt$. By assumption, $G$ is a strictly increasing, thrice continuously
differentiable function. It is also bounded: for $x>1$, we have
\[
0 < G(x)-G(1) = \int_1^x g(t)\,dt \leq \int_1^x dt/t^2 = 1,
\]
and similarly, for $x<-1$, we have $0 > G(x)-G(-1) \geq -1$.
It follows that the image of $G$ is some open interval $(A,B)$
and that $G^{-1}: (A,B) \to \RR$ is also thrice continuously differentiable.
Define $H: (A,B) \times (A,B) \to \RR$ by $H(x,y) = F(G^{-1}(x), G^{-1}(y))$;
it is twice continuously differentiable since $F$ and $G^{-1}$ are.
By our assumptions about $F$,
\begin{multline*}
\frac{\partial H}{\partial x} + \frac{\partial H}{\partial y} =
\frac{\partial F}{\partial x}(G^{-1}(x), G^{-1}(y))
\cdot \frac{1}{g(G^{-1}(x))}\\
+ \frac{\partial F}{\partial y}(G^{-1}(x), G^{-1}(y))
\cdot \frac{1}{g(G^{-1}(y))} = 0.
\end{multline*}
Therefore $H$ is constant along any line parallel to the vector $(1,1)$,
or equivalently, $H(x,y)$ depends only on $x-y$. We may thus write $H(x,y) =
h(x-y)$ for some function $h$ on $(-(B-A), B-A)$, and we then have $F(x,y)
= h(G(x) - G(y))$. Since $F(u,u) = 0$, we have $h(0) = 0$. Also, $h$
is twice continuously differentiable (since it can be written as $h(x)
= H(x,0)$), so $|h'|$ is bounded on the closed interval $[-(B-A)/2,
(B-A)/2]$, say by $M$.
Given $x_1,\dots,x_{n+1} \in \RR$ for some $n \geq 2$, the numbers
$G(x_1),\dots,G(x_{n+1})$ all belong to $(A,B)$, so we can choose indices
$i$ and $j$ so that $|G(x_i) - G(x_j)| \leq (B-A)/n \leq (B-A)/2$. By the
mean value theorem,
\[
|F(x_i, x_j)| = |h(G(x_i) - G(x_j))| \leq M \frac{B-A}{n},
\]
so the claim holds with $C = M(B-A)$.
\item[A6]
Choose some ordering $h_1,\dots, h_n$ of the elements of $G$ with $h_1 = e$.
Define an $n \times n$ matrix $M$
by settting $M_{ij} = 1/k$ if $h_j = h_i g$ for some $g \in \{g_1,\dots,g_k\}$ and $M_{ij} = 0$ otherwise.
Let $v$ denote the column vector $(1,0,\dots,0)$. The probability
that the product of $m$ random elements of $\{g_1,\dots,g_k\}$
equals $h_i$ can then be interpreted as the $i$-th component of the vector
$M^m v$.
Let $\hat{G}$ denote the dual group of $G$, i.e., the group of complex-valued characters of $G$.
Let $\hat{e} \in \hat{G}$ denote the trivial character.
For each $\chi \in \hat{G}$, the vector $v_\chi = (\chi(h_i))_{i=1}^n$ is an eigenvector of $M$
with eigenvalue $\lambda_\chi = (\chi(g_1) + \cdots + \chi(g_k))/k$.
In particular, $v_{\hat{e}}$ is the all-ones vector and $\lambda_{\hat{e}} = 1$.
Put
\[
b = \max\{|\lambda_\chi|: \chi \in \hat{G} - \{\hat{e}\}\};
\]
we show that $b \in (0,1)$ as follows.
First suppose $b=0$; then
\[
1 = \sum_{\chi \in \hat{G}} \lambda_\chi
= \frac{1}{k} \sum_{i=1}^k \sum_{\chi \in \hat{G}} \chi(g_i) = \frac{n}{k}
\]
because $\sum_{\chi \in \hat(G)} \chi(g_i)$ equals $n$ for $i=1$ and $0$ otherwise.
However, this contradicts the hypothesis that $\{g_1, \dots, g_k\}$ is not all of $G$.
Hence $b > 0$. Next suppose $b=1$, and choose $\chi \in \hat{G} - \{\hat{e}\}$ with $|\lambda_\chi| = 1$.
Since each of
$\chi(g_1), \dots, \chi(g_k)$ is a complex number of norm 1, the triangle inequality forces them all to be equal.
Since $\chi(g_1) = \chi(e) = 1$, $\chi$ must map each of $g_1,\dots, g_k$ to 1, but this is impossible because
$\chi$ is a nontrivial character and $g_1,\dots,g_k$ form a set of generators of $G$.
This contradiction yields $b < 1$.
Since $v = \frac{1}{n} \sum_{\chi \in \hat{G}} v_\chi$
and $M v_\chi = \lambda_\chi v_\chi$, we have
\[
M^m v - \frac{1}{n} v_{\hat{e}}
= \frac{1}{n} \sum_{\chi \in \hat{G} - \{\hat{e}\}} \lambda_\chi^m v_\chi.
\]
Since the vectors $v_\chi$ are pairwise orthogonal,
the limit we are interested in can be written as
\[
\lim_{m \to \infty} \frac{1}{b^{2m}} (M^m v - \frac{1}{n} v_{\hat{e}}) \cdot (M^m v - \frac{1}{n} v_{\hat{e}}).
\]
and then rewritten as
\[
\lim_{m \to \infty} \frac{1}{b^{2m}} \sum_{\chi \in \hat{G} - \{\hat{e}\}} |\lambda_\chi|^{2m}
= \#\{\chi \in \hat{G}: |\lambda_\chi| = b\}.
\]
By construction, this last quantity is nonzero and finite.
\textbf{Remark.}
It is easy to see that the result fails if we do not assume $g_1 = e$: take $G = \ZZ/2\ZZ$,
$n=1$, and $g_1 = 1$.
\textbf{Remark.}
Harm Derksen points out that a similar argument applies even if $G$ is not assumed to be abelian,
provided that the operator $g_1 + \cdots + g_k$ in the group algebra $\ZZ[G]$ is \emph{normal},
i.e., it commutes with the operator $g_1^{-1} + \cdots + g_k^{-1}$.
This includes the cases where the set $\{g_1,\dots,g_k\}$ is closed under taking inverses
and where it is a union of conjugacy classes (which in turn includes the case of $G$ abelian).
\textbf{Remark}.
The matrix $M$ used above has nonnegative entries with row sums equal to 1 (i.e., it corresponds to a Markov chain), and there exists a positive integer $m$ such that $M^m$ has positive entries. For any such matrix,
the Perron-Frobenius theorem implies that
the sequence of vectors $M^m v$ converges to a limit $w$, and there exists $b \in [0,1)$
such that
\[
\limsup_{m \to \infty} \frac{1}{b^{2m}} \sum_{i=1}^n ((M^m v - w)_i)^2
\]
is nonzero and finite. (The intended interpretation in case $b=0$ is that $M^m v = w$ for all large $m$.)
However, the limit need not exist in general.
\item[B1]
Since the rational numbers are dense in the reals, we can find positive integers $a,b$
such that
\[
\frac{3\epsilon}{hk} < \frac{b}{a}
< \frac{4\epsilon}{hk}.
\]
By multiplying $a$ and $b$ by a suitably large positive integer, we can also ensure that $3a^2 > b$. We then have
\[
\frac{\epsilon}{hk} < \frac{b}{3a} < \frac{b}{\sqrt{a^2+b} + a} = \sqrt{a^2+b} - a
\]
and
\[
\sqrt{a^2+b} - a = \frac{b}{\sqrt{a^2+b} + a} \leq \frac{b}{2a} < 2 \frac{\epsilon}{hk}.
\]
We may then take $m = k^2 (a^2+b), n = h^2 a^2$.
\item[B2]
Only the primes 2 and 5 appear seven or more times.
The fact that these primes appear is demonstrated by the examples
\[
(2,5,2), (2, 5, 3), (2, 7, 5), (2, 11, 5)
\]
and their reversals.
It remains to show that if either $\ell=3$ or $\ell$ is a prime greater than 5, then $\ell$ occurs at most six times
as an element of a triple in $S$.
Note that $(p,q,r) \in S$ if and only if
$q^2 - 4pr = a^2$ for some integer $a$; in particular, since $4pr \geq 16$, this forces $q \geq 5$.
In particular, $q$ is odd, as then is $a$, and so $q^2 \equiv a^2 \equiv 1 \pmod{8}$;
consequently, one of $p,r$ must equal 2. If $r=2$, then $8p = q^2-a^2 = (q+a)(q-a)$; since both factors
are of the same sign and their sum is the positive number $q$, both factors are positive.
Since they are also both even, we have $q+a \in \{2, 4, 2p, 4p\}$ and so
$q \in \{2p+1, p+2\}$. Similarly, if $p=2$, then $q \in \{2r+1, r+2\}$.
Consequently, $\ell$ occurs at most twice as many times as there are prime numbers in the list
\[
2\ell+1, \ell+2, \frac{\ell-1}{2}, \ell-2.
\]
For $\ell = 3$,$\ell-2= 1$ is not prime.
For $\ell \geq 7$, the numbers $\ell-2, \ell, \ell+2$ cannot all be prime,
since one of them is always a nontrivial multiple of 3.
\textbf{Remark.}
The above argument shows that the cases listed for 5 are the only ones that can occur. By contrast,
there are infinitely many cases where 2 occurs if either the twin prime conjecture holds or there are
infinitely many Sophie Germain primes (both of which are expected to be true).
\item[B3]
Yes, it follows that $f$ is differentiable.
\noindent
\textbf{First solution.}
Note first that in some neighborhood of $0$, $f/g$ and $g$ are both continuous, as then is their product $f$.
If $f(0) \neq 0$, then in some possibly smaller neighborhood of $0$, $f$ is either always positive or always
negative. We can thus choose $\epsilon \in \{\pm 1\}$ so that
$\epsilon f$ is the composition of the differentiable function
$(fg)\cdot (f/g)$ with the square root function. By the chain rule, $f$ is differentiable at $0$.
If $f(0) = 0$, then $(f/g)(0) = 0$, so we have
\[
(f/g)'(0) = \lim_{x \to 0}
\frac{f(x)}{x g(x)}.
\]
Since $g$ is continuous at 0, we may multiply limits to deduce that $\lim_{x \to 0} f(x)/x$ exists.
\noindent
\textbf{Second solution.}
Choose a neighborhood $N$ of $0$ on which $g(x) \neq 0$.
Define the following functions on $N \setminus \{0\}$:
$h_1(x) = \frac{f(x)g(x)-f(0)g(0)}{x}$;
$h_2(x) = \frac{f(x)g(0)-f(0)g(x)}{xg(0)g(x)}$;
$h_3(x) = g(0)g(x)$;
$h_4(x) = \frac{1}{g(x)+g(0)}$. Then
by assumption, $h_1,h_2,h_3,h_4$ all have limits as $x \to 0$. On the
other hand,
\[
\frac{f(x)-f(0)}{x} = (h_1(x)+h_2(x)h_3(x))h_4(x),
\]
and it follows that $\lim_{x\to 0} \frac{f(x)-f(0)}{x}$ exists, as desired.
\item[B4]
Number the games $1,\ldots,2011$, and let $A = (a_{jk})$ be the $2011 \times 2011$ matrix whose $jk$ entry is $1$ if player $k$ wins game $j$ and $i = \sqrt{-1}$ if player $k$ loses game $j$. Then $\overline{a_{hj}}a_{jk}$ is $1$ if players $h$ and $k$ tie in game $j$; $i$ if player $h$ wins and player $k$ loses in game $j$; and $-i$ if $h$ loses and $k$ wins. It follows that $T + i W = \overline{A}^T A$.
Now the determinant of $A$ is unchanged if we subtract the first row of $A$ from each of the other rows, producing a matrix whose rows, besides the first one, are $(1-i)$ times a row of integers. Thus we can write $\det A = (1-i)^{2010}(a+bi)$ for some integers $a,b$. But then
$\det(T+iW) = \det(\overline{A}^T A) = 2^{2010}(a^2+b^2)$ is a nonnegative integer multiple of $2^{2010}$, as desired.
\item[B5]
Define the function
\[
f(y) =
\int_{-\infty}^\infty \frac{dx}{(1+x^2)(1+(x+y)^2)}.
\]
For $y \geq 0$, in the range $-1 \leq x \leq 0$,
we have
\begin{align*}
(1+x^2)(1+(x+y)^2) &\leq (1+1)(1+(1+y)^2) = 2y^2+4y+4 \\
&\leq 2y^2+4+2(y^2+1) \leq 6+6y^2.
\end{align*}
We thus have the lower bound
\[
f(y) \geq \frac{1}{6(1+y^2)};
\]
the same bound is valid for $y \leq 0$ because $f(y) = f(-y)$.
The original hypothesis can be written as
\[
\sum_{i,j=1}^n f(a_i - a_j) \leq An
\]
and thus
implies that
\[
\sum_{i,j=1}^n \frac{1}{1 + (a_i-a_j)^2} \leq 6An.
\]
By the Cauchy-Schwarz inequality, this implies
\[
\sum_{i,j=1}^n (1 + (a_i-a_j)^2) \geq Bn^3
\]
for $B = 1/(6A)$.
\textbf{Remark.}
One can also compute explicitly (using partial fractions, Fourier transforms, or contour integration)
that $f(y) = \frac{2\pi}{4 + y^2}$.
\textbf{Remark.}
Praveen Venkataramana points out that the lower bound can be improved to $Bn^4$
as follows. For each $z \in \ZZ$, put $Q_{z,n} = \{i\in \{1,\dots,n\}: a_i \in [z, z+1)\}$ and $q_{z,n} = \#Q_{z,n}$. Then
$\sum_z q_{z,n} = n$ and
\[
6An \geq \sum_{i,j=1}^n \frac{1}{1 + (a_i-a_j)^2} \geq \sum_{z \in \ZZ} \frac{1}{2} q_{z,n}^2.
\]
If exactly $k$ of the $q_{z,n}$ are nonzero, then $\sum_{z \in \ZZ} q_{z,n}^2 \geq n^2/k$ by Jensen's inequality
(or various other methods), so we must have $k \geq n/(6A)$. Then
\begin{align*}
\sum_{i,j=1}^n (1 + (a_i-a_j)^2) &\geq
n^2 + \sum_{i,j=1}^k \max\{0, (|i-j|-1)^2\} \\
&\geq
n^2 + \frac{k^4}{6}-\frac{2k^3}{3}+\frac{5k^2}{6}-\frac{k}{3}.
\end{align*}
This is bounded below by $Bn^4$ for some $B>0$.
In the opposite direction, one can weaken the initial upper bound to $An^{4/3}$ and still
derive a lower bound of $Bn^3$. The argument is similar.
\item[B6]
In order to interpret the problem statement, one must choose a convention for the value of $0^0$; we will take
it to equal 1. (If one takes $0^0$ to be 0, then the problem fails for $p=3$.)
\textbf{First solution.}
By Wilson's theorem,
\[
k! (p-1-k)! \equiv (-1)^{k} (p-1)! \equiv (-1)^{k+1} \pmod{p},
\]
so we have a congruence of Laurent polynomials
\begin{align*}
\sum_{k=0}^{p-1} k! x^k &\equiv \sum_{k=0}^{p-1} \frac{(-1)^{k+1} x^k}{(p-1-k)!} \pmod{p} \\
&\equiv -x^{p-1} \sum_{k=0}^{p-1} \frac{(-x)^{-k}}{k!} \pmod{p}.
\end{align*}
Replacing $x$ with $-1/x$, we reduce the original problem to showing that the polynomial
\[
g(x) = \sum_{k=0}^{p-1} \frac{x^k}{k!}
\]
over $\FF_p$ has at most $(p-1)/2$ nonzero roots in $\FF_p$. To see this, write
\[
h(x) = x^p - x + g(x)
\]
and note that by Wilson's theorem again,
\[
h'(x) = 1 + \sum_{k=1}^{p-1} \frac{x^{k-1}}{(k-1)!} = x^{p-1} - 1 + g(x).
\]
If $z \in \FF_p$ is such that $g(z) = 0$, then $z \neq 0$ because $g(0) = 1$.
Therefore, $z^{p-1} = 1$, so $h(z) = h'(z) = 0$ and so $z$ is at least a double root of $h$.
Since $h$ is a polynomial of degree $p$, there can be at most $(p-1)/2$ zeroes of $g$ in $\FF_p$, as desired.
\noindent
\textbf{Second solution.} (By Noam Elkies)
Define the polynomial $f$ over $\FF_p$ by
\[
f(x) = \sum_{k=0}^{p-1} k! x^k.
\]
Put $t = (p-1)/2$; the problem statement is that $f$
has at most $t$ roots modulo $p$.
Suppose the contrary; since $f(0) = 1$,
this means that $f(x)$ is nonzero for at most $t-1$ values of $x \in \FF_p^*$.
Denote these values by $x_1,\dots,x_m$, where by assumption $m < t$, and define the polynomial
$Q$ over $\FF_p$ by
\[
Q(x) =
\prod_{k=1}^m (x - x_m) = \sum_{k=0}^{t-1} Q_k x^k.
\]
Then we can write
\[
f(x) = \frac{P(x)}{Q(x)} (1-x^{p-1})
\]
where $P(x)$ is some polynomial of degree at most $m$.
This means that the power series expansions of $f(x)$ and $P(x)/Q(x)$ coincide modulo $x^{p-1}$,
so the coefficients of $x^t, \dots, x^{2t-1}$ in $f(x) Q(x)$ vanish.
In other words, the product of the square matrix
\[
A = ((i+j+1)!)_{i,j=0}^{t-1}
\]
with the nonzero column vector $(Q_{t-1}, \dots, Q_0)$ is zero.
However, by the following lemma, $\det(A)$ is nonzero modulo $p$, a contradiction.
\begin{lemma}
For any nonnegative integer $m$ and any integer $n$,
\[
\det((i+j+n)!)_{i,j=0}^m = \prod_{k=0}^m k! (k+n)!.
\]
\end{lemma}
\begin{proof}
Define the $(m+1) \times (m+1)$ matrix $A_{m,n}$ by $(A_{m,n})_{i,j} = \binom{i+j+n}{i}$;
the desired result is then that $\det(A_{m,n}) = 1$. Note that
\[
(A_{m,n-1})_{ij} = \begin{cases}
(A_{m,n})_{ij} & i=0 \\
(A_{m,n})_{ij} - (A_{m,n})_{(i-1)j} & i > 0;
\end{cases}
\]
that is, $A_{m,n-1}$ can be obtained from $A_{m,n}$ by elementary row operations.
Therefore, $\det(A_{m,n}) = \det(A_{m,n-1})$, so $\det(A_{m,n})$ depends only on $m$.
The claim now follows by observing that $A_{0,0}$ is the $1 \times 1$ matrix with entry 1
and that $A_{m,-1}$ has the block representation $\begin{pmatrix} 1 & * \\ 0 & A_{m-1,0} \end{pmatrix}$.
\end{proof}
\noindent
\textbf{Remark.}
Elkies has given a more detailed discussion of the origins of this solution in the theory of orthogonal
polynomials; see \texttt{http://mathoverflow.net/questions/82648}.
\end{itemize}
\end{document}