\documentclass{article}
\newtheorem{problem}{Problem}
\title{CHM: A.4}
\author{Po-Shen Loh}
\date{9 April 2001}
\begin{document}
\maketitle

\begin{problem}[A.4]
  Call a polynomial \emph{positive reducible}\/ if it can be written
  as a product of two nonconstant polynomials with positive real
  coefficients.  Let $f(x)$ be a polynomial with $f(0) \neq 0$ such
  that $f(x^n)$ is positive reducible for some natural number $n$.
  Prove that $f(x)$ itself is positive reducible.
\end{problem}

\noindent \textbf{Solution:}

Let $a(x) = f(x^n)$.  That is, set the coefficient of $x^{nk}$ in
$a(x)$ equal to the coefficient of $x^k$ in $f(x^n)$, and set all
other coefficients in $a(x)$ to zero.  Since $a$ is positive
reducible, let it be the product $bc$.  Observe that since $a$ only
has coefficients for the $x^{kn}$ monomials, its only nonzero
derivatives at zero will be those of the $kn$-th orders.

Let's start taking derivatives (for brevity, we evaluate them all at
zero without writing that out): $0 = a' = b'c + bc'$.  Since we were
told that $f(0) \neq 0$, it follows that $b(0), c(0)$ are also
nonzero.  Yet they must be positive.  Therefore, $b'$ and $c'$ must
both be zero at zero.  Do it again: $0 = a'' = b''c + 2b'c' + bc''$.
From the previous step, the middle term is zero.  And we have the same
situation again so $b'' = c'' = 0$ at zero.

Keep going until we hit the $n$-th derivative.  Then $n!a_n = a^{(n)} =
b^{(n)}c + b^{(n-1)}c' + \cdots + bc^{(n)}$.  Everything in the middle
falls out again so we are left with the outer terms; the $n$-th
derivatives of $b$ and $c$ can be nonzero.  But all derivatives below
that had to be zero from our previous paragraph.

Continue the process.  We see that unless we are evaluating a
derivative of order divisible by $n$, the nonzero derivatives are
forced to multiply by derivatives that evaluate to zero.  It is only
when we are divisible by $n$ that they can align to multiply by each
other.  But for derivatives of order not divisible by $n$, we know
that they must be zero since $a$ only has terms $x^{kn}$.  Therefore,
we will find that for $b$ and $c$, the only nonzero derivatives at
zero are of order divisible by $n$; therefore, $b$ and $c$ are
polynomials in $x^n$ as well.  Therefore, we can let $y = x^n$ and
then have new polynomials $\beta(y)$ and $\gamma(y)$ that are equal to
$b$ and $c$; also, $f(y) = a(x)$, so we find that $f(y) = \beta(y)
\gamma(y)$.  Also, since we did not perturb the coefficients in going
from the latin to the greek, we still have positive reducibility.  And
we are done.

\end{document}

