Putnam 2013, Problem A-3

Suppose that the real numbers \( a_0, a_1, \ldots, a_n \) and \( x \), with \( 0 < x < 1 \), satisfy
\[
\frac{a_0}{1 - x} + \frac{a_1}{1 - x^2} + \cdots + \frac{a_n}{1 - x^{n+1}} = 0.
\]
Prove that there exists a real number \( y \) with \( 0 < y < 1 \) such that
\[
a_0 + a_1 y + \cdots + a_n y^n = 0.
\]

Solution:
Suppose on the contrary that $a_0 + a_1 y + \cdots + a_n y^n$ is nonzero for $0 < y < 1$. By the intermediate value theorem, this is only possible if $a_0 + a_1 y + \cdots + a_n y^n$ has the same sign for $0 < y < 1$; without loss of generality, we may assume that $a_0 + a_1 y + \cdots + a_n y^n > 0$ for $0 < y < 1$. For the given value of $x$, we then have

$$a_0 x^m + a_1 x^{2m} + \cdots + a_n x^{(n+1)m} \geq 0$$

for $m = 0, 1, \ldots$, with strict inequality for $m > 0$. Taking the sum over all $m$ is absolutely convergent and hence valid; this yields

$$\frac{a_0}{1-x} + \frac{a_1}{1-x^2} + \cdots + \frac{a_n}{1-x^{n+1}} > 0,$$

a contradiction.