\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Putnam 1972 B6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 62}
\maketitle

\section*{Problem}
Let $n_1 < n_2 < n_3 < \dotsb < n_k$ be a set of positive integers.
Prove that the polynomial $1+z^{n_1} + z^{n_2} + \dotsb + z^{n_k}$
has no roots inside the circle $|z| < (\sqrt 5 - 1)/2$.

\section*{Video}
\href{https://www.youtube.com/watch?v=iKfVuypCVos&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/iKfVuypCVos}}



\newpage

\section*{Solution}
Assume for contradiction $z$ is a root with $|z| < \frac{\sqrt5-1}{2}$,
and hence $|z| < 1$.
Let $z = x+yi$ and $|z| = r$ for brevity.

Note that
\[ -2\left( 1+z^{n_1}+\dots \right) +
  \underbrace{(z+z^2+z^3+\dots)}_{=\frac{z}{1-z}}
  = -2 \pm z \pm z^2 \pm \dots \]
Since $z$ is a root of the polynomial, we have
\begin{align*}
  \frac{z}{1-z} &= -2 \pm z \pm z^2 \pm \dots \\
  \iff \frac{2-z}{1-z} &= \pm z \pm z^2 \pm \dots \\
  \implies \frac{|2-z|}{|1-z|} &\le |z| + |z|^2 + \dots
    = \\
  \iff \frac{(2-z)(2-\ol z)}{(1-z)(1-\ol z)}
    &\le \frac{r^2}{(1-r)^2} \\
  \iff \frac{4 - 4x + r^2}{1 - 2x + r^2} &\le \frac{r^2}{(1-r)^2} \\
  \iff 2 + \frac{2 - r^2}{1 - 2x + r^2} &\le \frac{r^2}{(1-r)^2} \\
  \implies 2 + \frac{2 - r^2}{1 + 2r + r^2} &\le \frac{r^2}{(1-r)^2} \\
  \iff 2(1-r^2)^2 + (2-r^2)(1-r)^2 &\le r^2(1+r)^2 \\
  \iff 0 &\le 4r^2 + 4r - 4
\end{align*}
which implies $r \ge \frac{\sqrt5-1}{2}$, contradiction.

\begin{remark*}
  If $z = -\frac{\sqrt5-1}{2}$, then
  $z + z^3 + z^5 + \dots = -1$,
  so this is a sort of equality case for the problem.
\end{remark*}

\end{document}
