\documentclass[11pt]{scrartcl}
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\begin{document}
\title{HMIC 2015/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 49}
\maketitle

\section*{Problem}
Let $\omega = e^{\frac{2\pi i}{5}}$ be a primitive fifth root of unity.
Prove that there do not exist integers $a,b,c,d,k$ with $k>1$ such that
\[ (a+b\omega+c\omega^2+d\omega^3)^k=1+\omega. \]

\section*{Video}
\href{https://www.youtube.com/watch?v=DzEdbFcr4DU&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/DzEdbFcr4DU}}



\newpage

\section*{Solution}
Preliminary definitions:
\begin{itemize}
\ii Let $\varphi = \half(1+\sqrt5)$
  and recall that
  $\opname{Re} \omega = \cos 72\dg = \frac{\sqrt5-1}{4} = \half(\varphi-1)$.
\ii Let $K = \QQ(\omega)$ and $L = \QQ(\sqrt5)$,
  so $\OO_K = \ZZ[\omega]$ while $\OO_L = \ZZ[\varphi]$.
\ii The problem is essentially asking to show that $1+\omega$
  is not perfect power in $\OO_K$.
\end{itemize}

We consider the following norm map:
\[ N \colon \QQ(\omega) \to \QQ(\sqrt5) \text{ by }
  z \mapsto \left\lvert z \right\rvert^2. \]
It restricts to a multiplicative map from $N \colon \OO_K \to \OO_L$.
So we compute
\[ N(z)^k = N(z^k) = N(1+\omega)
  = (1+\omega)\left( 1+\frac{1}{\omega} \right)
  = 2 + 2\opname{Re} \omega = \varphi^2 \]
which is in fact a unit in $\OO_L$.
On the other hand, it is known that the units of $\OO_L$ are
\[ \OO_L^\times = \left< \varphi\right>. \]
So this already means $k \le 2$,
and the problem is reduced to proving the following claim.

\begin{claim*}
There is no $z \in \OO_K$
for which $N(z) = |z|^2 = \pm \varphi$.
\end{claim*}
\begin{proof}
Assume for contradiction there is such a $z$.
We pick a different basis than the one given in the original problem
statement, and assume instead that
\[ z = a\omega + b\omega^2 + c\omega^3 + d\omega^4 \in \OO_K \]
(which is nicer because it is more symmetric).
Then
\begin{align*}
  N(z) &= (a\omega + b\omega^2 + c\omega^3 + d\omega^4)
  (a\omega^{-1} + b\omega^{-2} + c\omega^{-3} + d\omega^{-4}) \\
  &= (a^2+b^2+c^2+d^2) + (ab+bc+cd) \varphi + (ac+bd+ad) \varphi^2 \\
  &= (a^2+b^2+c^2+d^2 + ac+bd+ad) \\
  &+ (ab+bc+cd+ac+bd+ad) \varphi
\end{align*}
However, this means
\[ 0 = a^2+b^2+c^2+d^2+ac+bd+ad
  = \frac{(a+c)^2}{2} + \frac{(b+d)^2}{2} +
  \frac{(a+d)^2}{2} + \frac{b^2+c^2}{2} \]
and this means $a=b=c=d=0$, a clear impossibility.
\end{proof}

\end{document}
