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

\begin{document}
\title{CAMO 2021/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 62}
\maketitle

\section*{Problem}
Find all functions $f \colon \RR_{> 0} \to \RR_{\ge 0}$ satisfying
\[ \big(f(x)+f(y)\big)^3=xf(x)+yf(y)+3f(xy)\big(f(x)+f(y)\big) \]
for all positive real numbers $x$ and $y$.

\section*{Video}
\href{https://www.youtube.com/watch?v=AlIvbT2FCI4&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/AlIvbT2FCI4}}

\section*{External Link}
\url{https://aops.com/community/p20150150}

\newpage

\section*{Solution}
The answer is any function of the form
\[
   f(x)
   = \begin{cases}
     0 & x \notin S \\
     \sqrt x & x \in S
   \end{cases}
\]
where $S$ is a subset of $\RR_{>0}$ which is closed under multiplication and division
(i.e.\ a multiplicative subgroup).
The verification is left to the reader.

Let $k = f(1) \ge 0$, plug in $x=y=1$ to get
\[ (2k)^3 = 2k + 6k^2 \implies k \in \{0,1\}. \]
\begin{claim*}
  If $k = 0$, then $f$ is identically zero.
\end{claim*}
\begin{proof}
  Plugging in $y=1$ gives
  \[ f(x)^3 = xf(x) + 3f(x)^2
    \implies f(x) \in \left\{ 0, \frac{3 + \sqrt{9+4x}}{2} \right\}. \]
  We say $t$ is bad if $f(t) \neq 0$.
  By plugging in $x=y=t$, we find that if $t$ is bad then $t^2$ is bad
  and actually
  \[ 8 \cdot \left( \frac{3+\sqrt{9+4t}}{2} \right)^3
    = 2t \cdot \frac{3+\sqrt{9+4t}}{2}
    + 3 \cdot \frac{3+\sqrt{9+4t^2}}{2} \cdot (3+\sqrt{9+4t}) \]
  Dividing out:
  \[ 2(3+\sqrt{9+4t})^2 = 2t + 3(3+\sqrt{9+4t^2}). \]
  This has finitely many solutions in $t$ though,
  while a single bad value of $t$ (other than $1$)
  should generate infinitely many.
  Contradiction.
\end{proof}
Otherwise, if $k=1$, then plugging in $y=1$ gives
\[ \left( f(x) + 1 \right)^3 = xf(x) + 1 + 3f(x) \left[ f(x) + 1 \right]. \]
and hence
\[ f(x)^3 = xf(x) \implies f(x) \in \left\{ 0, \sqrt x \right\}. \]
So, to finish, we only need to observe that if $f(x) = \sqrt x$
and $f(y) = \sqrt y$, then plugging in $x$ and $y$
shows $f(xy) = \sqrt{xy}$ as needed.

\end{document}
