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

\begin{document}
\title{Twitch 059.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 59}
\maketitle

\section*{Problem}
Find all functions $f \colon \ZZ_{\ge 0} \to \RR$ such that
\[ f(2x+3y) = 3f(x)+2f(y) \]
holds.

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



\newpage

\section*{Solution}
The only solution is $f \equiv 0$, which works.

By plugging in zeros,
we get $f(0)=0$, $f(2t)=3f(t)$, $f(3t)=2f(t)$.
Also, $x=y=t$ gives $f(5t)=5f(t)$.


Now, for any $t \in \RR$, let $y = f(t)$.
Then
\begin{align*}
  f(10t+3t) &= f(4t+9t) \\
  3f(5t)+2f(t) &= 3f(2t)+2f(3t) \\
  15y + 2y &= 9y + 4y
\end{align*}
Thus $f \equiv 0$.

\end{document}
