\documentclass[11pt]{scrartcl}
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\begin{document}
\title{Twitch 145.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 145}
\maketitle

\section*{Problem}
Find all functions $f \colon \mathbb R \rightarrow \mathbb R$
such that for all $x, y \in \mathbb R$, we have:
\[ f(x) = f(x + y + f(y)). \]

\section*{Video}
\href{https://www.youtube.com/watch?v=tc9C60_2-KM&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/tc9C60\_2-KM}}



\newpage

\section*{Solution}
Let $S$ be any $\ZZ$-submodule (additive subgroup) of $\RR$.
For each coset $T \in \RR/S$ we choose a representative $\rho(T)$ of it;
in other words we pick an arbitrary choice function
\[ f(x) = -\rho(x) \]
is a solution to the given equation, and moreover all solutions are of this form.

\begin{itemize}
  \ii To check these work, note that $y+f(y) = y-\rho(y) \in S$,
  hence $\rho(x) = \rho(x+y-\rho(y))$, so they work.

  \ii Conversely, given any function $f$ satisfying the given equation,
  if we let
  \[ S \coloneqq \left< y + f(y) \mid y \in \RR \right>_\ZZ \]
  then $f$ is invariant under shifting by $S$,
  i.e.\ $f(x+s) = f(x)$ for all $x \in \RR$ and $s \in S$.
  In other words, $f(x)$ only depends on $x \pmod S$.
  Since $x+f(x) \in S$ by construction, this finishes the proof.
\end{itemize}

\end{document}
