\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{DeuX MO 2020/I/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 43}
\maketitle
\section*{Problem}
Solve $f(f(x)f(y)) = xy-x-y$ over reals.
\section*{Video}
\href{https://www.youtube.com/watch?v=KXdg6Kv7d9k&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/KXdg6Kv7d9k}}
\newpage
\section*{Solution}
The only answer is $t \mapsto t-1$ which obviously works.
Let $P(x,y)$ denote the given assertion.
\begin{claim*}
$f$ is a bijection.
\end{claim*}
\begin{proof}
Follows from $P(x,0)$.
\end{proof}
\begin{claim*}
$f(1) = 0$ and $f(0) = -1$.
\end{claim*}
\begin{proof}
From $P(x,1)$ we have $f(f(x)f(1)) = -1$.
Since $f$ is bijection, this can only occur if $f(1)=0$.
\end{proof}
By $P(x,0)$ we have $f(-f(x)) = -x$.
So, if we let $g(x) = -f(x)$, then $g$ is an involution and
\[ g(g(x)g(y)) = x+y-xy \]
or equivalently
\begin{align*}
g(xy) &= g(x) + g(y) - g(x) - g(y) \\
\iff 1 - g(xy) &= \left( 1-g(x) \right)\left( 1-g(y) \right).
\end{align*}
Let's define \[ h = 1-g \]
so $h$ is multiplicative and bijective
with $h(0) = 0$, $h(1) = 1$, and hence $h(-1) = -1$.
Then the relation $g(g(x)) = x$ rewrites as
\[ h(1-h(x)) = 1-x. \qquad(\heartsuit) \]
\begin{claim*}
We have $h(h(x)) = x$.
\end{claim*}
\begin{proof}
Via
\begin{align*}
\frac{h(-1) \cdot (1-x)}{h(h(x))}
&= \frac{h(-1) \cdot h(1-h(x))}{h(h(x))}
= \frac{h(h(x)-1)}{h(h(x))} \\
&= h\left( 1 - \frac{1}{h(x)} \right)
= h\left( 1 - h\left( \frac 1x \right) \right) = 1 - \frac 1x.
\qedhere
\end{align*}
\end{proof}
Now $(\heartsuit)$ and the previous claim gives $1-h(x) = h(1-x)$;
and by taking $x=a/b$, we find that $h$ is an additive function.
Since $h$ is both additive and multiplicative on the real numbers
(and nonconstant), it must be the identity function.
This gives the set of solutions claimed earlier.
\end{document}