\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{JMO 2024/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 142}
\maketitle
\section*{Problem}
Solve over $\RR$ the functional equation $f(x^2-y)+2yf(x) = f(f(x))+f(y)$.
\section*{Video}
\href{https://www.youtube.com/watch?v=yLhlJHJDxvQ&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/yLhlJHJDxvQ}}
\section*{External Link}
\url{https://aops.com/community/p30227204}
\newpage
\section*{Solution}
The answer is $f(x) \equiv x^2$, $f(x) \equiv 0$, $f(x) \equiv -x^2$,
which obviously work.
Let $P(x,y)$ be the usual assertion.
\begin{claim*}
We have $f(0) = 0$ and $f$ even.
\end{claim*}
\begin{proof}
Combine $P(1,1/2)$ with $P(1,0)$ to get $f(0) = 0$.
Use $P(0,y)$ to deduce $f$ is even.
\end{proof}
\begin{claim*}
$f(x) \in \{-x^2, 0, x^2\}$ for every $x \in \RR$.
\end{claim*}
\begin{proof}
Note that $P(x,x^2/2)$ and $P(x,0)$ respectively give
\[ x^2f(x) = f(x^2) = f(f(x)). \]
Repeating this key identity several times gives
\begin{align*}
f(f(f(x))) &= f(f(x^2)) = f(x^4) = x^4 f(x^2) \\
&= f(x)^2 \cdot f(f(x)) = f(x)^2 f(x^2) = f(x)^3 x^2
\end{align*}
Suppose $t \neq 0$ is such that $f(t^2) \neq 0$.
Then the above equalities imply
\[ t^4 f(t^2) = f(t)^2 f(t^2) \implies f(t) = \pm t^2 \]
and then
\[ f(t)^2 f(t^2) = f(t)^3 t^2 \implies f(t^2) = \pm t^2. \]
Together with $f$ even, we get the desired result.
\end{proof}
\begin{remark*}
Another proof is possible here that doesn't use as iterations of $f$:
the idea is to ``show $f$ is injective up to sign outside its kernel''.
Specifically, if $f(a) = f(b) \neq 0$,
then $a^2f(a) = f(f(a)) = f(f(b)) = b^2f(b) \implies a^2=b^2$.
But we also have $f(f(x))=f(x^2)$,
so we are done except in the case $f(f(x))=f(x^2)=0$.
That would imply $x^2f(x) = 0$, so the claim follows.
\end{remark*}
Now, note that $P(1,y)$ gives
\[ f(1-y) + 2y \cdot f(1) = f(1) + f(y). \]
We consider cases on $f(1)$ and show that $f$ matches the desired form.
\begin{itemize}
\ii If $f(1) = 1$, then $f(1-y) + (2y-1) = f(y)$.
Consider the nine possibilities that arise:
\[
\begin{array}{lll}
(1-y)^2 + (2y-1) = y^2 &
0 + (2y-1) = y^2 &
-(1-y)^2 + (2y-1) = y^2 \\
(1-y)^2 + (2y-1) = 0 &
0 + (2y-1) = 0 &
-(1-y)^2 + (2y-1) = 0 \\
(1-y)^2 + (2y-1) = -y^2 &
0 + (2y-1) = -y^2 &
-(1-y)^2 + (2y-1) = -y^2.
\end{array}
\]
Each of the last eight equations is a nontrivial polynomial equation.
Hence, there is some constant $C > 100$ such that the latter eight equations
are all false for any real number $y > C$.
Consequently, $f(y) = y^2$ for $y > C$.
Finally, for any real number $z > 0$, take $x,y > C$ such that $x^2-y=z$;
then $P(x,y)$ proves $f(z) = z^2$ too.
\ii Note that (as $f$ is even), $f$ works iff $-f$ works,
so the case $f(1) = -1$ is analogous.
\ii If $f(1) = 0$, then $f(1-y) = f(y)$.
Hence for any $y$ such that $|1-y| \neq |y|$, we conclude $f(y) = 0$.
Then take $P(2, 7/2) \implies f(1/2) = 0$.
\end{itemize}
\begin{remark*}
There is another clever symmetry approach possible after the main claim.
The idea is to write
\[ P(x,y^2) \implies f(x^2-y^2) + 2y^2f(x) = f(f(x)) + f(f(y)). \]
Since $f$ is even gives $f(x^2-y^2) = f(y^2-x^2)$,
one can swap the roles of $x$ and $y$ to get $2y^2f(x) = 2x^2f(y)$.
Set $y=1$ to finish.
\end{remark*}
\end{document}