\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{USEMO 2020/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 35}
\maketitle
\section*{Problem}
A function $f$ from the set of positive real numbers to itself satisfies
\[ f(x+f(y)+xy) = xf(y)+f(x+y) \]
for all positive real numbers $x$ and $y$.
Prove that $f(x) = x$ for all positive real numbers $x$.
\section*{Video}
\href{https://www.youtube.com/watch?v=5a_XCGKiXnI}{\texttt{https://youtu.be/uj93tNL8f7M}}
\newpage
\section*{Solution}
We present two solutions.
\paragraph{First solution (Nikolai Beluhov)}
We first begin with the following observation.
\begin{claim*}
We must have $f(y) \ge y$ for all $y > 0$.
\end{claim*}
\begin{proof}
Otherwise, choose $0 < x < 1$ satisfying that $f(y) = (1-x) \cdot y$.
Then plugging this $P(x,y)$ gives $xf(y) = 0$, contradiction.
\end{proof}
Now, we make the substitution $f(x) = x+g(x)$,
so that $g$ is a function $\RR_{>0} \to \RR_{\ge 0}$.
The given function equation reads
$g(x+xy+(y+g(y))) + x+(y+g(y)) = (xy+xg(y)) + (x+y+g(x+y))$, or
\[ g(x+y+xy+g(y)) = (x-1) g(y) + g(x+y). \qquad (\dagger) \]
We have to show that $g$ is the zero function from $(\dagger)$.
\begin{claim*}
[Injectivity for nonzero outputs]
If $g(a) = g(b)$ for $a \neq b$,
then we must actually have $g(a) = g(b) = 0$.
\end{claim*}
\begin{proof}
Setting $(a,b)$ and $(b,a)$ in $(\dagger)$ gives
$(a-1) g(b) = (b-1) g(a)$
which, since $a-1 \neq b-1$, forces $g(a) = g(b) = 0$.
\end{proof}
\begin{claim*}
[$g$ vanishes on $(1, \infty)$]
We have $g(t) = 0$ for $t > 1$.
\end{claim*}
\begin{proof}
If we set $x=1$ in $(\dagger)$ we obtain that
$ g(g(y) + 2y + 1) = g(1+y) $.
As the inputs are obviously unequal,
the previous claim gives $g(1+y) = 0$ for all $y > 0$.
\end{proof}
Now $x = 2$ in $(\dagger)$ to get $g(y) = 0$, as needed.
\paragraph{Second solution (from authors)}
We start with the same opening of showing $f(y) \ge y$,
defining $f(x) = x + g(x)$, so $g$ satisfies $(\dagger)$.
Here is another proof that $g \equiv 0$ from $(\dagger)$.
\begin{claim*}
If $g$ is not the zero function,
then for any constant $C$,
we have $g(t) > C$ for sufficiently large $t$.
\end{claim*}
\begin{proof}
In $(\dagger)$ fix $y$ to be any input for which $g(y) > 0$.
Then \[ g\left( (1+y)x + (y+g(y)) \right) \ge (x-1) g(y) \]
so for large $x$, we get the conclusion.
\end{proof}
\begin{remark*}
You could phrase the lemma succinctly as
``$\lim_{x \to \infty} g(x) = +\infty$''.
But I personally think it's a bit confusing to do so
because in practice we usually talk about limits of continuous
(or well-behaved) functions,
so a statement like this would have the wrong connotations,
even if technically correct and shorter.
\end{remark*}
On the other hand,
by choosing $x=1$ and $y=t-1$ for $t > 1$ in ($\dagger$), we get
\[ g(2t+g(z)-1) = g(t) \]
and hence one can generate an infinite sequence of fixed points:
start from $t_0 = 100$, and
define $t_n = 2t_{n-1} + g(t_{n-1}) - 2 > t_{n-1} + 98$ for $n \ge 1$ to get
\[ g(t_0) = g(t_1) = g(t_2) = \cdots \]
and since the $t_i$ are arbitrarily large,
this produces a contradiction if $g \not\equiv 0$.
\end{document}