Team Selection Test for JBMO 2023

All constant functions.

Let $a<b$ be two arbitrary numbers. Consider a sufficiently large $x$ so that $x>-a$ and $x>b-f(-a)$, thus $-x<a$ and $x+f(x)>b-f(-a)+f(-a)=b$ (here we used $f(x)\ge f(-a)$ since $x>-a$). Now $-x<a<b<x+f(x)$ while $f(-x)=f(x+f(x))$, hence $f(a)=f(b)$, so $f$ is constant.