KAME06 wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y$ \[f(x + y)=f(f(x)) + y + 2022\] Setting $x=0$, we get $f(y)=y+a$ for some $a=f(f(0))+2022$ Plugging this back in equation, we get $a=-2022$ and solution $\boxed{f(x)=x-2022\quad\forall x}$, which indeed fits