Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$ :

$$
f(f(f(x)+f(y)))=f(x)+y
$$
First, $f$ is surjective. Indeed, by taking $x=0$ and $y=z-f(0)$:

$$
\mathrm{f}(\mathrm{f}(\mathrm{f}(0)+\mathrm{f}(z-\mathrm{f}(0))))=\mathrm{f}(0)+z-\mathrm{f}(0)=z
$$

Let $x_{0}$ be a preimage of 0. We inject $x=x_{0}$:

$$
f(f(0+f(y)))=y
$$

We inject $y=x_{0}$:

$$
f(f(f(x)+0))=f(x)
$$

From (III.19) and (III.20), we conclude that $f(z)=z$ for all $z \in \mathbb{R}$.