Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy:

$$
\forall x, y \in \mathbb{R}, \quad f(f(x+y))=f(x)+y
$$
By exchanging $x$ and $y$, we obtain for all $x, y \in \mathbb{R}: f(x)+y=f(y)+x$. In particular, by taking $x=0$ we have for all $y \in \mathbb{R}, f(y)=y+f(0)$. Conversely, the only solution function is the identity.

It was also possible to notice that $f$ is bijective by taking $x=0$: for all $y \in \mathbb{R}$, $f(f(y))=y+f(0)$. For $y=0$, we have for all $x \in \mathbb{R}, f(f(x))=f(x)$, so $f(x)=x$ by injectivity. We indeed fall back on the same result.

Finally, taking $y=-x$ also allows us to conclude.