Find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ satisfying

$$
\forall x, y \in \mathbb{Q}, \quad f(x+f(y))=f(x)+y
$$
We start by substituting $x$ with 0. We obtain $f(f(y))=y+f(0)$

We apply the trick from Exercise 5, and show that

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

Then we replace $y$ with 0 and get

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

By combining the two previous identities, we arrive at

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

Taking the first identity we found, we then have

$$
f(f(y))=y
$$

We know that $f$ is surjective. We return to the initial equation. Let $t \in \mathbb{Q}$. We choose $y$ such that $t=f(y)$. Then we know that $f(t)=f(f(y))=y$. We deduce that for all $x$ and for all $t$,

$$
f(x+t)=f(x)+f(t)
$$

Thus, $f$ satisfies Cauchy's equation, and the solutions are necessarily linear. Let's verify them:

$$
f(x)=a x \Longrightarrow a(x+a y)=a x+y \Longrightarrow a^{2} y=y \Longrightarrow a= \pm 1
$$

Finally, the only solutions are $f(x)=x$ and $f(x)=-x$