We denote by $\mathcal{F}$ the set of functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that

$$
f(x+f(y))=f(x)+f(y)
$$

for all real numbers $x$ and $y$. Find all rational numbers $q$ such that, for every function $f \in \mathcal{F}$, there exists a real number $z$ such that $f(z)=q z$.
Let $f$ be a function belonging to $\mathcal{F}$. For all real numbers $x$ and $y$, it is immediately shown by induction that $f(x+k f(y))=f(x)+k f(y)$ for any integer $k \geqslant 0$. We denote this equality by $E_{k}(x, y)$.
If we set $x^{\prime}=x-k f(y)$, we also observe that $f\left(x^{\prime}\right)=f\left(x^{\prime}+k f(y)\right)-k f(y)=f(x)-k f(y)$. Thus, the equality $E_{k}(x, y)$ is valid even when $k \leqslant 0$.
In particular, if $k \neq 0$ and $z=k f(0)$, the equality $E_{k}(0,0)$ indicates that $f(z)=(k+1) z / k$. This means that all rational numbers of the form $q=1+1 / k$, where $k$ is a non-zero integer, are solutions to the problem.
Conversely, let $a / b$ be an irreducible fraction such that $a / b-1$ is not the reciprocal of an integer, i.e., such that $a \neq b \pm 1$. We then consider the function $f$ defined as follows: for any real number $x$ such that $0 \leqslant x<1$ and any integer $k$, we set

$$
f(k+x)= \begin{cases}k+1 & \text { if } a x \text { is an integer that divides } b-a ; \\ k & \text { otherwise. }\end{cases}
$$

Since $f$ takes integer values, we verify that

$$
f(k+x+f(\ell+y))=f(k+\ell+f(y)+x)=k+\ell+f(y)+f(x)=f(k+x)+f(\ell+x),
$$

which means that $f \in \mathcal{F}$.
Suppose there exists an integer $k$ and a real number $z$ such that $0 \leqslant z<1$ and $f(k+z)=a(k+z) / b$. If $a z$ is not an integer that divides $b-a$, then

$$
a k+a z \equiv b f(k+z) \equiv b k \equiv a k \quad(\bmod b-a),
$$

which is absurd. Thus, $a z$ is an integer that divides $b-a$, and

$$
a k \equiv a(k+z) \equiv b f(k+z) \equiv b(k+1) \equiv a(k+1) \quad(\bmod b-a) .
$$

But then $b-a$ divides $a$, and thus also divides $b$, and even $\operatorname{PGCD}(a, b)=1$, which contradicts the fact that $a-b \neq \pm 1$. Therefore, the fraction $a / b$ is not a solution to the problem.
In conclusion, the solutions to the problem are the fractions of the form $1+1 / k$, where $k$ is a non-zero integer.

Comment from the graders: The problem was difficult and divided into two parts. The first part was to identify the rational solutions by iterating the equation to obtain a formula of the form $f(x+k f(y))=f(x)+k f(y)$ for any integer $k$, which uses elementary reasoning on functional equations. The second part, showing that the rationals found are indeed the only ones, required significantly more ideas, notably constructing somewhat pathological functions that are solutions. This second part was barely touched upon by the students.
We should also note frequent and crossed errors that should no longer be found in a selection test copy:
$\triangleright$ Many students misread the statement. Whether they thought that $q$ had to satisfy the equality $f(z)=q z$ for all $z$ (instead of a single $z$) or that $q$ had to satisfy $f(z)=q z$ for a function $f$ (instead of all functions $f$), students should have realized their mistake by seeing the (very quick, or even too quick) conclusion they reached.
$\triangleright$ Many students claim that the only solutions are the functions $x \mapsto x+c$ with $c$ real. Once again, it was possible to avoid this pitfall by critically examining the final solution obtained (which did not use the rationality of $q$, for example). The sources of errors are often that students, after obtaining that $f(f(y))=f(y)+f(0)$, set $x=f(y)$, obtain $f(x)=x+c$, and conclude, forgetting that the variable $x$ does not necessarily cover $\mathbb{R}$, but only the image of $f$, which is not necessarily surjective. More daring students tried to differentiate the functional equation, which should be avoided when there are no regularity assumptions on $f$.