, 2023

Let $Z$ be the set of all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{R}$ satisfying $f(z) = qz$. Let further
$$S = \left\{ \frac{n+1}{n} : n \in \mathbb{Z}, n \neq 0 \right\}.$$
We prove that $Z = S$ by showing the two inclusions: $S \subseteq Z$ and $Z \subseteq S$.

We first prove that $S \subseteq Z$. Let $f \in \mathcal{F}$ and let $P(x, y)$ be the relation $f(x + f(y)) = f(x) + f(y)$. First note that $P(0, 0)$ gives $f(f(0)) = 2f(0)$. Then, $P(0, f(0))$ gives $f(2f(0)) = 3f(0)$. We claim that
$$f(kf(0)) = (k + 1)f(0)$$
for every integer $k \ge 1$. The claim can be proved by induction. The case $k = 1$ and $k = 2$ have already been established. Assume that $f(kf(0)) = (k+1)f(0)$ and consider $P(0, kf(0))$ which gives
$$f((k + 1)f(0)) = f(0) + f(kf(0)) = (k + 2)f(0).$$
This proves the claim. We conclude that $\frac{k+1}{k} \in Z$ for every integer $k \ge 1$. Note that $P(-f(0), 0)$ gives $f(-f(0)) = 0$. We now claim that
$$f(-kf(0)) = (-k + 1)f(0)$$
for every integer $k \ge 1$. The proof by induction is similar to the one above. We conclude that $\frac{-k+1}{-k} \in Z$ for every integer $k \ge 1$. This shows that $S \subseteq Z$.

We now prove that $Z \subseteq S$. Let $p$ be a rational number outside the set $S$. We want to prove that $p$ does not belong to $Z$. To that end, we construct a function $f \in \mathcal{F}$ such that $f(z) \ne pz$ for every $z \in \mathbb{R}$. The strategy is to first construct a function
$$g : [0, 1) \to \mathbb{Z}$$
and then define $f$ as $f(x) = g(\{x\}) + \lfloor x \rfloor$. This function $f$ belongs to $\mathcal{F}$. Indeed,
$$\begin{align*} f(x + f(y)) &= g(\{x + f(y)\}) + \lfloor x + f(y) \rfloor \\ &= g(\{x + g(\{y\}) + \lfloor y \rfloor\}) + \lfloor x + g(\{y\}) + \lfloor y \rfloor \rfloor \\ &= g(\{x\}) + \lfloor x \rfloor + g(\{y\}) + \lfloor y \rfloor \\ &= f(x) + f(y), \end{align*}$$
where we used that $g$ only takes integer values. We now introduce the following lemma.

Lemma 1. For every $\alpha \in [0, 1)$, there exists $m \in \mathbb{Z}$ such that
$$m + n \neq p(\alpha + n)$$
for every $n \in \mathbb{Z}$.

Proof. Note that if $p = 1$ the claim is trivial. If $p \neq 1$, then the claim is equivalent to the existence of an integer $m$ such that $\frac{m-p\alpha}{p-1}$ is never an integer. Assume the contrary. That would mean that both $\frac{m-p\alpha}{p-1}$ and $\frac{(m+1)-p\alpha}{p-1}$ are integers, and so is their difference. The latter is equal to $\frac{1}{p-1}$. Since we assumed $p \notin S$, $\frac{1}{p-1}$ is never an integer. This is a contradiction. $\square$

Define $g : [0, 1) \to \mathbb{Z}$ by $g(\alpha) = m$ for any integer $m$ that satisfies the conclusion of Lemma 1. Note that $f(z) \neq pz$ if and only if
$$g(z) + \lfloor z \rfloor \neq p(z + \lfloor z \rfloor).$$
The latter is guaranteed by the construction of the function $g$. We conclude that $p \notin Z$ as desired. This shows that $Z \subset S$.