Subfield: Algebra


Question: Let $\mathbb{R}$ be the set of real numbers. We denote by $\mathcal{F}$ the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

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

for every $x, y \in \mathbb{R}$. Find all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{R}$ satisfying $f(z)=q z$.

Solution:
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 $f(z)=q z$. 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 $f(f(0))=2 f(0)$. Then, $P(0, f(0))$ gives $f(2 f(0))=3 f(0)$. We claim that

$$
f(k f(0))=(k+1) f(0)
$$

for every integer $k \geqslant 1$. The claim can be proved by induction. The cases $k=1$ and $k=2$ have already been established. Assume that $f(k f(0))=(k+1) f(0)$ and consider $P(0, k f(0))$ which gives

$$
f((k+1) f(0))=f(0)+f(k f(0))=(k+2) f(0) \text {. }
$$

This proves the claim. We conclude that $\frac{k+1}{k} \in Z$ for every integer $k \geqslant 1$. Note that $P(-f(0), 0)$ gives $f(-f(0))=0$. We now claim that

$$
f(-k f(0))=(-k+1) f(0)
$$

for every integer $k \geqslant 1$. The proof by induction is similar to the one above. We conclude that $\frac{-k+1}{-k} \in Z$ for every integer $k \geqslant 1$. 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 $f(z) \neq p z$ for every $z \in \mathbb{R}$. The strategy is to first construct a function

$$
g:[0,1) \rightarrow \mathbb{Z}
$$

and then define $f$ as $f(x)=g(\{x\})+\lfloor x\rfloor$. This function $f$ belongs to $\mathcal{F}$. Indeed,

$$
\begin{aligned}
f(x+f(y)) & =g(\{x+f(y)\})+\lfloor x+f(y)\rfloor \\
& =g(\{x+g(\{y\})+\lfloor y\rfloor\})+\lfloor x+g(\{y\})+\lfloor y\rfloor\rfloor \\
& =g(\{x\})+\lfloor x\rfloor+g(\{y\})+\lfloor y\rfloor \\
& =f(x)+f(y),
\end{aligned}
$$

where we used that $g$ only takes integer values.

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

$$
\frac{m-p \alpha}{p-1} \text { and } \quad \frac{(m+1)-p \alpha}{p-1}
$$

are integers, and so is their difference. The latter is equal to

$$
\frac{1}{p-1}
$$

Since we assumed $p \notin S, 1 /(p-1)$ is never an integer. This is a contradiction.

Define $g:[0,1) \rightarrow \mathbb{Z}$ by $g(\alpha)=m$ for any integer $m$ that satisfies the conclusion of Lemma 1. Note that $f(z) \neq p z$ if and 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 $Z \subset S$.

Final answer: $\left\{\frac{n+1}{n}: n \in \mathbb{Z}, n \neq 0\right\}$