Problem 3. Determine the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ for which there exists a function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(x)+f(y)=[g(x+y)]
$$

for any real numbers $x, y$.

We have denoted by $[a]$ the integer part of the real number $a$.
Solution. For $x \rightarrow(x+y), y \rightarrow 0$ in the given relation, we deduce

$$
f(x+y)+f(0)=[g(x+y)]=f(x)+f(y)
$$

Denoting $h$.

Denoting $h(x)=f(x)-f(0)$, we obtain $h(x+y)=h(x)+h(y)$ (Cauchy's equation) ...... 1p

We deduce that, for any $x_{0} \in \mathbb{R}$, chosen arbitrarily, and any natural $n$, $h\left(n x_{0}\right)=n h\left(x_{0}\right)$.

It follows that $h\left(x_{0}\right)=h\left(n \cdot \frac{x_{0}}{n}\right) \Leftrightarrow h\left(\frac{x_{0}}{n}\right)=\frac{h\left(x_{0}\right)}{n}$ for any $n \in \mathbb{N}^{*} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots 2 \mathbf{2}$

Since $2 f(x)=[g(2 x)] \in \mathbb{Z} \Rightarrow 2 h(x) \in \mathbb{Z} \Rightarrow \frac{2 h\left(x_{0}\right)}{n} \in \mathbb{Z}$ for any $n \in \mathbb{N}^{*}$. Choosing $n>\left|2 h\left(x_{0}\right)\right|$ we obtain that

$$
0 \leq\left|2 \cdot h\left(\frac{x_{0}}{n}\right)\right|=\left|\frac{2 h\left(x_{0}\right)}{n}\right|<1
$$

from which $h\left(\frac{x_{0}}{n}\right)=0$, hence $h\left(x_{0}\right)=0$. Since $x_{0}$ was chosen arbitrarily, we obtain $h(x)=0$ for any real $x$. We deduce that $f(x)=f(0)$ and since $2 f(x)=[g(2 x)] \in \mathbb{Z}$, it follows that $f(x)=\frac{k}{2}$ for any real $x$, where $k$ is a fixed integer.

$3 p$