Problem 1036

Set $x=0$ in the equation, we have $f(0)=f(0)\lfloor f(y)\rfloor$ for all real $y$;

- If $\mathrm{f}(0) \neq 0$, then $\mathrm{f}(\mathrm{y}) \in[1,2[$ for all real $y$. With $y=0$ in the equation, we get $f(x)=f(0)$ for all $x \in \mathbb{R}$, so $f$ is a constant function with value $v \in[1,2[$.
- If $f(0)=0$, we show that $f$ is the zero function. If for some $z \in[0,1[$, $f(z) \neq 0$, we get with $x=z: 0=f(0)=f(z)\lfloor f(y)]$ so for all real $y, f(y) \in[0,1[$. Take the initial equation with $x=1$ and $y=z$: we get $f(z)=0$, which is a contradiction. Therefore, $f$ is zero on $\left[0,1\left[\right.\right.$. If now $z \in \mathbb{R}$, there exists an integer $n \in \mathbb{Z}$ such that $\frac{z}{n} \in[0,1[$. We get with $x=n$ and $y=\frac{z}{n}$ :

$$\mathrm{f}(z)=\mathrm{f}(\mathfrak{n})\left\lfloor\mathrm{f}\left(\frac{z}{\mathfrak{n}}\right)\right\rfloor=0$$

so $f$ is indeed the zero function.

Conversely, these functions satisfy the equation.