Find all functions $f$ such that for all real numbers $x, y$ we have:

$$
f(x\lfloor y\rfloor)=\lfloor f(x)\rfloor f(y)
$$

where $\lfloor x\rfloor$ denotes the greatest integer $n$ such that $n \leq x$.

## 2 Solutions

## Thursday, August 19
On a

$$
f(x\lfloor y\rfloor)=f(x\lfloor\lfloor y\rfloor\rfloor)=\lfloor f(x)\rfloor f(\lfloor y\rfloor)
$$

So we have $(f(\lfloor y\rfloor)-f(y))\lfloor f(x)\rfloor=0$ for all real $x$ and $y$.

- If $\lfloor f(x)\rfloor=0$ for all $x$, then $f(y \times 1)=f(y)\lfloor f(1)\rfloor=0$ for all real $y$.
- Otherwise, there exists $x_{0}$ such that $\left\lfloor f\left(x_{0}\right)\right\rfloor \neq 0$, so for all real $y$ we have $f(\lfloor y\rfloor)=f(y)$. Therefore, we immediately have $f(x)=f(\lfloor x\rfloor 1)=f(x)\lfloor f(1)\rfloor$ and thus $f(x)(1-\lfloor f(1)\rfloor)=0$. For $x=x_{0}$, we have $\lfloor f(1)\rfloor=1$.
- If $\lfloor f(0)\rfloor=0$, then we have:

$$
1 \leq f(x) \leq f\left(2 \cdot \frac{1}{2}\right)=f(2)\left\lfloor f\left(\frac{1}{2}\right)\right\rfloor=f(2)\left\lfloor f\left(\left\lfloor\frac{1}{2}\right\rfloor\right)\right\rfloor=f(2)\lfloor f(0)\rfloor=0
$$

which is impossible.

- If $\lfloor f(0)\rfloor \neq 0$, using the original functional equation, we have $f(0)=f(x)\lfloor f(0)\rfloor$ for all $x$. Therefore, $f(x)=c$ where $c$ is a constant, and we have $\lfloor f(1)\rfloor=\lfloor c\rfloor=1$.

Conversely, the constant functions $f=c$ with $c \in[1,2[$.

## 4 Tests

## 1 Basic Strategy Test (3h)

## Problems