A1. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality
$$
f([x] y)=f(x)[f(y)] .
$$
holds for all $x, y \in \mathbb{R}$. Here, by $[x]$ we denote the greatest integer not exceeding $x$.
Answer. $f(x)=$ const $=C$, where $C=0$ or $1 \leq C<2$.
Solution 1. First, setting $x=0$ in (1) we get
$$
f(0)=f(0)[f(y)]
$$
for all $y \in \mathbb{R}$. Now, two cases are possible.
Case 1. Assume that $f(0) \neq 0$. Then from (2) we conclude that $[f(y)]=1$ for all $y \in \mathbb{R}$. Therefore, equation (1) becomes $f([x] y)=f(x)$, and substituting $y=0$ we have $f(x)=f(0)=C \neq 0$. Finally, from $[f(y)]=1=[C]$ we obtain that $1 \leq C<2$.
Case 2. Now we have $f(0)=0$. Here we consider two subcases.
Subcase 2a. Suppose that there exists $0<\alpha<1$ such that $f(\alpha) \neq 0$. Then setting $x=\alpha$ in (1) we obtain $0=f(0)=f(\alpha)[f(y)]$ for all $y \in \mathbb{R}$. Hence, $[f(y)]=0$ for all $y \in \mathbb{R}$. Finally, substituting $x=1$ in (1) provides $f(y)=0$ for all $y \in \mathbb{R}$, thus contradicting the condition $f(\alpha) \neq 0$.

Subcase 2b. Conversely, we have $f(\alpha)=0$ for all $0 \leq \alpha<1$. Consider any real $z$; there exists an integer $N$ such that $\alpha=\frac{z}{N} \in[0,1)$ (one may set $N=[z]+1$ if $z \geq 0$ and $N=[z]-1$ otherwise). Now, from (1) we get $f(z)=f([N] \alpha)=f(N)[f(\alpha)]=0$ for all $z \in \mathbb{R}$.
Finally, a straightforward check shows that all the obtained functions satisfy (1).

Solution 2. Assume that $[f(y)]=0$ for some $y$; then the substitution $x=1$ provides $f(y)=f(1)[f(y)]=0$. Hence, if $[f(y)]=0$ for all $y$, then $f(y)=0$ for all $y$. This function obviously satisfies the problem conditions.
So we are left to consider the case when $[f(a)] \neq 0$ for some $a$. Then we have
$$
f([x] a)=f(x)[f(a)], \quad \text { or } \quad f(x)=\frac{f([x] a)}{[f(a)]} .
$$

This means that $f\left(x_{1}\right)=f\left(x_{2}\right)$ whenever $\left[x_{1}\right]=\left[x_{2}\right]$, hence $f(x)=f([x])$, and we may assume that $a$ is an integer.
Now we have
$$
f(a)=f\left(2 a \cdot \frac{1}{2}\right)=f(2 a)\left[f\left(\frac{1}{2}\right)\right]=f(2 a)[f(0)] ;
$$
this implies $[f(0)] \neq 0$, so we may even assume that $a=0$. Therefore equation (3) provides
$$
f(x)=\frac{f(0)}{[f(0)]}=C \neq 0
$$
for each $x$. Now, condition (1) becomes equivalent to the equation $C=C[C]$ which holds exactly when $[C]=1$.