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 $x$. (France) Answer. $f(x)=$ const $=C$, where $C=0$ or $1 \leq C<2$.
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).