IMO 2010: f(⌊x⌋y) = f(x)⌊f(y)⌋

Case 1: $f(0)\ne0$. Setting $x=0$: $f(0)=f(0)\lfloor f(y)\rfloor$, so $\lfloor f(y)\rfloor=1$ for every $y$. The equation becomes $f(\lfloor x\rfloor y)=f(x)$ for all $x,y$; with $y=0$ we get $f(0)=f(x)$, so $f$ is a constant $c$ with $\lfloor c\rfloor=1$, i.e. $c\in[1,2)$. These constants satisfy the equation: both sides equal $c$.

Case 2: $f(0)=0$.

*Step 1: $f=0$ on $(0,1)$.* For $0<x<1$, $\lfloor x\rfloor=0$ and the equation reads $0=f(x)\lfloor f(y)\rfloor$ for all $y$. If $f(x)\ne0$ for some $x\in(0,1)$, then $\lfloor f(y)\rfloor=0$ for all $y$. Now put $x=1$: $f(y)=f(1)\lfloor f(y)\rfloor=0$ for all $y$, contradicting $f(x)\ne0$. Hence $f=0$ on $(0,1)$.

*Step 2: $f=0$ on $(0,\infty)$.* Take $z>0$ and an integer $n\ge1$ with $z/n<1$. Putting $x=n$ and $y=z/n$: $f(z)=f(n)\lfloor f(z/n)\rfloor=f(n)\lfloor0\rfloor=0$.

*Step 3: $f=0$ on $(-\infty,0)$.* Put $x=-1$: $f(-y)=f(-1)\lfloor f(y)\rfloor$. For $y\in(0,1)$, $f(y)=0$ so $f(-y)=0$: $f$ vanishes on $(-1,0)$. For $z<0$, take an integer $m\ge1$ with $|z|/m<1$, and put $x=-m$ and $y=z/(-m)\in(0,1)$. Then $\lfloor x\rfloor y=z$ and $f(z)=f(-m)\lfloor f(y)\rfloor=f(-m)\lfloor0\rfloor=0$.

So $f\equiv0$.

Answer. $f\equiv0$, or $f\equiv c$ for a constant $c\in[1,2)$.

Lesson. Case-split on the "degenerate" value ($f(0)=0$ or not) first, then propagate zeros from a small interval outward using the scaling built into the equation.