Find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the following functional equation: $$f(\lfloor x\rfloor y)=f(x)\lfloor f(y)\rfloor\quad\quad\text{for all }x,y\in\mathbb{R},$$ where $\lfloor\cdot\rfloor$ is the floor function (largest integer less than or equal to its argument).
(Source: IMO 2010 Shortlist, question A1.)
Some observations:
If there exists a number $x_0$ such as $\lfloor x_0 \rfloor \ne 0$ and $f(x_0)\ne 0$, we can write any number $z=\lfloor x_0 \rfloor y$ for some $y$. Now we get that $f(z)=f(\lfloor x_0 \rfloor y)=f(x_0)\lfloor f(y) \rfloor$. Since $\lfloor f(y) \rfloor$ is always an integer regardless of $y$, that means that all values of $f$ are integer multiples of some number $q=f(x_0)$. Assume that $f$ takes at least two nonzero different values $n_1q$ and $n_2q$, let $f(z_1)=n_1q$ and $f(z_2)=n_2q$, where $|n_1|<|n_2|$ and $z_i$ lie outside of $[0, 1)$. Now we get $z_1=\lfloor z_2 \rfloor y$ for some y and write $n_1q = f(z_1) = f(\lfloor z_2 \rfloor y)=f(z_2)\lfloor f(y)\rfloor=n_2 q \lfloor f(y) \rfloor$, or $\lfloor f(y) \rfloor = \frac{n_1}{n_2}$. But the left side is an integer but the right is not (since $|n_1|<|n_2|$, so $0<|\frac{n_1}{n_2}|<1$).
So, we conclude that $f$ can take at most one nonzero value outside of $[0, 1)$. And if $x$ lies in $[0, 1)$, we have $\lfloor x \rfloor = 0$ and $f(0\cdot y)=f(x)\lfloor f(y) \rfloor$, or $\frac{f(0)}{f(x)}=\lfloor f(y) \rfloor$ for any $y$. That means that if exists such $0\leqslant x<1$ and $f(x)\ne 0$, then $f(x)$ must be constant outside of $[0, 1)$ (see above).
Update (next part)
If $f(x)$ is constant outside of $[0, 1)$, then necessary $\lfloor f(x) \rfloor = 1$ (for example, that's because $f(4)=f(2)$, but $f(4)=f(2\cdot 2)=f(2) \lfloor f(2) \rfloor$. Since a number $0\leqslant z<1$ can be written as $-1(-z)$, we have $f(z)=f(-1)\lfloor f(-z)\rfloor$, so $f$ must be constant on all $\mathbb{R}$ (because both $-1$ and $-z$ lie outside of $[0, 1)$.
On the other hand, let's assume that $f(x)=0$ for any $x$ in $[0,1)$. So, for any number $z$ we can pick an integer $n$ such as $0\leqslant y=\frac z n < 1$. So we get $f(z)=f(n\cdot y)=f(n) \lfloor f(y) \rfloor$. Since $f(y)$ is $0$, so is $f(z)$. So, $f$ is constant zero function.
Final answer:
The only solutions are constant functions $f(x)=c$, where either $c=0$ or $1\leqslant c <2$ (i.e. $\lfloor c\rfloor =1$).
I guess I took a scenic route to the solution! First,
$$f(1) = f(1\cdot 1) = f(\lfloor 1\rfloor \cdot 1) = f(1) \lfloor f(1) \rfloor,$$
so either
$f(1) = 0$ or $\lfloor f(1) \rfloor = 1.$
If
$f(1) =0$
then for all $y\in\mathbb{R}$,
$$f(y) = f(1\cdot y) = f(\lfloor 1\rfloor y) = f(1) \lfloor f(y)\rfloor = 0,$$
so
$f(y) = 0$ for all $y$.
So from now on assume instead that
$\lfloor f(1)\rfloor = 1$. In particular, $f(1)\neq 0$.
Let
$n\in\mathbb{Z}$ and $\delta\in [0,1)$. Then $$f(n) = f( \lfloor n+\delta \rfloor \cdot 1) = f(n+\delta) \lfloor f(1) \rfloor = f(n+\delta), \phantom{NN} (***)$$
so
$f$ is constant on the half-open interval $[n,n+1)$, for each $n\in \mathbb{Z}$. Thus $f$ is determined by its values on the integers.
Now let
$q\in\mathbb{Z}$, and suppose $q\neq 0$. Then $$f(1) = f\left(q\cdot \frac{1}{q}\right) = f\left(\lfloor q\rfloor \cdot \frac{1}{q}\right) = f(q) \cdot \left\lfloor f\left(\frac{1}{q}\right)\right\rfloor.$$
Notice that
$f(q)\neq 0$ and $\lfloor f(1/q)\rfloor \neq 0$, since our assumption from earlier guarantees that $f(1)\neq 0$.
Now,
if $q> 1$ then $0<1/q<1$, so $$f(1/q) = f(0+1/q) = f(0),$$ by equation $(***)$. In particular, $\lfloor f(0)\rfloor = \lfloor f(1/2) \rfloor \neq 0$, since we said $\lfloor f(1/q)\rfloor \neq 0$ for all $q\in\mathbb{Z}\setminus \{ 0\}$.
Similarly,
if $q<-1$ then $-1<1/q<0$, so $f(1/q) = f(-1)$; hence $\lfloor f(-1)\rfloor \neq 0$.
So,
for $q\in \mathbb{Z}\setminus \{-1,0,1\}$, we have $$f(q)=\begin{cases}f(1) / \lfloor f(0)\rfloor & \textrm{ if } q > 1\\f(1) / \lfloor f(-1)\rfloor & \textrm{ if } q < 1.\end{cases}$$ Thus $f$ is determined by its values at $0$, $1$, and $-1$.
Recall that
we assumed $\lfloor f(1)\rfloor = 1$, so $f(1) = 1+\epsilon$ for some $\epsilon \in [0,1)$. Similarly, let's write $f(0) = m+\gamma$, and $f(-1) = n+\delta$, where $m,n\in\mathbb{Z}$ and $\gamma,\delta\in [0,1)$. By the way, we already found that $m = \lfloor f(0) \rfloor \neq 0$, and similarly $n\neq 0$.
Now
$$m+\gamma = f(0) = f( 0 (-1)) = (m+\gamma)(n),$$ so $(m+\gamma)(n-1) = 0$. We can't have $m+\gamma = 0$ because we said $f(0) \neq 0$, so we conclude that $n=1$.
We're almost done.
$$m+\gamma = f\left( -1 \cdot 0\right) = (1 + \delta) (m) = m + m\delta,$$ so $\gamma = m\delta$. Similarly, $\gamma = m \epsilon$.
Finally
$m + \gamma = f(0\cdot 0) = (m+\gamma) (m)$, which (since $m+\gamma \neq 0$) implies $m=1$. Thus $m=n=1$, and $\gamma=\delta=\epsilon$.
Therefore
$f(1) = f(-1) = f(0) = 1 + \epsilon$. It now follows that $f(q) = 1+\epsilon$ for all $q\in \mathbb{Z}$, and indeed $f(x) = f(\lfloor x\rfloor) = 1+\epsilon$, for all $x\in \mathbb{R}$.
So all of the solutions must be
constant functions; either $f(x) = 0$, or $f(x) = 1+\epsilon$ for some fixed constant $\epsilon \in [0,1)$.
Conversely,
any of these functions certainly satisfies the functional equation, as is easy to check. $\Box$
Here's another approach, related to Culver Kwan's but not identical. (Also, unlike C.K. I hadn't seen the question before :-). For the avoidance of doubt, of course I didn't look at C.K.'s solution before finding mine.) I'll be a little less terse than Culver.
Suppose $\lfloor f\rfloor=0$ always. Then setting $x=1$ in the given equation yields $f(y)$ on the left and $0$ on the right, so $f$ is identically 0. This is indeed a solution; let's now suppose that $f$ is something else, so that $\lfloor f\rfloor$ isn't always 0.
Now
pick a $y$ so that $\lfloor f(y)\rfloor\neq0$. The LHS of our equation depends only on $\lfloor x\rfloor$, so the RHS does too, so $f(x)$ depends only on $\lfloor x\rfloor$.
Finally,
fix $x$ at a large (positive or negative) value and let $y$ vary from $0$ inclusive to $1$ exclusive. As it does, the LHS of our equation covers $f(t)$ for all $t$ between $0$ inclusive and $x$ exclusive, but the RHS is constant. Hence $f$ is constant (separately for values $\leq0$ and for values $\geq0$, but since $0$ is in both ranges $f$ must be constant everywhere).
And
if $t$ is the constant value, our original equation says $t=t\lfloor t\rfloor$ or $t(\lfloor t\rfloor-1)=0$, so either $t=0$ (a case we already mentioned) or $1\leq t<2$.
So
the solutions to our equation are the constant functions at 0 and at values between 1 inclusive and 2 exclusive.
We can easily check that the solutions are:
$f(x)=C$ where $C=0$ or $1\le C<2$
Firstly,
$x=0$ yields $f(0)\big(\lfloor f(y)\rfloor-1\big)=0$.
If
$f(0)\ne0$, $\lfloor f(y)\rfloor=1$. We sub $x=1$ in the original equation and have $f(y)=f(1)\lfloor f(y)\rfloor=f(1)$, which makes $f(x)=C$ for any constant $1\le C< 2$.
If
$f(0)=0$, then put $x=k$ where $0\le k\le1$. Then we have $f(k)\lfloor f(y)\rfloor=0$.
If for some possible values of $k$, $f(k)\ne0$, then we have $\lfloor f(y)\rfloor=0$ for all $y$, so we put $x=1$ in the original equation, which yields $f(y)=f(1)\lfloor f(y)\rfloor=0$, but this contradicts with the condition that there exist $0<k<1$ that $f(k)\ne0$.
So for all $k$, $f(k)=0$. For all $y\in\mathbb{R}$, we can always choose an integer $n$, such that $0\le \frac yn<1$. So putting $(n,\frac yn)$ for $(x,y)$, we have $f(y)=f(\lfloor n\rfloor\cdot\frac yn)=f(n)\lfloor f(y)\rfloor=0$.
Note: I know the question is IMO 2010 shortlist A1, and I have done it before. But I just need a while to recover my memories.
