Put $x=y=0$. Then $f(0)=0$ or $\lfloor f(0) \rfloor=1$. $\bullet$ If $\lfloor f(0) \rfloor=1$, putting $y=0$ we get $f(x)=f(0)$, that is f is constant. Substituing in the original equation we find $f(x)=0, \ \forall x \in \mathbb{R}$ or $f(x)=a, \ \forall x \in \mathbb{R}$, where $a \in [1,2)$. $\bullet$ If $f(0)=0$, putting $x=y=1$ we get $f(1)=0$ or $\lfloor f(1) \rfloor=1$. For $f(1)=0$, we set $x=1$ to find $f(y)=0 \ \forall y$, which is a solution. For $\lfloor f(1) \rfloor=1$, setting $y=1$ yields $f(\lfloor x \rfloor)=f(x), \ (*)$. Putting $x=2, y=\frac{1}{2}$ to the original we get $f(1)=f(2)\lfloor f(\frac{1}{2}) \rfloor$. However, from $(*)$ we have $f(\frac{1}{2})=f(0)=0$, so $f(1)=0$ which contradicts the fact $\lfloor f(1) \rfloor=1$. So, $f(x)=0, \ \forall x$ or $f(x)=a, \ \forall x, \ a \in [1,2)$. Hope I'm right.

Let $P(x,y)$ be the assertation of $f([x]y)=f(x)[f(y)]$ $P(0,0): f(0)=f(0)[f(0)]$ There are 2 cases: 1. $[f(0)]=1$. Then, $P(x,0): f(0)=f(x)$ Making $f(x)=c$ for some constant $c$. Substituting back to $P(x,y)$, we get: $c=c[c]$ or $[c]=1$. Then, $1\leq c<2$. 2. $f(0)=0$. $P(1,1): f(1)=f(1)[f(1)]$ Again, there are 2 subcases: a. $f(1)=0$. $P(1,y): f(y)=0$. After checking back to the f.q., we get $f(x)=0$ is also a solution. b. $[f(1)]=1$. $P(x,1): f(x)=f([x])$. Consider: $P(2010,\frac1{2010}): f(1)=f(2010)[f(\frac{1}{2010})]=f(2010)[f(0)]=0$. Contradiction, because $1=[f(1)]=[0]=0$. Then the function which satisfies $P(x,y)$ is: $f(x)=0\forall x\in\mathbb{R}$ $f(x)=c\forall x\in\mathbb{R}, 1\leq c<2$. @above: we almost have the same solution.. hope we're right..

Solution to Problem 1: $f([x]y)=f(x)[f(y)]$ (1) Substituting $y=0$ we have $f(0) = f(x) [f(0)]$. If $[f0)] \ne 0$ then $f(x) = \frac{f(0)}{[f(0)]}$. Then $f(x)$ is constant. Let $f(x)=c$. Then substituting that in (1) we have $c=c[c] \Rightarrow c(1-[c])=0 \Rightarrow c=0$, or $[c]=1$. Therefore $f(x)=c$ where $c=0$ or $c \in [1,2)$ If $[f(0)] = 0$ then $f(0)=0$. Now substituting $x=1$ we have $f(y)=f(1)[f(y)]$. If $f(1) \ne 0$ then $[f(y)] = \frac{f(y)}{f(1)}$ and substituting this in (1) we have $f([x]y)=\frac{f(x)f(y)}{f(1)}$. Then $f([x]y)=f(x[y])$. Substituting $x=1/2, y=2$ we get $f(0)=f(1)$. Then $f(1)=0$, which is a contradiction Therefore $f(1)=0$. and then $f(y)=0$ for all $R$ Then the only solutions are $f(x)=0$ or $f(x)=c$ where $c \in [1,2)$

The answers are $f(x) = c$ for all $x$ and constant $c$, where $c \in \{0\} \cup [1,2)$. Put in $x = 0$ to get $f(0) = f(0) \lfloor f(y) \rfloor$. This implies either (a) $\lfloor f(y) \rfloor = 1$ for all $y$, or (b) $f(0) = 0$. If (a) is true, then the original equation reduces to $f(\lfloor x \rfloor y) = f(x)$. Set $y = 0$ and we get $f(0) = f(x)$, or rather that $f$ is constant. We know $\lfloor f(0) \rfloor = 1$, and we can verify for any $a \in [1,2)$, $f(x) = a$ for all $x$ is a solution. Otherwise we have (b), $f(0) = 0$. Let $0 < x < 1$. Then $\lfloor x \rfloor = 0$ and the equation becomes $f(x) \lfloor f(y) \rfloor = 0$. Either (c) $\lfloor f(y) \rfloor = 0$ for all $y$, or (d) $f(x) = 0$ for all $0 < x < 1$. If (c) is true, then the original equation becomes $f(\lfloor x \rfloor y) = 0$, and $x = 1$ gives $f(y) = 0$ for all $y$. If (d) is true, then let $0 < y < 1$ in the original equation. We get $f( \lfloor x \rfloor y) = 0$. For any real $a$ with $\lfloor a \rfloor \neq 0$, take $x = 2a$ and $y = \frac{a}{\lfloor 2a \rfloor}$, and note that $0 < y < 1$. Then $f( \lfloor x \rfloor y) = f(a) = 0$, and so $f$ is constant at $0$. EDIT: Alternate finish since the end is a bit sloppy. Use $x = 2, y = 1/2$ to get $f(1) = 0$, then plug in $x = 1$ to the original equation.

good problem..hope all the indian members could solve it

Ohhhh i'm sure that the most of the bolivian team didn't do this problem, i think just Arran did it ^_^ the bolivian team was not tried tried

Clearly $f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\right\rfloor$, so $(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0$ for all $x,y\in\mathbb{R}$. If $\left\lfloor f(y)\right\rfloor = 0$ for all $y \in \mathbb{R}$, then by taking $x=1$ we get $f(y)=f(1)\left\lfloor f(y)\right\rfloor = 0$, so $f$ is identically null (which checks). If, contrariwise, $\left\lfloor f(y_0)\right\rfloor \neq 0$ for some $y_0 \in \mathbb{R}$, it follows $f(x) = f(\lfloor x \rfloor)$ for all $x \in \mathbb{R}$. Now it immediately follows $f(x) = f(\lfloor x \rfloor \cdot 1) = f(x)\lfloor f(1) \rfloor$, hence $f(x)(1 - \lfloor f(1) \rfloor) = 0$. For $x=y_0$ this implies $\lfloor f(1) \rfloor = 1$. Assume $\lfloor f(0) \rfloor=0$; then $1 \leq f(1) = f\left ( 2\cdot \dfrac {1} {2} \right ) = f(2)\left \lfloor f \left ( \dfrac {1} {2} \right ) \right \rfloor = f(2)\left \lfloor f \left ( \left \lfloor \dfrac {1} {2} \right \rfloor \right ) \right \rfloor = f(2)\left \lfloor f(0) \right \rfloor = 0$, absurd. Therefore $\lfloor f(0) \rfloor \neq 0$, and now $y=0$ in the given functional equation yields $f(0) = f(x)\lfloor f(0) \rfloor$ for all $x \in \mathbb{R}$, therefore $f(x) = c \neq 0$ constant, with $\lfloor c \rfloor = \lfloor f(1) \rfloor = 1$, i.e. $c \in [1,2)$ (which obviously checks).

That is a nice problem for number 1 of day 1. If i could solve it in 40 minutes, I hope everyone from Uzbekistan team managed to solve it =p

Yay, I solved an IMO problem. Is it just me or does it seem as though this year's #1 and #4 are much easier than normal? Solution We claim that either $f(x)=k$ such that $k \in [1,2)$ or $f(x)=0$. We can assume that $f(1) \neq 0$. Because otherwise, by the equality, we have that $f(y)=f(1) \lfloor f(y) \rfloor =0$ and the claim is satsfied. Let $f(1)=k \neq 0$ where $k \in [1,2)$. Now we will establish several results about $f(x)$. $(1)$ Let $x=y=1$. Then since $f(1) \neq 0$, it follows that $f(1) \lfloor f(1) \rfloor \quad \Rightarrow \quad \lfloor f(1) \rfloor = 1$. $(2)$ Let $x=1$. Then $f(y)=f(1) \lfloor f(y) \rfloor = k \lfloor f(y) \rfloor$. $(3)$ Let $y=1$. Then by (1), $f(\lfloor x \rfloor)=f(x) \lfloor f(1) \rfloor =f(x)$. Now we will divide into two cases. Case 1: $f(0) \neq 0$. Let $x=0$. Then it follows that $f(0)=f(0) \lfloor f(y) \rfloor \quad \Rightarrow \quad \lfloor f(y) \rfloor =1$. Combining this with (2) yields that $f(x)=k \lfloor f(x) \rfloor =k$ where $k \in [1,2)$ and the claim is satisfied. Case 2: $f(0)=0$. Let $y$ be such that $y \in [0,1)$. Then it follows that $f(\lfloor x \rfloor y)=f(x) \lfloor f(y) \rfloor$. Now note that by (3), we have that $f(y)=f(\lfloor y \rfloor)=f(0)=0$. Hence $f(\lfloor x \rfloor y)=f(x) \lfloor 0 \rfloor=0$. Now we will prove that for each real number $a$, we have $x$ and $y$ such that $y \in [0,1)$ and $a= \lfloor x \rfloor y$. If $a>0$, then let $x=a+1$ and $0 \le y=a/\lfloor a+1 \rfloor < 1$. It is clear that from here, we have that $a= \lfloor x \rfloor y$. If $a<0$, then let $x=a-1$ and $0 \le y = a/\lfloor a-1 \rfloor <1$. Hence $a= \lfloor x \rfloor y$ This yields that for all real $a$, it follows that $f(a)=0$ and the claim is satisfied. Now we will show that both of the claimed functions satisfy the equation for all real $x$ and $y$. $(1)$ $f(x)=k$ where $k \in [1,2) \quad \Rightarrow \quad f(\lfloor x \rfloor y)=k=f(x) \lfloor f(y) \rfloor$ $(2)$ $f(x)=0 \quad \Rightarrow \quad f(\lfloor x \rfloor y)=0=f(x) \lfloor f(y) \rfloor$ Therefore the claim has been proven.

My Solution : For : $x=0$ we get : $f(0)=f(0)\left \lfloor f(y) \right \rfloor$ So either $f(0)=0$ or $\left \lfloor f(y) \right \rfloor=1$ for all y $\left \lfloor f(y) \right \rfloor=1$ for all y is indeed a solution . Now if $f(0)=0$ : Letting $y=1$ and $x=n \in \mathbb N^*$ we get : $f(n)=f(n)\left \lfloor f(1) \right \rfloor$ if $f(n)=0$ for all n then for $x=n$ we must have $f(yn)=0$ for all y so $f(y)=0$which is indeed a solution . If $f(0)=0$ and $\left \lfloor f(1) \right \rfloor=1$ and for $x=n$ : $f(ny)=f(n)\left \lfloor f(y) \right \rfloor$ for $x=n+p$ where $n\in \mathbb Z$ and $p \in \left( 0,1 \right )$: $f(ny)=f(n+p)\left \lfloor f(y) \right \rfloor$ so either : $f(n)=f(n+p)$ for $p \in \left( 0,1 \right )$ or $\left \lfloor f(y) \right \rfloor=0$ for all y but $\left \lfloor f(1) \right \rfloor=1$ so : $f(n)=f(n+p) for p \in \left( 0,1 \right )$ Let n be an integer $n<0$ we have : $f(n+1)=f(\left \lfloor n \right \rfloor\frac{n+1}{n})=f(n)\left \lfloor f(1+\frac{1}{n}) \right \rfloor=f(n).\left \lfloor f(0) \right \rfloor=0$ So $f(x) = 0$ for all $x\leq-2$ Now let n be an integer $n>1$ : We have : $f(n+1)=f(\left \lfloor n \right \rfloor\frac{n+1}{n})=f(n)\left \lfloor f(1+\frac{1}{n}) \right \rfloor=f(n).\left \lfloor f(1) \right \rfloor=f(n)=c$ so $f(x)=c$for all $x\geq2$ for$x=-2$ and $y=-1$ we get $c=0$ Using the fact that $f(n)=f(n+p)$ for$p\in (0,1)$ we will have : $f(x)=0$for $x\in(-\infty,-1)\cup [0,1)\cup (2,+\infty)$ $f(x)=f(-1)$for $x\in [-1,0)$ $f(x)=f(1)$for$x \in [1,2)$ but we have : $f(1)=f( 2 .\frac{1}{2})=f(2).\left \lfloor f(\frac{1}{2} )\right \rfloor =0$ whish gives a contradiction . So the solutions are : $f(x)=0$ $f(x)=c , c \in [1,2[$ Hope it's correct !

$f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor$- Q case 1 : there is $a$ s.t. $\left\lfloor a\right\rfloor$ is not zero and $f(a)=0$. $f(\left\lfloor a\right\rfloor y)=f(a)\left\lfloor f(y)\right\rfloor$. so $f(\left\lfloor a\right\rfloor y)=0$. put $y=\frac{z}{\left\lfloor a\right\rfloor}$. so we get $f(z)=0$ for all $z$. case 2 : take the negation of case 1. so if $f(x)=0$ then $x$ belongs to $[0,1)$. case 2.1 : $f(0)$ is not equal to $0$. put $x=0$ to equation Q. so we get $\left\lfloor f(y)\right\rfloor=1$. thus equation Q gives $f(\left\lfloor x\right\rfloor y)=f(x)$. put $x=1$. then $f(y)=f(1)$ for each $y$. thus $f(x)=b$ for each $x$ where $b$ is a number in $[1,2)$. case 2.2 : $f(0)=0$. case 2.2.1 : there is $b$ in $[1,0)$ s.t. $f(b)$ is not zero. so $f(\left\lfloor b\right\rfloor y)=f(b)\left\lfloor f(y)\right\rfloor$. thus $\left\lfloor f(y)\right\rfloor=0$. by Q we get $f(\left\lfloor x\right\rfloor y)=0$. putting $x=1$ gives $f(y)=0$ for each $y$. case 2.2.2 : negation of case 2.2.1. put $y=0.5$ to Q. then since $f(0.5)=0$ we get $f(\frac{\left\lfloor x\right\rfloor}{2})=0$. putting $x=2$ will give a contradiction since $f(1)$ is not zero. thus the solutions are $f(x)=0$ for each $x$, $f(x)=c$ for each $x$ where $c$ is a constsnt in $[1,2)$. we can easily verify them. $Q.E.D$

I will try to do it differently, maybe it's wrong I dont know.

Dr N0 wrote: Let $k$ be a positive integer, and let $\epsilon>0$ and let $\epsilon$ be very small number. Then because of $f$ is continous we have: $f$ is not assumed to be continuous. However, you were able to end up your proof differently : Case $[f(1)]=1$ Taking $x=2$ and $y=\frac{1}{2}$ gives $f(1)=f(2)[f(\frac{1}{2})]$. (1) Taking $y=1$ gives $f([x])=f(x)[f(1)]$, therefore, $f([x])=f(x)$ because we are in case $[f(1)]=1$. Hence for all real x in $]0,1[, f(x)=f(0)$. (2) (1) and (2) implies that $f(1)=f(2)[f(0)]$. And because $[f(1)]=1 \neq 0$, then $[f(0)]\neq 0$, hence $f(0)\neq 0$. Now let $x=0$. Then $f(0)=f(0)[f(y)]$. Thus, $f(0)=0$ or $[f(y)]=1$. But $f(0)\neq 0$. So $[f(y)]=1$. In particular, $[f(0)]=1$. Let $y=0$. Then $f(0)=f(x)[f(0)]$, which implies $f(0)=f(x)$ because $[f(0)]=1$. Therefore, $f$ is constant. At that step, the conclusion follows easily.

If $f$ is constant, say $f(x)=c$ for all $x$ and some constant $c$; we get $c=c\lfloor c\rfloor$, so $c=0$ or $\lfloor c\rfloor=1$, and $c=0$ or $1\le c<2$. Plugging in $y=0$ gives $f(0)=f(x)\lfloor f(0)\rfloor$. If $f(0)\neq0$, we also have that the RHS is not 0, so $f(x)=\dfrac{f(0)}{\lfloor f(0)\rfloor}=c$, contradiction. So $f(0)=0$. Now, let $x$ be such that $f(x)=0$. Then, for all $y$, $f(\lfloor x\rfloor y)=0$ for all $y\in\mathbb{R}$. If $\lfloor x\rfloor\neq0$, this will give us $f(x)=0$ for all $x$, contradiction. Thus, $\lfloor x\rfloor=0$. Now, switching $x$ and $y$ gives $f(\lfloor y\rfloor x)=f(y)\lfloor f(x)\rfloor=0$. But if $x\neq0$, we can choose $y$ appropriately so that $\lfloor y\rfloor x>1$, which is a contradiction. Thus, $x=0$. Finally, plugging in $x=1/2$ gives $f(0)=f(1/2)\lfloor f(y)\rfloor$ for all $y$, giving $\lfloor f(y)\rfloor=0$ for all $y$. So plugging in $x=1$ into the original equation, $f(y)=0$ for all $y$, again a contradiction. Thus, $f$ is constant, and our only solutions are those above.

Here is my solution (The same idea as many solutions given here): Consider the two possible cases: 1-If $f\left(\frac{1}{2}\right)=0$: Then for any $x\in\mathbb{R}$, $f(x)=f\left(\left\lfloor-\frac{1}{2}\right\rfloor(-x)\right)=f\left(-\frac{1}{2}\right)\lfloor f(-x)\rfloor=f(-1)\left\lfloor f\left(\frac{1}{2}\right)\right\rfloor\lfloor f(-x)\rfloor=0$. Clearly, $f=0$ satisfies the equation. 2-If $f\left(\frac{1}{2}\right)\neq 0$: Then for any $x\in\mathbb{R}$, $f(0)=f\left(\frac{1}{2}\right)\lfloor f(x)\rfloor$ and $f(x)=f(1)\lfloor f(x)\rfloor$. Therefore $f(x)=\frac{f(0)f(1)}{f\left(\frac{1}{2}\right)}=c=f\left(\frac{1}{2}\right)\neq 0$ is constant. Using $c=f(1)=f(1)\lfloor f(1)\rfloor=c\lfloor c\rfloor$ we deduce that $\lfloor c\rfloor=1$. Clearly, $f=c$ with $c\in[1,2[$ satisfy the equation.

Let x=0 So f(0)=f(0)⌊f(y)⌋ So f(0)(1-⌊f(y) ⌋ )=0 F(0)=0 OR ⌊f(y) ⌋=1 Case 1: f(0)=0 Substituting y=1, the equation become ⌊f(x) ⌋=f(x)⌊f(1) ⌋ Substituting x=1 and y=1, we get ⌊f(1) ⌋=0 or 1. If ⌊f(1) ⌋=1 it’s on the second case. So ⌊f(1) ⌋=0 Substituting x=1, we get f(y)=f(1) ⌊f(y) ⌋ If f(y) positive number except 0 f(1) must be greater of same with 1, contradiction So f(y) must be negative or 0 for all y If f(y) is negative, so f(1) is positive (contradiction) So f(y)=0 Case 2: f(y)=x which 1≤x<2 The answer: f(y)=0 or f(y)=c which 1≤x<2 Please help me if this answer is true or not. Thank you

i agree that it's a nice functional equation.but wasn't it too easy?i hope that France wont provide such water problems!France has ox of god like Gabriel Dospinescu,why dont they provide some problems for imo?

Substituting $(0,0)$, we find $f(0) = f(0) \cdot \lfloor f(0)\rfloor$, so $f(0)=0$ or $\lfloor f(0)\rfloor = 1$. Case 1: $f(0)=0$. Substituting $(1,1)$ tells us $f(1)=f(1) \cdot \lfloor f(1)\rfloor$, so $f(1)=0$ or $\lfloor f(1)\rfloor=1$. $f(1)=0$: Substitute $(1,t)$ to get $f(t)=f(1) \cdot \lfloor f(t)\rfloor = 0$, so we get the solution $f(x)=0$. $\lfloor f(1)\rfloor=1$: Suppose we have a real number $\alpha \in (0,1)$. Then substituting $(\alpha, 1)$ tells us $$f(0) = f(\alpha) \lfloor f(1) \rfloor \implies f(\alpha)=0.$$ However, if we substitute $(t, \alpha)$, we find $$f(\lfloor t \rfloor \alpha) = f(t) \cdot \lfloor f(\alpha)\rfloor = 0.$$ Since $\lfloor t \rfloor \alpha$ can take on any real value, this means $f(x)=0$ for all $x$, which contradicts $\lfloor f(1)\rfloor=1$. ${\color{blue} \Box}$ Case 2: $\lfloor f(0)\rfloor=1$. Substitute $(0,t)$ to get $$f(0) = f(0) \cdot \lfloor f(t)\rfloor \implies \lfloor f(t)\rfloor=1.$$ Thus the values of $f$ in this case are bounded from $[1,2)$. We can prove they all must be the same value by substituting $(t,0)$ to find $$f(0) = f(t) \cdot \lfloor f(0)\rfloor = f(t). \quad {\color{blue} \Box}$$ Hence our final solution is $\boxed{f(x)=c, \quad c \in \{0\} \cup [1,2)}$, which we can test to work. $\blacksquare$

canada wrote: Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds $$f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor$$ where $\left\lfloor a\right\rfloor$ is greatest integer not greater than $a.$ Proposed by Pierre Bornsztein, France A Very easy question for international mathematics olimpiad.

$f\equiv 0$ and $f \equiv c$ where $c\in [1,2)$ is a constant. $P(x,0) \implies f(0) = f(x) \left\lfloor f(0) \right\rfloor$. If $f(0) \not\in [0,1)$, then $\left\lfloor f(0) \right\rfloor \neq 0 \implies f(x) = \dfrac{f(0)}{\left\lfloor f(0) \right\rfloor} = c$. So putting $x=0$, we get $c = c\cdot \left\lfloor c \right\rfloor\implies c \in [1,2)$ because $c \neq 0$ as $f(0) \not\in [0,1)$. Otherwise $f(0) = [0,1) \implies \left\lfloor f(0) \right\rfloor = 0 \implies f(0) = 0$. If there exists $\alpha \in [0,1)$ such that $f(\alpha) \neq 0$, then $P(\alpha, y) \implies 0 = f(\alpha) \left\lfloor f(y) \right\rfloor \implies 0 \le f(y) < 1$ for all $y \in \mathbb R$. So $f(\left\lfloor x \right\rfloor y) = f(x) \left\lfloor f(y) \right\rfloor = 0$. Now $x=1 \implies f(y) = 0$ for all $y\in\mathbb R$. Now putting $y=\alpha$ gives a contradiction. Thus we must have that $f(x) = 0$ for all $x\in [0,1)$. If $f(1) =0$, then $P(1,y) \implies f(y) = 0$ for all $y$. Otherwise $P(1,1) \implies \left\lfloor f(1) \right\rfloor=1\implies 1 \le f(1) \le 2$. But at the same time, $P\left(2,\frac{1}{2}\right) \implies f(1) = 0$, contradiction and we are done.

$P(0,0)$ to get that $f(0)=f(0)\lfloor f(0) \rfloor$, so $f(0)=0$ or $\lfloor f(0) \rfloor=1$ If $\lfloor f(0) \rfloor=1$, then $P(x,0)$ gives $f(0)=f(x)$, so $\boxed{f(x)=c}$ works, where $\boxed{1 \le c < 2}$ If $f(0)=0$, then $P(1,1)$ gives $f(1)=f(1)\lfloor f(1) \rfloor$, so $f(1)=0$ or $\lfloor f(1) \rfloor=1$ If $f(1)=0$, then $P(1,x)$ gives $\boxed{f(x)=0}$ If $\lfloor f(1) \rfloor=1$, then $P(x,1)$ gives $f(\lfloor x \rfloor) = f(x)$. $P(x,\frac{1}{\lfloor x \rfloor})$ gives $f(1)=f(x) \lfloor f(\frac{1}{\lfloor x \rfloor}) \rfloor = f(x) \lfloor f(0) \rfloor = 0$. This is a contradiction.

Firstly, let $y=0$. If $\left\lfloor f(0)\right\rfloor$ is not equal to $0$,then it is obvious that $f(x)$ is constant. $f(x)=c$. Substituting into the original equation gives us that $\left\lfloor f(y)\right\rfloor$ is equal to 1.( If $c$ is not $0$. That case is trivial anyways). So $\left\lfloor c\right\rfloor$ is 1 .As a result, $1\leq c<2$. Now, assume that $\left\lfloor f(0)\right\rfloor$ is $0$. So, $f(0)=0$ by the first substitution. Putting $x=y=1$ gives us that $f(1)=0$ or $\left\lfloor f(1)\right\rfloor=1$. The former case is trivial; we can finish it easily, as it yields that $f(x)=0$ for all $x$. Now look at the latter case. Let $y=1$ to get that $f(x)=f(\left\lfloor x\right\rfloor)$. However, $P(x,y)$, where $x$ is a large real number and $y=\frac{1}{\left\lfloor x\right\rfloor}$ gives us that $f(1)=0$ since $f(y)=f(\left\lfloor y\right\rfloor)=0$ for that $y$. Contradiction.

The answer is $f \equiv c$ for $c = 0$ or any $c$ such that $\lfloor c \rfloor = 1$. Denote the given assertion as $P(x, y)$. From $P(0, y)$ we have $f(0) = f(0) \lfloor y \rfloor$ for any real $y$. This rearranges to $f(0)(\lfloor f(y) \rfloor - 1) = 0$, so we have two cases to consider: Case 1: $f(0) = 0$. Take any $t$ such that $\lfloor t \rfloor = 0$. From $P(t, t) = 0$, we get $0 = f(t)\lfloor f(t) \rfloor$. This is enough to imply that $f(t) = 0$, as if $\lfloor f(t) \rfloor = 0$ then $P(1, t)$ yields $f(t) = 0$. Now, let $k$ be any real number. Then we can pick reals $r_0, t_0$ such that $\lfloor r_0 \rfloor t_0 = k$ and $\lfloor t_0 \rfloor = 1$ (choose any $r_0$ such that $\lfloor r_0 \rfloor$ has the same sign as $k$ and $| \lfloor r_0 \rfloor | > |k|$). From $P(r_0, t_0)$ we conclude that $f(k) = 0$, so $f \equiv 0$. Case 2: $\lfloor f(y) \rfloor = 1$ for all $y$. From $P(1, y)$ we then get $f(y) = f(1)$. This is enough to imply that $f \equiv c$ for some constant $c$ such that $\lfloor c \rfloor = 1$.

Our two solutions are $f(x) = 0$ and $f(x) = c \forall c \in [1, 2]$. We can plug in $(0, 0)$ to get $f(0) = f(0)\left\lfloor f(0)\right\rfloor$ which implies $f(0) = 0$ or $\left\lfloor f(0)\right\rfloor = 1$. If the latter is true then plug in $(t, 0)$ then plug in $f(0) = f(t) = c \in [1, 2]$. If $f(0) = 0$ we can plug in $(1, 1)$ to get $f(1) = f(1)\left\lfloor f(1)\right\rfloor \implies f(1) = 0 \veebar \left\lfloor f(1)\right\rfloor = 1$. If $f(1) = 0$ then $(1, 0)$ gives $f(y) = 0$, done. If $\left\lfloor f(1)\right\rfloor = 1$ then $(x, 1) \implies f(\left\lfloor x\right\rfloor) = f(x)$. Now plug in $(1434, \frac{1}{1434})$ to get $f(1) = f(1434)f\left(\left\lfloor \frac{1}{1434} \right\rfloor\right) = 0$, contradiction. So our solutions are $f(x) = 0$ and $\left\lfloor f(x)\right\rfloor = 1$.

We claim the only solutions are $f\equiv c$, where $c\in \{0\}\cup [1,2)$. It is easy to see all such functions work; we know prove they are the only ones. Let $P(x,y)$ denote the assertion. From $P(0,0)$, we have that $f(0) = f(0)\lfloor f(0) \rfloor$. Thus, $f(0) = 0$ or $1\le f(0) < 2$. We proceed by casework. Case 1: $f(0)=0$. Fix $x\in (0,1)$, and we have $f(0) = 0 = f(x)\lfloor f(y) \rfloor$. Suppose there exists some real $y$ such that $f(y) \notin [0,1)$. Then, $f(x)=0$ for all $x\in (0,1)$. Then, let $x$ vary over the reals and $y$ over $(0,1)$ to find that $f(\lfloor x \rfloor y) = 0$. But since $\lfloor x \rfloor y$ can be any real number, $f\equiv 0$, contradiction. Thus, we always have $f(x)\in [0,1)$. Thus, we conclude $f(\lfloor x \rfloor y) = 0$, which implies $f\equiv 0$, as desired. Case 2: $f(0)\in [1,2)$. If this is the case, then by $P(x,0)$, we have $f(0) = f(x)$, as desired, and we are done. $\blacksquare$

Put $\displaystyle x=y=0$. We have $\displaystyle f(0)=0$ or $\left\lfloor f(0)\right\rfloor=1$. We split into two cases. $(i)\left\lfloor f(0)\right\rfloor=1$ Putting $\displaystyle y=0$ gives $f\equiv f(0)$,so $f\equiv c$ where $1\le c \le2$ $\displaystyle (ii)f(0)=0$. Put $\displaystyle x=y=1$. We have $\displaystyle f(1)=0$ or $\left\lfloor f(1)\right\rfloor=1$. Claim. $\displaystyle f(1)=0$ $\emph{Proof.}$ We use proof by contradiction. Assume that $\left\lfloor f(1)\right\rfloor=1$. Puttting $\displaystyle y=1$ gives $f(\left\lfloor x\right\rfloor)=f(x)$ Now $f(1)=f(2)\left\lfloor f(\frac{1}{2})\right\rfloor=f(2)\left\lfloor f(0)\right\rfloor=0$ since $\left\lfloor \frac{1}{2}\right\rfloor=0$.$\blacksquare$ Put $\displaystyle x=1,y=t$. $\displaystyle f(t)=0$ Combining these cases together, we have $f\equiv c$, where $c\in \{0\}\cup [1,2)$

Let $P(x,y)$ denote the assertion. First, we have that $P(0,y)$ implies $f(0)=f(0)\lfloor f(y)\rfloor$, so either $f(0)=0$ or $\lfloor f(1)\rfloor = 1$ for all $y$. In the latter case, we have $f(\lfloor x\rfloor y)=f(x)\cdot 1$, so setting $x=1$, we get $f(y)=f(1)$. Plugging in $f(x)=c$ for all $x$ into the equation, with $c=f(1)$, we see that either $c=0$ or $\lfloor c\rfloor = 1$. Now, we deal with the case where $f(0)=0$. $P(1,1)$ gives us $f(1) = f(1)\lfloor f(1)\rfloor$, so either $f(1)=0$ or $\lfloor f(1)\rfloor = 1$. If $f(1)=0$, then $P(1,y)$ gives $f(y)=f(1)\lfloor f(y)\rfloor = 0$ for all $y$. Otherwise if $\lfloor f(1)\rfloor = 1$, then $P(x,1)$ gives $$f(\lfloor x\rfloor) = f(x)\lfloor f(1)\rfloor = f(x).$$ Then, for any $0\leq y<1$, we have $$f(\lfloor x\rfloor y) = f(x)\lfloor f(y)\rfloor = f(x) \lfloor f(\lfloor y\rfloor)\rfloor = f(x)\cdot 0 = 0,$$ so since for any $a$, we can select some $x>a+1$ and let $0\leq y=\frac{a}{\lfloor a+1\rfloor } < 1$, we have $f(a)=0$ for all $a$. Since we have addressed all our cases, the solutions are $f(x)=c$ for all $x$ where $c\in \{0\}\cup [1, 2)$, and it's easy to check that all such functions satisfy our equation.

Warideeb wrote: Let $P(x,y)$ be the assertion of $f(\lfloor x \rfloor y)=f(x) \lfloor f(y) \rfloor$ Our $Claim$ is that $f \equiv 0$ or $f \equiv c$ where $1\leq c < 2$ Now $f \equiv 0$ is clearly a solution. Now we assume $f \not\equiv 0$ Now $P(0,x)$ gives us $f(0)=f(0) \lfloor f(x) \rfloor$ Now if $f(0) \ne 0$ then this gives us $f \equiv c$ for some $1\leq c < 2$ Now lets assume $f(0)=0$ $P(1,1)$ gives us $f(1)=c$ where $1 \leq c < 2$ now $P(x,1)$ gives us $f(x)=f(\lfloor x \rfloor)$ define a number $0 \leq A <1$ Now $P(x,A)$ and subsituting $f(A)$ with $f(\lfloor A \rfloor)$ gives us $f(\lfloor x \rfloor A)=0$ and give is $f \equiv 0$ and here is our solution

bad formatting because i copied my solution from mathdash vary x along [n, n+1), then we see for all x in this interval we must either have f(y) in [0,1) (to make the floor zero), or if the floor is not zero we are forced that all x in the interval have the same output when f is applied. we can now divide into two cases: f(y) lies in [0,1) for all y. then we get f(floor(x)y) = 0, so letting x = 1 gives f(y) = 0, which trivially satisfies the fe. otherwise, take some value of y for which this is not true. then we get for all integers n, if a,b in [n, n+1), then f(a) = f(b). now e = n, 0 < f - e < 1. then consider varying y over [e,f], then the right side is constant, the left side must be as well, so f(floor(x)y) is constant for all e <= y <= f. with e = 0, f = (m - 1)/m for some large integer m, set x = m, we then get f(my) is constant for all 0 <= y <= m - 1, take m to infinity and we see f is constant over positives. then set x = -1, vary y over all positives and see f is constant over negatives. we show the constant value for postiives and negatives must be the same. firstly, if the positive value is zero set x = 1 and vary y to get negative constant is also zero, so we can dsicrard this, same for other case except do x = -1. now neither constant value is 0, so set the value for x to 0 and we then get 1 = floor(f(y)), so floor(f(0)) = 1. then the positive constant value is in [1,2), then substite negative x, y = 0, then we instantly get the negative constant is also the same as f(0), so the other solution is f = k for k in [1,2), which obviously satisfies the fe.

Let $P(x, y)$ denote the assertion. Note that if there is some $k$ such that $f(k)=0,$ $P(k, y) \implies f(x) = 0$ for all $x.$ Thus assume that $f(x)$ has no zeros. Thus $P(0, 0) \implies \lfloor f(0) \rfloor = 1,$ so $P(x, 0) \implies f(x) = f(0)=c.$ Clearly $1 \leq c < 2$ and we see easily that this solution works. Therefore, the only solutions are $f(x) = 0, c \in [1, 2).$

Let $P(x,y)$ be the assertion for $f([x]y) = f(x)[f(y)]$. By $P(0,0)$ we get: $f(0) = f(0)[f(0)]$. Here there are two cases: Case 1: $f(0) \neq 0$ So we can cross out $f(0)$ and have $[f(0)] = 1$ Now by $P(x,0)$ we have: $f(0) = f(x)$ so $f$ is constant. Also by $[f(0)] = 1$ we have that if $f(x)=c$ then $[c]$ should equal to 1. We can easily check that for any such $c$ the equality holds. Case 2: $f(0)=0$ Now by $P(1,1)$ we have $f(1) = f(1)[f(1)]$. If $f(1) = 0$ then $P(1,x)$ gives $f(x) = 0$ which we can check that it is a solution. So we continue by supposing $f(1) \neq 0$ to check the other case. As a result, we can cross out $f(1)$ and have that $[f(1)]=1$. Then by $P(x,1)$ we have $f([x])=f(x)$. Now $\forall x>2: P(x,\frac{1}{[x]})$ gives us $f(1) = f(x).[f(\frac{1}{[x]}]$ and as we know $[f(\frac{1}{[x]})] = [f([\frac{1}{[x]}])] = [f(0)] = 0$. so $f(1) = 0$, contradiction. To sum up, we proved that the only functions in which the given equality holds are $f(x) = c, \forall x \in \mathbb{R}, \forall c \in [1,2)$ and $f(x) = 0, \forall x \in \mathbb{R}$. And we can easily check that these two functions work.

ily >_< Lets take $x=0$. Then, $f(0)=f(0)\lfloor f(y)\rfloor$. If $f(0)\neq 0$, then $\lfloor f(y) \rfloor=1$. If so, $f(\lfloor x \rfloor y)=f(x)$. Fix $x>1434$. Then $y$ times a scalar not zero gives all reals, hence $f$ is constant. Now, if $f(0)=0$, then take $0\leq x<1$, so $0=f(x)\lfloor f(y)\rfloor$. We vary $y$ over all reals now. If at any point $\lfloor f(y)\rfloor\neq 0$, then pick that $y$, and thus $f(x)=0$ for all $0\leq x<1$. Now, in the initial equation let $0\leq y<1$, so $f(y)=0$, and $f(\lfloor x\rfloor y)=0$. Now say we want to achieve $z$ with $\lfloor x\rfloor y$. If $z>0$, then make $\lfloor x\rfloor>z$, and scale down by $y$. If $z<0$ then make $\lfloor x \rfloor < z$ and scale down by $y$. Of course $f(0)=0$ already so nothing is to be done at $0$. Hence $f$ is simply $0$. Else, $\lfloor f(y)\rfloor=0$ for all $y$. Then, $f(\lfloor x\rfloor y)=0$. Now, fix $x>1+4+3+4$, and vary $y$ to get all reals, so $f$ is zero again. Now, if $f$ was just constant, then we have $c=c\lfloor c\rfloor$. Thus, $c=0$ or $\lfloor c\rfloor =1$. This means that $f(x)=0$ or $f(x)=c$, where $1\leq c<2$.