IMO 2010 Problem 1

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134 posts

#1 h Jul 7, 2010, 4:32 PM • 25 Y

Y by MSTang, Davi-8191, anantmudgal09, bel.jad5, Ramanujan_math, Gerninza, itslumi, TheMathCruncher_007, samrocksnature, centslordm, HWenslawski, megarnie, Amoonguss, ImSh95, David-Vieta, Mogmog8, Adventure10, Mango247, seansean01347, aidan0626, cubres, and 4 other users

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

This post has been edited 1 time. Last edited by djmathman, Apr 27, 2015, 2:58 PM
Reason: formatting

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socrates

2105 posts

socrates

#2 h Jul 7, 2010, 5:27 PM • 52 Y

Y by mad, plwseven, Pythagorasauras, AdithyaBhaskar, droid347, Sx763_, JasperL, Wizard_32, Illuzion, Severus, MathbugAOPS, Promi, Pluto1708, A_Math_Lover, Minusonetwelfth, ayan_mathematics_king, green_leaf, Bassiskicking, OlympusHero, itslumi, Smathematician, Aurn0b, kirillnaval, Muaaz.SY, samrocksnature, centslordm, mijail, Sush0, megarnie, Amoonguss, myh2910, skyguy88, David-Vieta, Zaro23, Stuffybear, Adventure10, Mango247, Sedro, aidan0626, dxd29070501, AlexCenteno2007, cosdealfa, poirasss, EpicBird08, cubres, and 7 other users

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.

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wangsacl

182 posts

wangsacl

#3 h Jul 7, 2010, 5:35 PM • 9 Y

Y by Pythagorasauras, Illuzion, samrocksnature, centslordm, Amoonguss, Adventure10, cubres, and 2 other users

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..

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m.candales

186 posts

m.candales

#4 h Jul 7, 2010, 5:47 PM • 6 Y

Y by samrocksnature, Adventure10, Mango247, cubres, and 2 other users

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 $c \in [1,2)$

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Jorge Miranda

157 posts

Jorge Miranda

#5 h Jul 7, 2010, 7:04 PM • 6 Y

Y by samrocksnature, Adventure10, Mango247, cubres, and 2 other users

Easy, but nice

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MellowMelon

5850 posts

MellowMelon

#6 h Jul 7, 2010, 9:10 PM • 8 Y

Y by samrocksnature, ZETA_in_olympiad, Adventure10, Mango247, cubres, and 3 other users

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.

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Justanotherone

10 posts

Justanotherone

#7 h Jul 7, 2010, 10:02 PM • 6 Y

Y by samrocksnature, Dhruv777, Adventure10, cubres, and 2 other users

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

This post has been edited 3 times. Last edited by Justanotherone, Jul 8, 2010, 7:22 AM

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ralexamigos

5 posts

ralexamigos

#8 h Jul 8, 2010, 12:19 AM • 4 Y

Y by samrocksnature, Adventure10, Mango247, cubres

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

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mavropnevma

15142 posts

mavropnevma

#9 h Jul 8, 2010, 6:54 AM • 13 Y

Y by FlakeLCR, samrocksnature, DofL, ZETA_in_olympiad, Adventure10, AlexCenteno2007, cubres, and 6 other users

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).

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rustam

348 posts

rustam

#10 h Jul 8, 2010, 7:03 AM • 4 Y

Y by samrocksnature, Adventure10, cubres, and 1 other user

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

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abacadaea

2176 posts

abacadaea

#11 h Jul 8, 2010, 2:03 PM • 6 Y

Y by samrocksnature, Adventure10, Mango247, cubres, and 2 other users

hopefully correct

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SnowEverywhere

801 posts

SnowEverywhere

#12 h Jul 8, 2010, 5:30 PM • 6 Y

Y by samrocksnature, Adventure10, Mango247, AlexCenteno2007, cubres, and 1 other user

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.

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Sylphaen

151 posts

Sylphaen

#13 h Jul 9, 2010, 9:33 AM • 4 Y

Y by samrocksnature, Adventure10, Mango247, cubres

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

!

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stsamster

18 posts

stsamster

#14 h Jul 10, 2010, 8:04 AM • 4 Y

Y by samrocksnature, Adventure10, Mango247, cubres

$f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor$ -

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.
$Q.E.D$

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Dr N0

131 posts

Dr N0

#15 h Jul 10, 2010, 3:59 PM • 4 Y

Y by samrocksnature, Adventure10, Mango247, cubres

I will try to do it differently, maybe it's wrong I dont know.
Click to reveal hidden text

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ryanbear

1055 posts

ryanbear

#131 h Jan 11, 2024, 4:21 PM • 1 Y

Y by cubres

$P(0,0)$ 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.

This post has been edited 2 times. Last edited by ryanbear, Jan 14, 2025, 3:43 AM

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Tellocan

35 posts

Tellocan

#132 h Mar 7, 2024, 2:50 PM • 3 Y

Y by Ibrahim_K, Mirhabib, cubres

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.

This post has been edited 2 times. Last edited by Tellocan, Mar 8, 2024, 7:03 AM
Reason: Typo

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eibc

600 posts

eibc

#133 h Mar 11, 2024, 11:22 PM • 1 Y

Y by cubres

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 $\lfloor c \rfloor = 1$ .

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dolphinday

1324 posts

dolphinday

#134 h May 16, 2024, 3:52 AM • 1 Y

Y by cubres

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 $\left\lfloor f(x)\right\rfloor = 1$ .

This post has been edited 1 time. Last edited by dolphinday, May 16, 2024, 3:53 AM

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Sedro

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Sedro

#135 h Jun 16, 2024, 2:25 AM • 1 Y

Y by cubres

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. $\blacksquare$

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Bryan0224

53 posts

Bryan0224

#137 h Jul 20, 2024, 11:40 PM • 1 Y

Y by cubres

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 $c\in \{0\}\cup [1,2)$

This post has been edited 2 times. Last edited by Bryan0224, Oct 15, 2024, 7:49 AM

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Jndd

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Jndd

#138 h Jul 24, 2024, 10:58 PM • 1 Y

Y by cubres

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.

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Warideeb

59 posts

Warideeb

#142 h Aug 14, 2024, 2:19 PM • 1 Y

Y by cubres

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

This post has been edited 1 time. Last edited by Warideeb, Aug 15, 2024, 2:30 PM

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ezpotd

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ezpotd

#144 h Aug 31, 2024, 6:24 PM • 1 Y

Y by cubres

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.

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Maximilian113

574 posts

Maximilian113

#146 h Jan 7, 2025, 6:41 AM • 1 Y

Y by cubres

Let $P(x, y)$ denote the assertion. First, suppose that $f(0)=0.$ Then for all $0 \leq x < 1,$ we have that $f(x) \lfloor f(y) \rfloor = 0.$ If there is some $x$ in that range such that $f(x) \neq 0,$ we have that $\lfloor f(y) \rfloor = 0$ for all $y$ so $f(x) = 0$ for all $x.$ Otherwise, for all $0 \leq x < 1$ we have $f(x)=0.$ Then $P(2, 0.5) \implies f(1)=0 \implies P(1, x) \implies f(x)=0 \, \forall x \in \mathbb R.$
Now assume that $f(0) \neq 0.$ 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 $f(x) = 0, c \in [1, 2).$

This post has been edited 1 time. Last edited by Maximilian113, Mar 16, 2025, 4:51 PM
Reason: oops

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Bardia7003

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Bardia7003

#147 h Jan 12, 2025, 2:55 PM • 1 Y

Y by cubres

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.

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TestX01

340 posts

TestX01

#148 h Jan 14, 2025, 10:07 AM • 2 Y

Y by MihaiT, cubres

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 $1\leq c<2$ .

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blueprimes

344 posts

blueprimes

#149 h Feb 28, 2025, 1:22 PM • 1 Y

Y by cubres

We claim the only solutions are $f(x) \equiv c$ where $c = 0$ or $\lfloor c \rfloor = 1$ , which obviously work. Now we prove they are the only solutions, let $x = y = 0$ , we obtain $f(0) = f(0) \lfloor f(0) \rfloor$ . We now consider cases:

$\textbf{Case 1: }$ $\lfloor f(0) \rfloor = 1$
Consider $y = 0$ , we obtain $f(0) = f(x)$ so $f$ is constant. Clearly the only function that works is $f(x) \equiv c$ where $\lfloor c \rfloor = 1$ .

$\textbf{Case 2: }$ $f(0) = 0$ .
Asserting $x = y = 1$ yields $f(1) = f(1) \lfloor f(1) \rfloor$ . If $f(1) = 0$ , then $x = 1$ gives $f(y) = 0$ for all $y$ so $f(x) \equiv 0$ . On the other hand, if $\lfloor f(1) \rfloor = 1$ then $y = 1$ implies $f( \lfloor x \rfloor) = f(x)$ . So $(x, y) = \left(2, \dfrac{1}{2} \right)$ gives
$f(1) = f(2) f \left(\dfrac{1}{2} \right) = f(2) f(0) = 0$ which fails.

Altogether our only solutions are $f(x) \equiv c$ where $c = 0$ or $\lfloor c \rfloor = 1$ as wanted.

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Marcus_Zhang

980 posts

Marcus_Zhang

#150 h Mar 15, 2025, 9:47 PM • 1 Y

Y by cubres

Finished an IMO problem for IMO 2010 - the year I was born yay

Storage

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Ilikeminecraft

611 posts

Ilikeminecraft

#151 h Apr 25, 2025, 8:00 AM • 1 Y

Y by cubres

I claim the answer is $f(x) = c,$ where $c\in\{0\}\cup[0,1)$ is a constant.
Let $P(x, y)$ denote our assertion. From $P(0, x),$ we see that $f(0) = f(0)\lfloor f(y)\rfloor.$ There exist 2 cases from here, if $f(0) = 0$ or if $1\leq f(y) < 2\forall y\in\mathbb R.$

We handle $f(0) = 0$ first. From $P(x, y)$ where $\lfloor x \rfloor = 0,$ we see that $0 = f(x)\lfloor f(y) \rfloor.$ From here, notice that in both cases, $\lfloor f(x) \rfloor = 0.$ Now, take $P(x, y)$ where $0 < y < 1.$ We see that $f(\lfloor x \rfloor y) = 0,$ and hence, for all $x\in\mathbb R_{\geq0},$ $f(x) = 0.$ Finally, take $P(-1, y)$ for $y > 0$ to see that $f(-y) = 0.$

Now, look at when $1\leq f(y) < 2$ for all $y.$ From $P(1, y),$ we see that $f(y) = f(1)\lfloor f(y)\rfloor = f(1).$ Thus, we have that $f(x)$ is constant.

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