Problem 5 of Fourth round

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Problem 5 of Fourth round

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: IX International Festival of Young Mathematicians Sozopol, Theme for 10-12 grade

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

437 posts

Pinko

#1 h Sep 22, 2018, 5:20 PM • 1 Y

Y by Adventure10

Find all functions $f :[0, +\infty) \rightarrow [0, +\infty)$ for which

$f(f(x)+f(y)) = xy f (x+y)$

for every two non-negative real numbers $x$ and $y$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

437 posts

Pinko

#2 h Sep 22, 2018, 8:07 PM • 2 Y

Y by Adventure10, Mango247

Answer
Solution

This post has been edited 1 time. Last edited by Pinko, Sep 22, 2018, 8:10 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

470 posts

#3 h Sep 22, 2018, 8:15 PM • 3 Y

Y by Pinko, Adventure10, Mango247

Ukrainian mathematical olympiad 2018 10.4 and 11.3.
See also.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

437 posts

Pinko

#4 h Sep 22, 2018, 8:26 PM • 2 Y

Y by Adventure10, Mango247

Thanks! I didn't know that this problem was also given on another competition.

This post has been edited 2 times. Last edited by Pinko, Sep 22, 2018, 8:27 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1188 posts

#5 h Jan 8, 2022, 3:05 PM

Y by

@Pinko I do not think 2) in your solution is correct, you implicitly assume $f(0) = 0$ which is not given or proven before. Moreover, later on you prove that if $z$ is such that $f(z) \neq 0$ then $f(2z) = 0$ , so overall you have "for each $x$ we have $f(x) = 0$ or $f(2x) = 0$ " but this does not imply that $f(2x) = 0$ for all $x$ - e.g. the function $f(1) = 1$ , $f(4) = 100$ , $f(x) = 0$ otherwise satisfies this (and there are many more examples)

This post has been edited 1 time. Last edited by Assassino9931, Jan 8, 2022, 4:02 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1188 posts

#7 h Jan 8, 2022, 3:59 PM

Y by

Here's a full solution, fixing Pinko's gaps. As $f\equiv 0$ is a solution, assume, for contradiction, that there is some $b$ with $f(b) > 0$ . We firstly determine $f(0) = a$ . Denote the assertion by $P(x,y)$ .

$P(x,0)$ gives $f(f(x) + a) = 0$ . Hence $P(x+f(a), 0)$ gives $0 = f(f(x+f(a)) + a) = f(a)$ . Then $P(f(x), a)$ leads to $f(f(f(x))) = 0$ (using $f(f(x) + a) = 0$ ). In particular $f(f(f(0))) = 0$ and since $f(f(0)) = 0$ , we deduce $f(0) = 0$ and $b\neq 0$ .

Moving on, $P(x,0)$ yields $f(f(x)) = 0$ and hence $P(x, f(y))$ gives $xf(y)f(x+f(y)) = f(f(x) + f(f(y))) = f(f(x)) = 0$ . In other words, if $f(y) > 0$ , then $f(x+f(y)) = 0$ for all $x$ and hence $f(f(x) + f(y)) = 0$ for all $x$ . With the latter, $P(x,b)$ implies $0 = f(f(x) + f(b)) = xbf(x+b)$ and so $f(x+b) = 0$ for all $x>0$ , i.e. $f(x) = 0$ for all $x>b$ . In particular, $b$ must be unique (as if $b_1>b_2$ are such that $f(b_1) > 0$ , $f(b_2) > 0$ , then $x=b_1$ gives a contradiction). But then $f(\frac{b}{2}) = 0$ and $P(\frac{b}{2}, \frac{b}{2})$ yields $0 = f(2f(\frac{b}{2})) = \frac{b^2}{4}f(b)$ which with $b>0$ implies $f(b) = 0$ , contradiction! Therefore $f \equiv 0$ is the only solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11091 posts

#8 h Jan 8, 2022, 7:11 PM • 1 Y

Y by megarnie

Under revision.

This post has been edited 1 time. Last edited by jasperE3, Jan 8, 2022, 11:11 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1063 posts

#9 h Jan 8, 2022, 7:47 PM

Y by

jasperE3 wrote:

Here's a quick proof.
$P(1,-1)\Rightarrow f(f(1)+f(-1))+f(0)=0$ and since $f(x)\ge0$ equality holds and $f(0)=0$ .

Ummm

Pinko wrote:

Find all functions $f :[0, +\infty) \rightarrow [0, +\infty)$ for which

$f(f(x)+f(y)) = xy f (x+y)$

for every two non-negative real numbers $x$ and $y$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a