Functional equation over nonzero reals

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

Functional equation over nonzero reals

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: Spain MO 2025 P6

The post below has been deleted. Click to close.

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

34 posts

#1 h Mar 31, 2025, 1:15 PM

Y by

Let $\mathbb{R}_{\neq 0}$ be the set of nonzero real numbers. Find all functions $f:\mathbb{R}_{\neq 0}\rightarrow\mathbb{R}_{\neq 0}$ such that, for all $x,y\in\mathbb{R}_{\neq 0}$ , $(x-y)f(y^2)+f\left(xy\,f\left(\frac{x^2}{y}\right)\right)=f(y^2f(y)).$

Z K Y

The post below has been deleted. Click to close.

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

23497 posts

pco

#2 h Mar 31, 2025, 3:19 PM • 1 Y

Y by Stuttgarden

Stuttgarden wrote:

Let $\mathbb{R}_{\neq 0}$ be the set of nonzero real numbers. Find all functions $f:\mathbb{R}_{\neq 0}\rightarrow\mathbb{R}_{\neq 0}$ such that, for all $x,y\in\mathbb{R}_{\neq 0}$ , $(x-y)f(y^2)+f\left(xy\,f\left(\frac{x^2}{y}\right)\right)=f(y^2f(y)).$

Let $P(x,y)$ be the assertion $(x-y)f(y^2)+f(xyf(\frac{x^2}y))=f(y^2f(y))$

If $y+\frac{f(y^2f(y))}{f(y^2)}\ne 0$ , then $P(y+\frac{f(y^2f(y))}{f(y^2)},y)$ $\implies$ $f(\text{something})=0$ , impossible

So $f(y^2f(y))=-yf(y^2)$ $\forall y\ne 0$ and $P(x,y)$ becomes (replacing RHS) :
New assertion $Q(x,y)$ : $f(xyf(\frac{x^2}y))=-xf(y^2)$ $\forall x,y\ne 0$

$Q(xf(x),xf(x)^2)$ $\implies$ $f(x^2f(x)^4)=-xf(x)f(x^2f(x)^4)$

And so $\boxed{f(x)=-\frac 1x\quad\forall x\ne 0}$ , which indeed fits

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a