Find the function!

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

Find the function!

G H J

Reveal topic tags

algebra unsolved

L

G H BBookmark kLocked kLocked NReply

Source: Problem 3, Brazilian Olympic Revenge 2005

The post below has been deleted. Click to close.

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

Johann Peter Dirichlet

376 posts

Johann Peter Dirichlet

#1 h Jun 1, 2005, 6:24 PM • 2 Y

Y by Adventure10, Mango247

Find all functions $f: R \rightarrow R$ such that
$f(x+yf(x))+f(xf(y)-y)=f(x)-f(y)+2xy$
for all $x,y \in R$

Z K Y

The post below has been deleted. Click to close.

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

708 posts

#2 h Jun 2, 2005, 8:04 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

at least i have found these things maybe helfull

we define this: $A:$ $x \longrightarrow y$ means in $A$ put $y$ beside of $x$ .
we name $f(yf(x)+x)+f(xf(y)-y)=f(x)-f(y)+2xy$ (we name this $A$ )

1. $f(0)=0$
2. $f(-x)=-f(x)$
3. $f(xf(x)+x)+f(xf(x)-x)=2x^2$
4. $f(xf(x)+x)=f(x)+x^2$
5. $f(xf(x)-x)=x^2-f(x)$

and these are the proves of the steps
$1$ .

$A$ : $x,y \longrightarrow 0$ and $f(0)=0$

$2$ .

$A$ : $x \longrightarrow y, y \longrightarrow x$ and adding what we will have to $A$ we will have
$f(yf(x)+x)+f(yf(x)-x)+f(xf(y)+y)+f(xf(y)-y)=4xy$
now $y \longrightarrow0$ and using $f(0)=0$ we will have $f(-x)=-f(x)$

$3$

$A$ : $y \longrightarrow x$ and we will have
$f(xf(x)+x)+f(xf(x)-x)=2x^2$

$4$

$A$ : $x \longrightarrow -x$ and using $f(-x)=-f(x)$ we will have $f(yf(x)+x)+f(xf(y)+y)=f(x)+f(y)+2xy$ ( $B$ )
now $B$ : $y \longrightarrow x$ we will have $f(xf(x)+x)=f(x)+x^2$

$5$

from $4$ we have reach $B:(yf(x)+x)+f(xf(y)+y)=f(x)+f(y)+2xy$ now we have
$C: A+B=2f(yf(x)+x)+f(xf(y)+y)+f(xf(y)-y)=2f(x)+4xy$ and $C: y \longrightarrow x$ we will have
$3f(xf(x)+x)+f(xf(x)-x)=2f(x)+4x^2$
but from $4$ we have that $f(xf(x)+x)=f(x)+x^2$ so $3f(xf(x)+x)+f(xf(x)-x)=3f(x)+3x^2+f(xf(x)-x)=2f(x)+4x^2$
that will lead us to $f(xf(x)-x)=x^2-f(x)$

This post has been edited 6 times. Last edited by lomos_lupin, Jun 2, 2005, 5:00 PM

Z K Y

The post below has been deleted. Click to close.

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

98 posts

galeos

#3 h Jun 2, 2005, 9:52 AM • 1 Y

Y by Adventure10

you have made a mistake. In 3) you should have a $2$ on the RHS. Moreover, $f(x)=x$ is clearly a solution.

Z K Y

The post below has been deleted. Click to close.

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

708 posts

#4 h Jun 2, 2005, 10:50 AM • 2 Y

Y by Adventure10, Mango247

you are right then the whole thing is useless

Z K Y

The post below has been deleted. Click to close.

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

99 posts

Saul

#5 h Jun 4, 2005, 1:48 PM • 2 Y

Y by Adventure10, Mango247

Here's my partial solution.

$f(x+yf(x))+f(-y+xf(y))=f(x)-f(y)+2xy$ (1)

Putting $x=y=0$ , we get $f(0)=0$ easily.
Now putting only $x=0$ , we have
$f(0) + f(-y) = f(0) - f(y)$
so that
$f(-x) = -f(x)$ .

Next, we prove that $0$ is the only $x$ for which $f(x)=0$ . Suppose $z$ is such that $z \neq 0$ , $f(z)=0$ . Put $x=y=z$ . Then we get $2z^2=0$ out of (1), a contradiction.

Now, we prove that $f(x)$ satisfies the Cauchy equation.

Substitute $-y$ for $y$ in (1). Because $f(y)=-f(-y)$ , we get

$f(x-yf(x))+f(y-xf(y))=f(x)+f(y)-2xy$

Adding this to (1) gives

$f(x+yf(x))+f(x-yf(x))+(-f(y-xf(y)))+f(y-xf(y))=f(x)+f(x)+f(y)-f(y)+2xy-2xy$

so that

$f(x+yf(x))+f(x-yf(x))=2f(x)$

If $f(x) \neq 0$ (that is iff $x \neq 0$ ) we can fix $x$ and vary $y$ so this becomes

$f(x+a)+f(x-a)=2f(x)$ (2)

for $a \in R$ , $x \neq 0$ . But it works for $x=0$ as well, as we have proved above. Putting $x=a$ in (2) gives $f(2x) = 2f(x)$ . Putting $x=\frac{x+y}2$ and $a=\frac{x-y}2$ gives

$f(x)+f(y)=2f(\frac{x+y}2)=f(x+y)$ .

If $f(x)=kx$ , then $2k^2=2$ from (1) so $k=1$ or $-1$ . $f(x)=x$ works and $f(x)=-x$ also works.

But I don't know how to eliminate the wild solutions... bijectivity doesn't do it, does it?

Z K Y

The post below has been deleted. Click to close.

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

708 posts

#6 h Jun 5, 2005, 6:35 AM • 1 Y

Y by Adventure10

nice piece of work

Z K Y

The post below has been deleted. Click to close.

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

268 posts

ali

#7 h Jun 7, 2005, 2:11 PM • 1 Y

Y by Adventure10

The problem is just getting started after we get:
f(x+y)=f(x)+f(y) and f(xf(x))=x^2
Anyone?

Z K Y

The post below has been deleted. Click to close.

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

268 posts

ali

#8 h Jun 7, 2005, 8:16 PM • 1 Y

Y by Adventure10

In fact note that since f(xf(x))=x^2 (*)
we get that 2yx = f(xf(y)) + f(yf(x))
Also so replacing y by 1 we obtain that
2x= f(f(x)) + f(bx) where b=f(1)
But since f(xf(x))=x^2 we get that f(b)=1.
f(xf(x))=x^2 and f(x+y)=f(x)+f(y) imply that f(x) is rational only if x is rational.
(proof. if f(x) rational then f(xf(x)) = f(x)^2 =x^2 so x is rational)
so since b=1, f(1)^2 = 1 and so f(1)=1/-1.
Case1.
f(1)=1
hence f(Q) = Q
Also f(f(x)) + f(x) =2x.
so f is bijective and therefore: xf(1/x) +1/x f(x) =2

Z K Y

The post below has been deleted. Click to close.

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

268 posts

ali

#9 h Jun 8, 2005, 10:29 PM • 1 Y

Y by Adventure10

Well, no one is interested in finishing this problem

So I will do it

Observe that f(1)=1 or -1.
I will only solve for the case when f(1)=1, the other case is exactly the same.
f(f(x)) + f(x) =2x and additivity of f(x) imply that f(x) +x is one to one.
So Now replace in equation f(xf(x))=x^2 , x with f(x)
to observe together with injectivity of f(x)+x that f(x^2) = f(x)^2
This implies that f(x) is increasing. So f(x)=x...

-Ali

Z K Y

The post below has been deleted. Click to close.

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

1275 posts

#10 h Jun 11, 2005, 6:06 AM • 2 Y

Y by Adventure10, Mango247

I almost understood nothng how you finished your solution. (I think you supposed that $f$ is continious). But this is my solution

First if $x=y=0$ then $f(0)=0$ . Now if $x=0$ then $f(-y)=-f(y)$ . So $f$ is odd. Now replace $x$ with $-x$ in the problem, then: $f(x+yf(x))+f(y+xf(y))=f(x)+f(y)+2xy \ \ \ \ \ \ (*)$
Suppose $f(c)=0$ and $c\in\mathbb R$ . Then in relation $(*)$ if $x=y=c$ then $c^2=0$ and $c=0$ . Now in relation $(*)$ replace $y$ with $-y$ then: $f(x-yf(x))-f(y+xf(y)=f(x)-f(y)-2xy$
Adding the 2 last relations we obtain: $f(x+yf(x))+f(x-f(x))=2f(x)$ . Then if we suppose $a=x+yf(x),\ b=x-yf(x)$ then $\forall a,b \in\mathbb R \ \ f(a)+f(b)=2f(\frac{a+b}{2})$
If $b=0$ then $\displaystyle f(\frac{a}2)=\frac{f(a)}2$ . So $\forall a,b \in\mathbb R\ \ f(a)+f(b)=f(a+b) \ \ (**)$
Now noticing relations $(*)$ and $(**)$ : $f(x)+f(xf(y))+f(y)+f(yf(x))=f(x)+f(y)+2xy$
So $f(xf(y))+f(yf(x))=2xy\ \ (1)$
Also because if $f(a)=f(b)\Rightarrow f(a-b)=0\Rightarrow a=b$ so $f$ is injective.
Now we prove that, there exist $t\in\mathbb R\backslash\{0\}$ that $f(t)=t$ or there exist $t\in\mathbb R\backslash\{0\}$ that $f(t)=-t$ . For this in relation $(*)$ if $x=y=1$ and $x=-y=1$ then: $f(f(1)+1)=f(1)+1\mbox{ and } f(1-f(1))=f(1)-1$
And at least one of the numbers $f(1)+1$ and $1-f(1)$ is nonzero. WLOG Suppose there exist $t$ that $f(t)=t$ (for the other case the calculations are similiar)

Now we prove that $f(1)=1$ . In relation (1) if $x=1,y=t$ then $t+f(tf(1))=f(t)+f(tf(1))=2t$ . So $f(tf(1))=t=f(t)\Rightarrow t=tf(1)\Rightarrow f(1)=1$

Now if $y=1$ then $f(f(x)+x)=2x$ . So $f(f(x)-x)=f(f(x)+x)-2f(x)=2x-2f(x)$ . Now if $f(x)\neq x$ then if $u=f(x)-x$ then $u\neq0$ and $f(u)=-2u$ . So $=f(-2u^2)=f(uf(u))=u^2$ . So $f(2u^2)=-u^2$ . Now suppose $x=1$ and $y=2u^2$ then $f(f(2u^2)+2u^2)=4u^2=f(2u.f(2u))=f(4uf(u))$ . So $f(2u^2)=2u^2=4u.f(u)$ . So $-u^2+2u^2=4u.(-2u)=-8u^2$ . Therefore $u=0$ and $f(x)=x$ for each $u\in\mathbb R$ .
Also if there exist $t$ that $f(t)=-t$ then with similiar calculations $f(x)=-x$ for each $x$ and solutions are $f(x)=x$ and $f(x)=-x$

Z K Y

The post below has been deleted. Click to close.

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

268 posts

ali

#11 h Jun 11, 2005, 4:17 PM • 2 Y

Y by Adventure10, Mango247

My solution is perfectly valid...No extra assumption, such as continuty

Z K Y

The post below has been deleted. Click to close.

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

268 posts

ali

#12 h Jun 11, 2005, 4:21 PM • 2 Y

Y by Adventure10, Mango247

BE ALWAYS SPECIFIC..
JUST DON'T WRITE A CRAP FOR ME...
IF IT IS WRONG, SHOW WHY?

Z K Y

The post below has been deleted. Click to close.

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

1275 posts

#13 h Jun 12, 2005, 9:31 AM • 1 Y

Y by Adventure10

First I didn't say your solution is wrong

Second Why do you get impatient?

I just didn't understood how you proved $f(x)$ is increasing and how you said $f(x)=x$

I'll be pleased if you explain more.

Z K Y

The post below has been deleted. Click to close.

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

705 posts

#14 h Apr 5, 2025, 3:16 PM • 1 Y

Y by Primeniyazidayi

$f(x+yf(x))+f(xf(y)-y)=f(x)-f(y)+2xy$ Answers are $f(x)=x$ and $f(x)=-x$ which clearly work. Let $P(x,y)$ be the assertion.
Claim: $f(0)=0$ .
Proof: $P(0,0)$ gives $2f(0)=0$ or $f(0)=0$ .
Claim: $f$ is odd.
Proof: $P(0,y)$ implies $f(-y)=-f(y)$ .
Claim: $f(a)=0\iff a=0$ .
Proof: If $f(a)=0$ , then $f(-a)=0$ and $P(a,a)$ yields $0=2a^2$ so $a=0$ . The other direction has been proven.
Claim: $f$ is additive.
Proof: Summing $P(x,y)$ and $P(-x,y)$ up gives $f(xf(y)+y)+f(y-xf(y))=2f(y)$ or $f(x+y)+f(y-x)=2f(y)$ or $f(a)+f(b)=2f(\frac{a+b}{2})$ . Since $b=0$ implies $f(a)=2f(\frac{a}{2})$ , we get $f(a)+f(b)=2f(\frac{a+b}{2})=f(a+b)$ hence $f$ is additive.
Claim: $f$ is injective.
Proof: If $f(p)=f(q)$ , then $f(p-q)=0$ but this is impossible for $p\neq q$ .
Claim: $f(x)=x$ for all reals or $f(x)=-x$ for all reals.
Proof: The main equation gives $f(xf(y)+yf(x))=f(yf(x))+f(xf(y))=2xy$ by additivity. Replace $\frac{x}{y}$ with $x$ to get $f(x.\frac{f(y)}{y}+yf(\frac{x}{y}))=2x=f(x.\frac{f(z)}{z}+zf(\frac{x}{z}))$ thus, $\frac{xf(y)}{y}+yf(\frac{x}{y})=\frac{xf(z)}{z}+zf(\frac{x}{z})$ or $\frac{f(y)}{y}+\frac{y}{x}f(\frac{x}{y})=\frac{f(z)}{z}+\frac{z}{x}f(\frac{x}{z})$ and if $f(x)=xg(x)$ , then $g(y)+g(\frac{x}{y})=g(z)+g(\frac{x}{z})$ or $g(x)+g(y)=g(z)+g(\frac{xy}{z})$ . Thus, $ab=cd$ implies $g(a)+g(b)=g(c)+g(d)$ .
Note that $f(2xf(x))=2x^2$ or $f(xf(x))=x^2$ . We have $\frac{f(x)}{x}+\frac{f(y)}{y}=\frac{f(xy)}{xy}+f(1)$ and $x,yf(y)$ gives
$\frac{f(x)}{x}+\frac{y}{f(y)}=\frac{f(x)}{x}+\frac{f(yf(y))}{yf(y)}=\frac{f(xyf(y))}{xyf(y)}+f(1)$ $\frac{f(x)}{x}+\frac{f(y)}{y}=\frac{f(xy)}{xy}+f(1)$ Hence by their difference, we see that $\frac{f(y)}{y}-\frac{y}{f(y)}=\frac{f(xy)}{xy}-\frac{f(xyf(y))}{xyf(y)}$ or $\frac{f(y)}{y}-\frac{y}{f(y)}=\frac{f(x)}{x}-\frac{f(xf(y))}{xf(y)}$ . Now replace $yf(y)$ with $y$ to get $\frac{y}{f(y)}-\frac{f(y)}{y}=\frac{f(x)}{x}-\frac{f(xy^2)}{xy^2}$ . If $y=1$ , then $f(1)^2=1$ so $f(1)=\pm 1$ . Pick $x=\frac{1}{y}$ which implies $\frac{y}{f(y)}=\frac{f(\frac{1}{y})}{\frac{1}{y}}$ or $f(y)f(\frac{1}{y})=1$ or $g(x)g(\frac{1}{x})=1$ . If $g(1)=1$ , then $g(x)+g(\frac{1}{x})=2$ and $g(x)g(\frac{1}{x})=1$ so $(g(x)-1)^2=0$ or $g\equiv 1$ or $f(x)=x$ for all reals. If $g(1)=-1$ , then $g(x)+g(\frac{1}{x})=-2$ and $g(x)g(\frac{1}{x})=1$ thus, $(g(x)+1)^2=0$ or $g\equiv -1$ or $f(x)=-x$ for all reals as desired. $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a