FE with a surprise

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

FE with a surprise

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: India Postal Set 3 P2 2016

The post below has been deleted. Click to close.

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

3070 posts

#1 h Jan 18, 2017, 3:11 AM • 4 Y

Y by Amir Hossein, mathematicsy, Adventure10, Mango247

Determine all functions $f:\mathbb R\to\mathbb R$ such that for all $x, y \in \mathbb R$
$f(xf(y) - yf(x)) = f(xy) - xy.$

Z K Y

The post below has been deleted. Click to close.

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

6857 posts

#2 h Jan 18, 2017, 6:09 AM • 8 Y

Y by Ankoganit, anantmudgal09, PRO2000, v4913, mathematicsy, Danielzh, Adventure10, Mango247

Put $y=0$ to get either $f$ constant, which is absurd, or $f(0)=0$ .
Put $x=y=\sqrt t$ for $t\ge0$ to get $f(t)=t$ for $t\ge0$ .
Now let $c=f(-1)$ . Put $(x,y)=(t,-1)$ and $(x,y)=(-1,t)$ for $t<0$ to conclude $f(-t)=-t+|t+ct| = (|1+c|-1)t$ . Let $k=1-|1+c|$ .

So we just need to find all $k$ such that the function $f$ defined by $f(x)=x$ for $x \ge 0$ and $f(x) = kx$ for $x \le 0$ works. Clearly it suffices to check when $xy<0$ . By cases we see that this forces $|k^2-k|=|k-1|$ hence $k=\pm1$ both of which work.

Hence the answers are the identity function and the absolute value function.

Z K Y

The post below has been deleted. Click to close.

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

132 posts

#3 h Jan 18, 2017, 7:43 AM • 1 Y

Y by Adventure10

f(x) equals x or modx .

Z K Y

The post below has been deleted. Click to close.

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

1000 posts

#4 h Jan 18, 2017, 3:48 PM • 4 Y

Y by Amir Hossein, mathematicsy, Adventure10, Mango247

$\text{Claim 1:}$ $f(0)=0$
Proof:if $f(0)\neq 0$ $P(\frac{x}{f(0)},0)$ $\implies$ $f$ is constant. So Claim is proved.

$\text{Claim 2:}$ $f(x)=x$ for all positive $x$
Proof:
$f(x,x)$ $\implies$ $f(x^2)=x^2$ so for positive $x$ we have $f(x)=x$

$\text{Claim 3:}$ $f(x)=0$ $\iff$ $x=0$
Proof:
Since $f(0)=0$ there exists $a$ s.t $f(a)=0$ .
$P(a,1)$ $\implies$ $a=0$

Now let's select positive $x$ and substitute $P(-x,-1)$ $\implies$
$f(f(-x)-xf(-1))=0$ so $f(-x)=xf(-1)$
So solutions are $f(x)=x$ for positives $f(x)=-xf(-1)$ for negatives.

This post has been edited 1 time. Last edited by tenplusten, Jan 18, 2017, 3:49 PM

Z K Y

The post below has been deleted. Click to close.

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

6857 posts

#5 h Jan 19, 2017, 3:09 AM • 2 Y

Y by v4913, Adventure10

Murad.Aghazade wrote:

So solutions are $f(x)=x$ for positives $f(x)=-xf(-1)$ for negatives.

Unless I'm mistaken, not all these functions work?

Z K Y

The post below has been deleted. Click to close.

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

1000 posts

#6 h Jan 19, 2017, 11:45 AM • 1 Y

Y by Adventure10

v_Enhance wrote:

Murad.Aghazade wrote:

So solutions are $f(x)=x$ for positives $f(x)=-xf(-1)$ for negatives.

Unless I'm mistaken, not all these functions work?

I think it is true whatever $f(-1)$ is.

Edited:Actually we can find the value of $f(-1)$ looking at $P(5,-3)$ and $P(-3,5)$

This post has been edited 2 times. Last edited by tenplusten, Jan 19, 2017, 12:48 PM

Z K Y

The post below has been deleted. Click to close.

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

8 posts

#7 h Jan 20, 2017, 12:49 PM • 2 Y

Y by Adventure10, Mango247

$f(-x)=xf(-1)$ then set x=-a (a is positive) then $f(-1)=-1$

Z K Y

The post below has been deleted. Click to close.

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

8 posts

#8 h Jan 20, 2017, 12:50 PM • 2 Y

Y by Adventure10, Mango247

oh sorry you set x as positive

Z K Y

The post below has been deleted. Click to close.

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

39 posts

#9 h Apr 2, 2021, 1:39 AM • 3 Y

Y by Mango247, Mango247, Mango247

Can someone explain me the casework in post 2?
I don't understand the part where he got $|k^2-1|=|k-1|$ .

This post has been edited 1 time. Last edited by mathematicsy, Apr 2, 2021, 1:39 AM
Reason: Referenced the wrong post.

Z K Y

The post below has been deleted. Click to close.

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

11091 posts

#10 h Apr 26, 2021, 5:04 PM • 1 Y

Y by Abdullahil_Kafi

Let $P(x,y)$ be the assertion $f(xf(y)-yf(x))=f(xy)-xy$ .
$P(x,0)\Rightarrow f(xf(0))=f(0)$ , so either $f$ is constant (no such solutions) or $f(0)=0$ .
$P(x,x)\Rightarrow f\left(x^2\right)=x^2$ , so $f$ fixes the positive reals. From now on, let $x,y\in\mathbb R^+$ .
$P(-x,-y)\Rightarrow f(yf(-x)-xf(-y))=f(xy)-xy=0$
$P(yf(-x)-xf(-y),1)\Rightarrow yf(-x)-xf(-y)=0\Rightarrow\frac y{f(-y)}=\frac x{f(-x)}=c$ for some constant $c$ . Thus $f(-x)=cx$ for some constant $c$ . Testing this in the original equation, we find $c\in\{-1,1\}$ , hence the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=|x|}$ which both work.

This post has been edited 2 times. Last edited by jasperE3, Apr 26, 2021, 5:27 PM

Z K Y

The post below has been deleted. Click to close.

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

476 posts

#11 h May 13, 2023, 11:20 PM

Y by

Nice problem!

Plug in $(x,0)$ to get $f(xf(0))=f(0)$ . So for all $r \in \mathbb{R}$ , $f(\frac{r}{f(0)}*f(0))=f(r)=f(0)$ , if $f(0) \neq 0$ , meaning that function $f$ would be constant. However, plugging in any $x,y \neq 0$ would give $0 = -xy$ , a contradiction.

Therefore, we conclude that $f(0)=0$ . So by now we have tried every pair including $0$ . Now let's plug in $(x,x)$ to get $f(x^{2})=x^{2}$ . Hence, for all nonnegative $t$ , $f(t)=t$ .

Ok, seems like we're almost there. How can we take advantage of this? Well, to get a positive number you can either multiply two positive numbers or two negative numbers. Let's try testing two negative numbers $(-x,-y)$ , while keeping $x,y \ge 0$ (if we test (x,y) we get $f(0)=0$ which is already found).

We get $f(-xf(-y)+yf(-x))=0$ . But what happens if $-xf(-y)+yf(-x) < 0$ ?, we wouldn't know the value of $f(-xf(-y)+yf(-x))$ then!
Well be can plug in $(-y,-x)$ for the same values of $x,y$ to get $f(-yf(-x)+xf(-y))=0$ . Notice that $-yf(-x)+xf(-y) = -(-xf(-y)+yf(-x))$ , so at least one of them must be nonnegative. But we also know that $f(t)=t$ for all nonnegative integers $t$ !

Therefore one of $(-xf(-y)+yf(-x),-yf(-x)+xf(-y))$ must be equal to $0$ . WLOG let $-xf(-y)+yf(-x)=0$ . Then $xf(-y)=yf(-x)$ (recall that $x,y$ are both nonnegative). Well, what does this tell us?
We have simply "fix" a value, say $f(-1)$ and put $f(-r)$ in terms of $f(-1)$ for all nonnegative $r$ .

Plugging in $(x,1)$ , we get $f(-x)=f(-1)x$ for some constant $f(-1)$ . Hence, function $f$ is linear. Well, how do I figure out the value of $f(-1)$ ?

Plug in $(1,-1)$ and $(-1,1)$ to get $f(f(-1)+1)=f(-1)+1=f(-1-f(-1))$ . Once again, we see that one of $(f(-1)+1,-1-f(-1))$ must be nonnegative.

WLOG if $f(-1)+1$ were to be positive, then $-1-f(-1)$ is negative, but $f(f(-1)+1)=f(-1)+1=f(-1-f(-1))$ , so $f(-x)=x$ .

WLOG if $f(-1)+1$ were to be 0, then $f(-1)=-1$ , so $f(-x)=-x$ .

Hence, functions $f$ can either be $y=x$ or $y=|x|$ . It is trivial to show that they satisfy the conditions, so we are done. $\blacksquare{}$

Z K Y

The post below has been deleted. Click to close.

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

5531 posts

#12 h Sep 29, 2023, 10:01 PM

Y by

The only solutions are $\boxed{f(x) = x}$ and $\boxed{f(x) = |x|}$ , which work.

Let $P(x,y)$ denote the given assertion.

$P(x,0): f(xf(0)) = f(0)$ . If $f(0) \ne 0$ , then $f$ is constant, absurd. So $f(0) = 0$ .

$P(x,x): f(x^2) = x^2$ , so $f(x) = x\forall x\ge 0$ .

$P(x,-1): f(xf(-1) + f(x)) = f(-x) + x$ .

Claim: If $f(x) = 0$ , then $x = 0$ .
Proof: Suppose there existed $f(a) = 0$ for some $a\ne 0$ . $P(a,1): f(a) = -a$ , so $a = 0$ , contradiction. $\square$

If $x \le 0$ , then $P(x,-1)\implies f(x f(-1) + f(x)) = 0$ , so $f(x) = -xf(-1)$ for all $x\le 0$ .

$P(x, 1): f(x - f(x)) = f(x) - x$ . For any $x < 0$ , then $f(x + xf(-1)) = -x - xf(-1)$ . Since $f(k) = -k\implies k \le 0$ , we have that $x(f(-1) + 1) \le 0$ , so $f(-1) + 1 \ge 0$ . If $f(-1) + 1 > 0$ , then $x + x f(-1)$ for $x\le 0$ can take any real number less than or equal to $0$ , so $f(x) = -x = |x|\forall x\le 0$ . If $f(-1) + 1 = 0$ , then $f(-1) = -1$ , so $f(x) = -x \cdot -1 = x$ for all $x\le 0$ . This gives the desired solutions.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a