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Source: Baltic Way 2014, Problem 4

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#1 h Nov 11, 2014, 1:48 PM • 12 Y

Y by A-Thought-Of-God, FaThEr-SqUiRrEl, samrocksnature, mathematicsy, centslordm, megarnie, jhu08, tiendung2006, ImSh95, Adventure10, Mango247, ItsBesi

Find all functions $f$ defined on all real numbers and taking real values such that $f(f(y)) + f(x - y) = f(xf(y) - x),$ for all real numbers $x, y.$

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#2 h Nov 11, 2014, 2:54 PM • 20 Y

Y by socrates, LittelPerel, Abdollahpour, e_plus_pi, Wizard_32, Pluto1708, Beginner2004, A-Thought-Of-God, thczarif, FaThEr-SqUiRrEl, samrocksnature, mathematicsy, centslordm, jhu08, megarnie, Kanimet0, ImSh95, Adventure10, Mango247, Motaha313

socrates wrote:

Find all functions $f$ defined on all real numbers and taking real values such that $f(f(y)) + f(x - y) = f(xf(y) - x),$ for all real numbers $x, y.$

Let $P(x,y)$ be the assertion $f(f(y))+f(x-y)=f(xf(y)-x)$ . We get:
$P(0,0)\rightarrow f(f(0))=0$
$P(0,f(0)),P(f(0),f(0))\rightarrow f(0)=0$
$P(x,0)\rightarrow f(x)=f(-x)\Rightarrow P(x,-x),P(x,x)\rightarrow f(x)\equiv 0$
Hence $f(x)\equiv 0 \,\,\, \forall x\in \mathbb{R}$ .

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#3 h Nov 11, 2014, 2:56 PM • 10 Y

Y by socrates, john111111, starfire27, FaThEr-SqUiRrEl, samrocksnature, centslordm, jhu08, ImSh95, Adventure10, Mango247

Nice problem!!
Let $P(x,y)$ the given relation.
Let $y\in\mathbb{R}$ such that $f(y)\neq 2.$
Then $P(-\frac{y}{f(y)-2},y)$ gives $f(f(y))=0.$
So, if $f(y)\neq 2$ the initial gives: $f(x-y)=f(xf(y)-x)$
and for $x=0$ the last one gives $f(-y)=f(0)$
We conclude that either $f(y)=2$ or $f(-y)=f(0).$
But if $f(y)=2$ then $f(-y)=f(0)-f(2)$
So $f(y)=c_1$ for some $y$ and $f(y)=c_2$ for all the other and $c_1=2\neq 0.$
Take $y_0$ such that $f(y_0)=2.$ Then the initial gives: $f(2)+f(x-y_0)=f(x)$
Taking $x=2$ to the last one we get $f(2-y_0)=0$ so $c_2=0$ and $f(0)=f(2).$
So $f(y)=2$ for some $y$ and $f(y)=0$ for all the other.
Now take $x=y=y_0,$ then $f(2)+f(0)=f(y_0)=2.$ But $f(0)=f(2)$ so $f(0)=f(1)=1,$ this is a contradiction.
So $f(x)=0$ for all $x.$

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#4 h Nov 4, 2015, 1:26 PM • 6 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, mannshah1211

Problem (Baltic Way 2014):
Find all functions $f$ defined on all real numbers and taking real values such that $f(f(y)) + f(x - y) = f(xf(y) - x),$ for all real numbers $x, y.$

Solution :
Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$

$f \equiv 1$ isn't obviously a solution.
Thus there exists some $a \in \mathbb R$ such that $f(a) \neq 1$

$P(\frac{f(a)}{f(a)-1},a) \implies$
$f(\frac{f(a)}{f(a)-1}-a)=0$
$\implies$ there exists a $k$ such that $f(k)=0$

$P(x,k) \implies f(0)+f(x-k)=f(-x)$
$P(0.k) \implies f(-k)=0$
$P(k,k) \implies f(0)=0$

$\implies f(0)=0$

$P(0,y) \implies f(f(y)) + f(-y) = 0$
$P(\frac{-y}{f(y)-2},y) \implies f(y)=0$ for all $f(y) \neq 2$

Now let there exist some $t \in R$ such that $f(t)=2$
$f(f(t))= - f(t) = -2$
But we know that that for all $f(x) \neq 2$ , we have $f(x)=0$

A contradiction.

$\implies \boxed{f \equiv 0}$ which indeed is a solution.

This post has been edited 1 time. Last edited by utkarshgupta, Nov 4, 2015, 1:26 PM

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#6 h Apr 13, 2018, 9:33 AM • 6 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, Mango247

$x=y=0 f(f(0))=0$
$y=f(0) f(0)+f(x-f(0))=f(-x)$
$x=0 f(0)+f(-f(0))=f(0)$
$f(-f(0))=0$
$x=f(0) 2f(0)=f(-f(0))=0$
$f(0)=0$
$y=0 f(x)=f(-x)$
$x=y f(f(x))=f(xf(x)-x)$
$y=-x f(f(x))+f(2x)=f(xf(x)-x)$
$f(x)=0$

This post has been edited 4 times. Last edited by E.A.K, Apr 17, 2018, 6:27 PM

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#7 h May 8, 2018, 6:28 AM • 8 Y

Y by nhusanboev, A-Thought-Of-God, FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, Mango247

Sweet and Simple

.

Answer: We show that the only function satisfying the assertion is $\boxed{f(x) \equiv 0}$ .

Proof: Let (as usual) $P(x,y)$ denote the assertion $f(f(y)) + f(x-y) = f(xf(y) - x)$ .

Claim 1. $f(0) =0$ .
$\longrightarrow$ $P(0,0) \rightarrow f^2(0) + f(0) = f(0) \implies f^2(0) =0$

Now let $f(0) =c$ for some $c \in \mathbb{R}$ . Consider $P(0,c) \rightarrow f^2(c) + f(-c) = c \iff f(-c)=0$ .

Also consider $P(c,c) \rightarrow f^2(c) + c = f( c f(c) -c) \implies 2c = 0 \iff c= 0$ .

Claim 2. $f$ is even.
$\longrightarrow$ Consider $P(x,0) \rightarrow f(x) = f(-x)$

Claim 3. $f \equiv 0$ .
$\longrightarrow$ Consider the two equations : $P(x,x)$ and $P(x,-x)$ .

(1) $\rightarrow f^2(x) + f(2x) =f(xf(x) -x).$

(2) $\rightarrow f^2(x) = f(xf(x) -x).$

Compairing the two equations yields $f(2x) = 0 \forall x\in \mathbb{R}$ .

This post has been edited 1 time. Last edited by e_plus_pi, May 8, 2018, 6:34 AM

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#8 h Oct 22, 2018, 2:23 AM • 6 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, Mango247

Let $P(x,y)$ denote the assertion $f(f(y))+f(x-y)=f(xf(y)-x).$

We claim that the only solution is $\boxed{f(x)=0}.$

Clearly, this solution does work and we aim to prove that this is the only solution.

Claim 1: $f(f(0))=0.$
Note that $P(0,0)$ gives $f(f(0))+f(0)=f(0)$ implying the desired.

Let $f(0)=a$ so $f(a)=0$ so $f(f(0))=0$ .

Claim 2: The only value of $a$ that works is $a=0$ so $f(0)=0.$
Let $P(0,a)$ so $a+f(-a)=f(0)=a\implies f(-a)=0.$ Let $P(a,a)$ so $2a=f(-a)$ . Hence, $2a=f(-a)=0\implies a=0.$ Thus, $f(0)=0.$

Claim 3: $f$ is even.
Let $P(x,0)$ so $f(x)=f(-x)$ with the fact that $f(0)=0.$

Claim 4: $f(x+y)=f(x-y)$ for all real $x,y\in \mathbb{R}$ .
Consider $P(x,y)$ and $P(x,-y)$ which give, respectively, $\begin{align*} f(f(y))+f(x-y)=f(xf(y)-x)\\f(f(-y))+f(x+y)=f(xf(-y)-x). \end{align*}$ Rearranging and noting that $f$ is even gives $f(x-y)=f(xf(y)-x)-f(f(y))=f(xf(-y)-x)-f(f(-y))=f(x+y).$
Hence, we can set $x=y=\frac{x}{2}$ in $f(x-y)=f(x+y)$ so $f(x)=f(0)=0.$ $\blacksquare$

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#9 h Apr 26, 2019, 3:41 AM • 5 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10

Taking $x=y=0$ gives $f(f(0))=0$ . Taking $y=f(0)$ implies $f(0)+f(x-f(0))=f(-x)$ . Letting $x=f(0)/2$ in the previous equation yields $f(0)=0$ . Putting this back into the equation it was derived from, we see $f(x)=f(-x)$ or that $f$ is even. From the original FE, we have $f(x-y)=f(xf(y)-x)-f(f(y)).$ Swapping the sign on $y$ and noting evenness, $f(x+y)=f(xf(y)-x)-f(f(y))$ . Hence $f(x-y)=f(x+y)$ for all $x,y$ , which shows the function is constant. Hence, there is only one solution: $\boxed{f(x)=0}$ .

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#10 h Oct 23, 2020, 1:52 PM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95

Let $P(x,y)$ denote the assertion $f(f(y))+f(x-y)=f(xf(y)-x).$

$P(0,0)$ : $f(f(0))+f(0)=f(0)\implies f(f(0))=0$ .

$P(0,y)$ : $f(f(y))+f(-y)=f(0)$ .

$P(x,f(0))$ : $f(f(f(0)))+f(x-f(0))=f(xf(f(0))-x) \implies f(0)+f(x-f(0))=f(-x)$ .

Last two together give us that $f(f(x))=-f(x-f(0))$ , which means that when $x=f(0)$ , thus $f(f(f(0)))=-f(f(0)-f(0)) \implies f(0)=-f(0)\implies f(0)=0$
Hence, $-f(-x)=f(f(x))=-f(x)\implies f(x)=f(-x)$ .

$P(x,y)$ : $f(f(y))+f(x-y)=f(xf(y)-x)$
$P(y,x)$ : $f(f(x))+f(y-x)=f(yf(x)-y)$
Since $f(x-y)=f(y-x)$ , thus $f(y)-f(x)=f(f(x))-f(f(y))=f(xf(y)-x)-f(yf(x)-y)$ , where the first equality comes from knowing that $-f(x)=f(f(x))$ .
Now, set $x$ to be $0$ and we get that $f(y)-f(0)=f(0\cdot f(y)-0)-f(yf(0)-y)\implies f(y)=-f(-y)=-f(y) \implies f(y)=0\forall y \in \mathbb{R}$ .
We claim $\boxed{f(y)=0\forall y \in \mathbb{R}}$ to be solution to this problem.

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A-Thought-Of-God

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#12 h Nov 12, 2020, 6:58 PM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95

Nice Problem.

Solution

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quirtt

#13 h Feb 5, 2021, 11:06 AM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95

Nice problem.

solution

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#14 h Apr 7, 2021, 10:57 PM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95

The only solution is that $f \equiv 0$ .

By setting $x=y=0$ , we have that $f(f(0))=0$ .
For simplicity let's say that $c=f(0)$ .

Now let's plug into the assertion $y \rightarrow c$ , then we have that $c+f(x-c)=f(-x)$ .
Now let's plug into the assertion $y = 0$ , then we must have that $f(x)=f(xc-x)$

Plug in $x=0$ , then we must have that $f(f(y))=-f(-y)=-c-f(y-c)$
Set $y=0$ into the upper relation to get that $f(-c)=-c$

Into the original assertion plug in $y=-c$ , then we must have that $-c+f(x+c)=f(-xc-x)$ .
Into the upper relation plug in $x=0$ to get that $c=0$ .

This implies that $f(x)=f(-x)$ .

This easily implies that $f(x+y)=f(x-y)$ and then we just set $x=y$ , to get that $f(2x)=0$ .
Thus this implies that $f \equiv 0$ .

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#15 h Apr 8, 2021, 12:25 AM • 5 Y

Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, tony88, ImSh95

Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$ .

$P(0,0)\Rightarrow f(f(0))=0$
$P(0,f(0))\Rightarrow f(-f(0))=0$
$P(f(0),f(0))\Rightarrow 2f(0)=f(-f(0))\Rightarrow f(0)=0$

$P(x,0)\Rightarrow f(x)=f(-x)$

$P(x,x)\Rightarrow f(f(x))=f(xf(x)-x)$
$P(x,-x)\Rightarrow f(f(x))+f(2x)=f(xf(x)-x)$
Comparing these, $\boxed{f(x)=0}$ , which works.

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#16 h Apr 8, 2021, 10:43 AM • 2 Y

Y by centslordm, ImSh95

Same solution but anyways.
Answer: $f(x) = 0 \ \forall \ x$ .
Proof:
Let $P(x,y)$ denote the given assertion.

Claim: $f(0) = 0.$
Proof:
$P(0,0) \to f(f(0)) + f(0) = f(0) \implies \boxed{f(f(0)) = 0}$ .

$P(0 , f(0)) \to f(0) + f(-f(0)) = f(0) \implies \boxed{f(-f(0)) = 0}$ .

$P(f(0) , f(0)) \to f(0) + f(0) = f(-f(0)) = 0 \implies \boxed{f(0) = 0}$ .

Claim: $f$ is even.
Proof:
$P(x,0) \to f(f(0)) + f(x) = f(xf(0) - x) \implies \boxed{f(x) = f(-x)}$ .

$P \left(\frac x2 , \frac x2 \right) \to f \left( f \left (\frac x2 \right) \right) + f(0) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right) \implies f \left( f \left (\frac x2 \right) \right) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right)$

$P \left(\frac x2 , -\frac x2 \right) \to f \left( f \left ( - \frac x2 \right) \right) + f(x) = f \left( \frac x2 f \left( - \frac x2 \right) - \frac x2 \right)$

$\implies f \left( f \left ( \frac x2 \right) \right) + f(x) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right)$ (Using $f$ is even.)

$\implies f \left( f \left ( \frac x2 \right) \right) + f(x) = f \left( f \left ( \frac x2 \right) \right)$ (Equating $P \left(\frac x2 , \frac x2 \right)$ and $P \left(\frac x2 , -\frac x2 \right)$ .)

$\implies f(x) = 0$ .

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hakN

#17 h Apr 8, 2021, 11:00 AM • 3 Y

Y by centslordm, ImSh95, Mango247

Let $P(x,y)$ be the assertion.
$P(0,0)\implies f(f(0))=0$ .
$P(x,0)\implies f(x)=f(x(f(0)-1))$ .
If $c=f(0)-1 \neq 1$ , then comparing $P(x,y)$ with $P(x,yc)$ gives $f(x)=f(x+y(1-c))$ , which implies that $f$ is constant since $c\neq 1$ . So only solution is $f(x)=0\ \forall x\in \mathbb{R}$ in this case.
If $f(0)=2$ , then we have $f(2)=0$ . From $P(x,2)$ we get $2+f(x-2)=f(-x)$ , but plugging $x=4$ and $x=-2$ gives a contradiction. Thus our only solution is $\boxed{f(x)=0 \ \forall x\in \mathbb{R}}$ , which clearly fits.

This post has been edited 1 time. Last edited by hakN, Sep 17, 2023, 5:58 PM
Reason: fixed error

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#18 h Aug 13, 2021, 10:27 PM • 2 Y

Y by centslordm, ImSh95

Let $P(x,y)$ be the assertion of $f(f(y)) + f(x - y) = f(xf(y) - x)$ .

We have $P(0,0)\implies f(f(0))=0$ , thus $P(0,f(0))\implies f(-f(0))=0$ and therefore $P(f(0),f(0))\implies f(0)=0$ .

Now, $P(x,0)\implies f(x)=f(-x)$ and $P(0,x)\implies f(f(x))=-f(-x)=-f(x)$ .

This means that $-f(x)=f(f(x))=f(-f(x))=f(f(f(x)))=-f(f(x))=f(x)\implies f\equiv 0,$ this obviously works.

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#19 h Aug 30, 2021, 2:47 PM • 4 Y

Y by ImSh95, Mango247, Mango247, Mango247

Let $P(x,y)$ denote the assertion.
$P(0,0)\implies f(f(0))=0$
$P(1,0) \implies f(1)=0$
$P(0,1) \implies f(-1)=0$
$P(1,1) \implies f(0)=0$ ,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$

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#20 h Aug 30, 2021, 5:54 PM • 1 Y

Y by ImSh95

Sprites wrote:

Let $P(x,y)$ denote the assertion.
$P(0,0)\implies f(f(0))=0$
$P(1,0) \implies f(1)=0$
$P(0,1) \implies f(-1)=0$
$P(1,1) \implies f(0)=0$ ,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$

Will you tell me how does $P(1,0) \implies f(1)=0 ?$

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#21 h Aug 30, 2021, 6:06 PM • 1 Y

Y by ImSh95

Sprites wrote:

Let $P(x,y)$ denote the assertion.
$P(0,0)\implies f(f(0))=0$
$P(1,0) \implies f(1)=0$
$P(0,1) \implies f(-1)=0$
$P(1,1) \implies f(0)=0$ ,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$

How do you induct to prove over $\mathbb R$ ?

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#22 h Sep 12, 2021, 2:28 PM • 1 Y

Y by ImSh95

hakN wrote:

Let $P(x,y)$ be the assertion.
$P(0,0)\implies f(f(0))=0$ .
$P(x,0)\implies f(x)=f(xf(0))$ .
Plugging $x=1$ in above, we get $f(1)=0$ .
$P(0,x)\implies f(f(x))=f(0) - f(-x)$ .
Plugging $x=1$ in above, we get $f(-1)=0$ .
$P(x,1)\implies f(0) + f(x-1) = f(-x)$ .
Plugging $x=1$ in above, we get $f(0)=0$ .
So we get $\boxed{f(x)=0 \forall x\in \mathbb{R}}$ which clearly fits.

Can u explain how u received line 3
How f(x)=f(xf(0))$

This post has been edited 1 time. Last edited by Rahmanhabibi, Sep 12, 2021, 2:29 PM
Reason: Adding some thing

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naruto_uzumaki_06

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#23 h Sep 12, 2021, 3:14 PM • 1 Y

Y by ImSh95

Let $P(x,y)$ be the assertion.
$P(0,0): f(f(0))=0.$ Let $f(0)=a \Rightarrow f(a)=0$
$P(a,a):f(-a)=2a$
$P(0,a): 3a=f(0)=a\Rightarrow f(0)=0$
$P(x,0):f(-x)=f(x) \forall x \in \mathbb{R}$
$P(x,-y):f(f(y))+f(x+y)=f(xf(y)-x)\Rightarrow f(x-y)=f(x+y) \forall x,y (1)$
In (1), let $x=y$ , we get $f(x)=a \text{(a=const)}$
It is easy to show that $f(x)=0$ is the only function which works.

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#24 h Feb 12, 2022, 5:09 PM • 1 Y

Y by ImSh95

$x=0 \implies f(0)=f(f(y))+f(-y)\rightarrow (1),x=y \implies f(0)=f(xf(x)-x)-f(f(x)) \implies f(f(0))=0$ $x=y=f(0)\implies 2f(0)=f(-f(0)),y=f(0),x=0 \implies f(0)=f(f(f(0)))+f(-f(0))=f(0)+2f(0)=3f(0)\implies f(0)=0$ $y=0 \implies f(x)=f(-x),f(0)=0 \implies f(f(x))=-f(x)\text{from (1)},f(x)=-(-f(x))=-f(f(x))=f(f(f(x)))=f(-f(x))\implies f(-f(x))=f(x)$ Finally,
$x=f(x),y=0 \implies f(f(x))=f(-f(x)) \implies f(x)=-f(x) \implies f(x)=0$

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ZETA_in_olympiad

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ZETA_in_olympiad

#25 h Jul 14, 2022, 1:24 PM • 1 Y

Y by ImSh95

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

$P(0,0)$ gives $f(f(0))=0.$ Further, $P(0,f(0))$ gives $f(-f(0))=0$ and finally $P(f(0),f(0))$ gives $f(0)=0.$ Then $P(x,0)$ implies $f$ is even. Subtracting $P(x,x)$ from $P(x,-x)$ implies $f(2x)=0.$ And it's easy to verify the identity zero function works.

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ZETA_in_olympiad

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ZETA_in_olympiad

#26 h Jul 14, 2022, 1:27 PM • 1 Y

Y by ImSh95

Rahmanhabibi wrote:

Can u explain how u received line 3
How f(x)=f(xf(0))$

It's wrong.

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#27 h Sep 20, 2022, 4:36 PM • 3 Y

Y by ImSh95, Mango247, Mango247

The only solution is $\boxed{f\equiv 0}$ , which works.

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

Let $f(0) = a$ .
$P(0,0): f(a) = 0$ .

$P(0,a): a + f(-a) = a$ , so $f(-a) = 0$ .

$P(a,a): 2a = f(-a)$ , so $a=0$ .

$P(x,0): f(x) = f(-x)$ .

$P(0,x): f(f(x)) + f(x) = 0$ .

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

$P(x+y,-y): f(f(y)) + f(x+2y) = f(xf(y) - x)$ .

So $f(x) = f(x+2y)$ , which implies $f$ is constant, so $f\equiv 0$ .

This post has been edited 1 time. Last edited by megarnie, Sep 20, 2022, 4:36 PM

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MagicalToaster53

#28 h Jun 30, 2023, 4:21 PM • 1 Y

Y by ImSh95

I claim that the only solution is $f(x) \equiv 0$ , which works.

Consider the following:
$\begin{align*} P(0, 0): &\rightarrow f(f(0)) = 0 \\ P(0, x): &\rightarrow f(f(x)) + f(-x) = f(0) \\ P(x, x): &\rightarrow f(f(x)) + f(0) = f(xf(x)-x) \\ P(x, 0): &\rightarrow f(x) = f(-x) \\ \\ P(0, f(0)): &\rightarrow f(-f(0)) = 0 \\ P(f(0), f(0)): &\rightarrow f(0) = 0 \\ \\ P(-x, x) \wedge P(x, 0): &\rightarrow f(f(x)) + f(-2x) = f(xf(x) - x) \\ P(-x, x) \wedge P(x, x): &\rightarrow f(-2x) = 0. \end{align*}$
Thus we have that as $-2x$ is surjective, $f(x) \equiv 0$ for all $x \in \mathbb{R}$ . $\blacksquare$

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#29 h Jun 30, 2023, 5:52 PM • 1 Y

Y by ImSh95

We claim that the only solution is $f\equiv0$ , which clearly fits.
We refer the assertion $f(f(y))+f(x-y)=f(xf(y)-x)$ as $p(x,y)$ . Note that if $f(x)$ is a constant function then $f\equiv0$ . Assume that $f(x)$ is a non-constant function.
Note that $p(0,-y)$ gives $f(y)=f(f(-y))$ .
Since the function is non-constant, there exists $y$ such that $f(y)\ne2$ . Then, $p\left(\frac{y}{2-f(y)},y\right)$ gives $f(f(y))=0$ . But $f(f(y))=f(-y)$ . So, $f(-y)=0$ . Now, $p\left(-\frac{y}{2},-y\right)$ gives $f(f(-y))=0=f(y)$ . We conclude that either $f(y)\in\{0,2\}$ . Note that $p(0,0)$ gives $f(f(0))=0=f(0)$ . But $p(x,0)$ gives $f(x)=f(-x)$ . Take $y$ such that $f(y)=2$ . We have $2=f(y)=f(-y)=f(f(y))=f(2)$ . Then, $p(x+2,2)$ gives $2+f(x)=f(x+2)$ . It follows that $f(x)=0$ and $f(x+2)=2$ . There is no restriction on $x$ , so we can take any real and $f(x)=0$ . So, $f\equiv0$ , contradicting $f(2)=2$ .
So, there are no solutions to the given equation other than $f\equiv0$ . $\square$

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#30 h Oct 3, 2023, 3:32 AM

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

Note that if $f(y)\neq 2$ , $f(f(y))=0$ . Proof: Choose $x=-\frac{y}{f(y)-2}$ .

Now clearly exists a number $l$ such that $f(l)\neq 2$ , since $f(x)=2$ for all $x$ is not a solution. Hence $f(f(l))=0, f(f(f(l)))=f(0)=0$ and so $f(0)=0$ . Now suppose exists $k$ such that $f(k)=2$ ,

$P(k,k): f(2)+f(0)=f(k)=2$ so $0=f(0)=f(2)-2$ . So $f(2)=2$ .

Now $P(x,0): 0+f(x)=f(-x)$ , so $f(x)=f(-x)$ .

$P(0,k): 2+f(-k)=0$ . So $f(-k)=-2$ at the same time $f(k)=2$ and since $f(k)=f(-k)$ , $2=-2$ which is not okay

So there is no $k$ such that $f(k)=2$ . So $f(f(x))=0$ for all $x$ .

$P(0,y): f(-y)=f(0)$ so $f$ is constant. Now checking only $f(x)=0$ works.

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#34 h Mar 11, 2024, 8:19 PM

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My solution is similar to # $6,7$ but I'm gonna post it anyway
Solution

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#35 h Jun 24, 2024, 7:46 PM

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

$P(0,0)$ gives $f(f(0))+f(0)=f(0)$ so $f(f(0))=0$ .

$P(x, 0)$ gives $f(x)=f(x(f(0)-1))$ .

$P(1, y)$ gives $f(f(y))+f(1-y)=f(f(y)-1)$ so plugging $y=f(0)$ gives $f(0)+f(1-f(0))=f(-1)$ but $f(x)=f(x(f(0)-1))$ so $f(0)+f(-1)=f(-1)$ so $f(0)=0$ . Plugging this back into $f(x)=f(x(f(0)-1))$ gives $f(x)=f(-x)$ .

$P(0, y)$ gives $f(f(y))+f(-y)=0$ so $f(f(y))=-f(y)$ .

We claim that the only solution is $\boxed{f(x)=0}$ . Assume FTSOC that there exists some $c\ne0$ in the image of $f$ . Then, $f(c)=f(-c)=-c$ so $-c\ne 0$ is also in the image of $f$ . However, this means that $f(-c)=f(c)=c$ . Since $c\ne -c$ , this is a contradiction, as desired.

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NicoN9

#36 h Today at 8:37 AM

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The answer is $f(x)\equiv 0$ , which obviously works. We claim that $f$ is constant. Let $P(x, y)$ denote the assertion.

Now we have $\begin{align} P(0, 0)\Longrightarrow f(f(0)) &= 0,\\ P(0, y)\Longrightarrow f(f(y)) + f(-y) &= f(0),\\ P(x, f(0))\Longrightarrow f(0) + f(x-f(0)) &= f(-x). \end{align}$ But since , from ( $2$ ) and ( $3$ ), we have $f(f(x)) + f(-x) + f(x - f(0))=f(-x)$ , thus $f(f(x))+f(x-f(0))=0.$ Put $x=f(0)$ in above, we have $f(0) = 0$ . Thus $f(x)=f(-x)$ from ( $3$ ).

Now, comparing $P(x, y)$ , and $P(x, -y)$ gives $f(x+y)=f(x-y)$ . This implies $f$ is constant. because for any real $t$ , Letting $x=y=\frac{t}{2}$ to get $f(t) = f(x+y)=f(x-y)=f(0).$ And so we're done.

This post has been edited 1 time. Last edited by NicoN9, Today at 8:37 AM
Reason: typo fixed

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