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Source: India IMO Training Camp 2016, Practice test 2, Problem 2

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3070 posts

#1 h Jul 22, 2016, 9:04 AM • 4 Y

Y by doxuanlong15052000, Davi-8191, toilaDang, Adventure10

Find all functions $f:\mathbb R\to\mathbb R$ such that $f\left( x^2+xf(y)\right)=xf(x+y)$ for all reals $x,y$ .

This post has been edited 1 time. Last edited by Ankoganit, Jul 22, 2016, 9:04 AM

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#2 h Jul 22, 2016, 9:27 AM • 1 Y

Y by Adventure10

I found that $f(x)=-x$ is the solution, but I don't have time to write it now, it may also be wrong. Anyways, I analysed $P(x,0),P(x,-x)$ and $P(-x,x)$ .

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#3 h Jul 22, 2016, 2:50 PM • 1 Y

Y by Adventure10

Wrong solution.

Edit: Yes. Thanks pco.

This post has been edited 1 time. Last edited by Diamondhead, Jul 22, 2016, 2:57 PM
Reason: Wrong

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pco

#4 h Jul 22, 2016, 2:55 PM • 3 Y

Y by Diamondhead, Adventure10, Mango247

Diamondhead wrote:

...
$P\left(\frac{x}{f(k)},k-x\right)$ : $f\left(\left(\frac{x}{f(k)}\right)^2+\frac{x}{f(k)}f(k-x)\right)=x$ for all $x\in\mathbb{R}$ , so $f$ is a bijective function....

No, you can only conclude from this that $f(x)$ is surjective. This is not enough to prove injectivity (and you need injectivity for your final step)

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#5 h Jul 22, 2016, 5:03 PM • 2 Y

Y by Adventure10, Mango247

Put $x,y=0$ to get $f(0)=0$
also put $y=0$ to obtain $f(x^{2})=xf(x)$

Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$ .

Putting $y=k$ we obtain $f(x^{2})=xf(x)=xf(x+k)$ which means for $x\not= 0$ we have $f(x)=f(x+k)$ . Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.

Suppose no such $k$ exists.

Then putting $y=-x$ we get $f(x^2+xf(-x))=0$ $\implies x^2 +xf(-x)=0$ which gives $f(-x)=-x$ replacing $-x$ with $x$ we get $f(x)=x$

Hence $f(x)=x$ and $f(x)=0$ are the only solutions.

Edit- Wrong solution. Thanks @Ashutoshmaths for pointing it out.

This post has been edited 1 time. Last edited by darthsid, Jul 23, 2016, 8:49 AM
Reason: wrong sol

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#6 h Jul 22, 2016, 5:09 PM • 2 Y

Y by Adventure10, Mango247

darthsid wrote:

Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$ .

we have $f(x)=f(x+k)$ . Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.

This is wrong.
I don't think you can solve this problem without using surjectivity.

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413 posts

#7 h Jul 22, 2016, 5:23 PM • 2 Y

Y by Adventure10, Mango247

Ashutoshmaths wrote:

darthsid wrote:

Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$ .

we have $f(x)=f(x+k)$ . Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.

This is wrong.
I don't think you can solve this problem without using surjectivity.

I don't get it. Why is it wrong? Edit: I get it now.

This post has been edited 1 time. Last edited by darthsid, Jul 23, 2016, 11:45 AM

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3028 posts

#8 h Jul 22, 2016, 6:54 PM • 8 Y

Y by ThE-dArK-lOrD, darthsid, kapilpavase, Ankoganit, anantmudgal09, opptoinfinity, Adventure10, Mango247

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2487 posts

#9 h Jul 23, 2016, 1:44 AM • 2 Y

Y by sa2001, Adventure10

Note: This problem had a lot of fake solves at the Camp and was by far the most troll problem there. And btw only 4-5 students solved it correctly in the time limit (excluding fake solves which almost all others did).

This post has been edited 1 time. Last edited by WizardMath, Jul 23, 2016, 1:44 AM

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#10 h Jul 23, 2016, 11:05 AM • 2 Y

Y by Adventure10, Mango247

darthsid wrote:

Ashutoshmaths wrote:

darthsid wrote:

Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$ .

we have $f(x)=f(x+k)$ . Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.

This is wrong.
I don't think you can solve this problem without using surjectivity.

I don't get it. Why is it wrong?

$f(x+k)=f(x)$ does not mean $f(x)$ is constant, $f$ can be periodic with period $k$

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25 posts

#11 h Jul 24, 2016, 1:17 AM • 6 Y

Y by lebathanh, Excuse, babu2001, Timmy456, Adventure10, D_S

Let $E(x,y)$ be the statement of the functional equation.
From $E(0,0)$ we get $f(0)=0$ . Clearly $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is a solution of the equation, henceforth suppose that $f\not\equiv 0$ .
Let $a$ be such that $f(a)\neq 0$ . Then $E(x, a-x)$ implies $f(z)=xf(a)$ , where $z=x^2+xf(a-x)$ . Since $f(a)\neq 0$ , it follows that $f$ is surjective.

Suppose that $f(n)=f(m)$ for real numbers $n,m\in\mathbb{R}$ . From $E(x,n)$ and $E(x,m)$ we get $xf(x+n)=f(x^2+xf(n))=f(x^2+xf(m))=xf(x+m)$ , therefore $f(x+n)=f(x+m), \forall x\in\mathbb{R}$ . Set $k=m-n$ ; then with $z=x+n$ the previous equation can be written as $f(z)=f(z+k),\forall z\in\mathbb{R}\,\,(*)$ .
Choose $y$ such that $f(y)=-k$ , which exists because $f$ is surjective. Now using $E(k,y)$ we have
$\begin{align*} 0=f(k^2-k^2)=f(k^2+kf(y))=kf(k+y)\stackrel{*}{=}kf(y)=-k^2 \end{align*}$ hence $k=0$ , so $n=m$ . Therefore $f$ is injective.
$E(1,x)$ gives $f(1+f(x))=f(1+x),\forall x\in\mathbb{R}$ , which is just $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ because of injectivity.

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acegikmoqsuwy2000

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acegikmoqsuwy2000

#12 h Jan 2, 2017, 6:16 AM • 4 Y

Y by Ankoganit, sa2001, Adventure10, Mango247

Considerably different solution

This post has been edited 1 time. Last edited by acegikmoqsuwy2000, Jan 2, 2017, 7:30 AM
Reason: no confirmation needed

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#14 h Mar 9, 2017, 5:31 PM • 2 Y

Y by Adventure10, Mango247

Let $P(x,y)$ be the assertion.
If f is constant. we get $f(x)=0$
$P(0,0) \implies f(0)=0$
Suppose there 's $f(k)=0$ and $k\not=0$
$P(x,0) \implies f(x^2)=xf(x)$
set $x=k \implies f(k^2)=0$
$P(x,k^2)$ and $P(x,0) \implies f(x^2)=xf(x+k^2)=xf(x) \implies f(x)=f(x+k^2)$
$P(x,k)$ and $P(x,0) \implies f(x^2)=xf(x+k)=xf(x) \implies f(x)=f(x+k)$
$P(k,x) \implies f(k^2+kf(x))=kf(x+k) \implies f(kf(x))=kf(x)$ _(1)
And easy to see $f$ is bijections.
from $(1)$ we get $f(kx)=kx \implies f(x)=x$ contradictions if $x=k$
So $f(z)=0$ if only if $z=0$
$P(x,-x) \implies f(x^2+xf(-x))=0 so f(x)=x$

This post has been edited 1 time. Last edited by MonsterS, Mar 9, 2017, 5:31 PM

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#15 h Mar 11, 2017, 5:01 PM • 2 Y

Y by Adventure10, Mango247

guys it is very easy.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$ .
$p(0,0)\Rightarrow\ f(0)=0$ .If there is f( $\omega$ )=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$ (constant)
Let there is no real such $f(x)=0$ then $P(x,0)\Rightarrow\ f(x^2)=xf(x)$ and $P(x,-x)\Rightarrow\ f(x^2+xf(-x))=0\Rightarrow\ x^2=-xf(-x)$ wich is $\boxed {f(x)=x}$

This post has been edited 2 times. Last edited by TRYTOSOLVE, Mar 11, 2017, 5:17 PM
Reason: so so

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pco

#16 h Mar 11, 2017, 5:07 PM • 1 Y

Y by Adventure10

TRYTOSOLVE wrote:

guys it is very easy.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$ .
$p(0,0)\Rightarrow\ f(0)=0$ .If there is f( $\omega$ )=0 then $p(x,\omega)$ implies that $f(\omega)=f(\omega+y)$

For me, $p(x,\omega)$ implies $f(x^2)=xf(x+\omega)$ and not $f(\omega)=f(\omega+y)$

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#17 h Mar 11, 2017, 5:10 PM • 2 Y

Y by Adventure10, Mango247

ups oh yes I will сhange it now

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#18 h Mar 11, 2017, 5:18 PM • 2 Y

Y by Adventure10, Mango247

it is not easy like I thought.

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#19 h Mar 11, 2017, 5:19 PM • 2 Y

Y by Adventure10, Mango247

Is my solution true guys?

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#20 h Mar 11, 2017, 5:19 PM • 2 Y

Y by Adventure10, Mango247

.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$ .
$p(0,0)\Rightarrow\ f(0)=0$ .If there is f( $\omega$ )=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$ (constant)
Let there is no real such $f(x)=0$ then $P(x,0)\Rightarrow\ f(x^2)=xf(x)$ and $P(x,-x)\Rightarrow\ f(x^2+xf(-x))=0\Rightarrow\ x^2=-xf(-x)$ wich is $\boxed {f(x)=x}$

This post has been edited 3 times. Last edited by TRYTOSOLVE, Mar 12, 2017, 4:02 AM

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#21 h Mar 11, 2017, 5:26 PM • 2 Y

Y by Adventure10, Mango247

guys I can not find mistake in my solution.What about you?

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bobthesmartypants

4337 posts

bobthesmartypants

#22 h Mar 11, 2017, 5:33 PM • 1 Y

Y by Adventure10

TRYTOSOLVE wrote:

If there is f( $\omega$ )=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$ (constant)

And how might this show that $f(x)=0$ for all $x\in\mathbb{R}$ ?

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sa2001

#23 h Feb 9, 2018, 4:27 PM • 2 Y

Y by Adventure10, Mango247

........................................

This post has been edited 15 times. Last edited by sa2001, Feb 10, 2018, 7:37 AM
Reason: Incorrect solution

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sa2001

#24 h Feb 9, 2018, 4:56 PM • 2 Y

Y by Adventure10, Mango247

...............

This post has been edited 1 time. Last edited by sa2001, Feb 10, 2018, 7:37 AM

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Vrangr

#25 h Feb 9, 2018, 6:30 PM • 3 Y

Y by Arhaan, Adventure10, Mango247

Here's a solution in the vain of INMO 2015 P3. (proving injectivity at $0$ )
Solution

This post has been edited 3 times. Last edited by Vrangr, Feb 19, 2018, 1:48 PM

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Vrangr

#26 h Feb 9, 2018, 7:02 PM • 3 Y

Y by math-o-enthu, Adventure10, Mango247

@2above and @3above, your solution seems to be true but needlessly complicated in places. Great solution nevertheless.

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#28 h Feb 10, 2018, 6:24 PM • 2 Y

Y by Adventure10, Mango247

Can I say this: $f(x)=0$ is solution. Suppose that $\boxed{f\equiv 0}$ and $f(a)\neq 0$ . $P(x;a-x)$ then $f(x^2+xf(a-x))=xf(a)$ and $f$ is surjective

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sa2001

#29 h Feb 10, 2018, 7:24 PM • 1 Y

Y by Adventure10

abbosjon2002 wrote:

Can I say this: $f(x)=0$ is solution. Suppose that $\boxed{f\equiv 0}$ and $f(a)\neq 0$ . $P(x;a-x)$ then $f(x^2+xf(a-x))=xf(a)$ and $f$ is surjective

I think you mean $f \not\equiv 0$ . Then, yes you can.

This post has been edited 1 time. Last edited by sa2001, Feb 10, 2018, 7:24 PM

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#30 h Feb 19, 2018, 1:07 PM • 2 Y

Y by Adventure10, Mango247

Probably the silliest way to solve this problem:

As usual, let $P(x,y)$ denote $f(x^2+xf(y))=xf(x+y)$ for all $x,y\in \mathbb{R}$ .
$P(0,0)$ easily gives us $f(0)=0$ and then $P(x,0)$ gives $f(x^2)=xf(x)$ .

Suppose there exists non-zero real number $a$ that $f(a)=0$ .
$P(x,a)$ gives us $f(x^2)=xf(x+a)=xf(x)$ for all $x\in \mathbb{R}$ . So, $f(x)=f(x+a)$ for all $x\in \mathbb{R} -\{ 0\}$ .
Since $f(0)=f(a)=0$ , we conclude that $f(x)=f(x+a)$ for all $x\in \mathbb{R}$ .
We can easily prove by induction that $f(x)=f(x+na)$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}^+$ .
$P(x+a,y)$ gives us $f(x^2+2xa+a^2+xf(y)+af(y))=(x+a)f(x+a+y)=(x+a)f(x+y)$ for all $x,y \in \mathbb{R}$ .
So, $f\Big( x^2+xf(y) +a(2x+a+f(y))\Big) =xf(x+y)+af(x+y)$ for all $x,y \in \mathbb{R}$ .
Plugging in $x$ by $t=\frac{n-a-f(y)}{2}$ where $n$ is a positive integer gives us
$f(t^2+tf(y)+na)=f(t^2+tf(y))=tf(t+y)=tf(t+y)+af(t+y)\implies f(t+y)=0.$ Hence, we get that for any $a\neq 0$ that $f(a)=0$ , $f\Big( \frac{n-a-f(y)}{2}+y\Big) =0$ for all positive integer $n$ and real number $y$ .

Suppose there exists non-zero real number $a$ that $f(a)=0$ .
We've $f\Big( \frac{1-a-f(1)}{2}+1\Big) =0$ and $f\Big( \frac{2-a-f(1)}{2}+1\Big) =0$ .
From the preceding paragraph, we've $f(x)=f(x+\frac{1-a-f(1)}{2}+1 )=f(x+\frac{2-a-f(1)}{2}+1)$ for all $x\in \mathbb{R}$ .
So, $f(x)=f(x+\frac{1}{2})$ for all real number $x$ . From $f(0)=0$ , we can easily prove that $f(m)=0$ for all integer $m$ .
Hence, we've $f(x)=f(x+n)$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}$ .
Using the preceding paragraph again, we've $f\Big( \frac{n-m-f(y)}{2}+y\Big) =0$ for all $n\in \mathbb{Z}^+,m\in \mathbb{Z},y\in \mathbb{R}$ .
So, $f(y-\frac{f(y)}{2})=0$ for all real number $y$ . This gives $f(n^2+nf(y)-\frac{nf(n+y)}{2})=f(n^2+\frac{nf(y)}{2})=0$ for all $n\in \mathbb{Z}^+$ and $y\in \mathbb{R}$ .
So, $f(4+f(y))=f(f(y))=0\implies f(x+f(y))=f(x)$ for all $x,y\in \mathbb{R}$ .
And so $f(x+f(y))=f(x)=f(x+f(z))\implies f(f(y)-f(z))=0$ for all $x,y,z\in \mathbb{R}$ .

If there exists $k\in \mathbb{R}$ that $f(k)\neq 0$ .
For any $r\in \mathbb{R}$ , not hard to show that there exists $a,b,c\in \mathbb{R}$ that $b-a=\frac{r}{f(k)}$ and $a+b+c=k$ .
Also, there exists $u,v\in \mathbb{R}$ that $f(u)=af(a+b+c)$ and $f(v)=bf(a+b+c)$ .
From the preceding paragraph, we get $f\Big( (b-a)f(a+b+c)\Big) =f(r)=0$ . This means $f(x)=0$ for all $x\in \mathbb{R}$ , contradiction with the existence of $k$ .

Hence, we conclude that if there exists non-zero real number $a$ that $f(a)=0$ then $f(x)=0$ for all $x\in \mathbb{R}$ .
If there not exists such $a$ , $P(x,-x)$ gives $f(x^2+xf(-x))=0\implies x^2+xf(-x)=0$ for all $x\in \mathbb{R}$ , this easily gives $f(x)=x$ for all $x\in \mathbb{R}$ .

This post has been edited 2 times. Last edited by ThE-dArK-lOrD, Feb 19, 2018, 1:09 PM

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#31 h May 2, 2019, 11:11 AM • 3 Y

Y by HouseofTerror_43, Adventure10, Mango247

Good Problem.

Ankoganit wrote:

Find all functions $f:\mathbb R\to\mathbb R$ such that $f\left( x^2+xf(y)\right)=xf(x+y)$ for all reals $x,y$ .

We shall show that $f(x) \equiv x \ \forall x \in \mathbb{R}$ and $f \equiv 0$ are the only solutions to the given functional equation and both of them clearly work.
$\text{[math]}$
Now assume that $f$ is not identically $0$ and hence there exists at least one $\alpha \in \mathbb{R}$ such that $f(\alpha) \neq 0$ . Then,
$P(x, \alpha-x) : f(x^2+x \cdot f(\alpha)) = \underbrace{x \cdot f(\alpha)}_{\text{Spans all of} \ \mathbb{R}} \implies f$ is surjective.
$\text{[math]}$
Note that $P(0,0) \implies f(0) = 0$ and $P(x,0) : f(x^2) = x \cdot f(x) \forall x \ \in \mathbb{R} \dots (\star)$ .
Now, suppose there exists $\lambda \in \mathbb{R}$ such that $f(\lambda) = 0 , \lambda \neq 0$ . Also by surjectivity of $f$ there must exist $\mu \in \mathbb{R}$ such that $f(\mu) = 1$ .
$\text{[math]}$
$P(x , \lambda) : f(x^2) = \underbrace{x \cdot f(x + \lambda) =f(x)}_{\bigstar} \implies f$ is periodic.
$P(\lambda , \mu): f(\lambda^2 + \lambda) = \underbrace{\lambda \cdot f(\lambda + \mu) =\lambda \cdot f(\mu)}_{\text{Periodic}} = \lambda$ .
$\text{[math]}$
But, $f(\lambda^2 + \lambda ) = f(\lambda^2)= \lambda \cdot f(\lambda) = 0 \iff \lambda = 0$ . So , $f$ is injective at $0$ . Now, the problem easily breaks down as:
$\text{[math]}$
$P(-f(y), y) : f(y) \cdot f(y-f(y))=0 \implies f(y - f(y))=0 \forall y \neq 0 \iff y=f(y) \forall y \neq 0$ .
As $f(0) = 0$ , we conclude that $\boxed{f(x) \equiv x \forall x \ \in \mathbb{R}}$

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#33 h May 28, 2020, 11:45 AM • 1 Y

Y by ILOVEMYFAMILY

my short solution: :rotfl:

:rotfl:

claim(1): $f$ is surjective
proof:
$P(\frac{x}{f(y)},y-x) \implies f(something)=x$
$\blacksquare$
claim(2): $f$ is injective
proof:
suppose there's $a \neq b : f(a)=f(b)$ note that
$P(x,a) \& P(x,b) \implies f(x+a)=f(x+b)$
define $\mathbb{S}=\{a-b | f(a)=f(b) \}$ then $f(x)=f(x+s) \forall s \in \mathbb{S}$
now let $f(y_1)=1 ,f(y_2)=2$
$P(s,y_1) \& P(s,y_2) \implies f(s^2)=s=2s \implies s=0$
$\blacksquare$
now just put $P(1,y) \implies f(y)=y$
and we win

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#34 h May 28, 2020, 12:09 PM

Y by

Ankoganit wrote:

Find all functions $f:\mathbb R\to\mathbb R$ such that $f\left( x^2+xf(y)\right)=xf(x+y)$ for all reals $x,y$ .

Nice actually and pritty short

Let $P(x,y)$ be the assertion
$f\left( x^2+xf(y)\right)=xf(x+y)$ Notice that if $f \equiv c$ then $c=0$ .
Suppose we are searching for a non-constant solution. In other words, there exist $t \in \mathbb{R}$ with $f(t) \neq 0$ , then $P(x,t-x)$ gives
$f\left( x^2+xf(t-x)\right)=xf(t) \ \forall x \in \mathbb{R}$ Thus $f$ is surjective, and $P(0,0)$ gives $f(0)=0$ , and by $P(x,0)$ we get
$f\left( x^2 \right)=xf(x) \quad (1)$ Now for each $z \in \mathbb{R}^*$ , there exist $x_0 \in \mathbb{R}^*$ such that $f(x_0)=z \neq 0$ , thus $P(-f(x_0),x_0)$ gives
$0=f(0)=-f(x_0)f\left(x_0-f(x_0)\right)$ because $f(x_0) \neq 0$ , it is forced that $f\left(x_0-f(x_0)\right)=0$ .
Finally, $P(f(x_0),x_0-f(x_0)$ gives
$f\left( f(x_0)^2 \right)=f(x_0)^2$ from $(1)$ , we get
$f(x_0)f\left(f(x_0)\right) = f\left( f(x_0)^2 \right)=f(x_0)^2$ $\Rightarrow f(z) = f\left(f(x_0)\right) = f(x_0) = z \ \forall z \in \mathbb{R}^*$ Hence $f \equiv x$ for all $x \in \mathbb{R}$ and $f \equiv 0$ are the only solutions. $\blacksquare$

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oVlad

#36 h Mar 5, 2022, 11:57 AM • 1 Y

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Very cool problem! Let $P(x,y)$ be the given assertion. Observe that $P(0,y)$ yields $f(0)=0.$ Thud, $P(x,0)$ implies $f(x^2)=xf(x).$

Note that $f\equiv 0$ is a solution; assume this sin't the case. Then, there exists $c$ such that $f(c)\neq 0$ then by varying $x$ in $P(x,c-x)$ we get that $f$ is surjective.

Claim: If $f(a)=f(b)$ then $f$ is periodic with period $|a-b|.$

Proof: If $f(a)=f(b)$ then by combining $P(x,a)$ and $P(x,b)$ we get $xf(x+a)=f(x^2+xf(a))=f(x^2+xf(b))=xf(x+b).$ Therefore, for all $x\neq 0,$ we have $f(x+a)=f(x+b).$ But since $f(a)=f(b)$ then $f(x+a)=f(x+b)$ for $x=0$ as well, so $f$ is periodic with period $|a-b|. \ \square$

Case 1: Assume that $f$ is not injective. Then there exist $a\neq b$ such that $f(a)=f(b).$ According to our claim, $f$ is periodic with period $c:=|a-b|\neq 0.$

Now, recall that $f(x^2)=xf(x).$ Combining this with the periodicity of $f$ , we can infer that for any integer $n$ we have $\begin{align*}\frac{n-c}{2}\cdot f\bigg(\frac{n-c}{2}\bigg)&=f\bigg(\frac{n^2+c^2-2nc}{4}\bigg)=f\bigg(\frac{n^2+c^2-2nc}{4}+n\cdot c\bigg)=f\bigg(\frac{n^2+c^2+2nc}{4}\bigg) \\ &=\frac{n+c}{2}\cdot f\bigg(\frac{n+c}{2}\bigg) =\frac{n+c}{2}\cdot f\bigg(\frac{n+c}{2}-c\bigg)=\frac{n+c}{2}\cdot f\bigg(\frac{n-c}{2}\bigg)\end{align*}$ Since $c\neq 0$ then $(n-c)/2\neq(n+c)/2$ so we must have $f((n-c)/2)=0.$

But using our claim, since $f(0)=0=f((n-c)/2)$ then $f$ is periodic with period $(n-c)/2.$ In particular, this implies that $f$ is periodic with period $n-c.$ Recall that $n$ could be any integer, so $f$ has periods $1-c$ and $2-c,$ resulting in the fact that $f$ has period $1.$

However, $P(1,y)$ gives us $f(1+f(y))=f(1+y).$ Using the fact that $f$ has period $1,$ we actually have $f(f(y))=f(1+f(y))=f(1+y)=f(y).$ But since $f$ is surjective, $f(y)$ can take any real value, so we get that $f(x)=x$ for all $x$ , contradicting the fact that $f$ is not injective!

Case 2: Assume that $f$ is injective. Well $P(1,y)$ gives us $f(1+f(y))=f(1+y)$ resulting in $f(y)=y$ for all $y.$

Therefore, the only functions are $f\equiv 0$ and $f\equiv\text{id}.$

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trying_to_solve_br

#37 h Mar 7, 2022, 4:44 PM

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This is one of the best FE's I've seen in a while. Let $P(x,y)$ be the assertion.

$P(0,0)$ implies $f(0)=0$ .

First, notice $f$ is surjective by taking $y=k-x$ and $x=L/f(k)$ (as we vary $L$ , $f$ goes through all reals).

Then, $y=0$ implies $f(x^2)=xf(x)=-xf(-x)$ and thus $f$ is odd. Now notice that $y=-x$ implies $f(x^2-f(x^2))=0$ , and hence if we prove injectivity at 0 we're done, as we'd prove that $f(x^2)=x^2$ and to extend to the negative reals just use it is odd.

To prove injectivity at 0, let $S=\{a_1,...\}$ the set of all real numbers that satisfy $f(a_i)=0$ .

But notice that if $a \in S$ , then $a^2 \in S$ , because of $x=a,y=0$ .

Now, to finish, just notice that: $P(x,a_i)$ implies $f(x^2)=xf(x)=xf(x+a_i)$ . Thus the function has period $a_i$ ( $f(x)=f(x+a_i)$ . Now, by surjectivity, take $c$ so that $f(c)=1$ , and $P(a_i,c) \implies f(a_i^2+(1.a_i))=f(a_i^2)=a_i.f(a_i+c)=a_i$ , because of $f(k+a_i)=f(k)$ . Thus $f(a_i^2)=0=a_i$ , and we're done. $\blacksquare$

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CT17

#38 h Mar 7, 2022, 5:58 PM

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Note that $f(x)\equiv 0$ and $f(x)\equiv x$ are both solutions. From now on, we can assume there exist (not necessarily distinct) $u$ and $v$ such that $f(u)\neq 0$ and $f(v)\neq v$ . Let $c = |v| - f(|v|)$ .

$P(x,u-x)\implies f$ is surjective.

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

$P(x,0)\implies f(x^2) = xf(x)$ . In particular $f$ is odd, so $f(|v|)\neq |v|$ .

$P\left(-\sqrt{|v|},\sqrt{|v|}\right)\implies f\left(|v| - \sqrt{|v|}f\left(\sqrt{|v|}\right)\right) = 0\implies f(c) = 0$ .

$P(x,c)\implies f(x^2) = xf(x+c)\implies f(x) = f(x + c)$ for $x\neq 0$ . But $0 = f(0) = f(c)$ , so $f(x) = f(x+c)$ for all $x$ .

$P(c,y)\implies f(c^2 + cf(y)) = cf(c+y) = cf(y)$ . Since $c\neq 0$ and $f$ is surjective the funtion $cf(y)$ is surjective, so $f(c^2 + x) = x$ for all $x$ . In particular $f$ is linear with slope $1$ , but plugging back in yields no new solutions.

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#39 h Mar 7, 2022, 7:35 PM

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

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

$P(x,0): f(x^2)=xf(x)$ .

Clearly $\boxed{f\equiv 0}$ is a solution. Henceforth assume there exists a $k$ with $f(k)\ne 0$ .

Claim: $f$ is surjective.
Proof: $P(x,k-x): f(x^2+xf(k-x))=xf(k)$ . Since $f(k)\ne 0$ , $xf(k)$ can take on any value, so $f$ is surjective $\blacksquare$ .

Claim: If $f(k)=0$ , then $k=0$ .
Proof: Suppose there exists a $k\ne 0$ with $f(k)=0$ .

$P(x,k): f(x^2)=xf(x+k)$ .

So $xf(x)=xf(x+k)\implies f(x)=f(x+k)$ .

Let $a$ satisfy $f(a)=-k$ .

$P(k,a): f(k^2-k^2)=0=kf(a+k)=kf(a)=-k^2$ , a contradiction. $\blacksquare$

$P(x,-x): f(x^2+xf(-x))=0\implies x^2=-xf(-x)$ . If $x\ne 0$ , then $-f(-x)=x\implies \boxed{f(x)=x}$ .

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#40 h Apr 9, 2022, 2:55 AM

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We prove that $f(x)=x$ and $f\equiv 0$ are the only solutions.
Let $P(x,y)$ be the assertion in $f\left( x^2+xf(y)\right)=xf(x+y)$ .

It can be easily checked that the only constant function is $\boxed{f\equiv0}$ , which indeed works and is our first solution.
Let $f$ be non-constant.

Claim : $f$ is injective.
Proof :
Assume possible $f(y_1)=f(y_2)$ and $y_1\neq y_2$ .
$P(x,y_1)-P(x,y_2) : f(x+y_1)=f(x+y_2) \forall x$ .
Hence, $f$ is periodic with period $|y_2-y_1|$ .
Hence, $f$ is bounded above.

But, we can choose very large $x$ and $y$ such that $f(x+y)\neq0$ (Note we can select such $y$ because $f\not\equiv 0$ ) : $f\left( x^2+xf(y)\right)<xf(x+y)$ .
Contradiction!
Hence, $f(y_1)=f(y_2) \implies y_1=y_2$ .
Hence proved claim!

$P(0,y) : f(0)=0$ .
$P(-x,x) : f(x^2-xf(x))=0$ .

Hence, $f(x^2-xf(x))=f(0) \overset{\text{injective}}{\implies} x^2-xf(x)=0 \implies f(x)=x \forall x\neq0$ .
Combining with $f(0)=0$ , we get $\boxed{f(x)=x },\forall x$ , which indeed satisfies given equation and is our second solution.

Hence, we are done

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ThisNameIsNotAvailable

#42 h Apr 9, 2022, 4:01 PM

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Ankoganit wrote:

Find all functions $f:\mathbb R\to\mathbb R$ such that $f\left( x^2+xf(y)\right)=xf(x+y)$ for all reals $x,y$ .

My solution, without using the injectivity.
Let $P(x,y)$ be the assertion that $f\left( x^2+xf(y)\right)=xf(x+y)$ .
$P(0;y) \implies f(0)=0$
$P(x;0) \implies f(x^2)=xf(x)$ , which leads to $f$ is odd.
Clearly $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is a solution, hence assume that there exist $k$ such that $f(k)\ne 0$ .
$P(x;k-x) \implies f(\cdots)=xf(k)$ , it follows that $f$ is surjective, so there exists $a \in \mathbb R$ such that $f(a)=0$ .

$P(x;a) \implies f(x+a)=f(x) \quad \forall x \in \mathbb R \implies f(x+na)=f(x) \quad \forall n \in \mathbb Z$
$P(x;-x) \implies f(x^2-xf(x))=0 \implies f(a^2)=0 \quad (*)$
$P(x;a^2) \implies f(x+a^2)=f(x) \quad \forall x \in \mathbb R$

Put $x \longrightarrow x+a$ into $f(x^2)=xf(x)$ , we have:
$f(x^2+2ax+a^2)=(x+a)f(x+a) \implies f(n^2)=(n+a)f(n)=nf(n) \implies af(n)=0 \quad \forall n \in \mathbb Z$
If $a=0$ , from $(*)$ we have $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ is also a solution.

If $f(n)=0 \quad \forall n \in \mathbb Z$ :
$P(x;n) \implies f(x+n)=f(x) \quad \forall n \in \mathbb Z$
$P(x+n;y) \implies f(x^2+2nx+n^2+(x+n)f(y))=(x+n)f(x+y+n)$
$\implies f(x^2+2nx+(x+n)f(y))=(x+n)f(x+y) \quad \forall n \in \mathbb Z$

Because of the surjectivity, for all $x \in \mathbb R, n \in \mathbb Z$ , there exists $t$ such that $f(t)=n-x$ .
Put $y \longrightarrow t$ into the above equation, we have:
$f(2nx)=(x+n)f(x+t) \quad \forall n \in \mathbb Z$
Put $x \longrightarrow x-n$ into the above equation:
$xf(x+t)=f(2nx) \implies nf(x+t)=0 \implies f(x+t)=0 \quad \forall n \in \mathbb Z$
So all the solutions are $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ or $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ .

This post has been edited 1 time. Last edited by ThisNameIsNotAvailable, Apr 10, 2022, 4:38 AM
Reason: Fix a typo

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hood09

#43 h Apr 10, 2022, 12:00 AM

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let $P(x,y)$ be the natural assertion.
$P(0,0) : f(0)=0$
$p(x,0) : f(x^2)=xf(x)$
if $\exists k \neq 0$ such that $f(k)=0$ then by $P(x,k) : f(x^2)=xf(x+k) \implies f(x)=f(x+k)$ then $f$ is $k-$ periodic .
and we have by $P(x+k,-k ) : f((x+k)^2)=(x+k)f(x) \implies f(x^2)=xf(x)+kf(x) \implies k=0$ or $\boxed{\forall x : f(x)=0}$ is a solution
if $k=0$ then $\forall z\neq 0 : f(z) \neq 0$ :
$P(-x,x) : f(x^2-xf(x))=0 \implies xf(x)=x^2 \implies f(x)=x$ so we have the second solution $\boxed{\forall x : f(x)=x}$

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#44 h Apr 22, 2022, 4:47 AM

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Oly grind back maybe?
Let $P(x,y)$ the assertion of the F.E. clearly if $f$ is constant then $f(x)=0$ works so lets work when $f$ is not constant.
$P(0,x)$
$f(0)=0$ $P(x,0)$
$f(x^2)=xf(x) \implies f \; \text{odd}$ $P(f(x),-x)$
$f(f(x)-x)=0$ $P(x,f(x)-x)$
$xf(x)=f(x^2)=xf(f(x)) \implies f(x)=f(f(x))$ As $f$ is non-cero there exists $d$ with $f(d) \ne 0$ so by $P(x,d-x)$
$f(x^2+xf(d-x))=xf(d) \implies f \; \text{surjective}$ Since $f$ is surjective we use that on the last equation by setting $t=f(x)$ to get
$f(t)=t \; \forall t \in \mathbb R$ Hence $\boxed{f(x)=0,x \; \forall x \in \mathbb R}$ work thus we are done

This post has been edited 1 time. Last edited by MathLuis, Jun 4, 2022, 2:39 PM

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VZH

#45 h Apr 22, 2022, 8:24 AM

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Let $P(x,y)$ be the assertion.
$P(0,0)$ : $f(0)=0$
$P(x,0)$ : $f(x^2)=xf(x)$

Case 1: $f(x)=0 \iff x=0$
$P(-f(x),x)$ : $f(x)f(x-f(x))=0 \implies f(x)=x, \forall x \in\mathbb{R}$ , which is a solution.

Case 2: There exists $t\neq0$ such that $f(t)=0$
$P(x,t)$ : $f(x^2)=xf(x+t) \implies xf(x+t)=xf(x) \implies f(x)=f(x+t), \forall x \in\mathbb{R}$ . This means $f(d)=0 \implies f$ has period $d$ .
$f(t^2)=tf(t)=0$ , so $f$ has period $t^2$ .
Now, $f(x^2+2tx+t^2)=f((x+t)^2)=(x+t)f(x+t)=xf(x)+tf(x)=f(x^2)+tf(x)$ . Let $x$ be an integer, then $f(x^2+2tx+t^2)=f(x^2) \implies tf(x)=0 \implies\forall x \in\mathbb{Z}, f(x)=0$ , so $f$ has period $1$ .
Take a positive integer $k$ that is not a perfect square, $f(k)=\sqrt{k}f(\sqrt{k})=0 \implies f(\sqrt{k})=0$ , where $\sqrt{k}$ is irrational.
Thus, $f$ has periods $1$ and $\sqrt{k}$ .
We aim to prove $f(x) \equiv 0$ . Only need to prove $f(x)=0$ for all $x \in [0,1)$ . For fixed irrational number $\alpha$ , let $n$ span all integers, $\{n\alpha\}$ is dense in $[0,1)$ . Hence $0=f(0)=f(n\sqrt{k}-[n\sqrt{k}])$ , and $n\sqrt{k}-[n\sqrt{k}]$ can be any number in $[0,1)$ $\implies f(x)=0, \forall x \in [0,1)$ . So $f(x) \equiv 0$ , which indeed is a solution.
Conclusion: Solutions are $f(x)=x, \forall x \in\mathbb{R}$ or $f(x) \equiv 0$ .

Weird solution I just came up with (any mistakes?).

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ZETA_in_olympiad

#46 h Jun 4, 2022, 2:13 PM

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

$P(0,0)$ gives $f(0)=0.$ If constant then $f\equiv 0.$ If not then $P(x,y-x)$ gives $f$ is surjective. Now let $f(u)=f(v).$ Comparing $P(z,u)$ and $P(z,v)$ gives $f(z)=f(z+w)$ where $w=u-v.$

Take $f(m)=0$ and $f(n)=1.$ Comparing $P(w,m)$ and $P(w,n)$ forces $w=0.$ So $f$ is injective and $P(1,x)$ gives $f(x)=x.$ Both work.

This post has been edited 1 time. Last edited by ZETA_in_olympiad, Jun 4, 2022, 4:12 PM
Reason: Typo

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ZETA_in_olympiad

#47 h Jun 4, 2022, 4:11 PM

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Ali3085 wrote:

my short solution: :rotfl:

:rotfl:

claim(1): $f$ is surjective
proof:
$P(\frac{x}{f(y)},y-x) \implies f(something)=x$
$\blacksquare$
claim(2): $f$ is injective
proof:
suppose there's $a \neq b : f(a)=f(b)$ note that
$P(x,a) \& P(x,b) \implies f(x+a)=f(x+b)$
define $\mathbb{S}=\{a-b | f(a)=f(b) \}$ then $f(x)=f(x+s) \forall s \in \mathbb{S}$
now let $f(y_1)=1 ,f(y_2)=2$
$P(s,y_1) \& P(s,y_2) \implies f(s^2)=s=2s \implies s=0$
$\blacksquare$
now just put $P(1,y) \implies f(y)=y$
and we win

Your solution needs a fix since surjectivity holds $\forall f(x)\neq 0.$

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#48 h Jul 12, 2023, 5:31 AM

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2016 India IMO Training T2P2 wrote:

Find all functions $f:\mathbb R\to\mathbb R$ such that $f\left( x^2+xf(y)\right)=xf(x+y),\forall x,y\in \mathbb{R}^{(1)}$

When $f$ is a constant function, we have $\boxed{f(x)=0,\forall x\in \mathbb{R}}$

Now consider the case when $f$ is not a constant function, then $\exists a \in \mathbb{R}$ such that $f(a)\ne 0$

$(1). P(x,a-x)\Rightarrow f(x^2+xf(a-x))=xf(a),$ subtitute $x\rightarrow \dfrac{x}{f(a)}\Rightarrow f(\text{something})=x\Rightarrow f$ is surjective

$(1).P(0,0)\Rightarrow f(0)=0; (1).P(x,0)\Rightarrow f(x^2)=xf(x)$

Assume that there is $c\ne 0$ such that $f(c)=0$

$(1).P(x,c)\Rightarrow f(x^2)=xf(x+c)\Rightarrow f(x)=f(x+c),\forall x\in \mathbb{R}$ (as we already have $f(x^2)=xf(x),\forall x\in \mathbb{R}$ )

$(1).P(c,y)\Rightarrow f(c^2+cf(y))=cf(y+c)=cf(y)\Rightarrow f(xc+c^2)=xc\Rightarrow f(x+c^2)=x$

Subtitute $x\rightarrow c-c^2\Rightarrow c-c^2=0\Rightarrow c=1\Rightarrow f(x+1)=x\Rightarrow f(0)=-1,$ a contradiction.

Hence $f(x)=0\Leftrightarrow x=0; (1).P(x,-x)\Rightarrow f(x^2+xf(-x))=0\Rightarrow x^2=-xf(-x)\Rightarrow \boxed{f(x)=x,\forall x\in \mathbb{R}}$

This post has been edited 9 times. Last edited by trinhquockhanh, Jul 12, 2023, 5:45 AM

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#49 h Apr 3, 2025, 7:06 AM

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Let $P(x,y)$ be the assertion $f\left(x^2+xf(y)\right)=xf(x+y)$ . Note that $\boxed{f(x)=0}$ is a solution, otherwise there is some $j$ with $f(j)\ne0$ .

$P(0,0)\Rightarrow f(0)=0$
$P(-f(x),x)\Rightarrow f(x)f(x-f(x))=0\Rightarrow f(x-f(x))=0$ if $f(x)\ne0$ , still obviously true when $f(x)=0$
$P(f(x),0)\Rightarrow f\left(f(x)^2\right)=f(x)f(f(x))$
$P(f(x),x-f(x))\Rightarrow f\left(f(x)^2\right)=f(x)^2$
Comparing the last two of these, we get $f(f(x))=f(x)$ if $f(x)\ne0$ , but still true when $f(x)=0$ .
$P(x,j-x)\Rightarrow f\left(x^2+xf(j-x)\right)=xf(j)$ and so $f$ is surjective. From $f(f(x))=f(x)$ we get $\boxed{f(x)=x}$ which fits.

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