Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $$f(f(x)-9f(y))=(x+3y)^2f(x-3y)$$for all $x,y\in \mathbb{R}$.
Problem
Source: 2022 Korea Winter Program Practice Test
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Jalil_Huseynov
24.09.2022 16:05
Bump....
pco
26.09.2022 13:11
F_Xavier1203 wrote: Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $$f(f(x)-9f(y))=(x+3y)^2f(x-3y)$$for all $x,y\in \mathbb{R}$. $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ fits. So let us from now look only for non allzero solutions. Let $P(x,y)$ be the assertion $f(f(x)-9f(y))=(x+3y)^2f(x-3y)$ Let $U=f^{-1}(\{0\})$ ($U\ne\mathbb R$) Let $a=f(0)$ $P(0,0)$ $\implies$ $f(-8a)=0$ and so $-8a\in U$ 1) Difficult part (maybe to improve) $U=\{0\}$
Let $u,v\in U$ (maybe identical)
$P(u,-\frac {x-v}3)$ $\implies$ $f(-9f(-\frac {x-v}3))=(x-u-v)^2f(x+u-v)$
$P(v,-\frac {x-v}3)$ $\implies$ $f(-9f(-\frac {x-v}3))=(x-2v)^2f(x)$
And so, subtracting, $(x-u-v)^2f(x+u-v)=(x-2v)^2f(x)$
And $f(x+u-v)=f(x)\left(\frac{x-2v}{x-u-v}\right)^2$ $\forall x\ne u+v$
This implies $f(x+2u-2v)=f(x+u-v)\left(\frac{x+u-3v}{x-2v}\right)^2$ $\forall x\ne 2u+2v$
And so $f(x+2u-2v)=f(x)\left(\frac{x-2v}{x-u-v}\right)^2\left(\frac{x+u-3v}{x-2v}\right)^2$ $\forall x\notin\{u+v,2u+2v\}$
Which is $f(x+2u-2v)=f(x)\left(\frac{x+u-3v}{x-u-v}\right)^2$ $\forall x\notin\{u+v,2u+2v\}$
This implies $f(x+3u-3v)=f(x+2u-2v)\left(\frac{x+2u-4v}{x+u-3v}\right)^2$ $\forall x\ne 3u+3v$
And so $f(x+3u-3v)=f(x)\left(\frac{x+u-3v}{x-u-v}\right)^2\left(\frac{x+2u-4v}{x+u-3v}\right)^2$ $\forall x\notin\{u+v,2u+2v, 3u+3v\}$
Which is $f(x+3u-3v)=f(x)\left(\frac{x+2u-4v}{x-u-v}\right)^2$ $\forall x\notin\{u+v,2u+2v, 3u+3v\}$
Then :
$P(x+3u,u)$ $\implies$ $f(f(x+3u))=(x+6u)^2f(x)$
$P(x+3u,v)$ $\implies$ $f(f(x+3u))=(x+3u+3v)^2f(x+3u-3v)$
And so, subtracting, $(x+6u)^2f(x)=(x+3u+3v)^2f(x+3u-3v)$
And $f(x+3u-3v)=f(x)\left(\frac{x+6u}{x+3u+3v}\right)^2$ $\forall x\ne -3u-3v$
And so, comparing the two values we got for $f(x+3u-3v)$ :
$f(x)\left(\frac{x+2u-4v}{x-u-v}\right)^2=f(x)\left(\frac{x+6u}{x+3u+3v}\right)^2$ $\forall x\notin\{-3u-3v,u+v,2u+2v, 3u+3v\}$
And so $\left(\frac{x+2u-4v}{x-u-v}\right)^2=\left(\frac{x+6u}{x+3u+3v}\right)^2$ $\forall x\notin\{-3u-3v,u+v,2u+2v, 3u+3v\}\cup U$
If $\frac{x+2u-4v}{x-u-v}=\frac{x+6u}{x+3u+3v}$, we get $u^2=v^2$
If $u^2\ne v^2$, we need $\frac{x+2u-4v}{x-u-v}=-\frac{x+6u}{x+3u+3v}$
Which is $x^2+(5u-v)x-6uv-6v^2=0$
And so also $x^2+(5v-u)x-6uv-6u^2=0$
Subtracting, we get $x=-u-v$ and plugging back in the equations, we get $u+v=0$, impossible here
And so $u^2=v^2$ and so $U=\{0\}$ or $U=\{u,-u\}$ for some $u\ne 0$
But, back to property "$(x-u-v)^2f(x+u-v)=(x-2v)^2f(x)$" found some lines above, setting there $v=-u\ne 0$ and $x=u$, we get $u^2f(3u)=0$ and so $3u\in U$, impossible too.
So $U=\{0\}$ and $a=0$ (since $-8a\in U$)
Q.E.D.
2) Solutions $P(3x,x)$ $\implies$ $f(f(3x)-9f(x))=0$ and so $f(3x)=9f(x)$ $P(-3x,x)$ $\implies$ $f(f(-3x)-9f(x))=0$ and so $f(-3x)=9f(x)=f(3x)$ and $f(x)$ is an even function So $P(3x,y)$ is $f(f(3x)-f(3y))=(3x+3y)^2f(3x-3y)$ and so New assertion $Q(x,y)$ : $f(f(x)-f(y))=(x+y)^2f(x-y)$ Comparing $Q(x,y)$ with $Q(x,-y)$ and since even, we get $(x+y)^2f(x-y)=(x-y)^2f(x+y)$ And so $x^2f(y)=y^2f(x)$ And so $f(x)=f(1)x^2$ Plugging this in original equation, we get $f(1)^3=f(1)$ and so $\boxed{\text{S2 : }f(x)=-x\quad\forall x}$, which indeed fits $\boxed{\text{S3 : }f(x)=x\quad\forall x}$, which indeed fits
idkk
10.10.2024 07:52
Solved with math_comb01. We consider the kernel of $f$ . and prove the kernel has only the element $0$ . First we prove it has at max 1 element. Say it has two elements $a,b$ in the kernel. Then $P(a,\frac{-(x-b)}{3}):$ u have $f(-9f(-\frac{x-b}{3}))=(x-a-b)^2f(a+x-b)$ U also have $ f(-9f(\frac{b-x}{3})) = (x-2b)^2f(x)$ so $f(x+a-b)=f(x)(x-\frac{2b}{x}-a-b)^2$. Keep adding $a-b$ So we have $f(x+3a-3b)=f(x)(x+\frac{6a}{x}+3a+3b)^2$ Now consider $P(x+3a,b)$ and $P(x+3b,a)$ and u have $(x+\frac{6a}{x}+3a+3b)^2=(x-\frac{2a}{x}+a-b)^2$ from here get $a^2=b^2$. so $a+b=0$ as they are distinct but as $a+3a-3b$ is also in kernel we conclude $a+3a-3b=-b$ so $a=\frac{b}{2}$ from here $b=0$ and hence $a=0$ . Contradiction. Easy to see kernel is non empty. now if $e$ is any element in kernel which in non zero consider $P(3y+e,y)$ to conclude $f(3y+e)-9f(y)$ is also in kernel. so $f(4e)$ is in kernel. consider $x=-3e$ and $y=e$ to conclude $f(-3e)=e$ now just consider $P(x,0)$ to get $16e^2=9e^2$ so $e=0$. So kernel is $\{0\}$ Now from here $f(f(x))=x^2f(x)$ to conclue if $f(a)=f(b)$ then $a^2=b^2$ Easy to see by comparign $x=-3y$ and $x=3y$ to see $f$ is even. and $f(3y)=9f(y)$ hence the equation converts to $f(f(x)-f(y))=(x+y)^2f(x-y)$ so $(x+y)^2f(x-y)=(x-y)^2f(x+y)$ rest is trivial