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Source: 2024 Turkey TST P2

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egxa

#1 h Mar 18, 2024, 8:34 PM • 4 Y

Y by Amir Hossein, ItsBesi, cubres, farhad.fritl

Find all $f:\mathbb{R}\to\mathbb{R}$ functions such that
$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$ for all real numbers $x,y$

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#2 h Mar 18, 2024, 9:28 PM • 3 Y

Y by VicKmath7, Amir Hossein, cubres

Here is my solution.
Let $P(x,y)$ be the assertion. $P(0,0)$ gives $f(0)=0$ . $P(x,0)$ implies $f(x)^3=x\,f(x^2)$ and $P(0,x)$ implies $f(x)^3=f(f(x))\,x^2$ for every $x\in\mathbb{R}$ . So, $f(x)^3=x\,f(x^2)=x^2f(f(x))$ . Since it is already true for $x=0$ , we can say that $f(x^2)=x\,f(f(x))$ for every $x\in\mathbb{R}$ and replace in the original equation to get $f(x+y)^3=(x^2+2xy)f(f(x))+(x^2+3xy+y^2)f(f(y))$ Reversing the $x,y$ order, we get $f(x+y)^3=(x^2+3xy+y^2)f(f(x))+(2xy+y^2)f(f(y))$ which means $f(f(x))(x+y)y=f(f(y))(x+y)x$ setting $y=1$ and letting $f(f(1))=k$ , we have $f(f(x))=kx$ for every $x\in\mathbb{R}-\{-1\}$ . So, for $x,y\in\mathbb{R}-\{-1\}$ , $f(x+y)^3=k(x^3+3x^2y+3xy^2+y^3)$ , and $f(x+y)=(x+y)t$ where $t=\sqrt[3]{k}$ . This means that $f(x)\equiv tx$ for a constant $t$ . Putting this into $f(x^2)=x\,f(f(x))$ , we get $t=t^2$ , so $t=0$ or $t=1$ . Therefore, the solutions are $f(x)=0$ and $f(x)=x$ .

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#4 h Mar 19, 2024, 2:02 PM • 2 Y

Y by ehuseyinyigit, cubres

$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$ $P(0,0)\implies f(0)^3=0\iff f(0)=0$
$P(0,y)\implies f(y)^3=f(f(y)).y^2$
$P(x,0)\implies f(x)^3=xf(x^2)$
Hence $f(f(x)).x^2=f(x)^3=xf(x^2)\iff x.f(f(x))=f(x^2)$
$P(x,y)=P(y,x)\implies (x+2y)f(x^2)+(x^2+3xy+y^2).f(f(y))=(y+2x)f(y^2)+(x^2+3xy+y^2).f(f(x))$
$(x+y)(f(x^2)-f(y^2))+yf(x^2)-xf(y^2)+(x^2+3xy+y^2)(f(f(y))-f(f(x)))=0$
$P(-2y,y)\implies f(-y)^3=f(f(y)).(4y^2-6y^2+y^2)=f(f(y)).-y^2$
Also $-f(f(y)).y^2=f(-y)^3=-y.f(y^2)\implies f(y^2)=y.f(f(y))$
Thus $\boxed{f(f(x))=\frac{f(x^2)}{x}}$
$f(x+y)^3=(x+2y).f(x^2)+\frac{f(y^2)}{y}(x^2+3xy+y^2)=(x+2y).f(x^2)+(2x+y).f(y^2)+(x+\frac{x^2}{y}).f(y^2)$ $f(x+y)^3-(x+2y).f(x^2)-(y+2x).f(y^2)=(x+\frac{x^2}{y}).f(y^2)$ While swapping $x\leftrightarrow y$ , $LHS$ doesn't change so $(x+\frac{x^2}{y}).f(y^2)=(y+\frac{y^2}{x}).f(x^2)$
$\iff (x^2y+x^3).f(y^2)=(xy^2+y^3).f(x^2)$
$\iff (x+y).x^2f(y^2)=(x+y).y^2f(x^2)$
$\iff (x+y)(\frac{f(x^2)}{x^2}-\frac{f(y^2)}{y^2})=0$
$x+y\neq 0\implies \frac{f(x^2)}{x^2}=\frac{f(y^2)}{y^2}$
Taking $x,y>0$ gives that $\boxed{f(x^2)=cx^2}$
Also by taking $x+y<0,x>0$ we get that $c=\frac{f(x^2)}{x^2}=\frac{f(y^2)}{y^2}$
So $f(x^2)=cx^2$ for all $x\in R$ .
$c^3x^6=f(x^2)^3=x^2.f(x^4)=cx^6\implies c=c^3\iff \boxed{c=-1,0,1}$

$f(f(x^2))=\frac{f(x^4)}{x^2}=cx^2$
If $c=-1,$ then $x^2=-x^2$ gives contradiction.
If $c=0\iff f\equiv 0$ for all $x>0,$ then $f(x+y)^3=0\implies \boxed{f\equiv 0, \forall x\in R}$
If $c=1 \iff f(x)=x$ for all $x>0,$ then $f(x)^3=x.f(x^2)=x^3\implies \boxed{f(x)=x, \forall x\in R}$

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#5 h Mar 19, 2024, 4:58 PM • 2 Y

Y by cubres, farhad.fritl

Let $P(x,y)$ denote the assertion $f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$
First, by $P(0,0)$ we have $f(0)=0$
By $P(0,x)$ and by $P(x,0)$ we can also deduce that $f(x)^3=xf(x^2)=f(f(x))\cdot x^2$ . Plugging in $x\rightarrow -x$ we have $f$ is odd. Also let's keep in mind that $f(f(x))=\frac{f(x^2)}{x}$ for any $x\neq 0$
Swapping x and y we get that $xf(y^2)+x^2\cdot f(f(y))=y^2\cdot f(f(x))+yf(x^2)$ and using $f(f(x))=\frac{f(x^2)}{x}$ with $x,y\neq 0$ we have
$xf(y^2)+\frac{f(y^2)\cdot x^2}{y}=yf(x^2)+\frac{f(x^2)\cdot y^2}{x}$ .
Plugging $x=1$ in the last one for $y\neq0,-1$ we get that $f(y^2)=y^2f(1)$ so $f(x)=xf(1)$ for any $x\geq 0$ . Using f odd we get $f(x)=xf(1)$ for any $x\in \mathbb R$ and we are done

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Aoxz

#6 h Mar 19, 2024, 5:31 PM • 1 Y

Y by cubres

My solution is same with #4

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#7 h Apr 3, 2024, 6:08 PM • 1 Y

Y by cubres

Here is my solution
Let $P(x,y)$ denote the assertion. $P(0,0)$ gives $f(0)=0$ . $P(x,0)$ gives $f(x)^3=xf(x^2)$ and $P(-x,0)$ gives $f(-x)^3=(-x)f(x^2)$ , so $f$ is odd function.
$P(x,-x)$ gives $f(x^2)=xf(f(x))$ . Now $P(x,y)$ and $P(y,x)$ gives that $(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)=f(x+y)^3=(y+2x)f(y^2)+f(f(x))(x^2+3xy+y^2)$ and if we put $xf(f(x))$ instead of $f(x^2)$ we get that
$(x+2y)xf(f(x)) +f(f(y))(x^2+3xy+y^2)= (y+2x)yf(f(y))+f(f(x))(x^2+3xy+y^2)$ , so we get $f(f(x))y(x+y)=f(f(y))x(x+y)$ , so $\frac {f(f(x))}{x}$ is constant for all $x\in\mathbb{R}$ , so $f(f(x))=cx$ and $f(x^2)=cx^2$ from $f(x^2)=xf(f(x))$ . And we put this in the condition then we get that $f(x)=ax$ . Again we put this in the condition then we get that $a=0$ or $a=1$ . Therefore, the solutions are $f(x)=0$ and $f(x)=x$ .

This post has been edited 1 time. Last edited by NuMBeRaToRiC, Apr 3, 2024, 6:09 PM

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#8 h Apr 27, 2024, 3:30 PM • 1 Y

Y by cubres

Let the given assertion be $P(x,y)$ . The only solutions are $\boxed{f(x)=0, f(x)=x}$ which can both be easily verified.

Claim: $f(0)=0$
This is immediate from $P(0,0)$

Claim: $f(x)$ is odd
Below are the assertions $P(-2y,y)$ and $P(0,y)$ , $f(-y)^3=-y^2f(f(y))$ $f(y^3)=y^2f(f(y)$ The result follows.

Claim: $f(x^2)=xf(f(x))$
This follows from $P(x,-x)$ combined with the fact that $f(x)$ is odd.

Claim: $f(x)=cx$ for all $x\in \mathbb{R}^{+}$
Consider $P(x,y)$ and $P(y,x)$ we get, $(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)=(y+2x)f(y^2)+f(f(x))(x^2+3xy+y^2)$ Multiplying by $xy$ and then using the reduction shown in the previous claim we get, $(xy^2+y^3)f(x^2)=(yx^2+x^3)f(y^2)$ If for any $x>0$ we have $f(x)=0$ then $f(x)=0$ for all $x\in \mathbb{R}^{+}$ . Otherwise we get $\frac{f(x^2)}{x^2}=\frac{f(y^2)}{y^2}$ for all $x,y\in \mathbb{R}^{+}$ which implies the result.

Claim: Either $f(x)=x$ or $f(x)=0$
Given that $f$ is odd and that $f(0)=0$ the previous claim implies that $f(x)=cx$ for all $x$ . As $f(x^2)=xf(f(x))$ we must have $c=c^2$ so either $c=0$ or $c=1$ .

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#9 h Aug 29, 2024, 4:00 PM • 2 Y

Y by Amy_Chen, cubres

$P(0,0) \implies f(0) = 0, P(x,0) \implies f(x)^3 = xf(x^2)$ * and $x \rightarrow -x \implies -f(x) = f(-x)$ . $P(-2y,y) \implies f(f(y)) = f(y)^3/y^2 \overset{*} = f(y^2)/y$ . Thus,
$f(x+y)^3 = xf(x^2) + 2yf(x^2) + \frac{x^2}{y} f(y^2) + 3xf(y^2) + yf(y^2) = xf(x^2) + 2xf(y^2) +\frac{y^2}{x} f(x^2) + 3yf(x^2) + yf(y^2) \implies$ $xf(y^2) + \frac{x^2}{y} f(y^2) = yf(x^2) + \frac{y^2}{x} f(x^2) \overset{y \rightarrow 1} \implies f(x^2) = f(1) x^2$ Because $f$ is odd, $f(x) = cx$ and plugging back gives $f \equiv x$ or $f \equiv 0$ .

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#10 h Oct 27, 2024, 1:40 PM • 1 Y

Y by cubres

Solution sketch :
$P(0,0)$ gives $f(0)=0$
$P(0,x)$ and $P(x,0)$ gives $f(x)^3=x^2f(f(x))=xf(x^2)$ .
$P(1,y)-P(y,1)$ : $f(f(y))+\frac{f(f(y))}{y}=c+cy$ where $f(f(1))=c$ $\Rightarrow f(f(y))=cy$ .
Thus $f(y)=c^{\frac{1}{3}}y$
We find then that $c=0,1$ which finishes the problem.

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#11 h Apr 12, 2025, 10:18 PM • 1 Y

Y by cubres

thanks to @HoRI_DA_GRe8 for suggesting this supaar bootiful problem (cuz i cant do hard FEs)

I'm not proving the claims that other's proved because i don't wanna spam the same solution a million times (just like i always do on other contests), so im just assuming those claims are true and doing the unique part in my proof

$\textbf{Proof:}$
let the given assertion be $P(x,y)$ and the only solutions are $\boxed{f(x)=0, f(x)=x}$ which can be easily verified, now to proof that these are the ONLY solutions,

$\bullet$ $P(0,0) \Rightarrow f(0) = 0$
$\bullet$ $P(-2y,y)$ and $P(0,y) \Rightarrow f$ is odd
$\bullet$ $f(f(x))x^{2}=f(x)^{3}$
$\bullet$ $f(f(x))=\frac{f(x^{2})}{x}$

$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$
$f(x+y)^3=(x+2y)f(x^2)+\frac{f(y^2)}{y}(x^2+3xy+y^2)$
$P(x,x) \Rightarrow f(2x)^3 = 3xf(x^2)+ \frac{f(x^2)}{x}(2x^2+3x^2)$
$\Rightarrow f(2x)^3 = 3xf(x^2) + f(x^2)5x$
$\Rightarrow f(2x)^3 =8xf(x^2)$
$\Rightarrow f(2x)^3 = 8x^2f(f(x))$
$\Rightarrow f(2x)^3= 8f(x)^3 \text{ because } f(f(x))x^{2}=f(x)^{3}$
$\Rightarrow f(2x)=2f(x)$
Claim: $f(x)=cx$ give that $f$ is odd and that $f(0)=0$ the previous claim implies that $f(x)=cx$ for some $c \in \mathbb{R} \forall x \in \mathbb{R}$

from $\bullet$ $f(f(x))x^{2}=f(x)^{3}$
$c^2x^3 = c^3 x^3$ $\Rightarrow c=0, \text{ or } c=1$ $\therefore \boxed{f(x)=0, f(x)=x}$ $\blacksquare$

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200th post!!!

This post has been edited 3 times. Last edited by Levieee, Apr 13, 2025, 6:46 AM

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#12 h Apr 14, 2025, 4:52 PM

Y by

egxa wrote:

Find all $f:\mathbb{R}\to\mathbb{R}$ functions such that
$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$ for all real numbers $x,y$

Let $P(x,y)$ be the assertion $f(x+y)^3=(x+2y)f\left(x^2\right)+f(f(y))\left(x^2+3xy+y^2\right)$ .
First, we show that $f(x)^3=xf\left(x^2\right)=x^2f(f(x))$ , which in particular implies $f(f(1))=f(1)$ :
$P(0,0)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(x)^3=xf\left(x^2\right)$
$P(0,x)\Rightarrow f(x)^3=x^2f(f(x))$

Because of this, we can rewrite $P(x,y)$ as:
$f(x+y)^3=f(x)^3+f(y)^3+2yf\left(x^2\right)+(x^2+3xy)f(f(y)).$ Swapping $x,y$ and comparing, this becomes:
$2yf\left(x^2\right)+(x^2+3xy)f(f(y))=2xf\left(y^2\right)+(y^2+3xy)f(f(x)),$ and setting $y=1$ and simplifying:
$2x^2f\left(x^2\right)+(x^4+3x^3)f(f(1))=2x^3f(1)+(3x^3+x^2)f(f(x))$ $2xf(x)^3+(x^4+3x^3)f(1)=2x^3f(1)+(3x+1)f(x)^3$ which gives $f(x)^3=x^3f(1)$ for all $x\ne-1$ .

Then, $f(1)^3=f(1)$ , so $f(1)\in\{-1,0,1\}$ .
If $f(1)=-1$ then $f(-1)^3=-f(1)=1$ (using $f(x)^3=xf\left(x^2\right)$ ) so $f(-1)=1$ , giving the solution $\boxed{f(x)=-x}$ .
If $f(1)=0$ then $f(-1)^3=-f(1)=0$ so $f(-1)=0$ , giving the solution $\boxed{f(x)=0}$ .
If $f(1)=1$ then $f(-1)^3=-f(1)=-1$ so $f(-1)=-1$ , giving the solution $\boxed{f(x)=x}$ .
Easy to verify that these all satisfy the equation.

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