Functional Equation:

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Functional Equation:

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#1 h Apr 9, 2025, 12:55 PM • 3 Y

Y by Nuran2010, TunarHasanzade, TDVOLIMPTEAM

Find all functions $f:\mathbb {R}\rightarrow \mathbb {R}$ such that:
$f(x^2f(x)+f(y))=(f(x))^3+y$

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#2 h Apr 9, 2025, 1:01 PM • 3 Y

Y by Omerking, TunarHasanzade, TDVOLIMPTEAM

Since $y$ takes all real values, we get $f$ is surjective.Additionally, take some $a$ and $b$ where $f(a)=f(b)$ .We substitute $y=a,y=b$ ,so we get the function bijective.Since function is surjective,we have some $a$ such that $f(a)=0$ .Substitute $x=a$ to get $f(f(y))=y$ . $y=0$ gives $f(f(0))=0$ .Since injective,we get $f(0)=0$ .Then,just substitute $x=f(x)$ and $y=0$ in the original to get $f(x^3)=x^3$ ,so $f(x)=x$ as the final solution.

This post has been edited 1 time. Last edited by Nuran2010, Apr 9, 2025, 1:02 PM

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#5 h Apr 9, 2025, 7:08 PM

Y by

Nuran2010 wrote:

Since $y$ takes all real values, we get $f$ is surjective.Additionally, take some $a$ and $b$ where $f(a)=f(b)$ .We substitute $y=a,y=b$ ,so we get the function bijective.Since function is surjective,we have some $a$ such that $f(a)=0$ .Substitute $x=a$ to get $f(f(y))=y$ . $y=0$ gives $f(f(0))=0$ .Since injective,we get $f(0)=0$ .Then,just substitute $x=f(x)$ and $y=0$ in the original to get $f(x^3)=x^3$ ,so $f(x)=x$ as the final solution.

On substituting $x\mapsto f(x)$ and $y\mapsto0$ I got $f(f(x)^2f(f(x)))=f(f(x))^3$ or $f(xf(x)^2)=x^3$ .

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#6 h Apr 15, 2025, 8:41 AM • 2 Y

Y by TunarHasanzade, TDVOLIMPTEAM

jasperE3 wrote:

Nuran2010 wrote:

Since $y$ takes all real values, we get $f$ is surjective.Additionally, take some $a$ and $b$ where $f(a)=f(b)$ .We substitute $y=a,y=b$ ,so we get the function bijective.Since function is surjective,we have some $a$ such that $f(a)=0$ .Substitute $x=a$ to get $f(f(y))=y$ . $y=0$ gives $f(f(0))=0$ .Since injective,we get $f(0)=0$ .Then,just substitute $x=f(x)$ and $y=0$ in the original to get $f(x^3)=x^3$ ,so $f(x)=x$ as the final solution.

On substituting $x\mapsto f(x)$ and $y\mapsto0$ I got $f(f(x)^2f(f(x)))=f(f(x))^3$ or $f(xf(x)^2)=x^3$ .

Sorry, calculation error! :oops_sign:

:oops_sign:

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