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#1 h Mar 26, 2025, 8:45 AM

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Find all $f:\mathbb{R}\to\mathbb{R}$ and
$f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}$

This post has been edited 1 time. Last edited by Ruji2018252, Mar 26, 2025, 9:30 AM
Reason: Sori

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#2 h Mar 26, 2025, 9:05 AM • 1 Y

Y by pco

$P(0,0)\implies f(0)=0$
$P(0,x)\implies f(f(x))=x$
$P(x,0)\implies f(x^2)=x$ for all x
but it leads contradiction

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Haris1

#3 h Mar 26, 2025, 9:16 AM

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If u can solve this then u can also do $IMO$ $2017$ $P2$ .

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#4 h Mar 26, 2025, 9:30 AM

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Mhremath wrote:

$P(0,0)\implies f(0)=0$
$P(0,x)\implies f(f(x))=x$
$P(x,0)\implies f(x^2)=x$ for all x
but it leads contradiction

Me update new problem bro!

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pco

#6 h Mar 26, 2025, 11:25 AM • 1 Y

Y by MS_asdfgzxcvb

Ruji2018252 wrote:

Find all $f:\mathbb{R}\to\mathbb{R}$ and
$f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}$

Let $P(x,y)$ be the assertion $f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y$
Let $a=f(0)$

$P(0,x)$ $\implies$ $f(f(x))=x+c$ where $c=f(a)^2+a$ and so $f(x)$ is bijective
$P(x,y)$ may be written as new assertion $Q(x,y)$ : $f(x+y)+f(x^2+f(y))=(x+c)^2+f(x)+f(y)+y$

1) $c=0$ and so $f(f(x))=x$
Proof

2) $\boxed{f(x)=x\quad\forall x}$
Proof

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