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#1 h Apr 18, 2025, 3:38 PM

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Find all functions $f:\mathbb{R}->\mathbb{R}$ such that,
$\begin{align} f(x)^2+f(y)=f(x^2-y)+2yf(x), \end{align}$ for all $x,y\in\mathbb{R}$ .

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#2 h Apr 18, 2025, 4:42 PM • 1 Y

Y by AlexCenteno2007

TUAN2k8 wrote:

Find all functions $f:\mathbb{R}->\mathbb{R}$ such that,
$\begin{align} f(x)^2+f(y)=f(x^2-y)+2yf(x), \end{align}$ for all $x,y\in\mathbb{R}$ .

Let $P(x,y)$ be the assertion $f(x)^2+f(y)=f\left(x^2-y\right)+2yf(x)$ .
$P\left(x,\frac12x^2\right)\Rightarrow f(x)^2=x^2f(x)\Rightarrow f(x)\in\left\{0,x^2\right\}$

Suppose there are some $a,b\ne0$ with $f(a)=0$ and $f(b)=b^2$ .
$P(a,b)\Rightarrow f\left(a^2-b\right)=b^2$
If $f\left(a^2-b\right)=0$ then $b^2=0$ , contradiction, so $f\left(a^2-b\right)=\left(a^2-b\right)^2$ and this simplifies to $a^2=2b$ .
$P(b,a)\Rightarrow f\left(b^2-a\right)=b^4-2ab^2$
If $f\left(b^2-a\right)=\left(b^2-a\right)^2$ then $a^2=0$ , contradiction, so $f\left(b^2-a\right)=0$ and this simplifies to $b^2=2a$ .
From these two equations we have (since $a\ne b$ ) $a+b=2$ . Now:
$P(0,0)\Rightarrow f(0)=0$
$P(a,0)\Rightarrow f\left(a^2\right)=0$ and repeating this process we must have $a^2+b=2$ , so $a\in\{0,1\}$
$P(b,0)\Rightarrow f\left(b^2\right)=b^4$ and repeating this process we must have $a+b^2=2$ , so $b\in\{0,1\}$
It's impossible to have $a\in\{0,1\}$ and $b\in\{0,1\}$ with $a+b=2$ .

Therefore either $\boxed{f(x)=0}$ for all $x$ or $\boxed{f(x)=x^2}$ for all $x$ , both satisfy the equation.

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