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$f:\mathbb{R}\to\mathbb{R},$ such that $f(|f(x)-f(y)|)=f(f(x))-2x^2f(y)+f(y^2)\,\,\forall \, x,y\in\mathbb{R}.$ Find all such $f(x).$

My Working:

Let $P(x,y): f(|f(x)-f(y)|)=f(f(x))-2x^2f(y)+f(y^2)\,\,\forall \, x,y\in\mathbb{R}$

$P(0,0):f(f(0))=0;$ Let $f(0)=a\implies f(a)=0$

$P(a,a): f(|f(a)-f(a)|)=f(f(a))-2a^2f(a)+f(a^2)\implies f(a^2)=0$

$P(0,x):f(|f(0)-f(x)|)=f(x^2)$ ---(1)

$P(x,0): f(|f(x)-f(0)|)=f(f(x))-2ax^2+a$ ---(2)

$P(x, x):a=f(f(x))-2x^2f(x)+f(x^2)$ ---(3)

From (1) and (2), we get $f(f(x))=f(x^2)+2ax^2-a$ ---(4)

$P(x,a): f(|f(x)|)=f(f(x))\text{ and }P(a,x): f(|f(x)|)=a-2a^2f(x)+f(x^2)$

$\implies f(f(x)=a-2a^2f(x)+f(x^2)$ ---(5)

From (4) and (5) we get, $2a^2f(x)=2a(1-x^2)$

$\implies a=0\text{ or } f(x)=\frac1a(1-x^2)$

Considering, $f(x)=\frac1a(1-x^2)\text{ and plugging it in parent we get two solutions, } f(x)=1-x^2,\,\, f(x)=x^2-1$

Now consider $a=0,\text{ i.e. } f(0)=0$

$\implies f(f(x))=f(x^2)=x^2f(x)$

I guessed $f(x)=0,\pm\, x^2$

Makar
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    If $a=0$, then (5) gives $f(x^2)=f(f(x))$. Plug that to $P(x,y)$ and consider left side symmetric in $x,y$, hence by right side you must have $-2x^2f(y)=-2y^2f(x)$. Then it should be easy to finish as $f(x)/x^2$ must be constant on $\mathbb{R} \setminus {0}$. – Sil Nov 05 '22 at 05:44
  • thanks a lot Now I got it – Makar Nov 05 '22 at 19:54
  • Once you make $a=0$, we have that the quotient is constant, and then you show that the constant must be either $1$ or $-1$ by plugging in the function into the condition you have on f. – Lorena Nov 06 '22 at 17:21
  • Adjusting my comment a bit, from $-2x^2f(y)=-2y^2f(x)$ we can avoid any division and $\mathbb{R} \setminus {0}$, simply put $y=1$ to get for example $x^2f(1)=f(x)$. – Sil Nov 21 '22 at 17:46

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