<s>[INST] A quadratic polynomial $f(x)$ satisfy $ f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ for all real $x$. then $f(x)$ is A quadratic polynomial $f(x)$ satisfy $\displaystyle f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ for all real $x$. then $f(x)$ is 
Let $f(x) = ax^2+bx+c$, then substitute $x=0$ in functional equation
$\displaystyle f(0) = \left(\frac{f(1)-f(1)}{2}\right) = 0$ So $c=0$
now $\displaystyle f'(x) = 2\left[\frac{f(x+1)-f((x-1)^2)}{2}\right]\cdot \left(\frac{f'(x+1)-f((x-1)^2)2(x-1)}{2}\right)$, 
then substitute $x=0$
So we have $b=0$
could some help me to find value of $a$, thanks 
 [/INST] With $f(x)=ax^2$ and $f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ we get
$ax^2=\left[\frac{a(x+1)^2-a(x-1)^4}{2}\right]^2$.
Now take $x=1$ and look what happens....
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