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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

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    Leave $a$ unknown and substitute. Since you already know $b$ and $c$, how bad could it be? – 2017-02-09
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    Actually, you must have a typo or something. There's no way that identity can be satisfied for a quadratic polynomial $f(x)$. The degrees of LHS and RHS can't be made equal. More precisely, the way you currently have it expressed, the LHS expands to a polynomial of degree $2$, while the RHS expands to a polynomial of degree $8$. – 2017-02-09

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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....