Number of real polynomials $p(x)$ ,$ p(x)>x^2$ 
Find the number of real polynomials satisfying $p(0)=0,p(x)>x^2$ for all $x$ not zero and
$p''(0)=\frac{1}{2}$

I started out with taking $$f(x)=p(x)-x^2$$ :thus $f(x)>0$ for $x$ not $0$.
Also $$f''(0)=\frac{-3}{2}$$.
I tried graphing and it makes sense that there are $0$ such polynomials.But how do i prove it rigorously ,probably a continuation of my method.
Source:KVPY 2018
 A: There are no such polynomials. To see this, we use the fact that the second derivative can be written as a limit  $$\boxed{f''(0) = \lim_{h \to 0} \frac{f(h) + f(-h) -2f(0)}{h^2}}$$ Now, if this equals $\frac {-3}2$, then as $f(0) =0$ we get :
$$
\frac{-3}{2} = \lim_{h \to 0} \frac{f(h) + f(-h)}{h^2}
$$
however if $f$ is positive everywhere except at $0$ , then this RHS limit must at least be non-negative, which is not the case. Of course, this shows analogously that if $p(x) > x^2$ everywhere except at $0$ with $p(0)=0$ then $p''(0) \geq 2$.
A: Here is another way.
Note that $f(x) = ax - {3 \over 2} x^2+ x^3 q(x)$ for some other polynomial $q$. We have $f(x) \ge 0$ for all $x$, so taking limits we see that $f'(0) = a = 0$.
We can write $f(x) = x^2 (- {3 \over 2} + x q(x))$.
Since $f(x) \ge 0$ we must have $- {3 \over 2} + x q(x) \ge 0$ for all $x \neq 0$ (and hence all $x$ of course, but that doesn't matter here).
However, we see that for small $x$ the quantity $- {3 \over 2} + x q(x) < 0$ which
is a contradiction.
