Find $f''(x)$ if $f\circ f'(x) = 4x^2 + 3$ Can you tell me the solution of this question?

If: $f\circ f'(x)=4 x^2 +3$
then what is $f''(x)$?

This was a question in math test which I just took yesterday.
One function satisfying the equation above is $f(x)=x^2+3$, for which $f'(x)=2x$ and therefore $f''(x)=2$.
We can also see that $f'(x)$ is monotonic in $[0,+\infty)$, and in $(-\infty,0]$.
What other analytic solutions $f(x)$ exist ?
Can we express all analytic solutions $f(x)$ with a few parameters ?
 A: I think I have an (incomplete and not rigorous) heuristic which suggests that $f(x)=x^2+3$ is the only answer. Perhaps somebody can turn this into a proper/rigorous answer.
It starts with the assumption that $f$ is analytic:
$$f(x)=\sum_{i=0}^\infty a_ix^i,$$
so that
$$f'(x)=\sum_{i=0}^\infty ia_ix^{i-1}.$$
If we write an expression for $f(f'(x))$ we get something like
$$\begin{align}
f(f'(x))&=a_0+a_1(a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots)
\\&\phantom{=}+a_2(a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots)^2
\\&\phantom{=}+a_3(a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots)^3+\cdots
\end{align}$$
If we can argue at this point (this is the big hole!) that all of the $a_n=0$ for $n\geq 3$ we are basically there. If this is true we have that $f(x)=a_0+a_1x+a_2x^2$. Just by plugging in we see that $a_2\neq 0$ and if $a_2\neq 0$ then $a_1=0$. 
We plug $f(x)=a_0+a_2x^2$ into the relation and we are left with $a_0=3$ and $a_2^3=1\Rightarrow a_2=1$.
A: Note that the function defined by 


*

*$f(x)=3+x^2$ if $x\le 0$

*$f(x)=3-x^2$ if $x\ge 0$


is also continuous, and $f(f'(x))=3+4x^2$ also. But then 


*

*$f''(x)=2$ if $x<0$

*$f''(x)=-2$ if $x>0$

*$f''$ is not defined for $x=0$

A: Together with the 2 answers already given here , for completeness I add this relevant link
Can $f(g(x))$ be a polynomial?
That should make everything clear.
