$F$ is a polynomial, $\deg F = 3$, and $(x^2 - 1)(x^2 - 2) | F(F(x)) - x$. Prove that $F$ exists $F$ is a polynomial, $\deg(F) = 3$, and $(x^2 - 1)(x^2 - 2)  | F(F(x)) - x$. Prove that:
a) $F$ exists 
b) There are at least 10 such polynomials
What I've tried to do:
$(x^2 - 1)(x^2 - 2) \mid F(F(x)) - x\implies \begin{cases}F(F(1)) = 1 \\ F(F(-1)) = -1 \\ F\left(F\left(\sqrt{2}\right)\right) = \sqrt{2}\\ F\left(F\left(-\sqrt{2}\right)\right) = -\sqrt{2}\end{cases}  $
Then I tried to apply interpolation somehow but it didn't help. 
Thanks in advance!

This is a partial answer.
Let $K(x)$ be the factor $\frac{F(F(x))-x}{(x^2-1)(x^2-2)}$.
One can simplify the search significantly by restricting our search of
$F(x)$ to be an odd polynomial in $x$, i.e.
$$F(x) = (ax^2 - b)x,\quad\text{ with }\; a \ne 0$$
There are already 4 pairs of real solutions given by
$$(a,b) = \begin{cases}
\pm (2, 3),\\
\pm (\frac{\sqrt{5}-1}{2}, \sqrt{5} ),\\
\pm (\frac{\sqrt{5}+1}{2}, \sqrt{5} ),\\
\pm (\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}} )
\end{cases}
\quad\longrightarrow\quad
\frac{K(x)}{x}
 = \begin{cases}
16 x^4- 24x^2+4\\
\\\frac12\left[ (7-3\sqrt{5})x^4 - (9-3\sqrt{5})x^2 + 4\right]\\
\\\frac12\left[ (7+3\sqrt{5})x^4 - (9+3\sqrt{5})x^2 + 4\right]\\
\\\frac14\left[ x^4 - 6x^2 + 7\right]
\end{cases}
$$
If one allow complex coefficients for $F(x)$, there are 4 more complex solutions
and we are done.
$$(a,b) = \pm (\frac{1+i}{2}, \frac{3+i}{2}) \quad\text{ and }\quad \pm (\frac{1-i}{2}, \frac{3-i}{2})$$
If $F(x)$ need to be real, then I don't have any good idea how to get the remaining two real solutions easily.
