What is my special quadratic? Start with $f(x)=x^2+bx+c$.
Then, attempt to solve for $x$ in $f(x)=x$.
It is easily found that $x=\frac{1-b\pm\sqrt{(b-1)^2-4c}}{2}$.
Then, start again with $f(x)=x$ and apply the function $f$ to both sides:
$$f(f(x))=f(x)$$
We see that the right side is equivalent to the original equation's left side, hence we have
$$f(f(x))=f(x)=x$$
Which we will remove the middle to get:
$$f(f(x))=x$$
Now, solving this problem is not equivalent to the original problem because we have now 'picked up' new solutions when we applied the function $f$.
However, we have not lost solutions.
Also, this, with manipulation, becomes
$$f(f(x))-x=0$$
Given that there are two solutions found through
$$f(x)-x=0$$
Because we have not lost any of the original solutions.
Since $f(x)-x$ contains solutions to the problem $f(f(x))-x$, is a polynomial, and does not contain any other solutions than the ones for $f(f(x))-x$, it is safe to assume that $f(x)-x$ is a factor of $f(f(x))-x$, which allows us to find the other two solutions that we 'picked up.'
And since it is a factor, we may divide to find the other factors:
$$\frac{f(f(x))-x}{f(x)-x}=P_1$$
Now, $P_1$ is a special polynomial that I have found to be:
$$P_1=x^2+(b+1)x+c+b+1$$
This is quite indeed an interesting quadratic in my opinion, and helpful as well, for it is the second quadratic factor of $f(f(x))-x$.
Furthermore, $f(f(f(x)))-x$ has factors $f(f(x))-x$ and $f(x)-x$ by a sort of recursive definition.
And if we proceed to find the unknown new roots we picked up, we will use division.
$$\frac{f(f(f(x)))-x}{[f(f(x))-x]\cdot[f(x)-x]}=P_2$$
Interestingly, $P_2$ is also a quadratic.
And if we proceed in this manner, we will find $P_n$ to be defined as:
$$P_n=\frac{f^n(x)-x}{\Pi_{i=1}^{n-1}\left(f^i(x)-x\right)}$$
Where we have $f^n(x)=f(f(f(\dots f(x)\dots)))$ where we have $n$ iterations of $f$.
Is there anything you can find that is special about $P_n$?  Is there any easier method by which I can find what $P_n$ is?
I note that the definition is somewhat recursive because we have:
$$\frac{f(f(f(x)))-x}{[f(f(x))-x]\cdot[f(x)-x]}=\frac{f(f(f(x)))-x}{[(f(x)-x)(P_1)]\cdot[f(x)-x]}=\frac{f(f(f(x)))-x}{P_1[f(x)-x]^2}=P_2$$
A similar definition can be made for $P_n$, but it is very difficult for me to find it.
So three questions:

Is there anything you can find that is special about $P_n$?  Is there any easier method by which I can find what $P_n$ is?  Can you simplify the recursive formula for $P_n$?

 A: Considering a quadratic map: $z_{n+1}=z_{n}^{2}+c$ with $z_{0}=c$.  For $c\in \mathbb{C}$ such that $z_{n+p}=z_{n}$, we say $c$ is a critical point with period $p$.  In particular $n=0$, $c$ is attractive.  Otherwise, it's called Misiurewicz point.  Of course $\{c:z_{n+p}=z_{n}\} \subset \{c:z_{nk+pm}=z_{nk}\}$.  For example,
$z_{1}=z_{0} \implies c^{2}+c=c \implies c^{2}=0$
$z_{2}=z_{1} \implies (c^{2}+c)^{2}+c=c^{2}+c
             \implies c^{3}(c+2)=0$
$z_{2}=z_{0} \implies (c^{2}+c)^{2}+c=c \implies c^{2}(c+1)^{2}=0$
$z_{3}=z_{2} \implies ((c^{2}+c)^{2}+c)^{2}+c=(c^{2}+c)^{2}+c
             \implies c^{4}(c+2)(c^{3}+2c^{2}+2c+2)=0$
$z_{3}=z_{1} \implies ((c^{2}+c)^{2}+c)^{2}+c=c^{2}+c
             \implies c^{3}(c+1)^{2}(c+2)(c^{2}+1)=0$
$z_{3}=z_{0} \implies ((c^{2}+c)^{2}+c)^{2}+c=c
             \implies c^{2}(c^{3}+2c^{2}+c+1)=0$
In fact, $\displaystyle M=\{c\in \mathbb{C}: \sup_{n\in \mathbb{N}} z_{n} < \infty \}$ is the famous Mandelbrot Set which is a fractal.

For the period doubling properties, see https://en.wikipedia.org/wiki/Feigenbaum_constants
A: Let $f(x) = x^2$, then $f^n(x) = x^{2^n}$.
$$f^n(x) - x = x^{2^n}-x = x\left(x^{2^n-1} - 1\right)$$
Thus the roots of $f^n(x) - x$ consist of $0$ and the $(2^n - 1)$th roots of unity. In particular, non-zero roots of


*

*$f^1(x) - x$ are just $1$.

*$f^2(x) - x$ are the $3$rd roots of unity $1, e^{i2\pi/3}, e^{i4\pi/3}$

*$f^3(x) - x$ are the $7$th roots of unity $1, e^{i2\pi/7}, e^{i4\pi/7}, e^{i6\pi/7}, e^{i8\pi/7}, e^{i10\pi/7}, e^{i12\pi/7}$.


By your claim 

Furthermore, $f(f(f(x)))−x$ has factors $f(f(x))−x$ and $f(x)−x$ by a sort
  of recursive definition.

the roots of $f^2(x) - x$ should be included among the roots of $f^3(x) - x$, but this is evidently not the case. 
For general $f$, letting $x_0$ be a root of $f^n(x) - x$, we see that $$f^{n+1}(x_0) = f(f^n(x_0)) = f(x_0).$$ Thus $x_0$ is only a root of $f^{n+1}(x) - x$ if it is also a root of $f(x) - x$.
I think there is still an interesting recursion here, but it is not the one shown. Instead, ask what happens if $P_1$ takes the place of $f$. That is:
$Q_0(x)$ is any quadratic, and for $n \in \Bbb N$,
$$Q_{n+1}(x) = \frac{Q_n(Q_n(x)) - x}{Q_n(x) - x}$$
Addendum since it turns out  $Q_{n+1}(x) = Q_n(x) + x + b + 1$ for all $n$, this version isn't all that interesting after all. Oh well.
A: Trivial solution of $f(f(x)) \equiv x$ is $f(x) \equiv \pm x$ and there's no other polynomial solution otherwise the power will be raised.  However, there's bilinear solution: $\displaystyle f(x) \equiv \frac{ax+b}{cx \pm a}$ where $bc \neq -a^{2}$.  In particular $a=0$ gives $\displaystyle f(x) \equiv \frac{b}{cx}$ where $b, c, x\neq 0$.
