Are these the only real fixed points of the polynomials $(f_n)_{n\ge 1}$ given by $f_1=X^2-1$ and $f_{n+1}=f_n^2-1$? Define a sequence of polynomials with integer coefficients $(f_n)_{n\ge 1}$ by $f_1 = X^2-1$ and
$$f_{n+1}=f_n^2-1.$$
Now, for any $n$, what are the $x\in\mathbb{R}$ with $f_n(x)=x$?
There are some obvious ones. Both roots of $X^2-X-1$ are fixed points of all $f_n$, and when $n$ is even, $0$ and $-1$ are as well. Computations with some (very) small values of $n$ suggest these are the only ones; is this correct? What if in the recursion we take a negative real number other than $-1$?
Edit: I believe that if we have a 'cycle' of $k$ distinct values $\alpha_1,\ldots,\alpha_k$ with $\alpha_{i+1}=\alpha_i^2-1$ and $\alpha_k^2-1=\alpha_1$, then the polynomial $g:=\prod_{i=1}^k(X-\alpha_i)$ satisfies $g(X^2-1)=(-1)^kg(X)g(-X)$. In particular, $g(X)\mid g(X^2-1)$, which might be useful.
 A: I'm going to define $f(x)=x^2-1$ so that your $f_n(x)$ are $f^{\circ n}(x) = f\left (f^{\circ(n-1)}(x)\right)$, the $n$-fold composition of $f$.
Now, for $0\leq x\lt \phi$ we have $f(x)\lt x$ and for $x\gt\phi$ we have $f(x)\gt x$, so by induction (also using the fact that $-1\lt f(x)\lt 0$ for $-1\lt x\lt 0$) the only positive fixed point of $f^{\circ n}$ is $\phi$. Since $f(x)\geq -1$ for all $x$, any remaining fixed points must be in $[-1, 0]$.
By the composition rule for derivatives, we have $\dfrac{d}{dx}f^{\circ n}(x)$ $=\dfrac{d}{dx}f\left(f^{\circ (n-1)}(x)\right)$ $=f'\left(f^{\circ(n-1)}(x)\right)\dfrac{d}{dx}f^{\circ(n-1)}(x)$; induction then gives that this is $\prod_{i=0}^{n-1}f'\left(f^{\circ i}(x)\right)$. Since $f'(x)=2x$, this is just $2^n\prod_{i=0}^{n-1}f^{\circ i}(x)$ . Now, if $n$ is odd then this product must be less than zero (each term in it is negative), so $\dfrac{d}{dx}\left(f^{\circ n}(x)-x\right)\lt 0$; this, combined with the values of $f^{\circ n}(-1)=0$ and $f^{\circ n}(0)=-1$ establishes that there's exactly one fixed point in the interval.
Finally, the case of $n$ even can be settled similarly to the case $x\gt 0$; for $-1\lt x\lt -\phi^{-1}$ we have $f^{\circ 2}(x)\lt x$ and for $-\phi^{-1}\lt x\lt 0$, $f^{\circ 2}(x)\gt x$, so induction yields $f^{\circ 2i}(x)=x$ only for $x=-\phi^{-1}$ in this range.
It's also pretty straightforward, using the above, to characterize the fixed points; both $-1$ and $0$ are attracting fixed points of $f^{\circ 2}()$ whereas $\phi$ and $-\phi^{-1}$ are repulsing fixed points of $f()$. In other words, for $-1\leq x\lt\phi$, iterating $f()$ converges to the cycle $\langle -1, 0\rangle$; for $x\gt\phi$, iterating $f()$ diverges. (The basin of attraction is somewhat larger than this, of course; in fact, the iterations converge to the cycle for exactly $x\in (-\phi, \phi)$. )

The case for general $c$ in $f_c(x)=x^2-c$ gets substantially more complicated, specifically as $c$ goes from $1$ to $2$. In fact, we can see that $c=1$ is a somewhat unusual point by studying convergence around the fixed cycle $\langle 0,1\rangle$. Consider a point $x=\epsilon$ for some small $\epsilon$; then after one iteration this goes to $\epsilon^2-1$ and after two this goes to $(\epsilon^2-1)^2-1=\epsilon^4-2\epsilon^2$. Since $\epsilon$ is small we can throw out the $\epsilon^4$ term and say that $f_1^{\circ 2}()$ maps $\epsilon$ to $-2\epsilon^2+o(\epsilon^2)$. But this is what's known as superlinear convergence; 'generically' near a fixed point, we would expect to have some constant $M$ such that $f_c^{\circ 2}()$ affects points around a fixed point $x_f$ by sending $x_f+\epsilon$ to $x_f+M\epsilon+o(\epsilon)$; the fixed point is then attracting if $|M|\lt 1$ and repelling if $|M|\gt 1$. The fact that our $M$ here is zero is a sign that we're not at a generic value of $c$. In fact, this $c$ is the exact center of the interval $(3/4, 5/4)$ where the dynamics of $f_c$ feature a period-2 cycle. For $c$ slightly larger than $5/4$, the cycle is instead period 4; e.g., if we look at $c=1.3$ then the iteration converges to a length-4 cycle roughly at $\langle -1.149, 0.019, -1.3, 0.389\rangle$.
These dynamics can be mapped to various features of the Mandelbrot set; the region $c\in(3/4, 5/4)$ where the iteration converges to a $2$-cycle corresponds exactly to the first circular bulb to the left of the main cardioid. They've also been studied even more extensively in another guise: by taking $x_n=ay_n+b$ we can write $ay_{n+1}+b$ $= x_{n+1}=x_n^2+c$ $=a^2y_n^2+2aby_n+b^2+c$, or on subtracting $b$ from both sides and dividing by $a$, $y_{n+1} = ay_n^2+2by_n+(b^2-b+c)/a$. Now, taking $b$ such that $b^2=b-c$ lets us eliminate the constant term and write $y_{n+1}=ay_n^2+2by_n$; further taking $a=-2b$ this can be written as $y_{n+1}=ay_n^2-ay_n$ $=ay_n(1-y_n)$. This reparametrization lets us look at the interval $y\in [0,1]$ specifically and iterations within this interval (as long as $a\leq 4$; for larger $a$ the map no longer carries the interval $[0,1]$ to a subset of itself and all bets are off). In this form, it's known as the Logistic Map and it's been very widely studied, including the famous 'bifurcation diagram' you may have seen that shows how the attracting fixed points/fixed cycles change as $a$ increases towards $4$ (or in your formulation of iterating $f_c=x^2-c$, as $c$ increases towards 2):

