Problem 1899

1) We will prove by induction on $n$ that the solutions of $g_n(x)$ are all distinct. With $n=1$, we have $g_1(x) = f(x)$ which has two solutions, $x=0, x=2$. Suppose the polynomial $g_n(x)$ has all $2^n$ solutions that are distinct, set as $x_1, x_2, \dots, x_{2^n}$, we write $g_n(x) = k(x - x_1)(x - x_2) \dots (x - x_{2^n})$. Hence,
$$g_{n+1}(x) = g_n(f(x)) = k(f(x) - x_1)(f(x) - x_2) \dots (f(x) - x_{2^n})$$
so with $i \neq j$, obviously the solution set of $f(x) - x_i$ and $f(x) - x_j$ will be disjoint. Also, if the equation $f(x) - x_i = 0$ has a double root, then that solution must be $1$. Next, we have
$$g_1(1) = f(1) = -5, \quad g_2(1) = f(-5) = 5(-5)(-7) = 175$$
so $g_n(1) > 0, \forall n \ge 2$. Hence, $x = 1$ cannot be a solution of $g_n(x), \forall n$. This shows that $g_{n+1}(x)$ also has a distinct solution $2^{n+1}$. The induction is completed.
Finally, from the above analysis, the solutions of $g_n(x)$ can be divided into $2^{n-1}$ disjoint pairs having sum equal $2$ so the sum of all the solutions will be $2^n$.

2) Note that for some parameter $d \in \mathbb{R}$, the equation
$$cx(x - 2) = d \iff x^2 - 2x = \frac{d}{c}$$
has two distinct solutions if and only if $\Delta' = 1 + \frac{d}{c} > 0$ or $c > -d$. (\heartsuit) It will then have the solutions
$$x = 1 + \sqrt{1 + \frac{d}{c}} \text{ and } x = 1 - \sqrt{1 + \frac{d}{c}}.$$
In order for $g_n(x)$ to have $2^n$, $g_{n-1}(x)$ must have enough $2^{n-1}$ distinct roots. Denote $r$ as one of those roots, we investigate the equation $f(x) = r$.
* If $r < 0$ then the equation has two positive solutions.
* If $r > 0$ then the equation has two solutions with opposite signs.
We see that (\heartsuit) generates a constraint for $c$ only when the parameter $d < 0$. Therefore, we are interested in pairs of opposite solutions generated from $r > 0$. Suppose that pair of solutions is $(r_1, r_2)$ with $r_1 < 0 < r_2$. We need $c > -r_1 = -(2 - r_2) = r_2 - 2$ so we take it back to the survey positive solution $r_2$ in that pair. Starting from an initial solution $r = 2$, we can build a sequence of numbers as follows: $u_1 = 2$ and $u_{n+1} = 1 + \sqrt{1 + \frac{u_n}{c}}$ with $n \ge 1$.
It is easy to see that $u_n$ is the solution of $g_n(x)$ and $u_2 > u_1$ so by induction, we can prove that the sequence $(u_n)$ increases strictly. Let $L$ be the solution of
$$L = 1 + \sqrt{1 + \frac{L}{c}} \text{ or } L = 2 + \frac{1}{c}.$$
By induction, we can also prove that $u_n < L, \forall n$, so it is clear that the sequence will converge to $L$. Note that we must always have $c > u_n - 2, \forall n$ so let $n \to +\infty$, we get $c \ge 2 + \frac{1}{c} - 2 = \frac{1}{c}$ or $c \ge 1$. $\square$