Saudi Arabian IMO Booklet

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 $i\ne j$, 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 $$g_1(1)=f(1)=-5,g_2(1)=f(-5)=5(-5)(-7)=175$$ 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 $$cx(x-2)=d⟺x^2-2x=\frac{d}{c}$$ has two distinct solutions if and only if ${\Delta }^{\prime }=1+\frac{d}{c}>0$ 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 $c\ge 1$. $\square$