IMC 2004 · Day 1 · P2

Put $P_n(x) = \underbrace{P(P(\dots(P}_{n}(x))\dots))$. As $P_1(x) \ge -1$, for each $x \in \mathbb{R}$, it must be that $P_{n+1}(x) = P_1(P_n(x)) \ge -1$, for each $n \in \mathbb{N}$ and each $x \in \mathbb{R}$. Therefore the equation $P_n(x) = a$, where $a < -1$ has no real solutions.

Let us prove that the equation $P_n(x) = a$, where $a > 0$, has exactly two distinct real solutions. To this end we use mathematical induction by $n$. If $n = 1$ the assertion follows directly. Assuming that the assertion holds for a $n \in \mathbb{N}$ we prove that it must also hold for $n+1$. Since $P_{n+1}(x) = a$ is equivalent to $P_1(P_n(x)) = a$, we conclude that $P_n(x) = \sqrt{a+1}$ or $P_n(x) = -\sqrt{a+1}$. The equation $P_n(x) = \sqrt{a+1}$, as $\sqrt{a+1} > 1$, has exactly two distinct real solutions by the inductive hypothesis, while the equation $P_n(x) = -\sqrt{a+1}$ has no real solutions (because $-\sqrt{a+1} < -1$). Hence the equation $P_{n+1}(x) = a$, has exactly two distinct real solutions.

Let us prove now that the equation $P_n(x) = 0$ has exactly $n+1$ distinct real solutions. Again we use mathematical induction. If $n = 1$ the solutions are $x = \pm 1$, and if $n = 2$ the solutions are $x = 0$ and $x = \pm\sqrt{2}$, so in both cases the number of solutions is equal to $n+1$. Suppose that the assertion holds for some $n \in \mathbb{N}$. Note that $P_{n+2}(x) = P_2(P_n(x)) = P_n^2(x)(P_n^2(x) - 2)$, so the set of all real solutions of the equation $P_{n+2} = 0$ is exactly the union of the sets of all real solutions of the equations $P_n(x) = 0$, $P_n(x) = \sqrt{2}$ and $P_n(x) = -\sqrt{2}$. By the inductive hypothesis the equation $P_n(x) = 0$ has exactly $n+1$ distinct real solutions, while the equations $P_n(x) = \sqrt{2}$ and $P_n(x) = -\sqrt{2}$ have two and no distinct real solutions, respectively. Hence, the sets above being pairwise disjoint, the equation $P_{n+2}(x) = 0$ has exactly $n+3$ distinct real solutions. Thus we have proved that, for each $n \in \mathbb{N}$, the equation $P_n(x) = 0$ has exactly $n+1$ distinct real solutions, so the answer to the question posed in this problem is 2005.