Thailand Mathematical Olympiad

Let $f(x) = x^2 + 2x$ and define $f^n = f \circ f \circ \dots \circ f$ ($n-1$ times). Let $a$ be a root of $p(x)$. From the second property, we see that $a, f(a), f^2(a), \dots$ are also roots of $p(x)$.

We subdivide the range of $a$ into four subintervals.

Case 1: If $a > 0$, then $0 < a < f(a)$. Since $f$ is strictly increasing over the interval $(0, \infty)$, we have $a < f(a) < f^2(a) < \dots$.

Case 2: If $-1 < a < 0$, then $0 > a > f(a) > -1$. Since $f$ is strictly increasing over the interval $(-1, 0)$, we have $a > f(a) > f^2(a) > \dots$.

Case 3: If $-2 < a < -1$, then $-1 < f(a) < 0$. Substituting for $a$ by $f(a)$ in Case 2, we get $f(a) > f^2(a) > \dots$.

Case 4: If $a < -2$, then $f(a) > 0$. Substituting for $a$ by $f(a)$ in Case 1, we get $f(a) < f^2(a) < \dots$.

From the four cases, we infer that if $a \notin \{-2, -1, 0\}$, then $p(x)$ has infinitely many distinct roots, which is impossible. Hence, $a \in \{-2, -1, 0\}$ and by direct checking all the possible $p(x)$ are
$$x,\ x+1,\ x(x+1),\ x(x+2),\ x(x+1)(x+2).$$