Suppose that $p(x) = 1/e(ax^{3} + bx^{2} + cx + d)$ where $a,b,c,d,e \in \mathbb{Z}$, $e > 0$. Then suppose that $m,M \in \mathbb{Z}$, $n,N \in \mathbb{N}$, the fractions $\frac{m}{n}$ , $\frac{M}{N}$ are reduced and $p(\frac{m}{n}) = \frac{M}{N}$. But $p(\frac{m}{n}) = \frac{am^{3} + bm^{2}n + cmn^{2} + dn^{3}}{en^{3}}$. Suppose $k = gcd(am^{3} + bm^{2}n + cmn^{2} + dn^{3},en^{3})$. Denote also $x = am^{3}$, $y = bm^{2}n + cmn^{2} + dn^{3}$. Then $k | x+y$ and $k | en^{3}$. Therefore $k | e(x^{3} + y^{3})$ ( because $x+y | x^{3} + y^{3} )$ , but $k | en^{3} | ey^{3}$, so $k | ex^{3} = ea^{3}m^{9}$ , and as before $k | en^{3}$. So, at the end, $k | ea^{3}m^{9} , ea^{3}n^{3}$. But $gcd(m^{9},n^{3}) = 1$, therefore $k | ea^{3}$. So we get that $gcd(am^{3} + bm^{2}n + cmn^{2} + dn^{3},en^{3}) | ea^{3}$, therefore $Mea^{3} \geqslant en^{3}$ , so $n \leqslant a \sqrt[3]{M}$ . Now suppose we have a sequence of rational numbers $q_{1},q_{2}, \ldots$ such that $q_{n} = p(q_{n+1})$. Write $q_{n} = \frac{a_{n}}{b_{n}}$ as a reduced fraction where $a_{n},b_{n} \in \mathbb{Z}$, $b_{n} > 0$. Then, as we saw before, $b_{n+1} \leqslant a \sqrt[3]{b_{n}}$. Therefore the sequence $\{ b_{n} \}$ is bounded. Also note that the sequence $\{ q_{n} \}$ is bounded. Indeed, there exists some $C > 0$ such that for every $x \in \mathbb{R}$, $|x| > C$ we have $|p(x)| > 2|x|$. Therefore it is impossible that $|q_{n}| > C$ for infinitely many $n$'s, because $q_{1}$ is bounded. So $\{ b_{n} \}$ is bounded and $\{ q_{n} \}$ is bounded, therefore $\{ a_{n} \}$ is bounded. Therefore the sequence $q_{n}$ takes only a finite number of values. Therefore there exists a rational number $q$, which appears in the sequence $\{ q_{n} \}$ infinitely many times. Because the number $q$ appears at least two times, there is $k \in \mathbb{N}$, such that $\underbrace{p \circ p \circ \ldots \circ p}_{k times} (q) = q$. Therefore $k$ is a period for $q_{n}$, because there exists $N$ as large as we want, such that $q_{N} = q$, so because of $q_{n-1} = p(q_{n})$, we get that $q_{n-k} = q_{n}$ for every $k+1 \leqslant n \leqslant N$. So because we can take $N$ to be as large as we want, we get that $q_{n-k} = q_{n}$ for every $k+1 \leqslant n$. Therefore $q_{n}$ is periodic.