9. While playing, Marcella finds by pure chance two polynomials $p(x)$ and $q(x)$, non-constant and with integer coefficients, satisfying the relation:

$$
p(q(x+1))=p\left(x^{3}\right) q(x+1)^{5} .
$$

What can we assert with certainty about the two polynomials found by Marcella?
(A) The leading coefficient of $p(x) q(x)$ is positive
(B) The polynomial $q(x)$ does not have integer roots $\begin{array}{llll}\text { (C) The degree of } q(x) \text { does not exceed that of } p(x) & \text { (D) The sum of the coefficients of } p(x)\end{array}$ is odd $\quad$ (E) The degree of $p(x) q(x)$ is a multiple of 8
9. The answer is (E). Let $m, n$ be the degrees of $p(x)$ and $q(x)$, respectively. They are positive integers. The relation between the polynomials given in the problem implies the following equation on the degrees, $m n = 3 m + 5 n$, which in turn is equivalent to $(m-5)(n-3)=15$. To find $(m, n)$, we must solve the systems

$$
\left\{\begin{array}{l}
m-5=k \\
n-3=\frac{15}{k}
\end{array}\right.
$$

as $k$ varies among the divisors of 15, that is, $k = \pm 1, \pm 3, \pm 5, \pm 15$. However, when $k$ is negative, at least one of $m$ and $n$ turns out to be negative or zero. Therefore, we discard these cases and consider only the remaining four systems, from which we obtain the following solutions:

| $m$ | 6 | 8 | 10 | 20 |
| :---: | :---: | :---: | :---: | :---: |
| $n$ | 18 | 8 | 6 | 4 |

Now we can observe that for each pair $(m, n)$, there actually exist two polynomials $p(x) = x^m, q(x) = (x-1)^n$ of degrees $m, n$ that satisfy the given relation. With these examples, we can already rule out answers (B) and (C). Regarding (A) and (D), it suffices to note that if $(p(x), q(x))$ satisfies Marcella's condition, the same holds for the polynomials $(\lambda p(x), q(x))$ for any integer $\lambda \neq 0$: hence $\left(-x^m, (x-1)^n\right)$ and $\left(2 x^m, (x-1)^n\right)$ are two easy counterexamples. Finally, the degree of $p(x) q(x)$ is $m+n$ and can only be 16 or 24, which confirms the correctness of (E).