5. $p, q$ are two distinct prime numbers, and the natural number $n \geqslant$ 3. Find all integers $a$, such that the polynomial $f(x)=x^{n}+$ $a x^{n-1}+p q$ can be factored into the product of two integer-coefficient polynomials, each of degree at least one. (Supplied by Huang Qiguo, Fudan University)
Let $f(x)=g(x) h(x)$.
Here
$$
\begin{array}{l}
g(x)=a_{1} x^{t}+\cdots+a_{1} x+a_{0}, \\
h(x)=b_{m} x^{m}+\cdots+b_{1} x+b_{0} .
\end{array}
$$

where $l, m$ are natural numbers, and the coefficients $a_{i}(0 \leqslant i \leqslant l), b_{0}(0 \leqslant \alpha \leqslant m)$ are all integers, and $a_{l} \neq 0, b_{m} \neq 0, l+m=n$. By comparing the coefficients of the highest terms on both sides of (1), we have $a_{t} b_{m}=1$. Thus, $a_{i}=b_{m}= \pm 1$. Without loss of generality, assume $a_{t}=b_{m}=1$. If $a_{t}=b_{m}=-1$, consider $f(x)=(-g(x)) \cdot(-h(x))$. By comparing the constant terms on both sides of (1), we know $a_{0} b_{0}=p q$. Since $p, q$ are two different primes, assume $p$ divides $b_{0}$. Given $b_{m}=1$, let $p \mid b_{\beta}(\beta=0,1, \cdots, m-1)$ and $p \nmid b_{r}(r \leqslant m)$. Using (1), the coefficient of $x^{r}$ in $f(x)$ is $c_{r}=a_{0} b_{r}+a_{1} b_{r-1}+a_{2} b_{r-2}+\cdots+a_{r} b_{0}$. Since $p \nmid a_{0}$, it follows that $p \nmid c_{r}$. But by the problem's condition, $c_{1}=c_{2}=\cdots=c_{n-2}=0$, which are all multiples of $p$, so the only possibility is $r=n-1$. Since $r \leqslant m \leqslant n-1$, it must be that $m=n-1, l=1$. Thus, $f(x)$ has a linear factor $g(x)=x+a_{0}$, and therefore $f\left(-a_{0}\right)=0$. Hence,
$$
\begin{aligned}
0 & =f\left(-a_{0}\right)^{\prime} \\
& =\left(-a_{0}\right)^{n}+a\left(-a_{0}\right)^{n-1}+p q .
\end{aligned}
$$

From $a_{0} b_{0}=p q$ and $p \nmid a_{0}$, we know that $a_{0}$ has only 4 possible values: 1, $-1, q, -q$.
When $a_{0}=-1$, by (2), we have $a=-(1+p q)$.
When $a_{0}=1$, by (2), we have
$$
a=1+(-1)^{n-2} p q.
$$

When $a_{0}= \pm q$, by (2), we have
$$
(\mp q)^{n}+a(\mp q)^{n-1}+p q=0,
$$

which implies $q^{n-2} \mid p$.
This contradicts the condition that $p, q$ are two different primes.