$3.77 p, q$ are two distinct prime numbers, and the positive integer $n \geqslant 3$. 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.
(China National Team Selection Examination, 1994)
【Sol】Let $f(x)=x^{n}+a x^{n-1}+p q=g(x) h(x)$.
We can assume that the leading coefficients of $g(x)$ and $h(x)$ are +1, otherwise we can consider $(-\dot{g}(x)) \cdot (-h(x))$. That is, we assume
$$
\begin{array}{c}
g(x)=x^{l}+a_{l-1} x^{l-1}+\cdots+a_{1} x+a_{0}, \\
h(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
a_{0} b_{0}=p q .
\end{array}
$$

Clearly,
$$
a_{0} b_{0}=p q \text {. }
$$

By $p, q$ being different primes and equation (4), we can assume $p \nmid a_{0}$, then $p \mid b_{0}$, let $p \mid b_{k}(\beta=0,1, \cdots, r-1), p \nmid b_{r}(r \leqslant m)$,
Let the coefficient of $x^{r}$ in $f(x)$ be $c_{r}$, then by (1), (2) and (3) we have
$$
c_{r}=a_{0} b_{r}+a_{1} b_{r-1}+\cdots+a_{r} b_{0} .
$$

By $p \nmid a_{0}$ we know $p \nmid c_{r}$, and by the given conditions
$$
c_{1}=c_{2}=\cdots=c_{n-2}=0 \text {, }
$$

they are all multiples of $p$.
Since $r \leqslant m \leqslant n-1$, it can only be that $r=n-1$, i.e.,
$$
m=n-1 \text {. }
$$
$h(x)$ is an $n-1$ degree polynomial, $g(x)$ is a linear polynomial. $g(x)=x+a_{0}$. Thus $f\left(-a_{0}\right)=0$, i.e., $0=\left(-a_{0}\right)^{n}+a\left(-a_{0}\right)^{n-1}+p q$.
Then $\left(-a_{0}\right)^{n-1}\left(a-a_{0}\right)=-p q$.
By $n-1 \geqslant 2$ and $p, q$ being different primes, we know from the above equation that $a_{0}= \pm 1$.
At this point, by equation (5) we have $a=1+(-1)^{n} p q$ or $a=-1-p q$.