210 Test all integers $n$ such that the polynomial $p(x)=x^{5}-n x^{2}-n-2$ can be expressed as the product of two non-constant polynomials with integer coefficients.
Solve for all integer values of $n$: $-2, -1, 10, 19, 34, 342$.
i) For $n=-2$, $p(x)=x^{5}+2 x=x\left(x^{4}+2\right)$; for $n=-1$, $p(x)=x^{5}+x-1=\left(x^{2}-x+1\right)\left(x^{3}+x^{2}-1\right)$; for $n=10$, $p(x)=x^{5}-10 x-12=(x-2)\left(x^{4}+2 x^{3}+4 x^{2}+8 x+6\right)$; for $n=19$, $p(x)=x^{5}-19 x-21=\left(x^{2}-x-3\right)\left(x^{3}+x^{2}+4 x+7\right)$; for $n=34$, $p(x)=x^{5}-34 x-36=(x+2)\left(x^{4}-2 x^{3}+4 x^{2}-8 x-18\right)$; for $n=342$, $p(x)=x^{3}-342 x-344=(x+1)\left(x^{4}-4 x^{3}+16 x^{2}-64 x-86\right)$.
ii) Polynomial: Suppose $p(x)=x^{5}-n x-n-2$ can be factored into the product of two non-constant integer polynomials, we will show: $n \in \{-2, -1, 10, 19, 34, 342\}$.
a) If the equation $p(x)=0$ has an integer root $\alpha$, then
$$
\begin{array}{l}
\alpha^{5}-n \alpha-n-2=0 \\
\Leftrightarrow \alpha^{5}-2=n(\alpha+1) \\
\Leftrightarrow\left\{\begin{array}{l}
\alpha \neq-1, \\
n=\frac{\alpha^{5}-2}{\alpha+1}.
\end{array}\right.
\end{array}
$$

Since $n \in \mathbf{Z}$, we know $(\alpha+1) \mid (\alpha-2)$. Also, $(\alpha+1) \mid (\alpha^{5}+1)$. Therefore, $(\alpha+1) \mid \left((\alpha^{5}+1)-(\alpha^{5}-2)\right)$, which means $(\alpha+1) \mid 3$. Thus, $\alpha \in \{-4, -2, 0, 2\}$, so $n \in \{-2, 10, 34, 342\}$.
b) If $p(x)=\left(x^{2}+a x+b\right)\left(x^{3}+c x^{2}+d x+e\right)$, where $a, b, c, d, e \in \mathbf{Z}$, then comparing the coefficients of corresponding terms, we get
$$
\left\{\begin{array}{l}
a+c=0, \\
a c+b+d=0, \\
a d+b c+e=0, \\
b d+a e=-n, \\
b e=-(n+2),
\end{array}\right.
$$

From (1), we get $c=-a$. Substituting into (2) and (5), we get $d=a^{2}-b$,
Combining (3), we get $e=-a d-b c=-a\left(a^{2}-b\right)-b(-a)=2 a b-a^{3}$,
From (5), (1), and (7), we get
$$
\begin{array}{l}
-n=b\left(a^{2}-b\right)+a\left(2 a b-a^{3}\right), \\
-(n+2)=b\left(2 a b-a^{3}\right),
\end{array}
$$
(8) minus (9) gives
$$
\begin{array}{l}
2=b\left(a^{2}-b\right)+a\left(2 a b-a^{3}\right)-b\left(2 a b-a^{3}\right) \\
\Leftrightarrow(2 a+1) b^{2}-\left(a^{3}+3 a^{2}\right) b+\left(a^{4}+2\right)=0 .
\end{array}
$$

Since $2 a+1 \neq 0$, we get $\Delta=\left(a^{3}+3 a^{2}\right)^{2}-4(2 a+1)\left(a^{4}+2\right)$ is a perfect square. That is, $\Delta=a^{6}-2 a^{5}+5 a^{4}-16 a-8=\left(a^{3}-a^{2}+2 a+2\right)^{2}-24 a-12$ is a perfect square.
When $a \geq 4$, it is clear that $\Delta < 0$.
$$
\begin{array}{l}
\left(a^{3}-a^{2}+2 a+2\right)^{2}>\Delta>\left(a^{3}-a^{2}+2 a+1\right)^{2},
\end{array}
$$

This contradicts $\Delta$ being a perfect square!
When $a \leq -3$, similarly, we get
$$
\left(a^{3}-a^{2}+2 a+2\right)^{2}<\Delta<\left(a^{3}-a^{2}+2 a+1\right)^{2}.
$$

This contradicts $\Delta$ being a perfect square!
In summary, $-2 \leq a \leq 3$. It is easy to verify that only when $a=-1$, $\Delta$ is a perfect square. Substituting into (1), we get $-b^{2}-2 b+3=0$, solving for $b=-1$ or 3, substituting into (8), we get $n=-1$ or 19.
Combining a) and b), we know (*) holds.
Combining i) and ii), we get $n=-2, -1, 10, 19, 34, 342$.