* Find all positive integers $n$, such that $n^{4}-4 n^{3}+22 n^{2}-36 n+18$ is a perfect square.
$$
\begin{aligned}
& n^{4}-4 n^{3}+22 n^{2}-36 n+18 \\
= & \left(n^{2}-2 n\right)^{2}+18\left(n^{2}-2 n\right)+18 \\
= & \left(n^{2}-2 n+9\right)^{2}-63
\end{aligned}
$$

Let $\left(n^{2}-2 n+9\right)^{2}-63=x^{2}$, then $\left(n^{2}-2 n+9+x\right)\left(n^{2}-2 n+9-x\right)=63$, thus
$$
\left\{\begin{array}{l}
n^{2}-2 n+9+x=63,21,9 \\
n^{2}-2 n+9-x=1,3,7,
\end{array}\right.
$$

Solving gives $n^{2}-2 n+9=32,12,8$, i.e., $(n-1)^{2}=24,4,0$.
Therefore, $n=1$ or 3.