1. Find all positive integers $n$, such that $n^{4}-4 n^{3}+22 n^{2}-$ $36 n+18$ is a perfect square.
1. Let $m \in \mathbf{N}_{+}$, such that $n^{4}-4 n^{3}+22 n^{2}-36 n+18=$ $m^{2}$, completing the square we get
$$
\left(n^{2}-2 n+9\right)^{2}-63=m^{2} \text {. }
$$

Thus, $63=\left(n^{2}-2 n+9\right)^{2}-m^{2}$
$$
=\left(n^{2}-2 n+9-m\right)\left(n^{2}-2 n+9+m\right) \text {. }
$$

Notice that, $n^{2}-2 n+9+m=(n-1)^{2}+m+8 \in$ $\mathbf{N}_{+}$, so we must have
$$
\left\{\begin{array}{l}
n^{2}-2 n+9-m=1,3,7, \\
n^{2}-2 n+9+m=63,21,9 .
\end{array}\right.
$$

And $n^{2}-2 n+9=32,12$ or 8, solving these respectively we get $n$ can only be 1 or 3 (corresponding to $m=1,9$).
Note: After obtaining equation (1), we can also use inequality estimation to handle it.