1 Find all positive integers $n$, such that
$$
n^{4}-4 n^{3}+22 n^{2}-36 n+18
$$

is a perfect square.
(Provided by Chengcheng)
We denote
$$
\begin{aligned}
A & =n^{4}-4 n^{3}+22 n^{2}-36 n+18 \\
& =\left(n^{2}-2 n\right)^{2}+18\left(n^{2}-2 n\right)+18 .
\end{aligned}
$$

Let \( n^{2}-2 n=x \) and \( A=y^{2} \), where \( y \) is a non-negative integer, then
$$
(x+9)^{2}-63=y^{2},
$$

which simplifies to
$$
(x+9-y)(x+9+y)=63.
$$

It follows that the only possible pairs \((x+9-y, x+9+y)\) are \((1,63)\), \((3,21)\), or \((7,9)\). Solving these, we get \((x, y)=(23,31)\), \((3,9)\), or \((-1,1)\). Among these, only \( x=3 \) and \( x=-1 \) yield positive integer solutions for \( n^{2}-2 n=x \), giving \( n=1 \) or \( n=3 \).

Therefore, the values of \( n \) that satisfy the condition are \( n=1 \) or \( n=3 \).