China Western Mathematical Olympiad

We write $$\begin{array}{rl}A& =n^4-4n^3+22n^2-36n+18\\ & =(n^2-2n{)}^2+18(n^2-2n)+18.\end{array}$$ Let $n^2-2n=x$, $A=y^2$, where $y$ is a nonnegative integer. Then $$(x+9{)}^2-63=y^2.$$ That is, $(x+9-y)(x+9+y)=63$. It can only be $(x+9-y,x+9+y)=(1,63)$, $(3,21)$ or $(7,9)$. We obtain, respectively, $(x,y)=(23,31)$, $(3,9)$ or $(-1,1)$. But only when $x=3$ and $-1$, $n^2-2n=x$ has solutions in positive integers and we get $n=1$ or 3. Therefore, $n=1$ or 3 satisfies the condition.