China Western Mathematical Olympiad

Suppose $A=n^4+6n^3+11n^2+3n+31$ is a perfect square. It means that $A=(n^2+3n+1{)}^2-3(n-10)$ is a perfect square.

If $n>10$, then $A<(n^2+3n+1{)}^2$, thus $A\le (n^2+3n{)}^2$.

Therefore $$(n^2+3n+1{)}^2-(n^2+3n{)}^2\le 3n-30,$$ or $$2n^2+3n+31\le 0,$$ which is impossible.

Case I $n\le -3$ or $0\le n<10$. Then $n^2+3n\ge 0$. Thus $$A\ge (n^2+3n+2{)}^2.$$ That is, $$2n^2+9n-27\le 0,$$ or $$-7<\frac{-3(\sqrt{33}+3)}{4}\le n\le \frac{3(\sqrt{33}-3)}{4}<3.$$ Therefore, $n=-6,-5,-4,-3,0,1,2$. For these values of $n$, the corresponding values of $A$ are $409,166,67,40,31,52,145$. All of them are not perfect squares.

Case II $n=-2,-1$. Then $A=37,34$ respectively, none of them is a perfect square.

If $n=10$, then $A=(10^2+3\times 10+1{)}^2=131^2$ is a perfect square.

Hence, only when $n=10$, $A$ is a perfect square.