Find all integers $n$ such that
$$
n^{4}+6 n^{3}+11 n^{2}+3 n+31
$$

is a perfect square.
(Xu Wandang)
Let $A=n^{4}+6 n^{3}+11 n^{2}+3 n+31$. Then
$$
A=\left(n^{2}+3 n+1\right)^{2}-3(n-10)
$$

is a perfect square.
When $n>10$, we have $A<\left(n^{2}+3 n+1\right)^{2}$.
If $n \leqslant-3$ or $n \geqslant 0$, then $n^{2}+3 n \geqslant 0$.
Thus, $A \geqslant\left(n^{2}+3 n+2\right)^{2}$.
Therefore, $2 n^{2}+9 n-27 \leqslant 0$.
Solving this, we get $-7<\frac{-3(\sqrt{33}+3)}{4} \leqslant n \leqslant \frac{3(\sqrt{33}-3)}{4}<3$.
So, $n=-6,-5,-4,-3,0,1,2$. In these cases, the corresponding values of $A=409,166,67,40,31,52,145$ are not perfect squares.

When $n=-2,-1$, the corresponding values of $A=37,34$ are also not perfect squares.
Therefore, only when $n=10$, $A$ is a perfect square.