1 Find all integers $n$, such that $n^{4}+6 n^{3}+11 n^{2}+3 n+31$ is a perfect square. (Xu Wanyi)
Let $A=n^{4}+6 n^{3}+11 n^{2}+3 n+31$ be a perfect square, i.e., $A=\left(n^{2}+3 n+1\right)^{2}-3(n-10)$ is a perfect square.
When $n>10$, $A<\left(n^{2}+3 n+1\right)^{2}$.
If $n \leqslant-3$, or $10>n \geqslant 0$, then $n^{2}+3 n \geqslant 0$. Therefore,
$$
\begin{array}{c}
A \geqslant\left(n^{2}+3 n+2\right)^{2}, \\
2 n^{2}+9 n-27 \leqslant 0, \\
-7<\frac{-3(\sqrt{33}+3)}{4} \leqslant n \leqslant \frac{3(\sqrt{33}-3)}{4}<3 .
\end{array}
$$

Thus, $n=-6,-5,-4,-3,0,1,2$, and the corresponding $A=409,166$, $67,40,31,52,145$ are not perfect squares.

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