19.1.9 * Find all integers $n$ such that $n^{4}+6 n^{3}+11 n^{2}+3 n+31$ is a perfect square.
Let $n^{4}+6 n^{3}+11 n^{2}+3 n+31=A$, then
$$
A=\left(n^{2}+3 n+1\right)^{2}-3(n-10).
$$

When $n=10$, $A$ is a perfect square.
When $n>10$, since
$$
\left(n^{2}+3 n+1\right)^{2}-\left(n^{2}+3 n\right)^{2}=2 n^{2}+6 n+1>3(n-10),
$$

so $\left(n^{2}+3 n\right)^{2}<A<\left(n^{2}+3 n+1\right)^{2}$.
Thus, $A$ lies between $\left(n^{2}+3 n+1\right)^{2}$ and $\left(n^{2}+3 n+2\right)^{2}$, and is not a perfect square.
When $n=-6,-5,-4,-3,-2,-1,0,1,2$, $A=409,166,67,40,37,31,31,52,145$, none of which are perfect squares.
Therefore, the required $n=10$.