10. (20 points) Let
$$
P=x^{4}+6 x^{3}+11 x^{2}+3 x+31 \text {. }
$$

Find the integer value(s) of $x$ that make $P$ a perfect square.
10. Since $P=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, therefore, when $x=10$, $P=131^{2}$ is a perfect square.

Next, we only need to prove: there are no other integer $x$ that satisfy the requirement.
(1) $x>10$.
If $P0$, then $P>\left(x^{2}+3 x\right)^{2}$.
Therefore, $\left(x^{2}+3 x\right)^{2}\left(x^{2}+3 x+1\right)^{2}$.
Let $P=y^{2}(y \in \mathbf{Z})$. Then
$$
|y|>\left|x^{2}+3 x+1\right| \text {, }
$$

which means $|y|-1 \geqslant \mid x^{2}+3 x+11$.
Thus, $y^{2}-2|y|+1 \geqslant\left(x^{2}+3 x+1\right)^{2}$, i.e., $-3(x-10)-2\left|x^{2}+3 x+1\right|+1 \geqslant 0$.
Solving this inequality yields the integer values of $x$ as
$$
\pm 2, \pm 1, 0, -3, -4, -5, -6 \text {. }
$$

However, the corresponding $P$ for these values are not perfect squares.
In conclusion, the integer value of $x$ that makes $P$ a perfect square is 10.