$2 \cdot 110$ To make $p=x^{4}+6 x^{3}+11 x^{2}+3 x+31$ a perfect square of an integer, the number of integer values of $x$ is
(A) 4.
(B) 3.
(C) 2.
(D) 1.
(E) 0.
(16th American High School Mathematics Examination, 1965)
[Solution] Let $p=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, and set $p=y^{2}$.
Therefore, when $x=10$, $p=\left(x^{2}+3 x+1\right)^{2}=131^{2}$.
Next, we prove that $p$ is a perfect square only when $x=10$.
Note: If $|N|>|M|$, where $N, M$ are integers, then
$$
|| N|-1| \geqslant|M| \text { or } N^{2}-M^{2} \geqslant 2|N|-1 \text {. }
$$
(1) If $x>10$, note that
$$
3(x-10)=\left(x^{2}+3 x+1\right)^{2}-y^{2} \geqslant 2\left|x^{2}+3 x+1\right|-1 \text {, }
$$

which cannot hold.
(2) If $x<10$, note that
$$
3(x-10)=\left(x^{2}+3 x+1\right)^{2}-y^{2} \leqslant 2\left|x^{2}+3 x+1\right|-1,
\end{aligned}
$$

this inequality holds only when $x=-6,-5,-4,-3,-2,-1,0,1,2$, but none of these values are perfect squares.
In conclusion, the answer is $(D)$.