Example 10. (19th All-Soviet Union Middle School Olympiad Problem) A quadratic trinomial $a x^{2}+b x+c$, where $a>100$. How many different integer points can there be at which the absolute value of its value does not exceed 50?
Assume there are three different integer points satisfying the given condition, then there must be two points located on the same side of the axis of symmetry $x=-\frac{b}{2 a}$ of the quadratic function $y=a x^{2} +b x+c$, or one point is located on this axis of symmetry. Without loss of generality, assume
$-\frac{b}{2 a}1\left(x_{1} \neq x_{2}\right.$ are integers $)$, then we have
$$
\begin{aligned}
100 & \geqslant\left|a x_{2}^{2}+b x_{2}+c\right|+\left|a x_{1}^{2}+b x_{1}+c\right| \\
& >\left|\left(a x_{2}^{2}+b x_{2}+c\right) \cdots\left(1 x_{1}^{2}+b x_{1}+c\right)\right| \\
& =a\left(x_{2}-x_{1}\right)\left(x_{2}+x_{1}+\frac{b}{a}\right) \geqslant a>100,
\end{aligned}
$$

which is a contradiction. Therefore, the number of points satisfying the given condition is at most two.