Example 1 Solve the system of equations
$$
\left\{\begin{array}{l}
a x^{2}+b x+c=y, \\
a y^{2}+b y+c=z, \\
a z^{2}+b z+c=x .
\end{array}\right.
$$

where $a \neq 0$, and $(b-1)^{2}=4 a c$.
(1979, Hubei Province High School Mathematics Competition)
We find that the parabola $y=a x^{2}+b x+c$ is tangent to the line $y=x$ at the point $\left(\frac{1-b}{2 a}, \frac{1-b}{2 a}\right)$. This leads us to think that $(x, y)=(p, p)$ is a solution to the first equation of the system (where $p=\frac{1-b}{2 a}$). Similarly, $(y, z)=(p, p)$ and $(z, x)=(p, p)$ are solutions to the second and third equations, respectively. Therefore, $(x, y, z)=(p, p, p)$ is a solution to the system of equations.

If the system of equations has a solution $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$, and $x^{\prime} \neq p$, then $y^{\prime}=a x^{\prime 2}+b x^{\prime}+c=a\left(x^{\prime}-p\right)^{2}+x^{\prime}$. When $a>0$, it can be deduced that $y^{\prime}>x^{\prime}$. Similarly, $x^{\prime}=a\left(z^{\prime}-p\right)^{2}+z^{\prime} \geqslant z^{\prime}$, $z^{\prime}=a\left(y^{\prime}-p\right)^{2}+y^{\prime} \geqslant y^{\prime}$, thus, $y^{\prime}>x^{\prime} \geqslant z^{\prime} \geqslant y^{\prime}$, leading to a contradiction; when $a<0$, similarly, it can be deduced that $y^{\prime}<x^{\prime} \leqslant z^{\prime} \leqslant y^{\prime}$, a contradiction. Therefore, $x^{\prime}=p$. Similarly, $y^{\prime}=z^{\prime}=p$. That is, the system of equations has a unique solution $(p, p, p)$.