Example 9 Consider a square on the complex plane, whose four vertices correspond to the four roots of a certain monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Analysis: After translating the origin of the complex plane to the center of a square, the vertices of the square are uniformly distributed on a circle. Therefore, in the original complex plane, the corresponding complex numbers of these vertices are the solutions to the equation $(x-a)^{4}=b$, where $a, b \in C$. Expanding and comparing coefficients, we can solve the problem.

Solution: According to the problem, the four vertices of the square correspond to the four roots of the equation $(x-a)^{4}=b(a, b \in C)$. Thus, $x^{4}+p x^{3}+q x^{2}+r x+s=(x-a)^{4}-b=x^{4}-4 a x^{3}+6 a^{2} x^{2}-4 a^{3} x+a^{4}-b$.

By comparing coefficients, we know that $-a=\frac{p}{4}$ is a rational number. Combining this with $-4 a^{3}=r$ being an integer, we conclude that $a$ is an integer. Therefore, since $s=a^{4}-b$ is an integer, $b$ is also an integer.

The above discussion shows that the complex numbers corresponding to the vertices of the square are the roots of the equation $(x-a)^{4}=b$ with integer coefficients, and the radius of the circumscribed circle $\sqrt[4]{|b|}$ is no less than 1. Thus, the area of the square is no less than $(\sqrt{2})^{2}=2$. The equation $x^{4}=1$ has four roots that correspond to the vertices of a square in the complex plane. Therefore, the minimum area of the square is 2.