5. 2.12 * Consider a square on the complex plane, whose four vertices correspond to the four roots of a 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.


Let the complex number corresponding to the center $A$ of a square be $a$. Then, after translating the origin of the complex plane to $A$, the vertices of the square are uniformly distributed on a circle, and the complex numbers corresponding to these vertices are the solutions of the equation $(x-a)^{4}=b$, where $b$ is some complex number. Thus,
\[
\begin{aligned}
& x^{4}+p x^{3}+q x^{2}+r x+s=(x-a)^{4}-b \\
= & x^{4}-4 a x^{3}+6 a^{2} x^{2}-4 a^{3} x+a^{4}-b .
\end{aligned}
\]

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, $b$ must also be an integer.

The above discussion shows that the complex numbers corresponding to the vertices of the square are the roots of the integer-coefficient equation $(x-a)^{4}=b$, 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 roots of the equation $x^{4}=1$ correspond to the vertices of a square in the complex plane. Therefore, the minimum area of the square is 2.

**Comment:** A deeper exploration of this problem is:
If the vertices of a regular $n$-sided polygon correspond to the $n$ complex roots of the integer-coefficient equation
\[
x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}=0,
\]
then the minimum area of the regular $n$-sided polygon is $\frac{n}{2} \sin \frac{2 \pi}{n}$.