3 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. (Xiong Bin)
Let the complex number corresponding to the center of the 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, i.e., they are the solutions to 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 the coefficients of each power of $x$, we find 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. This way, using $s=a^{4}-b$ being an integer, we know that $b$ is an integer.

The above discussion shows that the square is the root of the integer-coefficient equation $(x-a)^{4}=b$. Therefore, the radius of its circumscribed circle $(=\sqrt[4]{|b|})$ is no less than 1. Hence, the area of this square is no less than $(\sqrt{2})^{2}=2$. The equation $x^{4}=1$ has four roots in the complex plane corresponding to the vertices of a square. Therefore, the minimum area of the required square is 2.