Example 4 (2002 China Western Mathematical Olympiad) Consider a square on the complex plane, whose 4 vertices correspond to the 4 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.


According to the problem, we know that the 4 roots of the equation can only be in two scenarios: 2 real roots and 1 pair of conjugate complex roots; 2 pairs of conjugate complex roots.
(1) If the 4 roots of the equation are 2 real roots and 1 pair of conjugate complex roots, then we can set these 4 roots as \(\pm b, a \pm b \mathrm{i}\). Thus, the original equation is \((x-a)^{4}=b^{4}\),

which is \(x^{4}-4 a x^{3}+6 a^{2} x^{2}-4 a^{3} x+a^{4}-b^{4}=0\).
From \(-4 a \in \mathbf{Z}\) and \(4 a^{3} \in \mathbf{Z}\), we know \(a \in \mathbf{Z}\). From \(a^{4}-b^{4} \in \mathbf{Z}\), we know \(b^{4} \in \mathbf{Z}\), so \(b^{4} \geqslant 1, b^{2} \geqslant 1\). In this case, the area of the square is \(2 b^{2} \geqslant 2\), with equality holding when \(a \in \mathbf{Z}, b= \pm 1\).
(2) If the 4 roots of the equation are 2 pairs of conjugate complex roots, then we can set these 4 roots as \(a \pm b \mathrm{i}, a+2 b \pm b \mathrm{i}\). These 4 roots are the roots of the equation \((x-(a+b))^{4}=-4 b^{4}\). Similarly, we know \(4 b^{4} \in \mathbf{Z}\), thus \(4 b^{4} \geqslant 1, b^{2} \geqslant \frac{1}{2}\). Therefore, the area of the square is \(4 b^{2} \geqslant 2\). Equality holds when \(b= \pm \frac{\sqrt{2}}{2}, a+b \in \mathbf{Z}\).

In summary, the area of such a square is greater than or equal to 2. The equation \(x^{4}=1\) has 4 roots which are the 4 vertices of a square on the complex plane with an area of 2, so the minimum area of such a square is 2.