3. 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.


3. According to the problem, 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 $a \pm b, a \pm b \mathrm{i}$. 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.