China Western Mathematical Olympiad

Suppose the complex number corresponding to the center of the square is $a$. Then after translating the origin of the complex plane to $a$, the vertices of the square distribute evenly on the circumference. That is, they are the solutions of equation $(x-a{)}^4=b$, where $b$ is a complex number. Hence,

$$\begin{array}{rl}{x}^4+p{x}^3+q{x}^2+rx+s& =(x-a{)}^4-b\\ & =x^4-4a{x}^3+6a^2{x}^2-4a^{3}x+a^4-b.\end{array}$$

Comparing the coefficients of terms for $x$ with the same degree, we know that $-a=\frac{p}{4}$, and it is a rational number. Combining further that $-4a^3=r$ is an integer, we can see that $a$ is an integer. So by using the fact that $s=a^4-b$ is an integer, we can show that $b$ is also an integer.

The above discussion makes clear of a fact that the four numbers corresponding to the four vertices of this square are roots of integer coefficients equation $(x-a{)}^4=b$. Hence, the radius of its circumcircle ($=\sqrt[4]{|b|}$) is not less than $1$. Therefore, the area of this square is not less than $(\sqrt{2}{)}^2=2$. But the four roots of the equation $x^4=1$ are corresponding to the four vertices of a square on the complex plane. Hence, the minimum value of the area of the square is $2$.