Find the minimum area of a nth-side polygon whose vertices satisfy In the Argand diagram, a square has its vertices satisfying a 4th power equation  with coefficients being integers, i.e.
$$z^4+az^3+bz^2+cz+d=0, z\in \mathbb C, a,b,c,d \in \mathbb Z$$
Find the minimum area.
I suspect that the vertices statisfying $$z^4=1$$ will do, but I cannot find a way to prove it.
Apart from square, the question also asks me to find the minimum area of nth-side polygon. Again, I lack some insights to solve this problem.
 A: You can show that, for any fourth-order polynomial $P(z)$ whose zeroes lie on the vertices of a square, there must be complex numbers $z_0$ and $z_1$ such that
$$
P(z) = (z - z_0)^4 + z_1^4.
$$
The proof of that statement should suggest how to solve the rest of the problem.
A: \begin{align*}
  f(z) &= \left( z+\frac{a}{4} \right)^4+d-\frac{a^4}{256} \\
  &= z^4+az^3+\frac{3a^2}{8}z^2+\frac{a^3}{16}z+d \\
  \left| z_k+\frac{a}{4} \right| &= 
  \sqrt[4]{\left| d-\frac{a^4}{256} \right|} \\
  \text{area} &= 2\sqrt{\left| d-\frac{a^4}{256} \right|}
\end{align*}

Case I: $a=0$
$$\text{area}=2\sqrt{d}$$ where $d\in \mathbb{Z}[i]\backslash \{ 0 \}$

Case II: $a\ne 0$
$$\dfrac{3a^2}{8},\, \dfrac{a^3}{16},\, d\in \mathbb{Z}[i]$$
$$\implies a=4m+4ni \quad \text{or} \quad a=(4m+2)+(4n+2)i$$
Without loss of generality, considering the cases for $a=4$ and $a=2+2i$ only.
The area is given by
$$2\sqrt{\left| d-1 \right|} \quad \text{and}
\quad 2\sqrt{\left| d+\frac{1}{4} \right|} \quad \text{respectively}$$


Hence, the minimum area is $1$.

