Not really a difference, is there? Guess it matters in how you present the answer, but you can still find which square it is.
The four roots of the equation are $ \alpha + \beta, \alpha + i \beta, \alpha - \beta, \alpha - i \beta$ for some complex $ \alpha, \beta$ with $ \beta \neq 0$. We want the minimum of $ |\beta|$. Let $ P(x)$ be the polynomial given (so it has the roots $ \alpha + \beta$ etc.), then $ P(x) = x^4 - 4\alpha x^3 + 6\alpha^2 x^2 - 4 \alpha^3 x + \alpha^4 - \beta^4$ by Vieta. Since $ -4\alpha$ is an integer $ \alpha$ is clearly rational. But $ -4\alpha^3$ is also an integer, so $ \alpha$ has to be an integer as well. WLOG $ \alpha = 0$. $ x^4 - \beta^4$ has to have integer coefficients, and $ |\beta| = 1$ is the minimum.
So the minimum size square is the one inscribed in the unit circle.