A square $$ABCD$$ has all its vertices on the curve $$x^2 y^2 = 1$$. The midpoints of its sides also lie on the same curve. Then, the square of area of $$ABCD$$ is ________.

The curve is $$x^2 y^2 = 1$$, which can be written as $$|xy| = 1$$, i.e., $$xy = 1$$ or $$xy = -1$$. This consists of four hyperbolic branches in the four quadrants.

Since the square has all vertices on this curve and the midpoints of its sides also lie on the curve, we use the symmetry of the curve. The curve is symmetric about both axes and about the lines $$y = x$$ and $$y = -x$$.

Let the square be centered at the origin with vertices at $$(a, b)$$, $$(-b, a)$$, $$(-a, -b)$$, $$(b, -a)$$ (a square rotated by some angle). Since each vertex lies on $$x^2 y^2 = 1$$, we need $$a^2 b^2 = 1$$, so $$|ab| = 1$$.

The midpoint of the side joining $$(a, b)$$ and $$(-b, a)$$ is $$\left(\frac{a-b}{2}, \frac{a+b}{2}\right)$$. This midpoint must also lie on $$x^2 y^2 = 1$$, so $$\left(\frac{a-b}{2}\right)^2 \left(\frac{a+b}{2}\right)^2 = 1$$.

This gives $$\frac{(a-b)^2(a+b)^2}{16} = 1$$, so $$(a^2 - b^2)^2 = 16$$, meaning $$a^2 - b^2 = \pm 4$$.

We also have $$ab = \pm 1$$. Let us take $$ab = 1$$ (and $$a, b > 0$$). Then $$b = \frac{1}{a}$$.

Substituting into $$a^2 - b^2 = 4$$: $$a^2 - \frac{1}{a^2} = 4$$. Let $$u = a^2$$, so $$u - \frac{1}{u} = 4$$, giving $$u^2 - 4u - 1 = 0$$. Solving: $$u = \frac{4 \pm \sqrt{16+4}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$$. Since $$u > 0$$, we take $$u = 2 + \sqrt{5}$$.

The side length of the square: the distance between $$(a,b)$$ and $$(-b,a)$$ is $$\sqrt{(a+b)^2 + (b-a)^2} = \sqrt{2(a^2+b^2)}$$.

Now $$a^2 + b^2 = a^2 + \frac{1}{a^2} = u + \frac{1}{u}$$. Since $$u = 2+\sqrt{5}$$, we get $$\frac{1}{u} = \frac{1}{2+\sqrt{5}} = \frac{2-\sqrt{5}}{(2+\sqrt{5})(2-\sqrt{5})} = \frac{2-\sqrt{5}}{4-5} = \sqrt{5}-2$$. So $$a^2 + b^2 = (2+\sqrt{5}) + (\sqrt{5}-2) = 2\sqrt{5}$$.

The side length squared is $$2(a^2 + b^2) = 2 \cdot 2\sqrt{5} = 4\sqrt{5}$$.

The area of the square is $$(\text{side length})^2 = 4\sqrt{5}$$.

The square of the area is $$(4\sqrt{5})^2 = 16 \times 5 = 80$$.