Cubic Square

A very simplified proof (it is just a lot of bash):

Let a cubic which satisfies be called $f(x) = x^3 + a_2 x^2 + a_1 x + a_0$ . We will first perform some transformations to make the cubic symmetric about the $y$ -axis.

Perform a translation $x \to x - \frac {a_2}{3}$ of this cubic so the coefficient of the second term is 0. This cubic still isn't symmetric about the $y$ -axis, though it's current rule is $f(x)=x^3 + b_1 x + b_0$ . We perform another translation, $y \to y - b_0$ , so the constant term is cancelled out. Thus, we are left with $f(x)=x^3 - ax$ , so we have shown a successful mapping to this expression from any cubic.

WLOG $f(x)=x^3-ax$ , with $a>0$ . Thus, $f$ is an odd function, which implies the square has a rotational centre about the origin, with an order at least 2, so the centre of the square is $(0, 0)$ .

Also, squares have rotational symmetry of order 4, so the coordinates of the points of the square are $(r, s), (-s, r), (-r, -s), (s, -r)$ . Thus, $f(x)$ this square will be exactly the same if $f(x)$ is rotated $90^{\circ}$ (call this function $f'(y)$ ). Thus,

$\begin{aligned} f'(y) &= -y^3 + ay\\ &= x(x^3 - ax)^3 - a(x^3 - ax)\\ &= x (x^2 - r^2)^2 (x^2 - s^2)^2\\ \end{aligned}$

Bashing this out, which is left as an exercise for the reader, we get the area of this square would be $\sqrt{72}$ .