13.6.23 ** Given a real number $a$ satisfies: there is exactly one square whose four vertices all lie on the curve $y=$ $x^{3}+a x$. Try to find: the side length of this square.
Let the four vertices of the square be $A, B, C, D$. We first prove that the center of the square $ABCD$ is the origin $O$. Otherwise, since $y=x^3 + ax$ is an odd function, the points $A', B', C', D'$, which are the reflections of $A, B, C, D$ about the origin $O$, are also on the curve, and $A'B'C'D'$ is also a square, contradicting the condition that there is only one square.

Therefore, we can set $A(x_0, y_0), B(-y_0, x_0), C(-x_0, -y_0), D(y_0, -x_0)$, where $x_0 > 0, y_0 > 0$. Thus, we have $y_0 = x_0^3 + ax_0$, (1) and $-x_0 = y_0^3 + ay_0$.
(1) $\times x_0 +$ (2) $\times y_0$ gives $x_0^4 + y_0^4 + a(x_0^2 + y_0^2) = 0$.
(3) From (3), we know $a = -\frac{x_0^4 + y_0^4}{x_0^2 + y_0^2} < 0, \theta \in \left(0, \frac{\pi}{2}\right)$. From (3) and (4), we get
$$
\left\{\begin{array}{l}
a = -r^2 \left(1 - 2 \sin^2 \theta \cos^2 \theta\right) = -r^2 \left(1 - \frac{1}{2} \sin^2 2\theta\right), \\
r^2 = \frac{4}{\sin 4\theta}
\end{array}\right.
$$

Eliminating $r^2$ from these equations, we get $a \sin 4\theta = -4 + 2 \sin^2 2\theta$. Simplifying, we have $\left(1 + a^2\right) \left(\sin^2 2\theta\right)^2 - (4 + a^2) \sin^2 2\theta + 4 = 0$, which is a quadratic equation in $\sin^2 2\theta$. By the problem's condition, this equation has only one root in $(0,1)$, so $\Delta = \left(a^2 + 4\right)^2 - 16\left(1 + a^2\right) = a^4 - 8a^2 = 0$. Since $a < 0$, we have $a = -2\sqrt{2}$. Consequently, $\sin 2\theta = \frac{\sqrt{6}}{3}$, $\sin 4\theta = \frac{2\sqrt{2}}{3}$. Therefore, $r = \sqrt[4]{18}$. Thus, the side length of the square is $\sqrt{2} r = \sqrt[4]{72}$.