7. Given a real number $a$ satisfies: there is one and only one square, whose 4 vertices all lie on the curve $y=x^{3}+a x$. Try to find the side length of the square.

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7. Given a real number $a$ satisfies: there is one and only one square, whose 4 vertices all lie on the curve $y=x^{3}+a x$. Try to find the side length of the square.
7. Let the 4 vertices of the square be $A, B, C, D$. The center of the square must be the origin $O$, otherwise, because $y=x^{3}+a x$ is an odd function, the points $A, B, C, D$ would have their symmetric points $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ with respect to $O$ on the curve. $1 \mathrm{~L} A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ would also form a square, contradicting the given conditions. Let the 4 points be $A\left(x_{0}, y_{0}\right), B\left(-y_{0}, x_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(y_{0},-x_{11}\right)$, where $x_{0}>0, y_{0}>0$, then $y_{0}=x_{0}^{3}+a x_{0}$ (1), $-x_{0}=y_{0}^{3}+a y_{0}$ (2). By multiplying (1) by $x_{0}$ and (2) by $y_{0}$, we get $x_{0}^{4}+y_{0}^{4}+a\left(x_{0}^{2}+y_{0}^{2}\right)=0 \quad$ (3). By multiplying (1) by $y_{0}$ and (2) by $x_{0}$, we get $x_{0}^{2}+y_{0}^{2}=x_{0} y_{0}\left(x_{0}^{2}-y_{0}^{2}\right) \quad$ (4). Let $x_{0}=r \cos \theta, y_{0}=r \sin \theta, r>0, \theta \in\left(0, \frac{\pi}{2}\right)$, substituting into (3) and (4) gives $a=-r^{2}\left(\cos ^{4} \theta+\sin ^{4} \theta\right)=-r^{2}\left(1-2 \cos ^{2} \theta \sin ^{2} \theta\right), r^{2}=\frac{2}{\sin 2 \theta+\cos 2 \theta} \quad$ (5). Eliminating $r^{2}$, we get $\left(1+a^{2}\right)\left(\sin ^{2} 2 \theta\right)^{2}-\left(4+a^{2}\right) \sin ^{2} 2 \theta+4=0$. Since $\sin ^{2} 2 \theta$ has only one real root in $(0,1)$, we have $\Delta=\left(a^{2}+4\right)^{2}-16\left(1+a^{2}\right)=a^{4}-8 a^{2}=0$, so $a=-2 \sqrt{2}$ (since (3) implies $a<0$), $\sin 2 \theta=\frac{\sqrt{6}}{3}, \cos 2 \theta=\frac{\sqrt{3}}{3}$. Substituting into (5) gives $r=\sqrt[4]{18}$, hence the side length of the square is $\sqrt{2} r=\sqrt[4]{72}$.