II If one side of the square $A B C D$ lies on the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$. Then the minimum value of the area of the square is $\qquad$ .
Let the side $AB$ of a square lie on the line $y=2x-17$, and the coordinates of the two vertices on the parabola be $C\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$. Then the equation of the line $l$ on which $CD$ lies is $y=2x+b$. By solving the system of equations formed by the line $l$ and the parabola, we get $x^{2}=2x+b \Rightarrow x_{1,2}=1 \pm \sqrt{b+1}$.

Let the side length of the square be $a$, then
$$
\begin{aligned}
a^{2} & =\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=5\left(x_{1}-x_{2}\right)^{2} \\
& =20(b+1) .
\end{aligned}
$$

Take any point $(6,-5)$ on the line $y=2x-17$, and the distance from this point to the line $y=2x+b$ is $a$, so
$$
a=\frac{|17+b|}{\sqrt{5}} .
$$
Solving (1) and (2) together, we get $b_{1}=3, b_{2}=63$, so $a^{2}=80$, or $a^{2}=1280$. Therefore, $a_{\min }^{2}=80$.