16. Given that 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 $\qquad$ .
16. 80 .

From the problem, we know that the distance from a point $\left(x, x^{2}\right)$ on the parabola to the line $l$ is
$$
d=\frac{\left|2 x-x^{2}-17\right|}{\sqrt{5}}>0 .
$$

When $x=1$, $d$ reaches its minimum value $\frac{16}{\sqrt{5}}$.
To minimize the area of the square, and by the symmetry of the square, the distances from the two points on the parabola to $(1,1)$ should be equal. Thus, we can assume these two points are $\left(a, a^{2}\right),\left(2-a,(2-a)^{2}\right)(a<1)$.
Therefore, $\sqrt{[a-(2-a)]^{2}+\left[a^{2}-(2-a)^{2}\right]^{2}}$ $=\frac{\left|a^{2}-2 a-17\right|}{\sqrt{5}}$
$\Rightarrow a=-1$ or -7 .
Thus, the coordinates of the two points are $(-1,1),(3, a)$ or $(-7,49),(9,8)$.
Therefore, the minimum value of the required area is
$$
(3+1)^{2}+(9-1)^{2}=80 \text {. }
$$