Three, (16 points) It is known that on the parabola $y=x^{2}$, there are three vertices $A, B$, and $C$ of a square. Find the minimum value of the area of such a square.

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Let's assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis), and let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, with the slope of $BC$ being $k (k>0)$. Then we have
$$
\begin{array}{c}
y_{3}-y_{2}=k\left(x_{3}-x_{2}\right), \\
y_{1}-y_{2}=-\frac{1}{k}\left(x_{1}-\right. \\
\left.x_{2}\right) .
\end{array}
$$

Since points $A$, $B$, and $C$ lie on the parabola, we have $y_{1}=x_{1}^{2}$, $y_{2}=x_{2}^{2}$, and $y_{3}=x_{3}^{2}$. Substituting these into the above equations, we get
$$
x_{3}=k-x_{2}, x_{1}=-\frac{1}{k}-x_{2} \text {. }
$$

Since $|AB|=|BC|$, we have
$$
\begin{array}{l}
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \\
=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}, \\
\therefore \sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right) . \\
x_{2}-x_{1}=k\left(x_{3}-x_{2}\right) . \\
\frac{1}{k}+2 x_{2}=k\left(k-2 x_{2}\right), \\
k^{2}-\frac{1}{k}=(2 k+2) x_{2} \geqslant 0, k^{3} \geqslant 1, k \geqslant 1 .
\end{array}
$$

Thus, $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$. The side length of the square is
$$
\begin{array}{l}
\sqrt{k^{2}+1}\left(x_{3}-x_{2}\right)=\sqrt{k^{2}+1}\left(k-2 x_{2}\right) \\
=\sqrt{k^{2}+1}\left(k-\frac{k^{3}-1}{k(k+1)}\right) \\
=\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \\
\geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k+1)^{2}}}{k+1}=\sqrt{2} .
\end{array}
$$

The equality holds if and only if $k=1$, and point $B$ is at the origin.