Example 5 Given that a square has three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
(1998, Shanghai High School Mathematics Competition)
Solution: As shown in Figure 3, without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis). 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, and the slope of $BC$ be $k$ $(k>0)$. Then we have
$$
\begin{aligned}
y_{3}-y_{2} & =k\left(x_{3}-x_{2}\right), \\
y_{1}-y_{2} & =-\frac{1}{k}\left(x_{1}-x_{2}\right).
\end{aligned}
$$

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
$$
x_{3}=k-x_{2}, \quad x_{1}=-\frac{1}{k}-x_{2}.
$$

Since $|AB|=|BC|$,
$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} = \sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}},
$$

we have $\sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)$. Therefore, $x_{2}-x_{1}=k\left(x_{3}-x_{2}\right)$. From equation (1), we get
$$
\frac{1}{k}+2 x_{2}=k\left(k-2 x_{2}\right).
$$

Thus, $k^{2}-\frac{1}{k}=(2 k+2) x_{2} \geqslant 0$. Solving this, we get $k \geqslant 1$, and $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$. Therefore, the side length of the square is
$$
\begin{array}{l}
\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)=\sqrt{1+k^{2}}\left(k-2 x_{2}\right) \\
=\sqrt{1+k^{2}}\left[k-\frac{k^{3}-1}{k(k+1)}\right] \\
=\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \geqslant \frac{2 k}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \\
\geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k+1)^{2}}}{k+1}=\sqrt{2}.
\end{array}
$$

The equality holds if and only if $k=1$, i.e., when point $B$ is at the origin. Therefore, the minimum area of the square is 2.