Example 14 (1998 Shanghai Competition Problem) Given that there are three vertices $A, B, C$ of a square on the parabola $y=x^{2}$, find the minimum value of the area of such a square.

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Let's assume $A$ and $B$ are on the right side of the $y$-axis (including the $y$-axis), and the coordinates of $A, B, C$ are $\left(x_{1}, x_{1}^{2}\right)$, $\left(x_{2}, x_{2}^{2}\right), \left(x_{3}, x_{3}^{2}\right) \left(x_{3} \geq 0\right)$, then
$$
\left\{\begin{array}{l}
x_{1}^{2}-x_{2}^{2}=k\left(x_{1}-x_{2}\right), \\
x_{3}^{2}-x_{2}^{2}=-\frac{1}{k}\left(x_{3}-x_{2}\right),
\end{array}\right.
$$

Solving, we get $x_{1}=k-x_{2}, x_{3}=-\frac{1}{k}-x_{2}$. From $|A B|^{2}=|B C|^{2}$, we have
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}^{2}-x_{2}^{2}\right)^{2}=\left(x_{3}-x_{2}\right)^{2}+\left(x_{3}^{2}-x_{2}^{2}\right)^{2},
$$

which simplifies to $\left(1+k^{2}\right)\left(x_{1}-x_{2}\right)^{2}=\left(1+\frac{1}{k^{2}}\right)\left(x_{2}-x_{3}\right)^{2}$,
thus $x_{2}-x_{3}=k\left(x_{1}-x_{2}\right)$. Substituting $x_{1}=k-x_{2}, x_{3}=-\frac{1}{k}-x_{2}$,
we get $2 x_{2}+\frac{1}{k}=k\left(k-2 x_{2}\right)$, hence $k^{2}-\frac{1}{k}=(2 k+2) x_{2} \geqslant 0$,
solving for $k \geqslant 1$, and $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$, so
$$
\begin{aligned}
|A B| & =\sqrt{1+k^{2}}\left(x_{1}-x_{2}\right)=\sqrt{1+k^{2}}\left(k-2 x_{2}\right)=\sqrt{1+k^{2}}\left[k-\frac{k^{3}-1}{k(k+1)}\right] \\
& =\sqrt{1+k^{2}} \frac{k^{2}+1}{k(k+1)} \\
& =\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k+1)^{2}}}{k+1}=\sqrt{2},
\end{aligned}
$$

Equality holds if and only if $k=1$, in which case $x_{2}=0$, and $B$ coincides with the origin.
Therefore, the minimum value of the area of the square is 2.