SKET $k$ Given: On the parabola $y=x^{2}$, there are three vertices $A, B, C$ of a square. Find: The minimum value of the area of such a square.

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Assume that among the three vertices of the square on the parabola, two are on the right side of the $y$-axis (including the $y$-axis), and let the coordinates of points $A$, $B$, and $C$ be $\left(x_{i}, y_{i}\right), i=1,2,3$. The slope of $BC$ is $k (k > 0)$. Since the points lie on the parabola $y=x^{2}$, we have $y_{1}=x_{1}^{2}, y_{2}=x_{2}^{2}, y_{3}=x_{3}^{2}$. Substituting these into the above equations, we get $x_{3}=k-x_{2}, x_{1}=-\frac{1}{k}-x_{2}$.

Since $|A B|=|B C|$, we have $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}$, so $\sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)$, which simplifies to $x_{2}-x_{1}=k\left(x_{3}-x_{2}\right)$. Therefore, $\frac{1}{k}+2 x_{2}=k\left(k-2 x_{2}\right)$. From $k^{2}-\frac{1}{k}-(2 k+2) x_{2} \geqslant 0$, we know $k^{3} \geqslant 1$, hence $k>1$. Thus, $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$, and the side length of the square is
$$
\begin{aligned}
\sqrt{k^{2}+1}\left(x_{3}-x_{2}\right) & =\sqrt{k^{2}+1}\left(k-2 x_{2}\right)=\sqrt{k^{2}+1}\left[k-\frac{k^{3}-1}{k(k+1)}\right] \\
& =\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \geqslant \frac{2 k}{k} \cdot \frac{\frac{k+1}{\sqrt{2}}}{k+1}=\sqrt{2},
\end{aligned}
$$

with equality holding if and only if $k=1$, i.e., when $B$ is at the origin. Therefore, $S_{\max }=(\sqrt{2})^{2}=2$.