Question 173, Given that a square has three vertices on the parabola $y=x^{2}$, find the minimum value of the area of such a square.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. 

(Note: The note itself is not part of the translation task and should not be included in the final output.)

Question 173, Given that a square has three vertices on the parabola $y=x^{2}$, find the minimum value of the area of such a square.
Question 173, Solution: Let the square be $ABCD$, and $A\left(a, a^{2}\right), B\left(b, b^{2}\right), D\left(d, d^{2}\right)$ lie on the parabola $y=x^{2}$. Suppose the slope of line $AB$ is $k (k>0)$, then $b^{2}-a^{2}=k(b-a)$, hence $b=k-a$. Similarly, $d=-\frac{1}{k}-a$. Also,
$$
\begin{array}{l}
|AB|=\sqrt{1+k^{2}}(b-a)=\sqrt{1+k^{2}}(k-2a) \\
|AD|=\sqrt{1+\frac{1}{k^{2}}}(a-d)=\sqrt{1+\frac{1}{k^{2}}}\left(2a+\frac{1}{k}\right)=\sqrt{1+k^{2}} \cdot \frac{2ak+1}{k^{2}}
\end{array}
$$

Noting that $|AB|=|AD|$, we have
$$
\begin{array}{l}
\sqrt{1+k^{2}}(k-2a)=\sqrt{1+k^{2}} \cdot \frac{2ak+1}{k^{2}} \\
\Rightarrow (k-2a)k^{2}=2ak+1 \\
\Rightarrow a=\frac{k^{3}-1}{2(k^{2}+k)}
\end{array}
$$

Thus, $|AB|=\sqrt{1+k^{2}}\left(k-\frac{k^{3}-1}{k^{2}+k}\right)=\frac{(k^{2}+1)\sqrt{1+k^{2}}}{k(k+1)} \geq \frac{2k \frac{k+1}{\sqrt{2}}}{k(k+1)}=\sqrt{2}$, with equality holding if and only if $k=1$. Therefore, the area of the square is no less than 2, and when $A(0,0), B(1,1), D(-1,1)$, the area of square $ABCD$ is exactly 2.