设 $A$ 的坐标为 $(t_1,t_1^2)$，$B$ 的坐标为 $(t_2,t_2^2)$，显然，由 $AD\varparallel BC$，得 $1=\frac{t_2^2-t_1^2}{t_2-t_1}$，即 $t_1+t_2=1$．
设正方形 $ABCD$ 的边长为 $d$，则$$\begin{aligned}d^2&=|AB|^2=(t_1-t_2)^2+(t_1^2-t_2^2)^2=(t_1-t_2)^2(1+(t_1+t_2)^2)\\
&=2((t_1+t_2)^2-4t_1t_2)=2(1-4t_1t_2),\\
\end{aligned}$$即$$t_1t_2=\frac{1}{8}(2-d^2).~~~~ ① $$又有$$d=|AD|=\frac{|t_1-t_1^2-4|}{\sqrt{2}}=\frac{|t_1(1-t_1)-4|}{\sqrt{2}}=\frac{|t_1t_2-4|}{\sqrt{2}}.$$于是$$2d^2=(t_1t_2-4)^2.~~~~ ② $$将式 ① 代入式 ②，整理得$$d^4-68d^2+900=0.$$解得 $d=\pm 3\sqrt{2}$ 或 $\pm 5\sqrt{2}$．舍去负值，即知正方形 $ABCD$ 的边长为 $3\sqrt{2}$ 或 $5\sqrt{2}$．