14. Given a square $A B C D$ with two vertices $A$ and $B$ on the parabola $y=x^{2}$, and points $C$ and $D$ on the line $y=x-4$. Find the side length $d$ of the square.
14. Let $A\left(t_{1}, t_{1}^{2}\right) 、 B\left(t_{2}, t_{2}^{2}\right)$, and $t_{1} \neq t_{2}$.

If $A B / / D C$, then $1=\frac{t_{2}^{2}-t_{1}^{2}}{t_{2}-t_{1}}$, which means $t_{1}+t_{2}=1$.
- Therefore,
$$
\begin{array}{l}
d^{2}=|A B|^{2}=\left(t_{1}-t_{2}\right)^{2}+\left(t_{1}^{2}-t_{2}^{2}\right)^{2} \\
=\left(t_{1}-t_{2}\right)^{2}\left[1+\left(t_{1}+t_{2}\right)^{2}\right] \\
=2\left[\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}\right]=2\left(1-4 t_{1} t_{2}\right) .
\end{array}
$$

Thus, $t_{1} t_{2}=\frac{1}{8}\left(2-d^{2}\right)$.
On the other hand,
$$
\begin{aligned}
d & =A D=\frac{\left|t_{1}-t_{1}^{2}-4\right|}{\sqrt{2}}=\frac{\left|t_{1}\left(1-t_{1}\right)-4\right|}{\sqrt{2}} \\
& =\frac{\left|t_{1} t_{2}-4\right|}{\sqrt{2}},
\end{aligned}
$$

we have $2 d^{2}=\left(t_{1} t_{2}-4\right)^{2}$.
Substituting equation (1) into equation (2) gives $d^{4}-68 d^{2}+900=0$.
Solving this, we get $d= \pm 3 \sqrt{2}$ or $d= \pm 5 \sqrt{2}$.
Discarding the negative values, the side length of the square is $3 \sqrt{2}$ or $5 \sqrt{2}$.