13 Let $A, B$ be two distinct points on the parabola $y^{2}=2 p x(p>0)$, then $\mid \overrightarrow{O A}+$ $\left.\overrightarrow{O B}\right|^{2}-|\overrightarrow{A B}|^{2}$ the minimum value is
(13 $-4 p^{2}$ Solution 1 Let $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$, then
$$
\begin{array}{c}
|\overrightarrow{O A}+\overrightarrow{O B}|^{2}=\left(x_{A}+x_{B}\right)^{2}+\left(y_{A}+y_{B}\right)^{2}, \\
|\overrightarrow{A B}|^{2}=\left(x_{A}-x_{B}\right)^{2}+\left(y_{A}-y_{B}\right)^{2}, \\
|\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2}=4\left(x_{A} \cdot x_{B}+y_{A} \cdot y_{B}\right) .
\end{array}
$$

Let the line $A B$ intersect the $x$-axis at point $P(a, 0)$. If the slope of line $A B$ exists, let it be $m$, then the equation of line $A B$ is $y=m(x-a)$, substituting it into the parabola equation gives
$$
m^{2} x^{2}-2\left(a m^{2}+p\right) x+m^{2} a^{2}=0 .
$$

By the relationship between the roots and coefficients of a quadratic equation, we get $x_{A} x_{B}=a^{2}$, from which we obtain
$$
y_{A} y_{B}=m^{2}\left(x_{A}-a\right)\left(x_{B}-a\right)=-2 a p \text {. }
$$

Therefore,
$$
\begin{aligned}
& |\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2} \\
= & 4\left(x_{A} \cdot x_{B}+y_{A} \cdot y_{B}\right) \\
= & 4\left[(a-p)^{2}-p^{2}\right]>-4 p^{2} .
\end{aligned}
$$

When the slope of line $A B$ does not exist, we have $x_{A}=x_{B}=a, y_{A}=-y_{B}=$ $\sqrt{2 a p}$. Therefore, we still have
$$
\begin{aligned}
& |\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2} \\
= & 4\left(x_{A} \cdot x_{B}+y_{A} \cdot y_{B}\right) \\
= & 4\left[(a-p)^{2}-p^{2}\right] \geqslant-4 p^{2} .
\end{aligned}
$$

Clearly, when and only when $a=p$, that is, when the slope of line $A B$ does not exist, the equality holds, and $|\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2}$ has the minimum value $-4 p^{2}$.
Solution 2 Let $A\left(\frac{y_{A}^{2}}{2 p}, y_{A}\right), B\left(\frac{y_{B}^{2}}{2 p}, y_{B}\right)$, then
$$
\begin{aligned}
|\overrightarrow{O A}+\overrightarrow{O B}|^{2} & =\left(\frac{y_{A}^{2}+y_{B}^{2}}{2 p}\right)^{2}+\left(y_{A}+y_{B}\right)^{2}, \\
|\overrightarrow{A B}|^{2} & =\left(\frac{y_{A}^{2}-y_{B}^{2}}{2 p}\right)^{2}+\left(y_{A}-y_{B}\right)^{2} .
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
& |\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2} \\
= & 4\left(\frac{y_{A}^{2} \cdot y_{B}^{2}}{4 p^{2}}+y_{A} \cdot y_{B}\right) \\
= & 4\left[\left(\frac{y_{A} \cdot y_{B}}{2 p}+p\right)^{2}-p^{2}\right] \geqslant-4 p^{2} .
\end{aligned}
$$

When $y_{A} y_{B}=-2 p^{2}$, $|\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2}$ takes the minimum value $-4 p^{2}$.