One. (20 points) Given the equations about $x$: $4 x^{2}-8 n x -3 n=2$ and $x^{2}-(n+3) x-2 n^{2}+2=0$. Does there exist a value of $n$ such that the square of the difference of the two real roots of the first equation equals an integer root of the second equation? If it exists, find such $n$ values; if not, explain the reason.
$$
\begin{aligned}
-\Delta_{1} & =(-8 n)^{2}-4 \times 4 \times(-3 n-2) \\
& =(8 n+3)^{2}+23>0 .
\end{aligned}
$$

Then $n$ is any real number, the first equation always has real roots.
Let the two roots of the first equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} . \\
\therefore(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=4 n^{2}+3 n+2 .
\end{array}
$$

From the second equation, we get
$$
[x-(2 n+2)][x+(n-1)]=0 \text {. }
$$

Solving, we get $x_{1}=2 n+2, x_{2}=1-n$.
If $x_{1}$ is an integer, then
$$
4 n^{2}+3 n+2=2 n+2 \text {. }
$$

Solving, we get $n_{1}=0, n_{2}=-\frac{1}{4}$.
When $n=0$, $x_{1}=2$;
When $n=-\frac{1}{4}$, $x_{1}=\frac{3}{2}$ is not an integer. Discard.
If $x_{2}$ is an integer, then $4 n^{2}+3 n+2=1-n$.
Solving, we get $n_{3}=n_{4}=-\frac{1}{2}$.
When $n=-\frac{1}{2}$, $x_{2}=\frac{3}{2}$ is not an integer. Discard.
In summary, when $n=0$, the square of the difference of the two real roots of the first equation equals the integer root of the second equation.