Three. (25 points) Given the equation about $x$
$$
4 x^{2}-8 n x-3 n-2=0
$$

and $x^{2}-(n+3) x-2 n^{2}+2=0$.
Question: Is there such a value of $n$ that the square of the difference of the two real roots of equation (1) equals an integer root of equation (2)? If it exists, find such $n$ values; if not, explain the reason.
Three, because the discriminant of equation (1) is
$$
\begin{array}{l}
\Delta_{1}=(-8 n)^{2}-4 \times 4(-3 n-2) . \\
=(8 n+3)^{2}+23>0,
\end{array}
$$

Therefore, equation (1) has two distinct real roots $\alpha_{1} 、 \beta_{1}$.
By Vieta's formulas, we get
$$
\begin{array}{l}
\left(\alpha_{1}-\beta_{1}\right)^{2}=\left(\alpha_{1}+\beta_{1}\right)^{2}-4 \alpha_{1} \beta_{1} \\
=4 n^{2}+3 n+2 .
\end{array}
$$

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

If $4 n^{2}+3 n+2$ is an integer root of equation (1), substituting it in, we get
$$
\begin{array}{l}
\left(4 n^{2}+n\right)\left(4 n^{2}+4 n+1\right)=0 \\
\Rightarrow n=0,-\frac{1}{4},-\frac{1}{2} .
\end{array}
$$

Upon verification, only when $n=0$, it satisfies the problem's conditions.
In summary, $n=0$.