Example 14 Given the equations in $x$: $4 x^{2}-8 n x- 3 n=2$ 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.

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Solution: From $\Delta_{1}=(-8 n)^{2}-4 \times 4 \times(-3 n-2)$ $=(8 n+3)^{2}+23>0$, we know that for any real number $n$, equation (1) always has real roots.
Let the two roots of the first equation be $\alpha, \beta$. Then
$$
\alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} \text {. }
$$

Thus, $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$
$$
=4 n^{2}+3 n+2 \text {. }
$$

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

Solving for the two roots, we get
$$
x_{1}=2 n+2, x_{2}=-n+1 \text {. }
$$

If $x_{1}$ is an integer, then $4 n^{2}+3 n+2=2 n+2$.
Solving gives $n_{1}=0, n_{2}=-\frac{1}{4}$.
When $n=0$, $x_{1}=2$ is an integer; when $n=-\frac{1}{4}$, $x_{1}=\frac{3}{2}$ is not an integer, so it is discarded.
If $x_{2}$ is an integer, then $4 n^{2}+3 n+2=1-n$.
Solving gives $n_{3}=n_{4}=-\frac{1}{2}$. In this case, $x_{2}=\frac{3}{2}$ is not an integer, so it is discarded.

In summary, when $n=0$, the square of the difference of the two real roots of the first equation equals one integer root of the second equation.