16. 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.
16. $\Delta_{1}=(8 n+3)^{2}+23>0$.

Then $n$ is any real number, equation (1) always has real roots.
Let the two roots of the first equation be $\alpha, \beta$. Then
$$
\begin{array}{l}
\alpha+\beta=2 n, \alpha \beta=\frac{-3 n-2}{4} . \\
\therefore(\alpha-\beta)^{2}=4 n^{2}+3 n+2 .
\end{array}
$$

From the second equation, we get $[x-(2 n+2)][x+(n-1)]=0$,
solving for $x_{1}=2 n+2, x_{2}=1-n$.
If $x_{1}$ is an integer, then $4 n^{2}+3 n+2=2 n+2$,
thus $n_{1}=0, n_{2}=-\frac{1}{4}$.
When $n=0$, $x_{1}=2$ is an integer;
When $n=-\frac{1}{4}$, $x_{1}=\frac{3}{2}$ is not an integer, discard.
If $x_{2}$ is an integer, then $4 n^{2}+3 n+2=1-n$,
thus $n_{3}=n_{4}=-\frac{1}{2}$.
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 first integer root of the second equation.