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, $n=0$.