Three. (25 points) Find all real numbers $a$ such that the quadratic equation $a x^{2}+a x+a+2006=0$ has two distinct integer roots.

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Three. (25 points) Find all real numbers $a$ such that the quadratic equation $a x^{2}+a x+a+2006=0$ has two distinct integer roots.
If $a x^{2}+a x+a+2006=0$ has two distinct integer roots $x_{1}, x_{2}$, then
$$
a=\frac{-2006}{x_{1}^{2}+x_{1}+1}=-\frac{2006}{t^{2}+t+1}(t \in Z) .
$$

If $a=-\frac{2006}{t^{2}+t+1}$. Then $a x^{2}+a x+a+2006=0$ becomes $-\frac{2006}{t^{2}+t+1}\left(x^{2}+x+1\right)+2006=0$, which is
$$
x^{2}+x+1-\left(t^{2}+t+1\right)=0 .
$$

Thus, $(x-t)(x+t+1)=0$ indeed has two distinct integer roots $x=t, x=-t-1$.

In summary, the real number $a=-\frac{2006}{t^{2}+t+1}(t=0, \pm 1, \pm 2, \pm 3, \cdots)$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006)