3. Given that $a$ and $b$ are real numbers, and $b-a=2$006. If the roots of the quadratic equation $x^{2}+a x+b=0$ are all integers, then the number of sets of roots for this equation is ( ).
(A) 4
(B) 6
(C) 8
(D) 10
3.B.

By Vieta's formulas, we have $x_{1}+x_{2}=-a, x_{1} x_{2}=b$. Then $x_{1}+x_{2}+x_{1} x_{2}=2006$.
Therefore, $\left(x_{1}+1\right)\left(x_{2}+1\right)=2007=9 \times 223$
$$
\begin{array}{l}
=-9 \times(-223)=3 \times 669=-3 \times(-669) \\
=1 \times 2007=(-1) \times(-2007) .
\end{array}
$$

It is easy to see that the equation has 6 solutions.