8. Given $a, b \in \mathbf{Z}$, and $a+b$ is a root of the equation
$$
x^{2}+a x+b=0
$$

Then the maximum possible value of $b$ is $\qquad$
8. 9 .

From the given, we have
$$
\begin{array}{l}
(a+b)^{2}+a(a+b)+b=0 \\
\Rightarrow 2 a^{2}+3 a b+b^{2}+b=0 .
\end{array}
$$

Since $a, b \in \mathbf{Z}$, we have
$$
\Delta=9 b^{2}-8\left(b^{2}+b\right)=b^{2}-8 b
$$

must be a perfect square.
Let $b^{2}-8 b=m^{2}(m \in \mathbf{N})$. Then
$$
\begin{array}{l}
(b-4)^{2}-m^{2}=16 \\
\Rightarrow(b-4+m)(b-4-m)=16 \\
\Rightarrow\left\{\begin{array} { l } 
{ b - 4 + m = 8 , } \\
{ b - 4 - m = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
b-4+m=4, \\
b-4-m=4
\end{array}\right.\right. \\
\quad \text { or }\left\{\begin{array} { l } 
{ b - 4 + m = - 2 , } \\
{ b - 4 - m = - 8 }
\end{array} \text { or } \left\{\begin{array}{l}
b-4+m=-4, \\
b-4-m=-4
\end{array}\right.\right. \\
\Rightarrow b=9,8,-1,0 .
\end{array}
$$

Therefore, the maximum possible value of $b$ is 9.