3. If three integers $a, b, c (a \neq 0)$ make the equation $a x^{2}$ $+b x+c=0$ have two roots $a$ and $b$, then $a+b+c$ equals. $\qquad$
3. 18 .

From the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
a+b=-\frac{b}{a}, \\
a b=\frac{c}{a} .
\end{array}\right.
$$

From (1), we have $a^{2}+a b+b=0, (a+1) b=-a^{2}$.
Since $a \neq 0$, then $a+1 \neq 0$, $b \neq 0$. Thus,
$$
b=-\frac{a^{2}}{a+1}=-\frac{(a+1-1)^{2}}{a+1}=-a+1-\frac{1}{a+1} \text {, }
$$

Since $b$ is an integer, it follows that $a+1$ divides 1. Since $a+1 \neq 1$ (because $a \neq 0$), then $a+1=-1, a=-2$. Therefore, $b=4, c=a^{2} b$ $=16$. Hence, $a+b+c=18$.