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

From (1), we get \(a^{2}+a \cdot b+b=0\), \((a+1) \cdot b=-a^{2}\). Since \(a \neq 0\), it follows that \(a+1 \neq 0\),
$$
b=-\frac{a^{2}}{a+1}=-\frac{(a+1-1)^{2}}{a+1}=-(a+1)+2-\frac{1}{a-1} .
$$

Since \(b\) is an integer,
$$
a+1=-1, a=-2 . b=4, c=a^{2} b=16 . a+b+c=18 .
$$