Example 1 Given the quadratic equation in $x$
$$
x^{2}+c x+a=0
$$

has two integer roots which are exactly 1 more than the roots of the equation
$$
x^{2}+a x+b=0
$$

Find the value of $a+b+c$.
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition)
【Analysis】The problem gives the relationship between the roots of two quadratic equations. Let's assume the factored form of these two equations to find the coefficients of each term.

Solution: Let $\alpha, \beta (\alpha, \beta \in \mathbf{Z}, \alpha < \beta)$ be the two integer roots of equation (1). Then $\alpha-1, \beta-1$ are the two integer roots of equation (2).

Let $x^{2} + c x + a = (x - \alpha)(x - \beta)$.
Clearly, $a = \alpha \beta$.
Let $x^{2} + a x + b = (x - \alpha + 1)(x - \beta + 1)$.
Then $a = -(\alpha - 1) - (\beta - 1) = 2 - \alpha - \beta$
$$
\begin{array}{l}
\Rightarrow \alpha \beta = 2 - \alpha - \beta \\
\Rightarrow \alpha \beta + \alpha + \beta + 1 = 3 \\
\Rightarrow (\alpha + 1)(\beta + 1) = 3 .
\end{array}
$$

Since $\alpha, \beta \in \mathbf{Z}$, we have
$$
\left\{\begin{array}{l}
\alpha = 0, \\
\beta = 2
\end{array} \text { or } \left\{\begin{array}{l}
\alpha = -4, \\
\beta = -2 .
\end{array}\right.\right.
$$

Thus, $x^{2} + c x + a = x(x - 2)$
$\Rightarrow a = 0, c = -2$
$\Rightarrow x^{2} + a x + b = (x - 1)(x + 1)$
$\Rightarrow b = -1$;
$x^{2} + c x + a = (x + 4)(x + 2)$
$\Rightarrow a = 8, c = 6$
$\Rightarrow x^{2} + a x + b = (x + 5)(x + 3)$
$\Rightarrow b = 15$.
In summary, $a + b + c = -3$ or 29.