1092. Solve the equation

$$
a x^{2}+b x+b=0
$$

for all possible values of the parameters $a$ and $b$, given that all its roots are integers.
$\triangle$ Let's consider several cases.

1) Let $a \neq 0$ and $b \neq 0$.

Denote the roots of the quadratic equation by $x_{1}$ and $x_{2}$. By Vieta's formulas, we get:

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a} \\
x_{1} x_{2}=\frac{b}{a}
\end{array}\right.
$$

Add the equations of this system:

$$
x_{1}+x_{2}+x_{1} x_{2}=0
$$

Solve the last equation. For this, express $x_{2}$ in terms of $x_{1}$:

$$
x_{2}=-\frac{x_{1}}{x_{1}+1}=-\frac{\left(x_{1}+1\right)-1}{x_{1}+1}=-1+\frac{1}{x_{1}+1}
$$

Then either $x_{1}+1=1$, i.e., $x_{1}=0$, or $x_{1}+1=-1$, i.e., $x_{1}=-2$. The value $x_{1}=0$ should be discarded, otherwise $b=0$. For $x_{1}=-2$, find $x_{2}: x_{2}=-2$.

Since

$$
\frac{b}{a}=x_{1} \cdot x_{2}=(-2) \cdot(-2)=4
$$

for this solution to exist, it is necessary that $b=4a$. If $b \neq 4a$, it is not true that both roots of the quadratic equation are integers.

2) Let $a=0, b \neq 0$.

In this case, the given equation becomes linear:

$$
b x+b=0
$$

Solve for $x: x=-1$.

3) Let $a \neq 0, b=0$.

Then the equation has a single root: $x=0$.

4) Let $a=0, b=0$.

The equation takes the form:

$$
0 \cdot x^{2}+0 \cdot x+0=0
$$

It is satisfied by any value of $x$, including non-integers, so this case does not satisfy the condition of the problem.

Answer: if $a b \neq 0$ and $b=4 a$, then $x_{1}=x_{2}=-2$; if $a b \neq 0$ and $b \neq 4 a$ or $a=b=0$, then there are no solutions; if $a=0, b \neq 0$, then $x=-1$; if $a \neq 0, b=0$, then $x=0$.