2. [4 points] Find all integer parameter triples $a, b$, and $c$, for each of which the system of equations

$$
\left\{\begin{array}{l}
a x+2 y+c z=c \\
3 x+b y+4 z=4 b
\end{array}\right.
$$

has no solutions.
Answer: $(-6, -1, -8), (-3, -2, -4), (3, 2, 4)$.

Solution. A system of two linear equations has no solutions if and only if the coefficients of the variables in the equations are proportional to each other, but not proportional to the free terms. It is also impossible for the coefficients of one of the variables to be zero in both equations ${ }^{1}$. We get

$$
\frac{a}{3}=\frac{2}{b}=\frac{c}{4} \neq \frac{c}{4 b}
$$

From the first equality, it follows that $a b=6$, and from the second, that $b c=8$. Since the numbers $a, b$, and $c$ are integers, it follows that the number $b$ is a divisor of two. Thus, there are four possible cases:

$$
\begin{aligned}
& b=1 \Rightarrow a=6, c=8 \\
& b=-1 \Rightarrow a=-6, c=-8 \\
& b=-2 \Rightarrow a=-3, c=-4
\end{aligned}
$$[^2]$b=2 \Rightarrow a=3, c=4$.

Of these, all except the first one violate the inequality in (??). Therefore, the condition of the problem is satisfied by three triples of integers: $(-6, -1, -8), (-3, -2, -4), (3, 2, 4)$.