What are the necessary and sufficient conditions for the following system of equations:

$$
\begin{aligned}
& x=b y+c z \\
& y=c z+a x \\
& z=a x+b y
\end{aligned}
$$

to have a non-zero solution?
Let's substitute $z$ from equation (3) into equations (1) and (2) and rearrange them in terms of $x$ and $y$:

$$
\begin{aligned}
& (1-a c) x-(b+b c) y=0 \\
& (a+a c) x-(1-b c) y=0
\end{aligned}
$$

These two equations do not uniquely determine $x$ and $y$, but only their ratio. However, they generally give two different values. $x=y=0$ is a solution in this case as well (then $z$ is also 0). Other solutions exist if the two equations give the same result for the ratio of $x$ and $y$. Multiply the first equation by $(a+a c)$ and the second by $(1-a c)$:

$$
\begin{aligned}
& (1-a c)(a+a c) x-(a+a c)(b+b c) y=0 \\
& (1-a c)(a+a c) x-(1-a c)(1-b c) y=0
\end{aligned}
$$

Here, the coefficients of $y$ must also be equal for a non-zero solution to exist, so it is necessary that

$$
(a+a c)(b+b c)=(1-a c)(1-b c)
$$

If this condition is satisfied, then, for example, $x$ can be chosen arbitrarily. We can calculate $y$ from equation (4) or (5) and $z$ from equation (3). (If in equation (4) or (5) all terms on the left side cancel out, we calculate $y$ from the other equation; if all terms on the left side of both equations are zero, we can choose $y$ arbitrarily.)

With this choice of variables, equations (1) and (2) are also satisfied, which follows immediately if we subtract the $c$-multiple of $z$, which we calculated from equation (3), from equations (4) and (5).

Thus, the system of equations has non-zero solutions if and only if equation (6) is satisfied, which can be written in the following form:

$$
2 a b c+a b+b c+c a-1=0
$$