We have a system of equations for variables A, x, y, and z, where A is a constant.
$x + y = Az$,
$z + x = Ay$,
$y + z = Ax$.
Obviously, this system of equations is unsolvable if there were no coefficient A, but with the introduction of this coefficient I am aware that there may be multiple solutions, if there are any. If there are multiple, how can they be expressed? And if there are none, why?
Adding these equations together, we get
$$2x+2y+2z = A(x+y+z)$$
From this we either get that $A=2$ or $x+y+z=0$. In the first case:
$$z+x=2y$$
$$z = 2y-x$$
and from the first equation:
$$x+y = 2z = 2(2y-x)$$
$$x+y = 4y-2x$$
$$3x = 3y$$
$$x = y$$
A similar process can be used to show that $x=z$, so if $A=2, x=y=z$, and it is easy to verify that any $x=y=z$ works with $A=2$. In the second case:
$$x+y+z=0$$
$$z = -x-y$$
$$x+y = A(-x-y)$$
$$(x+y)(A+1) = 0$$
$$x+y = 0$$
OR
$$A = -1$$
If $x+y = 0$:
$$y = -x$$
$$z = 0$$
but then
$$0+x = A(-x)$$
and $x = 0$ or $A=-1$, so $x=y=z=0$ and $A$ is any real number is also one solution. If $A=-1$, we can easily verify that any $x, y, z$ so that $x+y+z=0$ works, providing a complete characterization of solutions. In summary:
$A = 2, x=y=z$
$x=y=z=0$
$A = -1, x+y+z=0$