TITLE: Find all the solutions of this equation:$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $ in reals.
QUESTION [8 upvotes]: Find all the solutions of this equation
$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $
My attempt:In the hint of the question it was written that show $x,y,z \in [-1,1]$ then add all equations and conclude $(x,y,z)=(-1,-1,-1),(0,0,0),(1,1,1)$ are the solutions.I can show that $x,y,z \in [-1,1]$like below:
By symmetry consider $x \ge y \ge z$ then if $x>1$ we have:
$4x^3-4x>0 \Rightarrow 4x^3-3x>x \Rightarrow z>x$
which is wrong.Now consider $x<-1$ we have:
$x<-1 \Rightarrow y<-1 \Rightarrow 4y^3-4y<0 \Rightarrow 4y^3-3y<y \Rightarrow x<y$
By the same way we can get $x,y,z \in [-1,1]$.By adding the equations we can get:
$x+y+z=x^3+y^3+z^3$
But what should I do now?

REPLY [2 votes]: I am assuming that you want integer solutions.  First note that if any of $x$, $y$, and $z$ is zero, then all of them are.  Suppose now that they are all nonzero.  Then, you have relations $y\mid x$, $z\mid y$, and $x\mid z$, whence $|x|=|y|=|z|=:t$.  Then, $x+3y=4y^3$ implies that $$4t\geq |x+3y|=\left|4y^3\right|=4t^3\,,$$
making $t=1$ the only possibility.  The rest is straightforward, and the only solutions are $(x,y,z)=(\alpha,\alpha,\alpha)$ with $\alpha\in\{-1,0,+1\}$.