Working with a Cyclic System of Equations I recently found the following problem:

Find all non-negative reals $(x,y,z,w)$ such that:
$$\begin{cases}x^2 = 3y + 10z + 5w\\
y^2 = 3z + 10w + 5x\\
z^2 = 3w + 10x + 5y\\
w^2 = 3x + 10y + 5z\end{cases}$$

After looking at previous MSE forums regarding cyclic systems, the main piece of advice I got was to try adding everything up. That yields: $x^2 + y^2 + z^2 + w^2 = 18(x+y+z+w)$ but I don't know how to continue on from there.
After entering the problem into WolframAlpha, I figured out that the only solutions that actually work are $(0,0,0,0)$ and $(18,18,18,18).$ That makes me want to find a way to prove that $x=y=z=w$ but I don't really get how to.
Any help would be greatly appreciated!

Just a partial answer :
suppose we have $x \leq y \leq z \leq w  $,
then  $x^2 \leq y^2 \leq z^2 \leq w^2 $ which we can use to write
$$3y + 10z + 5w \leq 3z + 10w + 5x \iff 3(y-z)+10(z-w)+5(w-x) \leq 0 $$
$$ 3z + 10w + 5x \leq 3w + 10x + 5y \iff 3(z-w)+10(w-x)+5(x-y) \leq 0 $$
$$3w + 10x + 5y \leq 3x + 10y + 5z \iff 3(w-x)+10(x-y)+5(y-z) \leq 0$$
Summing all the eq. we obtain :
$$3(y-x)+10(z-y)+5(w-z) \leq 0 $$
but if at least one of the three terms is strictly greater than $0$ (e.g. $y>x$) we can notice that
$$3(y-x)+10(z-y)+5(w-z) > 0 $$
which is a contradiction.

I don't think this method can be extended to other possible cases.
