Equation with huge number of nested square roots: Find all real roots of equation:
$$
\underbrace{\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}}}_\text{2018 roots}=x
$$

I know that the answers are 0 and 3. What is common about those values is that each time, the square root is applied, the result is exactly $x$, therefore the number of iterations doesn't really matter.
To clarify that, declare the sequence of $a_i(x)$ as:
$$
a_0(x)=x;\quad a_n(x)=\sqrt{x+2a_{n-1}(x)}
$$
So, the equation of the problem can be written as $a_{2018}(x)=x$.
If it could be proven that for given $x$ the sign of $a_n-a_{n-1}$ is same for any $n$ (i.e. the sequence is monotonously increasing, decreasing or constant) then, since $a_{2018}(x)=a_{0}(x)$ the sequence must be constant, in particular $a_1(x)=a_0(x)$, ie. $\sqrt{3x}=x$, from which it follows that $x$ must be either $0$ or $3$. And indeed, it can be easily checked that the sequence is constant for those values.
To my disappointment, I can't find a way to prove that sign of $a_n-a_{n-1}$ is same. It may be really wrong.

Hint:
$\displaystyle\;\; a_n-a_{n-1}=\sqrt{x+2a_{n-1}}-\sqrt{x+2a_{n-2}}=\frac{2(a_{n-1}-a_{n-2})}{\sqrt{x+2a_{n-1}}+\sqrt{x+2a_{n-2}}}\,$, so the differences between consecutive terms have the same sign.
