Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$ 
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$

I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y = 5.$ (I found $5$ as a solution of the equation by hit and trial method).
Therefore, $x = 0$ or $3.$
I'm wondering if there's any  another method to solve it as the repeated squaring step seems to be somewhat absurd.
 A: It's not so hard to imagine someone guessing the solutions $0$ and $3$. The solution $0$ is something one can see from the positions of all of the "$x$". The solution $3$ might be inspired by asking what would make $\sqrt{3x}$ rational.
Now the idea is to prove that there can be no more than two solutions by showing that the left side is concave down. If $f(x)$ is positive and concave down, then first of all $2\sqrt{x+f(x)}$ is also positive. But:
$$\begin{align}
\frac{d^2}{dx^2}2\sqrt{x+f(x)}
&=\frac{d}{dx}\frac{1+f'(x)}{\sqrt{x+f(x)}}\\
&=\frac{f''(x)\sqrt{x+f(x)}-\frac{(1+f'(x))^2}{2\sqrt{x+f(x)}}}{(x+f(x))}\\
&=\frac{f''(x)(x+f(x))-\frac12(1+f'(x))^2}{(x+f(x))\sqrt{x+f(x)}}
\end{align}$$
By assumption, $f''(x)$ is negative and $f(x)$ is positive, so this expression is also negative. So $2\sqrt{x+f(x)}$ is also concave down.
So since $2\sqrt{3x}$ is positive and concave down, so is  $2\sqrt{x+2\sqrt{3x}}$. And so is $2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}$. And so is $2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}$.
Since that last expression $F(x)$ is concave down, there can be at most two solutions to $\frac12F(x)=x$.
A: We have,
$$\begin{align}&\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x≥0\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=2x\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=3x\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=\sqrt{3x}\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=2\sqrt{3x}\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=x+2\sqrt{3x}\\ 
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}} = \sqrt{x+2\sqrt{3x}}\end{align}$$
Let, $\sqrt{x+2\sqrt{3x}}=u,\thinspace u≥0$ then we get,
$$\begin{align}\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}=u\end{align}$$
Then, we see that,
$$\begin{align}&\sqrt{x+2u} = u\\
\iff &\sqrt{x+2\sqrt{x+2u}} = u\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}=u\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}= u\end{align}$$
Therefore, we have
$$\begin{align}&x+2u= x+2\sqrt{3x}\\
\iff &u^2=3x\\
\iff &x+2\sqrt{3x}=3x\\
\iff &\sqrt{3x}=x\\
\iff &x\in\left\{0,3\right\}.\end{align}$$
A: We want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x    $
So, we want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2x}}}}=x    \quad \quad (E)$

*

*We see that if $\sqrt{x+2x} =x$, then, $x$ is a solution of $(E)$.
Therefore, $0$ and $3$ are two solutions of $(E)$.

*Suppose that $x$ is a solution of $(E)$ and $x\neq 0$. Then $x>0$ and $F(x)=1$ with
$F(x)=\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{ \dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}}}}}$

*$F$ is strictly decreasing on $(0,+\infty)$, so, the equation $F(x)=1$ has at most one solution.

We can conclude that $(E)$ has exactly two solutions, $0$ and $3$.
A: Set $F(x,y) = \sqrt{x+2y}$ for $x,y\geq 0$. Note that the function $F(x,y)$ is strictly increasing in $y$, i.e., if $y_1<y_2$, then $F(x,y_1)<F(x,y_2)$.
Set
$$
y_0:=x;
$$
$$
y_1:=F(x,y_0);
$$
$$
y_2:=F(x,y_1);
$$
$$
y_3:=F(x,y_2);
$$
$$
y_4:=F(x,y_3),
$$
and then, the equation is that $y_4 = y_0$.
Assume that $y_0<y_1$. Then $F(x,y_0) < F(x, y_1)$, i.e., $y_1<y_2$. Similarly, we get $y_2<y_3$ and $y_3<y_4$. I.e., in this case, we get that $y_0<y_1<y_2<y_3<y_4$, so the equality $y_4 = y_0$ cannot be satisfied.
Similarly, if we assume that $y_0>y_1$, then we get $y_0>y_1>y_2>y_3>y_4$, and again, the equality $y_4 = y_0$ cannot be satisfied.
Therefore, for the equation to be satisfied, $y_1$ has to be equal to $y_0$, i.e., $\sqrt{3x}=x$, which means that $x$ can be either $0$ or $3$. Substituting these values to the original equation, we see that both of them satisfy it.
