<s>[INST] 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.
 [/INST] 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 $\frac12F(x)=x$.
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