Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$

I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?

Let $a = \sqrt{3}$, so $x = \sqrt{a-\sqrt{a+x}}$. Let's call this equation $(*)$.
Let $y = \sqrt{a+x}$ so $x = \sqrt{a-y}$. Therefore

*

*$x^2 = a-y$

*$y^2 = a+x$
Subtracting 2. from 1. we have $x^2 - y^2 = -(x+y)$. Let's consider the following cases:

*

*If $x+y = 0$, then $x = -\sqrt{a+x}\le 0$, but from $(*)$ we have $x\ge 0$, so we must have $x = 0$. It's easy to see that this doesn't work.


*If $x+y \neq 0$, we have $\begin{aligned}\frac{x^2-y^2}{x+y} = -1 \therefore x-y = -1\end{aligned}$, i.e., $x+1 = \sqrt{a+x}$, so $(x+1)^2 = x^2+2x+1 = a+x$, therefore $x^2+x+1-a = 0$, which can be solved as a quadratic equation to get $\begin{aligned}x = \frac{-1+ \sqrt{1-4(1-a)}}{2} = \frac{-1+\sqrt{4a-3}}{2}\end{aligned}$ (notice that this is the only positive solution).
