An equation with infinite nested square roots

Question

Find all values of $x$ that satisfy the following equation:

$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}} =x\sqrt{x-\sqrt{x-\sqrt{x-\ldots}}}$

Answer 1

I think that the only valid solutions are

$x=0$ and $x=2$.

Reasoning

For the following purposes, we'll assume that the square root operator is a mapping from the non-negative real numbers to the non-negative reals. $x=0$ is a trivial solution so we'll assume, additionally, that $x>0$. In particular both sides of the equation are positive.
Let $A$ represent the left hand side and $xB$ the right hand side ($A$ and $B>0$).
Then $A$ satisfies $$A^2 - x = A$$ In other words $$ A = \frac{1 + \sqrt{1+4x}}{2}$$ where we've taken the positive sign in the solution to ensure $A>0$.
$B$ satisfies the equation $$x- B^2 = B$$ which means that $$B = \frac{-1 + \sqrt{1+4x}}{2} $$ Then overall, we have $$ \frac{1 + \sqrt{1+4x}}{2} = x \left( \frac{-1 + \sqrt{1+4x}}{2} \right)$$ or simplifiying $$\sqrt{1+4x} = \frac{x+1}{x-1} $$.
Squaring both sides and multiplying across by $(x-1)^2$ we find that some terms cancel and we end up with $$4x^2 (x-2) = 0$$ which gives $x=2$ as the only other solution.
We must check this works with the equation above and, plugging it back in clearly gives a valid solution.
We must also check that $x=1$ is not a solution since we may be multiplying across by zero above but a quick check ensures us that this does not work.

Answer 2

One of the possible solutions is:

$x=0$