# An equation with infinite nested square roots

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

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

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.

• Very nicely solved and super quick! Jul 5 at 15:10
• There are no additional solutions in the complex domain, as long as we assume the $\sqrt{}$ symbol means any consistent function satisfying $(\sqrt{z})^2 = z$ for all $z$, and if $z$ is a positive real number, so is $\sqrt{z}$. Jul 6 at 17:02

One of the possible solutions is:

$$x=0$$