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

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


2 Answers 2


I think that the only valid solutions are

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


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.

  • $\begingroup$ Very nicely solved and super quick! $\endgroup$ Jul 5 at 15:10
  • $\begingroup$ 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 $\sqrt{z}$. $\endgroup$
    – aschepler
    Jul 6 at 17:02

One of the possible solutions is:



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