TITLE: Proving $\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$
QUESTION [10 upvotes]: How can I prove this equality? 

$$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$

REPLY [11 votes]: Assuming convergence of both expressions, you have $A^2=x+A$ and $B^2=x-B$, so
$$
(A-B)(A+B)=A^2-B^2=(A+B).
$$
We conclude that $A-B=1$.  Then $$A^2+B^2=2x+(A-B)=2x+1,$$ and therefore
$$
AB=\frac{1}{2}(A^2+B^2-(A-B)^2)=\frac{1}{2}(2x+1-1)=x.
$$