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

How can I prove this equality?

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

Solution 1:

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 $$ AB=\frac{1}{2}(A^2+B^2-(A-B)^2)=\frac{1}{2}(2x+1-1)=x. $$

Solution 2:

Let $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=y\implies x+y=y^2\implies y^2-y-x=0\implies y=\dfrac{1\pm\sqrt{1+4x}}2$

As $x>1,y>1\implies y=\dfrac{1+\sqrt{1+4x}}2$

$\sqrt{x-\sqrt{x-\sqrt{x-\cdots}}}=z\implies x-z=z^2\iff z^2+z-x=0\implies z=\dfrac{-1\pm\sqrt{1+4x}}2$

As $x>1, z<1\implies z=\dfrac{-1+\sqrt{1+4x}}2$