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How can I prove this equality?

$$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$ if $$x>1$$

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

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    $\begingroup$ +1, though as is always the case with these problems, OP will need to show that $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ converges for $x > 1$ (and of course similarly $\sqrt{x-\sqrt{x-\sqrt{x-\cdots}}}$ converges for $x > 1$). $\endgroup$ – Alex Wertheim Dec 4 '14 at 17:36
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    $\begingroup$ @AWertheim, Adding math.stackexchange.com/questions/61048/… for the proof of convergence $\endgroup$ – lab bhattacharjee Dec 4 '14 at 17:40
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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 $$ AB=\frac{1}{2}(A^2+B^2-(A-B)^2)=\frac{1}{2}(2x+1-1)=x. $$

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  • $\begingroup$ +1 nicely done - not a radical in sight $\endgroup$ – nbubis Dec 4 '14 at 18:40

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