How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$ How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?

If we consider an infinite chain.
Suppose $x = \sqrt{20 +\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}}}$
$x = \sqrt{20 +x}\\
x^2 = 20 + x\\
x^2 - x - 20 = 0\\
(x-5)(x+4) = 0$
$x$ must be greater than $0, x = 5$
and $y = \sqrt{20 -\sqrt{20-\sqrt{20-\sqrt{20-\sqrt{\cdots}}}}}$
$y = \sqrt{20 - y}\\
y^2 + y - 20=0\\
y = 4$
$x-y = 1$
As we add more terms under those square roots 
$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}$
converges toward 1.
