Nested roots

$\frac{1}{\sqrt{2011+\sqrt{2011^{2}-1}}}=\frac{1}{\sqrt{1006+2\sqrt{1006 \times 1005}+1005}}$

= $\frac{1}{\sqrt{(\sqrt{1006}+\sqrt{1005})^2}}$

= $\frac{1}{\sqrt{1006}+\sqrt{1005}}=\frac{\sqrt{1006}-\sqrt{1005}}{1006-1005}$

= $\sqrt{1006}-\sqrt{1005}$

Let $x=\sqrt{a+\sqrt{a^2-1}}$ and $y=\sqrt{a-\sqrt{a^2-1}}$ . Then

$\begin{aligned} xy &= \sqrt{a+\sqrt{a^2-1}} \sqrt{a-\sqrt{a^2-1}}\\ &= \sqrt{a^2-(a^2-1)}\\ &= 1 \end{aligned}$ We want to express $x^{-1}=y$ as $\sqrt{m}-\sqrt{n}$ for integers $m$ and $n$ . Note that $\begin{aligned} (x+y)^2 &= x^2+2xy+y^2\\ &= a+\sqrt{a^2-1} +2 + a-\sqrt{a^2-1} \\ &= 2a+2 \end{aligned}$ Thus $x+y=\sqrt{2a+2}$ . Similarly $x-y=\sqrt{2a-2}$ . So

$\begin{aligned} y &= \frac{1}{2} (\sqrt{2a+2}-\sqrt{2a-2})\\ &= \sqrt{\frac{2a+2}{4}}-\sqrt{\frac{2a-2}{4}} \\ &= \sqrt{\frac{a+1}{2}}-\sqrt{\frac{a-1}{2}}\\ &= \sqrt{m}-\sqrt{n} \end{aligned}$ The sum $m+n=a$ . Therefore when $a=2011$ , the solution is $\boxed{2011}$ .

Let $m = \frac{x+1}2, n = \frac{x-1}2.$ Then $x-\sqrt{x^2-1} = m+n-2\sqrt{mn},$ so $\frac1{x+\sqrt{x^2-1}} = m+n-2\sqrt{mn}$ , and taking the square root of both sides gives us the desired result. In this case, $x = 2011, m = 1006, n = 1005, m+n = x = \fbox{2011}$ .

This demonstrates that the solution works; I got it by looking at the first equation and setting $m+n = x$ and $4mn = x^2-1$ , and solving for $m$ and $n$ .