Squares and Rationalization I was solving this question, and I'm hitting a wall.

If ${1\over{\sqrt{2011+\sqrt{2011^2-1}}}}=\sqrt{m}-\sqrt{n}$, where $m$ and $n$ are positive integers, what is the value of $m+n$?

I tried to solve this question with two approaches:


${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$
    $={\sqrt{2011+\sqrt{2011^2-1}}\over{2011+\sqrt{2011^2-1}}}$
    $={({\sqrt{2011+\sqrt{2011^2-1}})(\sqrt{2011^2-1}-2011)}\over{2011^2-2012}}$
Squaring,
$m+n-2\sqrt{mn}$
    $={{{{(2011+\sqrt{2011^2-1}})\{4022(2011-\sqrt{2011^2-1})-1\}}}\over{2011^4+2012^2-2\cdot{2011^2}\cdot{2012}}}$
This goes nowhere


And


${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$
Squaring
${m+n-2\sqrt{mn}}$
    $={{\sqrt{2011^2-1}-2011}\over{(2011)(2010)-1}}$
Which again, probably goes nowhere.


Can anyone help?

HINT:
$$(a+\sqrt{a^2-1})(a-\sqrt{a^2-1})=a^2-(a^2-1)=?$$
$$\implies\sqrt{a+\sqrt{a^2-1}}\cdot\sqrt{a-\sqrt{a^2-1}}=?$$
