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 $m+n$?
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> 
> 

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?