Squares and Rationalization

Question

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:

${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$ $={\sqrt{2011+\sqrt{2011^2-1}}\over{2011+\sqrt{2011^2-1}}}$ $={({\sqrt{2011+\sqrt{2011^2-1}})(\sqrt{2011^2-1}-2011)}\over{2011^2-2012}}$

<p>Squaring,</p>

<p>$m+n-2\sqrt{mn}$
$={{{{(2011+\sqrt{2011^2-1}})\{4022(2011-\sqrt{2011^2-1})-1\}}}\over{2011^4+2012^2-2\cdot{2011^2}\cdot{2012}}}$</p>

<p>This goes nowhere</p>

And

${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$

<p>Squaring</p>

<p>${m+n-2\sqrt{mn}}$
$={{\sqrt{2011^2-1}-2011}\over{(2011)(2010)-1}}$</p>

<p>Which again, probably goes nowhere.</p>

Can anyone help?

Answer 1

Hint

$\dfrac1{\sqrt{2011+\sqrt{2011^2-1}}}$ $=\dfrac1{\sqrt{2011+2\sqrt{1005*1006}}}$ $=\dfrac1{\sqrt{\sqrt{1005}+\sqrt{1006}}^2}$

Answer 2

after my hint we have $$\sqrt{2011-\sqrt{2011^2-1}}=\sqrt{m}-\sqrt{n}$$ after squaring we get $$2011-2\sqrt{1011030}=m+n-s\sqrt{mn}$$ we set $$m+n=2011$$ $$nm=1011030$$ solving this System we get $$m_1=1005$$ $$m_2=1006$$ can you finish?

Answer 3

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

$$\implies\sqrt{a+\sqrt{a^2-1}}\cdot\sqrt{a-\sqrt{a^2-1}}=?$$