Flash back of the year 2011

Number Theory Level 3

I f 1 2011 + 201 1 2 − 1 = m − n , If \frac{1}{\sqrt{2011+\sqrt{2011^{2}-1}}} =\sqrt{m} - \sqrt{n} , where m and n are positive integers , what is the value of m + n proof is needed compulsary. according to me the poster's of solution are the solvers


The answer is 2011.

Sudoku Subbu by sudoku subbu , 19, India

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1 solution

Chew-Seong Cheong
Jan 21, 2015

It is given that:

1 2011 + 201 1 2 − 1 = m − n \quad \quad \quad \dfrac {1}{\sqrt{2011+\sqrt{2011^2-1}}} = \sqrt{m}-\sqrt{n}

( 1 2011 + 201 1 2 − 1 ) 2 = ( m − n ) 2 \quad \quad \quad \left( \dfrac {1}{\sqrt{2011+\sqrt{2011^2-1}}} \right)^2 = \left( \sqrt{m}-\sqrt{n} \right)^2

1 2011 + 201 1 2 − 1 = m + n − 2 m n \quad \quad \quad \dfrac {1}{2011+\sqrt{2011^2-1}} = m + n - 2\sqrt{mn}

2011 − 201 1 2 − 1 ( 2011 + 201 1 2 − 1 ) ( 2011 − 201 1 2 − 1 ) = m + n − 2 m n \quad \quad \quad \dfrac {2011-\sqrt{2011^2-1}}{(2011+\sqrt{2011^2-1})(2011-\sqrt{2011^2-1})} = m + n - 2\sqrt{mn}

2011 − 201 1 2 − 1 201 1 2 − 201 1 2 + 1 = m + n − 2 m n \quad \quad \quad \dfrac {2011-\sqrt{2011^2-1}}{2011^2- 2011^2+1} = m + n - 2\sqrt{mn}

⇒ 2011 − 201 1 2 − 1 = m + n − 2 m n \quad \quad \Rightarrow 2011-\sqrt{2011^2-1} = m + n - 2\sqrt{mn}

⇒ m + n = 2011 ⇒ 4 m n = 201 1 2 − 1 \quad \quad \Rightarrow m + n = \boxed{2011} \quad \Rightarrow 4mn = 2011^2 - 1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

absolutely right you are the real solver my friend

sudoku subbu - 6 years, 4 months ago

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