An algebra problem by swastik p (3)

Algebra Level 2

If 1 2011 + 201 1 2 1 = m n \dfrac{1}{\sqrt{2011+\sqrt{2011^2-1}}}=\sqrt{m}-\sqrt{n} , where m , n Z + m,n\in\mathbb{Z^+} , what is the value of m + n m+n ?


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1234 1009 2011 2012

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2 solutions

Rab Gani
May 25, 2018

square both sides, 1/(2011+√(〖2011〗^2-1)) = m+n - 2√mn 1/(2011+√(〖2011〗^2-1)) . (2011-√(〖2011〗^2-1))/(2011-√(〖2011〗^2-1)) = 2011-√((2012)(2010)), m+n = 2011, and 4mn=(2012)(2010), or mn = 503 . 2010 = 1006 . 1005

X X
May 25, 2018

2011 + 201 1 2 1 = 2011 + 2010 × 2012 \sqrt{2011+\sqrt{2011^2-1}}=\sqrt{2011+\sqrt{2010\times2012}} = ( 1005 ) 2 + ( 1006 ) 2 + 2 1005 1006 =\sqrt{(\sqrt{1005})^2+(\sqrt{1006})^2+2\sqrt{1005}\sqrt{1006}} = ( 1005 + 1006 ) 2 =\sqrt{(\sqrt{1005}+\sqrt{1006})^2} = 1005 + 1006 =\sqrt{1005}+\sqrt{1006} 1 2011 + 201 1 2 1 = 1 1005 + 1006 = 1006 1005 ( 1006 + 1005 ) ( 1006 1005 ) \dfrac{1}{\sqrt{2011+\sqrt{2011^2-1}}}=\frac1{\sqrt{1005}+\sqrt{1006}}=\frac{\sqrt{1006}-\sqrt{1005}}{(\sqrt{1006}+\sqrt{1005})(\sqrt{1006}-\sqrt{1005})} = 1006 1005 = m n =\sqrt{1006}-\sqrt{1005}=\sqrt{m}-\sqrt{n} m + n = 1006 + 1005 = 2011 m+n=1006+1005=2011

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