An algebra problem by Lucas Nascimento

Algebra Level 2

[ 2 ( 3 + 1 ) ( 3 2 + 1 ) ( 3 4 + 1 ) ( 3 8 + 1 ) ⋯ ( 3 128 + 1 ) + 1 ] 1 / 128 = ? \left[2(3+1)(3^2+1)(3^4+1)(3^8+1)\cdots(3^{128}+1)+1\right]^{1/128}=\, ?


The answer is 9.

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Lucas Nascimento by Lucas Nascimento , 20, Brazil

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1 solution

Brian Moehring
Feb 8, 2017

Relevant wiki: Difference Of Squares

Write 2 = 3 − 1 2 = 3-1 so that

2 ( 3 + 1 ) ( 3 2 + 1 ) ( 3 4 + 1 ) … ( 3 128 + 1 ) = ( 3 − 1 ) ( 3 + 1 ) ( 3 2 + 1 ) ( 3 4 + 1 ) … ( 3 128 + 1 ) = ( 3 2 − 1 ) ( 3 2 + 1 ) ( 3 4 + 1 ) … ( 3 128 + 1 ) = ( 3 4 − 1 ) ( 3 4 + 1 ) … ( 3 128 + 1 ) ⋮ = ( 3 128 − 1 ) ( 3 128 + 1 ) \ = 3 256 − 1 \begin{aligned} 2(3+1)(3^2+1)(3^4+1)\ldots(3^{128}+1) &= (3-1)(3+1)(3^2+1)(3^4+1)\ldots(3^{128}+1) \\ &= (3^2-1)(3^2+1)(3^4+1)\ldots(3^{128}+1) \\ &= (3^4-1)(3^4+1)\ldots(3^{128}+1) \\ &\qquad \vdots \\ &= (3^{128}-1)(3^{128}+1) \\\ &= 3^{256}-1 \end{aligned}

Therefore, the entire expression simplifies to

[ ( 3 256 − 1 ) + 1 ] 1 / 128 = [ 3 256 ] 1 / 128 = 3 2 = 9 . \big[(3^{256}-1)+1\big]^{1/128} = \big[3^{256}\big]^{1/128} = 3^2 = \boxed{9}.

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