Just a simple question #7

Number Theory Level 3

2014 × 2015 × 2016 × 2017 + 1 = . . . . \sqrt{2014\times2015\times2016\times2017 + 1} = . . . .

201 5 2 + 2014 2015^{2} + 2014 201 7 2 + 2015 2017^2 + 2015 201 4 2 + 2015 2014^2 + 2015 201 4 2 + 2017 2014^2 + 2017 201 6 2 + 2015 2016^2 + 2015

Andronikus Lumembang by Andronikus Lumembang , 25, Indonesia

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1 solution

Asume 2015 2015 as M M

So we can rewrite the equation as :

= ( M − 1 ) × M × ( M + 1 ) × ( M + 2 ) + 1 = \sqrt{(M-1)\times M\times(M+1)\times(M+2)+1} = M × ( M + 1 ) × ( M − 1 ) × ( M + 2 ) + 1 = \sqrt{M\times(M+1)\times(M-1)\times(M+2)+1} = ( M 2 + M ) × ( M 2 + M − 2 ) + 1 = \sqrt{(M^2+M)\times(M^2+M-2)+1}

Asume M 2 + M M^2 + M as N N

Rewrite the last equation as :

= ( N ) × ( N − 2 ) + 1 = \sqrt{(N)\times(N-2)+1}

= N 2 − 2 N + 1 = \sqrt{N^2 - 2N + 1}

= ( N − 1 ) 2 = \sqrt{(N-1)^2}

= N − 1 = N-1 ,

change it back N − 1 = M 2 + M − 1 = 201 5 2 + 2015 − 1 N - 1 = M^2 + M - 1 = 2015^2 + 2015 - 1 = 201 5 2 + 2014 = \boxed{2015^2 + 2014}

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Andronikus Lumembang - 6 years ago

Nice solution, mine different with yours I try the small number instead the question i.e 1 × 2 × 3 × 4 + 1 \sqrt{1\times2\times3\times4 + 1} then I found out that the solution is 2 2 + 1 2^{2}+1

Resha Dwika Hefni Al-Fahsi - 4 years, 12 months ago

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