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Question Number 212513 by MrGaster last updated on 16/Oct/24
certificate: x^(2n) −1 Can always be x+1 be divisible
certificate:x2n−1Canalwaysbex+1bedivisible
Answered by A5T last updated on 16/Oct/24
x≡−1(mod x+1)⇒x^(2n) −1≡(−1)^(2n) −1(mod x+1) ⇒x^(2n) −1≡0(mod x+1)⇒x+1∣x^(2n) −1
x≡−1(modx+1)⇒x2n−1≡(−1)2n−1(modx+1)⇒x2n−1≡0(modx+1)⇒x+1∣x2n−1

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