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Question Number 212513 by MrGaster last updated on 16/Oct/24

$$\mathrm{certificate}:$$$$x^{2n}-1$$$$\mathrm{Can}\mathrm{always}\mathrm{be}x+1\mathrm{be}\mathrm{divisible}$$

Answered by A5T last updated on 16/Oct/24

$$x\equiv -1(modx+1)\Rightarrow x^{2n}-1\equiv {(-1)}^{2n}-1(modx+1)$$$$\Rightarrow x^{2n}-1\equiv 0(modx+1)\Rightarrow x+1\mid x^{2n}-1$$

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