Question Number 212513 by MrGaster last updated on 16/Oct/24
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}{n}} −\mathrm{1} \\ $$$$\mathrm{Can}\:\mathrm{always}\:\mathrm{be}\:{x}+\mathrm{1}\:\mathrm{be}\:\mathrm{divisible}\:\:\:\:\:\:\:\:\: \\ $$
Answered by A5T last updated on 16/Oct/24
$${x}\equiv−\mathrm{1}\left({mod}\:{x}+\mathrm{1}\right)\Rightarrow{x}^{\mathrm{2}{n}} −\mathrm{1}\equiv\left(−\mathrm{1}\right)^{\mathrm{2}{n}} −\mathrm{1}\left({mod}\:{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}{n}} −\mathrm{1}\equiv\mathrm{0}\left({mod}\:{x}+\mathrm{1}\right)\Rightarrow{x}+\mathrm{1}\mid{x}^{\mathrm{2}{n}} −\mathrm{1} \\ $$