Question Number 146863 by lyubita last updated on 16/Jul/21
Commented by MJS_new last updated on 16/Jul/21
$${question}\:\mathrm{144084} \\ $$
Answered by Olaf_Thorendsen last updated on 16/Jul/21
$${a}+\frac{\mathrm{1}}{{a}}\:=\:−\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{a}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${a}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:\mathrm{or}\:{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{1st}\:\mathrm{case}\::\:{a}\:=\:{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \\ $$$${n}\:=\:\mathrm{1234567891011}\:=\:\mathrm{3}{k},\:{k}\in\mathbb{Z} \\ $$$$\mathrm{because}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{1}\:=\:\mathrm{48}\:=\:\mathrm{3}{p} \\ $$$$\Rightarrow\:{a}^{{n}} \:=\:\left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}{k}} \:=\:{e}^{\mathrm{2}{ik}\pi} \:=\:\mathrm{1} \\ $$$${m}\:=\:\mathrm{1110987654321}\:=\:\mathrm{3}{k}',\:{k}'\in\mathbb{Z} \\ $$$$\mathrm{because}\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{0}+…+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}\:=\:\mathrm{48}\:=\:\mathrm{3}{p} \\ $$$$\frac{\mathrm{1}}{{a}^{{m}} }\:=\:\left({e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}{k}'} \:=\:{e}^{−\mathrm{2}{ik}'\pi} \:=\:\mathrm{1} \\ $$$$\Rightarrow\:{a}^{{n}} +\frac{\mathrm{1}}{{a}^{{m}} }\:=\:\mathrm{1}+\mathrm{1}\:=\:\mathrm{2} \\ $$$$\mathrm{Same}\:\mathrm{result}\:\mathrm{with}\:{a}\:=\:{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{3}}} \\ $$
Commented by lyubita last updated on 16/Jul/21
$${thank}\:{you}\:{so}\:{much}.\:{are}\:{you}\:{from}\:{french}? \\ $$