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Question Number 146863 by lyubita last updated on 16/Jul/21

Commented by MJS_new last updated on 16/Jul/21

question 144084

$${question}\:\mathrm{144084} \\ $$

Answered by Olaf_Thorendsen last updated on 16/Jul/21

a+(1/a) = −1 a^2 +a+1 = 0 a = ((−1±i(√3))/2) = e^((2iπ)/3) or e^((4iπ)/3) 1st case : a = e^((2iπ)/3) n = 1234567891011 = 3k, k∈Z because 1+2+3+4+...+1+0+1+1 = 48 = 3p ⇒ a^n = (e^((2iπ)/3) )^(3k) = e^(2ikπ) = 1 m = 1110987654321 = 3k′, k′∈Z because 1+1+1+0+...+4+3+2+1 = 48 = 3p (1/a^m ) = (e^(−((2iπ)/3)) )^(3k′) = e^(−2ik′π) = 1 ⇒ a^n +(1/a^m ) = 1+1 = 2 Same result with a = e^((4iπ)/3)

$${a}+\frac{\mathrm{1}}{{a}}\:=\:−\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{a}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${a}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:\mathrm{or}\:{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{1st}\:\mathrm{case}\::\:{a}\:=\:{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \\ $$$${n}\:=\:\mathrm{1234567891011}\:=\:\mathrm{3}{k},\:{k}\in\mathbb{Z} \\ $$$$\mathrm{because}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{1}\:=\:\mathrm{48}\:=\:\mathrm{3}{p} \\ $$$$\Rightarrow\:{a}^{{n}} \:=\:\left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}{k}} \:=\:{e}^{\mathrm{2}{ik}\pi} \:=\:\mathrm{1} \\ $$$${m}\:=\:\mathrm{1110987654321}\:=\:\mathrm{3}{k}',\:{k}'\in\mathbb{Z} \\ $$$$\mathrm{because}\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{0}+...+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}\:=\:\mathrm{48}\:=\:\mathrm{3}{p} \\ $$$$\frac{\mathrm{1}}{{a}^{{m}} }\:=\:\left({e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}{k}'} \:=\:{e}^{−\mathrm{2}{ik}'\pi} \:=\:\mathrm{1} \\ $$$$\Rightarrow\:{a}^{{n}} +\frac{\mathrm{1}}{{a}^{{m}} }\:=\:\mathrm{1}+\mathrm{1}\:=\:\mathrm{2} \\ $$$$\mathrm{Same}\:\mathrm{result}\:\mathrm{with}\:{a}\:=\:{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{3}}} \\ $$

Commented by lyubita last updated on 16/Jul/21

thank you so much. are you from french?

$${thank}\:{you}\:{so}\:{much}.\:{are}\:{you}\:{from}\:{french}? \\ $$

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