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Question Number 102771 by bramlex last updated on 11/Jul/20

x+(1/x) = −1 ⇒x^(1907) +(1/x^(1907) ) ?

$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:=\:−\mathrm{1}\:\Rightarrow\mathrm{x}^{\mathrm{1907}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{1907}} }\:?\: \\ $$

Answered by ajfour last updated on 11/Jul/20

x^2 +x+1=0 ⇒ x=ω, ω^2 ω^(1907) +(1/ω^(1907) )= (1/ω)+ω = ω^2 +ω=−1.

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}=\omega,\:\omega^{\mathrm{2}} \\ $$$$\:\:\omega^{\mathrm{1907}} +\frac{\mathrm{1}}{\omega^{\mathrm{1907}} }=\:\frac{\mathrm{1}}{\omega}+\omega\:=\:\omega^{\mathrm{2}} +\omega=−\mathrm{1}. \\ $$

Answered by floor(10²Eta[1]) last updated on 11/Jul/20

x^2 =−x−1 x^4 =x^2 +2x+1=(−x−1)+2x+1=x ⇒x^4 =x⇒x^(16) =x^4 =x x^(160) =x^(10) =x^4 .x^4 .x^2 =x.x.x^2 =x^4 =x ⇒x^(160) =x⇒x^(1600) =x^(10) =x ★x^(307) =(x^(10) )^(30) .x^7 =x^(30) .x^4 .x^3 =(x^(10) )^3 .x.x^3 =x^3 .x^4 =x^3 .x=x ⇒x^(1907) =x^(1600) .x^(307) =x.x=x^2 x^(1907) +(1/x^(1907) )=x^2 +(1/x^2 )=((x^4 +1)/x^2 )=((x+1)/(−x−1))=−1

$$\mathrm{x}^{\mathrm{2}} =−\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{4}} =\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}=\left(−\mathrm{x}−\mathrm{1}\right)+\mathrm{2x}+\mathrm{1}=\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{4}} =\mathrm{x}\Rightarrow\mathrm{x}^{\mathrm{16}} =\mathrm{x}^{\mathrm{4}} =\mathrm{x} \\ $$$$\mathrm{x}^{\mathrm{160}} =\mathrm{x}^{\mathrm{10}} =\mathrm{x}^{\mathrm{4}} .\mathrm{x}^{\mathrm{4}} .\mathrm{x}^{\mathrm{2}} =\mathrm{x}.\mathrm{x}.\mathrm{x}^{\mathrm{2}} =\mathrm{x}^{\mathrm{4}} =\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{160}} =\mathrm{x}\Rightarrow\mathrm{x}^{\mathrm{1600}} =\mathrm{x}^{\mathrm{10}} =\mathrm{x} \\ $$$$\bigstar\mathrm{x}^{\mathrm{307}} =\left(\mathrm{x}^{\mathrm{10}} \right)^{\mathrm{30}} .\mathrm{x}^{\mathrm{7}} =\mathrm{x}^{\mathrm{30}} .\mathrm{x}^{\mathrm{4}} .\mathrm{x}^{\mathrm{3}} =\left(\mathrm{x}^{\mathrm{10}} \right)^{\mathrm{3}} .\mathrm{x}.\mathrm{x}^{\mathrm{3}} \\ $$$$=\mathrm{x}^{\mathrm{3}} .\mathrm{x}^{\mathrm{4}} =\mathrm{x}^{\mathrm{3}} .\mathrm{x}=\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{1907}} =\mathrm{x}^{\mathrm{1600}} .\mathrm{x}^{\mathrm{307}} =\mathrm{x}.\mathrm{x}=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{1907}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{1907}} }=\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{x}+\mathrm{1}}{−\mathrm{x}−\mathrm{1}}=−\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Cool!

$${Cool}! \\ $$

Commented by bramlex last updated on 11/Jul/20

jooss...

$$\mathrm{jooss}... \\ $$

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