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

x+(1/x) = −1 ⇒x^(1907) +(1/x^(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.

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

Commented byRasheed.Sindhi last updated on 11/Jul/20

Cool!

Commented bybramlex last updated on 11/Jul/20

jooss...