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

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

Answered by ajfour last updated on 11/Jul/20

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

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

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

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

$$Cool!$$

Commented by bramlex last updated on 11/Jul/20

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

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