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Question Number 117045 by bobhans last updated on 09/Oct/20

$$\mathrm{If}\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{number}\mathrm{satisfying}$$$${\mathrm{x}}^2+\mathrm{x}+1=0,\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}$$$${\mathrm{x}}^{53}+{\mathrm{x}}^{52}+{\mathrm{x}}^{51}+{\mathrm{x}}^{50}+{\mathrm{x}}^{49}?$$

Answered by john santu last updated on 09/Oct/20

$$weknowthat{x}^2+x+1=\frac{x^3-1}{x-1},x\ne 1$$$$since{x}^2+x+1=0,impliesthat{x}^3=1$$$$andx\ne 1.Now{x}^{53}+x^{52}+x^{51}+x^{50}+x^{49}=$$$$x^{51}(x^2+x)+x^{49}(x^2+x+1)=$$$$x^{51}(-1)+x^{49}\left(0\right)=-x^{51}=-{\left(x^3\right)}^{17}$$$$=-{\left(1\right)}^{17}=-1$$

Answered by 1549442205PVT last updated on 09/Oct/20

$$\mathrm{P}={\mathrm{x}}^{53}+{\mathrm{x}}^{52}+{\mathrm{x}}^{51}+{\mathrm{x}}^{50}+{\mathrm{x}}^{49}$$$$={\mathrm{x}}^{53}+{\mathrm{x}}^{52}+{\mathrm{x}}^{51}+{\mathrm{x}}^{50}+{\mathrm{x}}^{49}+{\mathrm{x}}^{48}-{\mathrm{x}}^{48}$$$$={\mathrm{x}}^{51}({\mathrm{x}}^2+\mathrm{x}+1)+{\mathrm{x}}^{48}({\mathrm{x}}^2+\mathrm{x}+1)-{\mathrm{x}}^{48}$$$$=-{\mathrm{x}}^{48}(\mathrm{since}{\mathrm{x}}^2+\mathrm{x}+1=0\left(\mathrm{by}\mathrm{hypothesis}\right))$$$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}{\mathrm{x}}^2+\mathrm{x}+1=0$$$$\mathrm{we}\mathrm{infer}\mathrm{x}=\frac{-1\pm \mathrm{i}\sqrt{3}}{2}=\cos 120°\pm \mathrm{isin}120°$$$$\mathrm{Hence},\mathrm{by}{\mathrm{Mauvra}}^{\prime }\mathrm{s}\mathrm{formula}\mathrm{we}\mathrm{have}$$$${\mathrm{x}}^{48}=\cos (48.120°)\pm \mathrm{isin}(48.120°)$$$$=\cos (16.360°)\pm \mathrm{isin}(16.360°)$$$$=1\pm 0=1\Rightarrow \mathrm{P}=-{\mathrm{x}}^{48}=-1$$

Answered by floor(10²Eta[1]) last updated on 09/Oct/20

$${\mathrm{x}}^2+\mathrm{x}+1=0\Rightarrow {\mathrm{x}}^2=-\mathrm{x}-1$$$${\mathrm{x}}^3=-{\mathrm{x}}^2-\mathrm{x}=\mathrm{x}+1-\mathrm{x}=1$$$$\Rightarrow {\mathrm{x}}^{3\mathrm{k}}=1^{\mathrm{k}}=1\mathrm{\forall }\mathrm{k}\in \mathbb{N}$$$${\mathrm{x}}^{53}={\mathrm{x}}^{51}.{\mathrm{x}}^2={\mathrm{x}}^2$$$${\mathrm{x}}^{52}={\mathrm{x}}^{51}.\mathrm{x}=\mathrm{x}$$$${\mathrm{x}}^{51}=1$$$${\mathrm{x}}^{50}={\mathrm{x}}^{48}.{\mathrm{x}}^2={\mathrm{x}}^2$$$${\mathrm{x}}^{49}={\mathrm{x}}^{48}.\mathrm{x}=\mathrm{x}$$$$\Rightarrow {\mathrm{x}}^{53}+{\mathrm{x}}^{52}+{\mathrm{x}}^{51}+{\mathrm{x}}^{50}+{\mathrm{x}}^{49}$$$$=2({\mathrm{x}}^2+\mathrm{x})+1=-2+1=-1$$

Answered by Olaf last updated on 09/Oct/20

$$x^2+x+1=0=\frac{1-x^3}{1-x}\Rightarrow x^3=1$$$$x^2+x+1=0\Rightarrow x+\frac{1}{x}=-1$$$$x^{53}+x^{52}+x^{51}+x^{50}+x^{49}=$$$$x^{51}(x^2+x+1+\frac{1}{x}+\frac{1}{x^2})=$$$$x^{51}[{(x+\frac{1}{x})}^2-2+(x+\frac{1}{x})+1]=$$$$x^{51}(1-2-1+1)=-x^{51}=-{\left(x^3\right)}^{17}=-1$$

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