Question Number 1540 by Rasheed Soomro last updated on 17/Aug/15
$$\mathrm{Determine}\:\mathrm{three}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:,\:\beta\:,\gamma\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha=\beta^{\:\mathrm{2}} \:\:\:\:\:\:\:{but}\:\:\:\:\beta\:\neq\:\alpha^{\:\mathrm{2}} \\ $$$$\beta\:=\:\gamma^{\:\mathrm{2}} \:\:\:\:\:\:{but}\:\:\:\:\:\gamma\:\neq\:\beta^{\:\mathrm{2}} \\ $$$$\gamma\:=\:\alpha^{\:\mathrm{2}\:} \:\:\:\:\:{but}\:\:\:\:\:\alpha\:\neq\:\gamma^{\:\mathrm{2}} \\ $$
Answered by 123456 last updated on 17/Aug/15
![{ ((α=β^2 ∧β≠α^2 )),((β=γ^2 ∧γ≠β^2 )),((γ=α^2 ∧α≠γ^2 )) :}≡ { ((α=β^2 ∧β≠γ)),((β=γ^2 ∧γ≠α)),((γ=α^2 ∧α≠β)) :} α=β^2 =(γ^2 )^2 =[(α^2 )^2 ]^2 =α^(2×2×2) =α^8 α^8 −α=0 α(α^7 −1)=0 α=0∨α^7 =1 α=0⇒γ=0⇒γ^2 =α α=e^(((2π)/7)kı) ,k∈Z_7 γ=α^2 =e^(((4π)/7)kı) β=γ^2 =e^(((8π)/7)kı) (α,β,γ)=(e^(((2π)/7)kı) ,e^(((8π)/7)kı) ,e^(((4π)/7)kι) ) β≠α^2 ⇒e^(((8π)/7)kı) ≠e^(((4π)/7)kı) (k≠0) γ≠β^2 ⇒e^(((4π)/7)kı) ≠e^(((16π)/7)kı) =e^(((2π)/7)kı) (k≠0) α≠γ^2 ⇒e^(((2π)/7)kı) ≠e^(((8π)/7)kı) (k≠0) (α,β,γ)=(e^(((2π)/7)kı) ,e^(((8π)/7)kı) ,e^(((4π)/7)kı) ),k∈Z_7 \{0}](Q1541.png)
$$\begin{cases}{\alpha=\beta^{\mathrm{2}} \wedge\beta\neq\alpha^{\mathrm{2}} }\\{\beta=\gamma^{\mathrm{2}} \wedge\gamma\neq\beta^{\mathrm{2}} }\\{\gamma=\alpha^{\mathrm{2}} \wedge\alpha\neq\gamma^{\mathrm{2}} }\end{cases}\equiv\begin{cases}{\alpha=\beta^{\mathrm{2}} \wedge\beta\neq\gamma}\\{\beta=\gamma^{\mathrm{2}} \wedge\gamma\neq\alpha}\\{\gamma=\alpha^{\mathrm{2}} \wedge\alpha\neq\beta}\end{cases} \\ $$$$\alpha=\beta^{\mathrm{2}} =\left(\gamma^{\mathrm{2}} \right)^{\mathrm{2}} =\left[\left(\alpha^{\mathrm{2}} \right)^{\mathrm{2}} \right]^{\mathrm{2}} =\alpha^{\mathrm{2}×\mathrm{2}×\mathrm{2}} =\alpha^{\mathrm{8}} \\ $$$$\alpha^{\mathrm{8}} −\alpha=\mathrm{0} \\ $$$$\alpha\left(\alpha^{\mathrm{7}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\alpha=\mathrm{0}\vee\alpha^{\mathrm{7}} =\mathrm{1} \\ $$$$\alpha=\mathrm{0}\Rightarrow\gamma=\mathrm{0}\Rightarrow\gamma^{\mathrm{2}} =\alpha \\ $$$$\alpha={e}^{\frac{\mathrm{2}\pi}{\mathrm{7}}{k}\imath} ,{k}\in\mathbb{Z}_{\mathrm{7}} \\ $$$$\gamma=\alpha^{\mathrm{2}} ={e}^{\frac{\mathrm{4}\pi}{\mathrm{7}}{k}\imath} \\ $$$$\beta=\gamma^{\mathrm{2}} ={e}^{\frac{\mathrm{8}\pi}{\mathrm{7}}{k}\imath} \\ $$$$\left(\alpha,\beta,\gamma\right)=\left({e}^{\frac{\mathrm{2}\pi}{\mathrm{7}}{k}\imath} ,{e}^{\frac{\mathrm{8}\pi}{\mathrm{7}}{k}\imath} ,{e}^{\frac{\mathrm{4}\pi}{\mathrm{7}}{k}\iota} \right) \\ $$$$\beta\neq\alpha^{\mathrm{2}} \Rightarrow{e}^{\frac{\mathrm{8}\pi}{\mathrm{7}}{k}\imath} \neq{e}^{\frac{\mathrm{4}\pi}{\mathrm{7}}{k}\imath} \left({k}\neq\mathrm{0}\right) \\ $$$$\gamma\neq\beta^{\mathrm{2}} \Rightarrow{e}^{\frac{\mathrm{4}\pi}{\mathrm{7}}{k}\imath} \neq{e}^{\frac{\mathrm{16}\pi}{\mathrm{7}}{k}\imath} ={e}^{\frac{\mathrm{2}\pi}{\mathrm{7}}{k}\imath} \left({k}\neq\mathrm{0}\right) \\ $$$$\alpha\neq\gamma^{\mathrm{2}} \Rightarrow{e}^{\frac{\mathrm{2}\pi}{\mathrm{7}}{k}\imath} \neq{e}^{\frac{\mathrm{8}\pi}{\mathrm{7}}{k}\imath} \left({k}\neq\mathrm{0}\right) \\ $$$$\left(\alpha,\beta,\gamma\right)=\left({e}^{\frac{\mathrm{2}\pi}{\mathrm{7}}{k}\imath} ,{e}^{\frac{\mathrm{8}\pi}{\mathrm{7}}{k}\imath} ,{e}^{\frac{\mathrm{4}\pi}{\mathrm{7}}{k}\imath} \right),{k}\in\mathbb{Z}_{\mathrm{7}} \backslash\left\{\mathrm{0}\right\} \\ $$
Commented by 123456 last updated on 17/Aug/15
$$\left(\omega,\omega^{\mathrm{4}} ,\omega^{\mathrm{2}} \right) \\ $$$$\left(\omega^{\mathrm{2}} ,\omega,\omega^{\mathrm{4}} \right) \\ $$$$\left(\omega^{\mathrm{3}} ,\omega^{\mathrm{5}} ,\omega^{\mathrm{6}} \right) \\ $$$$\left(\omega^{\mathrm{4}} ,\omega^{\mathrm{2}} ,\omega\right) \\ $$$$\left(\omega^{\mathrm{5}} ,\omega^{\mathrm{6}} ,\omega^{\mathrm{3}} \right) \\ $$$$\left(\omega^{\mathrm{6}} ,\omega^{\mathrm{3}} ,\omega^{\mathrm{5}} \right) \\ $$