Question Number 1557 by 123456 last updated on 19/Aug/15
$$\mathrm{find}\:\alpha,\beta,\gamma\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha=\beta^{\mathrm{2}} \\ $$$$\beta^{\mathrm{3}} =\gamma^{\mathrm{4}} \\ $$$$\gamma^{\mathrm{5}} =\alpha^{\mathrm{6}} \\ $$
Commented by 123456 last updated on 19/Aug/15
![{ ((α=β^2 )),((β^3 =γ^4 )),((γ^5 =α^6 )) :}⇒ { ((α^(15) =β^(30) )),((β^(30) =γ^(40) )),((γ^(40) =α^(48) )) :} [(1,2),(3,4),(5,6)] [(3,6),(6,8),(5,6)] [(15,30),(30,40),(40,48)] α^(15) =α^(48) ⇒α^(15) (α^(33) −1)=0](https://www.tinkutara.com/question/Q1558.png)
$$\begin{cases}{\alpha=\beta^{\mathrm{2}} }\\{\beta^{\mathrm{3}} =\gamma^{\mathrm{4}} }\\{\gamma^{\mathrm{5}} =\alpha^{\mathrm{6}} }\end{cases}\Rightarrow\begin{cases}{\alpha^{\mathrm{15}} =\beta^{\mathrm{30}} }\\{\beta^{\mathrm{30}} =\gamma^{\mathrm{40}} }\\{\gamma^{\mathrm{40}} =\alpha^{\mathrm{48}} }\end{cases} \\ $$$$\left[\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{5},\mathrm{6}\right)\right] \\ $$$$\left[\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{6},\mathrm{8}\right),\left(\mathrm{5},\mathrm{6}\right)\right] \\ $$$$\left[\left(\mathrm{15},\mathrm{30}\right),\left(\mathrm{30},\mathrm{40}\right),\left(\mathrm{40},\mathrm{48}\right)\right] \\ $$$$\alpha^{\mathrm{15}} =\alpha^{\mathrm{48}} \Rightarrow\alpha^{\mathrm{15}} \left(\alpha^{\mathrm{33}} −\mathrm{1}\right)=\mathrm{0} \\ $$
Answered by Rasheed Soomro last updated on 20/Aug/15
![α=β^2 ⇒α^3 =β^6 ..............(I) β^3 =γ^4 ⇒β^6 =γ^8 ............(II) From (I) and (II) α^3 =γ^(8 ) ⇒α^6 =γ^(16) .....(III) γ^5 =α^6 ⇒α^6 =γ^5 .........................(IV) From (III) and (IV) γ^5 =γ^(16) ⇒γ^(16) −γ^5 =0⇒ γ^5 (γ^(11) −1)=0 γ=0 ∨ γ^(11) =1 , γ is 11th root of unity. [Above was the process of eliminating α and β. In the same way by eliminating β and γ and after that by α and γ we will get results about α and β respectively.] Results about α and β α=0 ∨ α^(33) =1 , α is 33rd root of unity. β=0 ∨ β^(33) =1 , β is 33rd root of unity. (α,β,γ)=(0,0,0) Solution of α or β: {0}∪{1,w,ω^2 ,...ω^(32) },ω=cos((2π)/(33))+ı sin((2π)/(33)) Solution of γ: {0}∪{1,ρ,ρ^2 ,...ρ^(10) },ρ=cos((2π)/(11))+ı sin((2π)/(11)) Let (α,β,γ)=(ω^p ,ω^q ,ρ^r ) fulfilling given conditions. w^p =(ω^q )^2 ⇒ p=2q................................I (ω^q )^3 =(ρ^r )^4 ⇒ω^(3q) =ρ^(4r) ⇒ω^(15q) =ρ^(20r) ...........II (ρ^r )^5 =(ω^p )^6 ⇒ρ^(5r) =ω^(6p) ⇒ρ^(20r) =ω^(24p) ..........III p=2q.......[from I] 24p=15q⇒8p=5q...............[from II and III] 8(2q)=5q⇒q=0⇒p=2(0)=0⇒p=0 ρ^(20r) =ω^(24p) =ω^(24(0)) =1=ρ^0 ⇒ 20r=0 ⇒r=0 Sol. Set : {(0,0,0),(1,1,1)}](https://www.tinkutara.com/question/Q1563.png)
$$\alpha=\beta^{\mathrm{2}} \:\Rightarrow\alpha^{\mathrm{3}} =\beta^{\mathrm{6}} …………..\left(\boldsymbol{{I}}\right) \\ $$$$\beta^{\mathrm{3}} =\gamma^{\mathrm{4}} \:\Rightarrow\beta^{\mathrm{6}} =\gamma^{\mathrm{8}} \:…………\left(\boldsymbol{{II}}\right) \\ $$$${From}\:\left(\boldsymbol{{I}}\right)\:\:\:{and}\:\:\:\left(\boldsymbol{{II}}\right)\:\:\:\:\:\alpha^{\mathrm{3}} =\gamma^{\mathrm{8}\:\:} \Rightarrow\alpha^{\mathrm{6}} =\gamma^{\mathrm{16}} \:…..\left(\boldsymbol{{III}}\right) \\ $$$$\gamma^{\mathrm{5}} =\alpha^{\mathrm{6}} \:\Rightarrow\alpha^{\mathrm{6}} =\gamma^{\mathrm{5}} …………………….\left(\boldsymbol{{IV}}\right) \\ $$$${From}\:\left(\boldsymbol{{III}}\right)\:\:{and}\:\:\:\:\left(\boldsymbol{{IV}}\right) \\ $$$$\gamma^{\mathrm{5}} =\gamma^{\mathrm{16}} \Rightarrow\gamma^{\mathrm{16}} −\gamma^{\mathrm{5}} =\mathrm{0}\Rightarrow\:\gamma^{\mathrm{5}} \left(\gamma^{\mathrm{11}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\gamma=\mathrm{0}\:\vee\:\gamma^{\mathrm{11}} =\mathrm{1}\:,\:\gamma\:\:{is}\:\mathrm{11}\boldsymbol{{th}}\:\:{root}\:{of}\:{unity}. \\ $$$$\left[{Above}\:{was}\:{the}\:{process}\:{of}\:{eliminating}\:\:\:\alpha\:\:{and}\:\:\:\beta.\right. \\ $$$${In}\:{the}\:{same}\:{way}\:\:{by}\:{eliminating}\:\:\:\beta\:\:\:\:{and}\:\:\:\:\gamma\:\:\:{and}\:\:{after}\:{that} \\ $$$${by}\:\:\:\:\alpha\:\:\:\:{and}\:\:\:\:\gamma\:\:\:\:\:{we}\:{will}\:{get}\:{results}\:{about}\:\:\:\alpha\:\:\:\:\:{and}\:\:\:\:\:\beta \\ $$$$\left.{respectively}.\right] \\ $$$${Results}\:{about}\:\alpha\:{and}\:\:\beta \\ $$$$\alpha=\mathrm{0}\:\vee\:\alpha^{\mathrm{33}} =\mathrm{1}\:,\:\alpha\:\:{is}\:\mathrm{33}\boldsymbol{{rd}}\:\:{root}\:{of}\:{unity}. \\ $$$$\beta=\mathrm{0}\:\vee\:\beta^{\mathrm{33}} =\mathrm{1}\:,\:\beta\:\:{is}\:\mathrm{33}\boldsymbol{{rd}}\:\:{root}\:{of}\:{unity}. \\ $$$$\left(\alpha,\beta,\gamma\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$${Solution}\:{of}\:\alpha\:{or}\:\beta:\:\:\left\{\mathrm{0}\right\}\cup\left\{\mathrm{1},{w},\omega^{\mathrm{2}} ,…\omega^{\mathrm{32}} \right\},\omega={cos}\frac{\mathrm{2}\pi}{\mathrm{33}}+\imath\:{sin}\frac{\mathrm{2}\pi}{\mathrm{33}} \\ $$$${Solution}\:{of}\:\:\gamma:\:\left\{\mathrm{0}\right\}\cup\left\{\mathrm{1},\rho,\rho^{\mathrm{2}} ,…\rho^{\mathrm{10}} \right\},\rho={cos}\frac{\mathrm{2}\pi}{\mathrm{11}}+\imath\:{sin}\frac{\mathrm{2}\pi}{\mathrm{11}} \\ $$$${Let}\:\left(\alpha,\beta,\gamma\right)=\left(\omega^{{p}} ,\omega^{{q}} ,\rho^{{r}} \right)\:{fulfilling}\:{given}\:{conditions}. \\ $$$${w}^{{p}} =\left(\omega^{{q}} \right)^{\mathrm{2}} \Rightarrow\:{p}=\mathrm{2}{q}…………………………..\boldsymbol{{I}} \\ $$$$\left(\omega^{{q}} \right)^{\mathrm{3}} =\left(\rho^{{r}} \right)^{\mathrm{4}} \Rightarrow\omega^{\mathrm{3}{q}} =\rho^{\mathrm{4}{r}} \Rightarrow\omega^{\mathrm{15}{q}} =\rho^{\mathrm{20}{r}} ………..\boldsymbol{{II}} \\ $$$$\left(\rho^{{r}} \right)^{\mathrm{5}} =\left(\omega^{{p}} \right)^{\mathrm{6}} \Rightarrow\rho^{\mathrm{5}{r}} =\omega^{\mathrm{6}{p}} \Rightarrow\rho^{\mathrm{20}{r}} =\omega^{\mathrm{24}{p}} ……….\boldsymbol{{III}} \\ $$$${p}=\mathrm{2}{q}…….\left[{from}\:\boldsymbol{{I}}\right] \\ $$$$\mathrm{24}{p}=\mathrm{15}{q}\Rightarrow\mathrm{8}{p}=\mathrm{5}{q}……………\left[{from}\:\boldsymbol{{II}}\:\:{and}\:\:\boldsymbol{{III}}\right] \\ $$$$\mathrm{8}\left(\mathrm{2}{q}\right)=\mathrm{5}{q}\Rightarrow{q}=\mathrm{0}\Rightarrow{p}=\mathrm{2}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{p}=\mathrm{0} \\ $$$$\rho^{\mathrm{20}{r}} =\omega^{\mathrm{24}{p}} =\omega^{\mathrm{24}\left(\mathrm{0}\right)} =\mathrm{1}=\rho^{\mathrm{0}} \:\Rightarrow\:\mathrm{20}{r}=\mathrm{0}\:\Rightarrow{r}=\mathrm{0} \\ $$$${Sol}.\:{Set}\::\:\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right),\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\right\} \\ $$