Question-132293

Question Number 132293 by Algoritm last updated on 13/Feb/21

Commented by mr W last updated on 13/Feb/21

x=1, y=0 x=0, y=1 the proof you may ask for is thinking.

$${x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{0},\:{y}=\mathrm{1} \\ $$$${the}\:{proof}\:{you}\:{may}\:{ask}\:{for}\:{is}\:{thinking}. \\ $$

Answered by MJS_new last updated on 13/Feb/21

{ ((x^5 +y^5 =1)),((x^3 +y^3 =1)) :} let x=u−v∧y=u+v { ((2u^5 +20u^3 v^2 +10uv^4 −1=0)),((2u^3 +6uv^2 −1=0 ⇒ v^2 =((1−2u^3 )/(6u)))) :} insert in (1) and transform ⇒ u^6 −(5/8)u^3 +(9/(32))u−(5/(64))=0 (u−(1/2))^3 (u^3 +(3/2)u^2 +(3/2)u+(5/8))=0 now use Cardano ⇒ u_(1, 2, 3) =(1/2) u_4 =−(1/2)+α−β u_5 =−(1/2)+ωα−ω^2 β u_6 =−(1/2)+ω^2 α−ωβ with α=(((−1+(√5))/(16)))^(1/3) ∧β=(((1+(√5))/(16)))^(1/3) ∧ω=−(1/2)+((√3)/2)i ⇒ 6 solutions x=0∧y=1 ∨ x=1∧y=0 x≈−.661093±.630705i∧y=x^− x≈−.791887±.755487∧y=−.0470208±.998894i

$$\begin{cases}{{x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{1}}\\{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{1}}\end{cases} \\ $$$$\mathrm{let}\:{x}={u}−{v}\wedge{y}={u}+{v} \\ $$$$\begin{cases}{\mathrm{2}{u}^{\mathrm{5}} +\mathrm{20}{u}^{\mathrm{3}} {v}^{\mathrm{2}} +\mathrm{10}{uv}^{\mathrm{4}} −\mathrm{1}=\mathrm{0}}\\{\mathrm{2}{u}^{\mathrm{3}} +\mathrm{6}{uv}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\:{v}^{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{2}{u}^{\mathrm{3}} }{\mathrm{6}{u}}}\end{cases} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\mathrm{transform}\:\Rightarrow \\ $$$${u}^{\mathrm{6}} −\frac{\mathrm{5}}{\mathrm{8}}{u}^{\mathrm{3}} +\frac{\mathrm{9}}{\mathrm{32}}{u}−\frac{\mathrm{5}}{\mathrm{64}}=\mathrm{0} \\ $$$$\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \left({u}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{u}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{u}+\frac{\mathrm{5}}{\mathrm{8}}\right)=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{Cardano} \\ $$$$\Rightarrow \\ $$$${u}_{\mathrm{1},\:\mathrm{2},\:\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${u}_{\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{2}}+\alpha−\beta \\ $$$${u}_{\mathrm{5}} =−\frac{\mathrm{1}}{\mathrm{2}}+\omega\alpha−\omega^{\mathrm{2}} \beta \\ $$$${u}_{\mathrm{6}} =−\frac{\mathrm{1}}{\mathrm{2}}+\omega^{\mathrm{2}} \alpha−\omega\beta \\ $$$$\mathrm{with}\:\alpha=\sqrt[{\mathrm{3}}]{\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{16}}}\wedge\beta=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{16}}}\wedge\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$$\mathrm{6}\:\mathrm{solutions} \\ $$$${x}=\mathrm{0}\wedge{y}=\mathrm{1}\:\vee\:{x}=\mathrm{1}\wedge{y}=\mathrm{0} \\ $$$${x}\approx−.\mathrm{661093}\pm.\mathrm{630705i}\wedge{y}=\overset{−} {{x}} \\ $$$${x}\approx−.\mathrm{791887}\pm.\mathrm{755487}\wedge{y}=−.\mathrm{0470208}\pm.\mathrm{998894i} \\ $$

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