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

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

$x = 1, y = 0$ $x = 0, y = 1$ $t h e p r o o f y o u m a y a s k f o r i s t h i n k i n g .$
	Answered by MJS_new last updated on 13/Feb/21
	${ ((x^5 +y^5 =1)),((x^3 +y^3 =1)) :} let x=u−v∧y=u+v { ((2u^5 +20u^3 v^2 +10uv^4 −1=0)),((2u^3 +6uv^2 −1=0 ⇒ v^2 =((1−2u^3 )/(6u)))) :} insert in (1) and transform ⇒ u^6 −(5/8)u^3 +(9/(32))u−(5/(64))=0 (u−(1/2))^3 (u^3 +(3/2)u^2 +(3/2)u+(5/8))=0 now use Cardano ⇒ u_(1, 2, 3) =(1/2) u_4 =−(1/2)+α−β u_5 =−(1/2)+ωα−ω^2 β u_6 =−(1/2)+ω^2 α−ωβ with α=(((−1+(√5))/(16)))^(1/3) ∧β=(((1+(√5))/(16)))^(1/3) ∧ω=−(1/2)+((√3)/2)i ⇒ 6 solutions x=0∧y=1 ∨ x=1∧y=0 x≈−.661093±.630705i∧y=x^− x≈−.791887±.755487∧y=−.0470208±.998894i$
	${\begin{cases} x^{5} + y^{5} = 1 \\ x^{3} + y^{3} = 1 \end{cases}$ $let x = u - v \land y = u + v$ ${\begin{cases} 2 u^{5} + 20 u^{3} v^{2} + 10 {u v}^{4} - 1 = 0 \\ 2 u^{3} + 6 {u v}^{2} - 1 = 0 \Rightarrow v^{2} = \frac{1 - 2 u^{3}}{6 u} \end{cases}$ $insert in (1) and transform \Rightarrow$ $u^{6} - \frac{5}{8} u^{3} + \frac{9}{32} u - \frac{5}{64} = 0$ ${(u - \frac{1}{2})}^{3} (u^{3} + \frac{3}{2} u^{2} + \frac{3}{2} u + \frac{5}{8}) = 0$ $now use Cardano$ $\Rightarrow$ $u_{1, 2, 3} = \frac{1}{2}$ $u_{4} = - \frac{1}{2} + α - β$ $u_{5} = - \frac{1}{2} + ω α - ω^{2} β$ $u_{6} = - \frac{1}{2} + ω^{2} α - ω β$ $with α = \sqrt[3]{\frac{- 1 + \sqrt{5}}{16}} \land β = \sqrt[3]{\frac{1 + \sqrt{5}}{16}} \land ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $\Rightarrow$ $6 solutions$ $x = 0 \land y = 1 \lor x = 1 \land y = 0$ $x \approx - . 661093 \pm . 630705 i \land y = \bar{x}$ $x \approx - . 791887 \pm . 755487 \land y = - . 0470208 \pm . 998894 i$
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