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Question Number 92084 by Power last updated on 04/May/20

	Answered by MJS last updated on 04/May/20
	$obviously x_0 =y_0 =0 let x≠0 ∧ y=px ∧ p≠0 { ((x((p^5 +1)x^4 −33p)=0)),((x(3p^2 x^4 −8(p+1))=0)) :} { ((x^4 =((33p)/(p^5 +1)))),((x^4 =((8(p+1))/(3p2)))) :} ⇒ p^6 +p^5 −((99)/8)p^3 +p+1=0 trying I found (p−(1/2))(p−2)(p^4 +(7/2)p^3 +((31)/4)p^2 +(7/2)p+1)=0 ⇒ p_1 =(1/2) p_2 =2 p_(3, 4) =−(7/8)+((√(−29+2(√(737))))/8)±(((√(43))/8)−((√(29+2(√(737))))/8))i p_(5, 6) =−(7/8)−((√(−29+2(√(737))))/8)±(((√(43))/8)+((√(29+2(√(737))))/8))i now be so kind and calculate the rest...$
	$obviously x_{0} = y_{0} = 0$ $let x \neq 0 \land y = p x \land p \neq 0$ ${\begin{cases} x ((p^{5} + 1) x^{4} - 33 p) = 0 \\ x (3 p^{2} x^{4} - 8 (p + 1)) = 0 \end{cases}$ ${\begin{cases} x^{4} = \frac{33 p}{p^{5} + 1} \\ x^{4} = \frac{8 (p + 1)}{3 p 2} \end{cases}$ $\Rightarrow$ $p^{6} + p^{5} - \frac{99}{8} p^{3} + p + 1 = 0$ $trying I found$ $(p - \frac{1}{2}) (p - 2) (p^{4} + \frac{7}{2} p^{3} + \frac{31}{4} p^{2} + \frac{7}{2} p + 1) = 0$ $\Rightarrow$ $p_{1} = \frac{1}{2}$ $p_{2} = 2$ $p_{3, 4} = - \frac{7}{8} + \frac{\sqrt{- 29 + 2 \sqrt{737}}}{8} \pm (\frac{\sqrt{43}}{8} - \frac{\sqrt{29 + 2 \sqrt{737}}}{8}) i$ $p_{5, 6} = - \frac{7}{8} - \frac{\sqrt{- 29 + 2 \sqrt{737}}}{8} \pm (\frac{\sqrt{43}}{8} + \frac{\sqrt{29 + 2 \sqrt{737}}}{8}) i$ $now be so kind and calculate the rest . . .$
	Commented by Power last updated on 05/May/20

	$thanks$
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