Prove that x^3 =y^6 +20 has no positive integer solution.


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Question Number 196650 by CrispyXYZ last updated on 28/Aug/23

$Prove that x^{3} = y^{6} + 20 has no positive integer solution .$
	Answered by deleteduser1 last updated on 28/Aug/23

	$x^{3} \equiv 1, - 1 o r 0 (m o d 9)$ $w h e n (y, 9) = 1; y^{6} \overset{9}{\equiv} 1 \Rightarrow y^{6} + 20 \equiv 3 (m o d 9)$ $w h e n y = 9 k; y^{6} \overset{9}{\equiv} 0 \Rightarrow y^{6} + 20 \equiv 2 (m o d 9)$ $T h e r e e x i s t s n o s o l u t i o n f o r x, y \in Z^{+}, s i n c e$ $y^{6} + 20 n e v e r e q u a l s - 1, 0, 1 (m o d 9)$
	Answered by MM42 last updated on 29/Aug/23
	$x^3 −y^6 =20⇒(x−y^2 )(x^2 +xy^2 +y^4 )=20 1) { ((x−y^2 =1)),((x^2 +xy^2 +y^4 =20)) :}⇒3y^2 ( y^2 +1)=19 × 2) { ((x−y^2 =2)),((x^2 +xy^2 +y^4 =10)) :}⇒y^2 ( y^2 +2)=2 × 3) { ((x−y^2 =4)),((x^2 +xy^2 +y^4 =5)) :}⇒3y^2 ( y^2 +4)=−1 ×$
	$x^{3} - y^{6} = 20 \Rightarrow (x - y^{2}) (x^{2} + {x y}^{2} + y^{4}) = 20$ $1) {\begin{cases} x - y^{2} = 1 \\ x^{2} + {x y}^{2} + y^{4} = 20 \end{cases} \Rightarrow 3 y^{2} (y^{2} + 1) = 19 \times$ $2) {\begin{cases} x - y^{2} = 2 \\ x^{2} + {x y}^{2} + y^{4} = 10 \end{cases} \Rightarrow y^{2} (y^{2} + 2) = 2 \times$ $3) {\begin{cases} x - y^{2} = 4 \\ x^{2} + {x y}^{2} + y^{4} = 5 \end{cases} \Rightarrow 3 y^{2} (y^{2} + 4) = - 1 \times$
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