find all pairs of integers (x,y) such that x^3 +y^3 =(x+y)^2


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Question Number 115906 by Eric002 last updated on 29/Sep/20

$f i n d a l l p a i r s o f i n t e g e r s (x, y) s u c h t h a t$ $x^{3} + y^{3} = {(x + y)}^{2}$
	Answered by mindispower last updated on 29/Sep/20
	$x^3 +y^3 =(x+y)(x^2 +y^2 −xy) ⇔(x+y)(x^2 +y^2 −xy−(x+y))=0 x=−y or x^2 +y^2 −xy−(x+y)=0 y^2 +y(−1−x)+x^2 −x=0 Δ=x^2 +2x+1−4x^2 +4x≥0 ⇒6x−4x^2 +1≥0 x∈[((−6+(√(48)))/(−8)),((6+(√(48)))/8)] 0≤x≤1 x=0⇒y^2 −y=0⇒y∈{0,1} x=1⇒y^2 −2y=0 y∈{0,2} S={(a,−a);(0,1);(0,2)}$
	$x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$ $\Leftrightarrow (x + y) (x^{2} + y^{2} - x y - (x + y)) = 0$ $x = - y$ $o r$ $x^{2} + y^{2} - x y - (x + y) = 0$ $y^{2} + y (- 1 - x) + x^{2} - x = 0$ $Δ = x^{2} + 2 x + 1 - 4 x^{2} + 4 x ⩾ 0$ $\Rightarrow 6 x - 4 x^{2} + 1 ⩾ 0$ $x \in [\frac{- 6 + \sqrt{48}}{- 8}, \frac{6 + \sqrt{48}}{8}]$ $0 ⩽ x ⩽ 1$ $x = 0 \Rightarrow y^{2} - y = 0 \Rightarrow y \in {0, 1}$ $x = 1 \Rightarrow y^{2} - 2 y = 0$ $y \in {0, 2}$ $S = {(a, - a); (0, 1); (0, 2)}$
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