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Question Number 160270 by alf123 last updated on 27/Nov/21

	Answered by Rasheed.Sindhi last updated on 27/Nov/21
	$(x^2 +y)(y^2 +x)=(x−y)^3 ⇒ { ((x^2 +y=x−y_((i)) ∧ y^2 +x=(x−y)^2 _((ii)) )),(( OR)),((x^2 +y=(x−y)^2 _((iii)) ∧ y^2 +x=x−y_((iv)) )) :} ⇒ { (((ii)/(i): ((y^2 +x)/(x^2 +y))=x−y.....(v))),(( OR)),(((iii)/(iv): ((x^2 +y)/(y^2 +x))=x−y.....(vi))) :} From (v) & (vi): ((y^2 +x)/(x^2 +y))=((x^2 +y)/(y^2 +x))⇒(x^2 +y)^2 =(y^2 +x)^2 ⇒ { ((x^2 +y=y^2 +x⇒x^2 −y^2 +y−x=0)),((x^2 +y=−y^2 −x⇒x^2 +y^2 +y+x=0)) :} ⇒ { (((x−y)(x+y−1)=0⇒x=y ∣ x+y=1)),((x^2 +y^2 +y+x=0⇒(x,y)=(−1,−1),(0,0))) :} x=y ∣ y=1−x x=y 2(x^2 +x)=0 x(x+1)=0 x=0⇒y=0 x=−1,y=−1 y=1−x (x^2 +1−x)( (1−x)^2 +x )=(x−(1−x) )^3 (x^2 −x+1)(x^2 −x+1)=(2x−1)^3 x^2 −x+1=(2x−1)^1 =(2x−1)^2 ...A (2x−1)^1 =(2x−1)^2 suggest: 2x−1=0,1 { ((x=1/2 doesn′t obey fully A(All the three sides are not equal))),((x=1(✓)⇒y=1−1=0)) :} x=1,y=0 (x,y)=(0,0),(−1,−1),(1,0)$
	$(x^{2} + y) (y^{2} + x) = {(x - y)}^{3}$ $\Rightarrow {\begin{cases} \underset{(i)}{\underset{⏟}{x^{2} + y = x - y}} \land \underset{(i i)}{\underset{⏟}{y^{2} + x = {(x - y)}^{2}}} \\ OR \\ \underset{(i i i)}{\underset{⏟}{x^{2} + y = {(x - y)}^{2}}} \land \underset{(i v)}{\underset{⏟}{y^{2} + x = x - y}} \end{cases}$ $\Rightarrow {\begin{cases} (i i) / (i) : \frac{y^{2} + x}{x^{2} + y} = x - y . . . . . (v) \\ OR \\ (i i i) / (i v) : \frac{x^{2} + y}{y^{2} + x} = x - y . . . . . (v i) \end{cases}$ $F r o m (v) & (v i) :$ $\frac{y^{2} + x}{x^{2} + y} = \frac{x^{2} + y}{y^{2} + x} \Rightarrow {(x^{2} + y)}^{2} = {(y^{2} + x)}^{2}$ $\Rightarrow {\begin{cases} x^{2} + y = y^{2} + x \Rightarrow x^{2} - y^{2} + y - x = 0 \\ x^{2} + y = - y^{2} - x \Rightarrow x^{2} + y^{2} + y + x = 0 \end{cases}$ $\Rightarrow {\begin{cases} (x - y) (x + y - 1) = 0 \Rightarrow x = y ∣ x + y = 1 \\ x^{2} + y^{2} + y + x = 0 \Rightarrow (x, y) = (- 1, - 1), (0, 0) \end{cases}$ $x = y ∣ y = 1 - x$ $\underset{―}{x = y}$ $2 (x^{2} + x) = 0$ $x (x + 1) = 0$ $x = 0 \Rightarrow y = 0$ $x = - 1, y = - 1$ $\underset{―}{y = 1 - x}$ $(x^{2} + 1 - x) ({(1 - x)}^{2} + x) = {(x - (1 - x))}^{3}$ $(x^{2} - x + 1) (x^{2} - x + 1) = {(2 x - 1)}^{3}$ $x^{2} - x + 1 = {(2 x - 1)}^{1} = {(2 x - 1)}^{2} . . . A$ ${(2 x - 1)}^{1} = {(2 x - 1)}^{2} s u g g e s t :$ $2 x - 1 = 0, 1$ ${\begin{cases} x = 1 / 2 {d o e s n}^{'} t o b e y f u l l y A (A l l t h e t h r e e s i d e s a r e n o t e q u a l) \\ x = 1 (✓) \Rightarrow y = 1 - 1 = 0 \end{cases}$ $x = 1, y = 0$ $(x, y) = (0, 0), (- 1, - 1), (1, 0)$
	Commented by mr W last updated on 27/Nov/21

	$(X, 0) w i t h X = a n y i n t e g e r$ $(- 1, - 1)$ $(8, - 10)$ $(9, - 21)$ $m a y b e t h e r e a r e m o r e .$
	Commented by Rasheed.Sindhi last updated on 27/Nov/21

	$T h a n X S i r, h a v e y o u s o l v e d i t ? I f y e s p l e a s e$ $s h a r e .$
	Answered by mr W last updated on 27/Nov/21

	$x^{2} y^{2} + x y + x^{3} + y^{3} = x^{3} - 3 x^{2} y + 3 {x y}^{2} - y^{3}$ $x^{2} y^{2} + x y + 2 y^{3} = - 3 x^{2} y + 3 {x y}^{2}$ $y = 0 \Rightarrow x = a n y i n t e g e r$ $x^{2} y + x + 2 y^{2} = - 3 x^{2} + 3 x y$ $(3 + y) x^{2} + (1 - 3 y) x + 2 y^{2} = 0$ $x = \frac{- 1 + 3 y \pm \sqrt{Δ}}{2 (3 + y)}$ $Δ = {(1 - 3 y)}^{2} - 8 y^{2} (3 + y)$ $Δ = {(1 + y)}^{2} (1 - 8 y) s h o u l d b e p e r f e c t s q u a r e!$ $1 - 8 y = n^{2}$ $y = - \frac{n^{2} - 1}{8} = \frac{(n - 1) (n + 1)}{8}$ $(n - 1) (n + 1) = 2 \times 4 = 8 \Rightarrow n = 3 \Rightarrow y = - 1 \Rightarrow x = - 1$ $(n - 1) (n + 1) = 4 \times 6 = 24 \Rightarrow n = 5 \Rightarrow y = - 3 \Rightarrow x \notin Z$ $(n - 1) (n + 1) = 6 \times 8 = 48 \Rightarrow n = 7 \Rightarrow y = - 6 \Rightarrow x \notin Z$ $(n - 1) (n + 1) = 8 \times 10 = 80 \Rightarrow n = 9 \Rightarrow y = - 10 \Rightarrow x = 8$ $. . .$ $(n - 1) (n + 1) = 12 \times 14 \Rightarrow n = 13 \Rightarrow y = - 21 \Rightarrow x = 9$ $. . . . . . m a y b e m o r e$
	Commented by Rasheed.Sindhi last updated on 27/Nov/21

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