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Question Number 156648 by MathSh last updated on 13/Oct/21

$Solve for integers :$ $x^{2} y^{2} - x^{2} - xy - y^{2} + x + y = 0$
	Answered by Rasheed.Sindhi last updated on 14/Oct/21
	$x^2 y^2 - x^2 - xy - y^2 + x + y = 0 x^2 y^2 - x^2 - xy+x - y^2 + y = 0 x^2 (y^2 −1)−x(y−1)−y(y−1)=0 (y−1)( x^2 (y+1)−x−y )=0 y=1_(∀x) ∣ x^2 (y+1)−x−y =0 ∣ x^2 y+x^2 −x−y =0 ∣ x^2 y−y+x^2 −x =0 y(x^2 −1)+x(x−1)=0 y(x−1)(x+1)+x(x−1)=0 (x−1){ y(x+1)+x}=0 x=1_(∀y) ∣ xy+y+x=0 y(x+1)=−x y=−(x/(x+1)) (1)0 is divided by any integer⇒x=0 (2)±1 divide any integer ⇒ x+1=1⇒x=0 or x+1=−1⇒x=−2 •x=0⇒y=−(x/(x+1))=−(0/(0+1))=0 x=0,y=0 •x=−2⇒y=−((−2)/(−2+1))=−2 x=−2,y=−2 determinant ((((x,y)=(1,y),(x,1),(0,0),(−2,−2))))$
	$x^{2} y^{2} - x^{2} - xy - y^{2} + x + y = 0$ $x^{2} y^{2} - x^{2} - xy + x - y^{2} + y = 0$ $x^{2} (y^{2} - 1) - x (y - 1) - y (y - 1) = 0$ $(y - 1) (x^{2} (y + 1) - x - y) = 0$ $\underset{\forall x}{y = 1} ∣ x^{2} (y + 1) - x - y = 0$ $∣ x^{2} y + x^{2} - x - y = 0$ $∣ x^{2} y - y + x^{2} - x = 0$ $y (x^{2} - 1) + x (x - 1) = 0$ $y (x - 1) (x + 1) + x (x - 1) = 0$ $(x - 1) {y (x + 1) + x} = 0$ $\underset{\forall y}{x = 1} ∣ xy + y + x = 0$ $y (x + 1) = - x$ $y = - \frac{x}{x + 1}$ $(1) 0 is divided by any integer \Rightarrow x = 0$ $(2) \pm 1 d i v i d e a n y i n t e g e r$ $\Rightarrow x + 1 = 1 \Rightarrow x = 0$ $or x + 1 = - 1 \Rightarrow x = - 2$ $∙ x = 0 \Rightarrow y = - \frac{x}{x + 1} = - \frac{0}{0 + 1} = 0$ $x = 0, y = 0$ $∙ x = - 2 \Rightarrow y = - \frac{- 2}{- 2 + 1} = - 2$ $x = - 2, y = - 2$ $\begin{matrix} (x, y) = (1, y), (x, 1), (0, 0), (- 2, - 2) \end{matrix}$
	Commented by MathSh last updated on 14/Oct/21

	$Perfcet dear S er, thank you$
	Answered by TheSupreme last updated on 13/Oct/21

	$x^{2} (y^{2} - 1) + x (1 - y) + y (1 - y) = 0$ $(y - 1) [x^{2} (y + 1) - x - y] =$ $= (y - 1) [y (x^{2} - 1) + x (x - 1)] =$ $= (y - 1) (x - 1) [y (x + 1) + y + x] =$ $= (y - 1) (x - 1) [x y + y + x + 1] = 0$ $(1, y) a n d (x, 1) a r e a l l t h e s o l u t i o s$
	Commented by TheSupreme last updated on 13/Oct/21

	$a n d a l s o$ $(y - 1) (x - 1) (x + 1) (y + 1)$
	Answered by GuruBelakangPadang last updated on 13/Oct/21
	$(x^2 −1)(y^2 −1)=(x−1)(y−1) x=1,y=1⇒(x+1)(y+1)=1 k∈z⇒(x,y)={(1,z),(z,1),(0,0),(-2,-2)}$
	$(x^{2} - 1) (y^{2} - 1) = (x - 1) (y - 1)$ $x = 1, y = 1 \Rightarrow (x + 1) (y + 1) = 1$ $k \in z \Rightarrow (x, y) = {(1, z), (z, 1), (0, 0), (- 2, - 2)}$
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