Solve-for-integers-x-2-y-2-x-2-xy-y-2-x-y-0-

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

\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 \Rightarrow (x + 1) (y + 1) = 1

k \in z \Rightarrow (x, y) = {(1, z), (z, 1), (0, 0), (- 2, - 2)}