How-many-integer-pairs-x-y-satisfy-x-2-4y-2-2xy-2x-4y-8-0-

Question Number 19413 by Tinkutara last updated on 10/Aug/17

How many integer pairs (x, y) satisfy

x^{2} + 4 y^{2} - 2 x y - 2 x - 4 y - 8 = 0 ?

Commented by mrW1 last updated on 11/Aug/17

(0, - 1)

(- 2, 0)

(4, 0)

(0, 2)

(6, 2)

(4, 3)

is this correct ?

Answered by mrW1 last updated on 11/Aug/17

x^{2} - 2 (y + 1) x - 4 (2 + y - y^{2}) = 0

\Rightarrow x = \frac{2 (y + 1) \pm \sqrt{4 {(y + 1)}^{2} + 16 (2 + y - y^{2})}}{2}

x = y + 1 \pm \sqrt{y^{2} + 2 y + 1 + 8 + 4 y - {4 y}^{2}}

x = y + 1 \pm \sqrt{3 (3 + 2 y - y^{2})}

x = y + 1 \pm \sqrt{3 [4 - {(y - 1)}^{2}]}

4 - {(y - 1)}^{2} ⩾ 0

{(y - 1)}^{2} ⩽ 4

since x is integer, 4 - {(y - 1)}^{2} = 0 or 3

\Rightarrow {(y - 1)}^{2} = 4 or 1

\Rightarrow y - 1 = \pm 2 or \pm 1

\Rightarrow y = 1 \pm 2 or 1 \pm 1

\Rightarrow y = 3, - 1, 2, 0

\Rightarrow x = 4, 0, 3 \pm 3, 1 \pm 3

\Rightarrow (x, y) = (4, 3) / (0, - 1) / (6, 2) / (0, 2) / (4, 0) / (- 2, 0)

Commented by Tinkutara last updated on 11/Aug/17

Thank you very much Sir!