# What is the number of ordered pair of integers (x,y) satisfying the equation x^2 + 6x + y^2 = 4?

Feb 23, 2018

$8$

#### Explanation:

$\text{Complete the square for x : }$

"(x+3)^2 + y^2 = 13

$\text{As both terms are positive, we know that}$

$- 4 < x + 3 < 4$
$\text{and}$
$- 4 < y < 4$

$y = \pm 3 \implies x + 3 = \pm 2 \implies x = - 5 \mathmr{and} - 1$
$y = \pm 2 \implies x + 3 = \pm 3 \implies x = - 6 \mathmr{and} 0$
$y = \pm 1 \text{ and " y = 0, " yield no perfect square}$

$\text{So we have 8 solutions : }$
$\left(- 5 , - 3\right) , \left(- 5 , 3\right) , \left(- 1 , - 3\right) , \left(- 1 , 3\right) ,$
$\left(- 6 , - 2\right) , \left(- 6 , 2\right) , \left(0 , - 2\right) , \left(0 , 2\right) .$