The number of order pairs of integers $(x, y)$ satisfying the equation $x^{2} + 6 x + y^{2} = 4$

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Solution

$x^{2} + 6 x + y^{2} = 4 \Rightarrow (x + 3)^{2} + y^{2} = 13 \Rightarrow x + 3 = \pm 2; y = \pm 3 x = - 5, - 1; y = - 3, 3$
Number of order pair $= 4$
and
$x + 3 = \pm 3; y = \pm 2 \Rightarrow x = 0, - 6; y = - 2, 2$
Number of order pair $= 4$
total number of order pairs $= 8$

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