How many integer pairs (x , y) satisfy x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0 ?

\[ x^2 + 4y^2 - 2xy - 2x - 4y - 8 = 0 \] \[ \Rightarrow x^2 - 2x(y+1) + 4y^2 - 4y - 8 = 0 \] \[ \Rightarrow x^2 - 2x(y+1) + y^2 + 2y + 1 + 3y^2 - 6y - 9 = 0 \] \[ \Rightarrow (x - (y+1))^2 + 3(y-1)^2 = 12 \] \[ \Rightarrow (x - y - 1)^2 + 3(y - 1)^2 = 12 \] Let \( X = x - y - 1 \) and \( Y = y - 1 \) \[ \Rightarrow X^2 + 3Y^2 = 12 \] Since \( x, y \in \mathbb{Z} \), it follows that \( X, Y \in \mathbb{Z} \). \[ (X^2, Y^2) \in \{(0,4), (9,1), (3,4)\} \] \[ (X, Y) \in \{(0,2), (0,-2), (3,1), (3,-1), (-3,1), (-3,-1)\} \] Evaluate \( (x, y) \) for each \( (X, Y) \): \[ X = x - y - 1, \quad Y = y - 1 \] \[ (x, y) \in \{(4,3), (0,-1), (6,2), (1,0), (0,2), (-2,0)\} \] **6 possible integer pairs.**