Given: $$x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}$$
Solutions to the equation are: $$\{(1,1), (1,-1), (-1,1), (-1,-1)\}$$
Among them, only \( (1, -1) \) satisfies \( x > y \).
Answer: $$\boxed{1}$$
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