Discussion
The number of elements in the set \(\{(x, y): 2x^{2} + 3y^{2} = 35, x, y \in \mathbb{Z}\}\) , where \(\mathbb{Z}\) is the set of all integers, is
Answer: E
The set has 8 elements.
Since \(3y^{2} \le 35\), \(y^{2}\) can only be 0, 1, 4 or 9. Check each:
- \(y^{2} = 0\): \(2x^{2} = 35\), not possible in integers.
- \(y^{2} = 1\): \(2x^{2} = 32\), \(x^{2} = 16\), \(x = \pm 4\).
- \(y^{2} = 4\): \(2x^{2} = 23\), not possible.
- \(y^{2} = 9\): \(2x^{2} = 8\), \(x^{2} = 4\), \(x = \pm 2\).
So \(x = \pm 4, y = \pm 1\) gives 4 pairs and \(x = \pm 2, y = \pm 3\) gives 4 more: 8 in all. Counting only positive values gives 2, and counting signs for one variable only gives 4.
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