Question

Let \( Z \) be the set of all integers, and consider the set \( X = \{(x, y): x^2 + 2y^2 = 3, x, y \in Z \) and \( Y = \{(x, y): x>y, x, y \in Z\} \). Then the number of elements in \( X \cap Y \) is:

• A. 2
• B. 1
• C. 3
• D. 4

Hint:

When solving such problems, start by finding all the elements of the set and then apply additional conditions.

Answer 1

The Correct Option is B

We are given two sets: 1. \( X = \{(x, y): x^2 + 2y^2 = 3, x, y \in Z\} \) 2. \( Y = \{(x, y): x>y, x, y \in Z\} \) 
Step 1: Find the elements of set \( X \).
We solve the equation \( x^2 + 2y^2 = 3 \) for integer values of \( x \) and \( y \):
If \( x = 1 \), then \( 1^2 + 2y^2 = 3 \) gives \( 2y^2 = 2 \), so \( y = \pm 1 \).
If \( x = -1 \), then \( (-1)^2 + 2y^2 = 3 \) gives \( 2y^2 = 2 \), so \( y = \pm 1 \).
If \( x = 2 \), then \( 2^2 + 2y^2 = 3 \) gives \( 4 + 2y^2 = 3 \), which has no integer solutions.
If \( x = -2 \), then \( (-2)^2 + 2y^2 = 3 \) gives \( 4 + 2y^2 = 3 \), which has no integer solutions.
Thus, the integer solutions for \( x^2 + 2y^2 = 3 \) are: \[ (x, y) = (1, 1), (1, -1), (-1, 1), (-1, -1). \] Step 2: Apply the condition \( x>y \) from set \( Y \).
We now check which of these solutions satisfy \( x>y \):
For \( (1, 1) \), \( x = 1 \) and \( y = 1 \), so \( x \not> y \).
For \( (1, -1) \), \( x = 1 \) and \( y = -1 \), so \( x>y \).
For \( (-1, 1) \), \( x = -1 \) and \( y = 1 \), so \( x \not> y \).
For \( (-1, -1) \), \( x = -1 \) and \( y = -1 \), so \( x \not> y \).
Thus, the only solution that satisfies both conditions is \( (1, -1) \). Therefore, the number of elements in \( X \cap Y \) is \( \boxed{1} \).