Question

Let \( C \) denote the set of all tuples \( (x, y) \) which satisfy \( x^2 = 2^y \), where \( x \) and \( y \) are natural numbers. What is the cardinality of \( C \)?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When solving equations involving powers, express one side as a power of the same base to find integer solutions.

Answer 1

The Correct Option is C

Step 1: Start with the given equation. We are given the equation: \[ x^2 = 2^y. \] We are tasked with finding the values of \( x \) and \( y \) where both \( x \) and \( y \) are natural numbers. 
Step 2: Solve for possible values of \( x \) and \( y \). For \( x^2 = 2^y \) to hold, \( x \) must be a power of 2. Let: \[ x = 2^k. \] Substitute into the equation: \[ (2^k)^2 = 2^y \quad \Rightarrow \quad 2^{2k} = 2^y \quad \Rightarrow \quad 2k = y. \] Thus, for every integer \( k \), we can find a corresponding \( y = 2k \). 
Step 3: Identify the valid tuples. Let’s check for small values of \( k \):
For \( k = 1 \), we have \( x = 2 \) and \( y = 2 \), which gives the tuple \( (2, 2) \).
For \( k = 2 \), we have \( x = 4 \) and \( y = 4 \), which gives the tuple \( (4, 4) \).
These are the only valid solutions, as higher values of \( k \) would give non-natural values for \( y \). 
Step 4: Conclude the result. Thus, the set \( C \) contains exactly 2 tuples: \( (2, 2) \) and \( (4, 4) \). The cardinality of \( C \) is: \[ \boxed{2}. \]