If x is integer and `x^2` is even, which of the following must be true?

Options

A. x is even
B. x is odd
C. `x^3` is odd
D. `x^3` is even

Answer

To solve the problem, we need to determine what must be true if \( x \) is an integer and \( x^2 \) is even.

### Step-by-Step Solution:

1. **Understanding the Condition**: We start with the condition that \( x^2 \) is even. Recall that a number is even if it can be expressed as \( 2k \), where \( k \) is an integer.

2. **Analyzing \( x^2 \)**: If \( x^2 \) is even, we can use the property of squares. The square of an integer is even if and only if the integer itself is even.

3. **Using Contradiction**: Let's assume \( x \) is odd. An odd integer can be expressed in the form \( x = 2k + 1 \) for some integer \( k \). Squaring this gives:
\[ x^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \]
This shows that \( x^2 \) is odd, which contradicts our initial condition that \( x^2 \) is even.

4. **Conclusion**: Since assuming \( x \) is odd leads to a contradiction, we conclude that \( x \) must be even.

5. **Final Statement**: Therefore, if \( x^2 \) is even, it must be true that \( x \) is even.