Prove or disprove the following: “If x is an odd integer and y is an even integer then x² + y² is an odd integer.”

We will analyze whether the given statement is true or false.

### Step 1: Define Odd and Even Integers
- An **odd integer** can be written as \( x = 2k + 1 \), where \( k \) is an integer.
- An **even integer** can be written as \( y = 2m \), where \( m \) is an integer.

### Step 2: Compute \( x^2 + y^2 \)
- Squaring \( x \):

  \[
  x^2 = (2k+1)^2 = 4k^2 + 4k + 1
  \]

  Since \( 4k^2 + 4k \) is always even, \( x^2 \) is **odd**.

- Squaring \( y \):

  \[
  y^2 = (2m)^2 = 4m^2
  \]

  Since \( 4m^2 \) is always divisible by 2, \( y^2 \) is **even**.

### Step 3: Sum of \( x^2 \) and \( y^2 \)

\[
x^2 + y^2 = (\text{odd}) + (\text{even}) = \text{odd}
\]

### Conclusion:
Since an **odd + even = odd**, the sum \( x^2 + y^2 \) is always odd. 

Thus, the given statement is **true**.

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Do you want more details or have any questions? Here are some related questions to explore:

1. What happens if both \( x \) and \( y \) are even?
2. What happens if both \( x \) and \( y \) are odd?
3. Can we generalize this result to other exponents, like \( x^3 + y^3 \)?
4. How does this result apply to modular arithmetic?
5. Can we extend this idea to non-integer real numbers?

**Tip:** When dealing with odd and even numbers, always express them in terms of \( 2k \) and \( 2k+1 \) to simplify proofs.

Explore the proof that the sum of the square of an odd integer and the square of an even integer is always odd. This problem involves understanding the parity of integers and applying basic algebraic properties. Suitable for students in Grades 7-9.

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