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.

---

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.