x and y are integers, 0 < x < y, and x² + y^2 is even. Which of the following integers must be even?

Rationale

x + y must be even.

For the sum \( x^2 + y^2 \) to be even, both \( x \) and \( y \) must either be both even or both odd. Since the sum of two odd integers is even, this means that \( x + y \), which is the sum of \( x \) and \( y \), is also even.

A (xy) YOUR ANSWER

The product \( xy \) can be odd if both \( x \) and \( y \) are odd and will only be even if at least one of them is even. Since both \( x \) and \( y \) can be odd, \( xy \) does not have to be even and therefore cannot be guaranteed as such.

B (x + y) CORRECT

As stated, if both \( x \) and \( y \) are either even or odd, their sum \( x + y \) must be even. This is a direct consequence of the properties of even and odd integers, confirming that \( x + y \) is necessarily even.

C (y - x) YOUR ANSWER

The difference \( y - x \) depends on the parity of \( x \) and \( y \). If both are odd, \( y - x \) is even; if both are even, \( y - x \) is also even. However, if one is odd and the other is even (which is not allowed in this case), the difference could be odd. Thus, \( y - x \) cannot be guaranteed to be even.

D (x² + y) YOUR ANSWER

The term \( x^2 + y \) can be either even or odd depending on the individual parities of \( x \) and \( y \). If \( x \) is odd, \( x^2 \) is odd and adding \( y \) (which could be odd or even) may yield an odd result. Therefore, \( x^2 + y \) does not have to be even.

Conclusion

In summary, when \( x \) and \( y \) are both integers such that \( x^2 + y^2 \) is even, it necessarily implies that both integers are either both even or both odd. Consequently, their sum \( x + y \) must be even, while the other choices do not hold this property under the given conditions. This logical deduction is crucial for understanding integer properties in mathematical contexts.

x and y are integers, 0 < x < y, and x² + y^2 is even. Which of the following integers must be even?

x + y must be even.

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