Answer
With this, we can assume that if a sum of x and y (both integers) is ODD, then one is Even, and the other is Odd.
Work Step by Step
This question is wrong, it didn't meant this
Let's specify all the combinations to make it easy to understand why the book made a mistake in this question.
Imagine that $x$ and $y$ are odd numbers, and $S$ is the Sum of both:
$x = 2k + 1$ ODD
$y = 2r + 1$ ODD
$\begin{split}
S & = x + y \\
& = 2k + 1 + 2r + 1 \\
& = 2k + 2r + 2 \\
& = 2(k + r + 1) \\
\end{split}$
$S$ is even. So we have $2$ odd numbers resulting in an even number.
This already shows that the assumption Susanna Epp. makes in the book isn't correct, but let's continue to cover all cases.
Now, imagine that $x$ and $y$ are even numbers, and $S$ is the Sum of both:
$x = 2k$ EVEN
$y = 2r$ EVEN
$\begin{split}
S & = x + y \\
& = 2k + 2r \\
& = 2(k + r) \\
\end{split}$
$S$ is even. So we have $2$ even numbers resulting in an even number.
Now, imagine that $x$ is even and $y$ is odd, and $S$ is the Sum of both:
$x = 2k$ EVEN
$y = 2r + 1$ ODD
$\begin{split}
S & = x + y \\
& = 2k + 2r + 1 \\
& = 2(k + r) + 1\\
\end{split}$
$S$ is odd. So we have $1$ odd and $1$ even, resulting in an even number.
Now, imagine that $x$ is odd and $y$ is even, and $S$ is the Sum of both:
(This case is similar to the case above)
$x = 2k + 1$ ODD
$y = 2r$ EVEN
$\begin{split}
S & = x + y \\
& = 2k + 1 + 2r \\
& = 2k + 2r + 1 \\
& = 2(k + r) + 1\\
\end{split}$
$S$ is odd. So we have $1$ odd and $1$ even (again), resulting in an even number.
The results we saw:
Even + Odd or Odd + Even = ODD
Even + Even or Odd + Odd = Even
With this, we can assume that if a sum of x and y (both integers) is ODD, then one is Even, and the other is Odd.
Also, if a sum of x and y (both integers) is Even, or they are both Even, or Odd.
