Answer
$k^2 + m^2$ is a even number if $k$ is odd, and $m$ is even, because $m$ can be write as $2r+1$, and $m$ can be rewrite as $2s$
$k^2 + m^2 = 2(2r^2 + 2r + 2s^2) + 1$ , that is a odd integer, by the definition of odd numbers.
Work Step by Step
If $k$ is any odd integer, and $m$ is any even integer, $k$ and $m$ can be writen as:
$k = 2r + 1$
$m = 2s$
(with $k$ and $r$ being two integers)
$r, s \in \mathbb{Z}$
Summing $k^2$ and $m^2$:
$\begin{split}
k^2 + m^2 & = (2r + 1)^2 + (2s)^2 \\
& = (2r)^2 + 2(2r.1) + 1^2 + (2s)^2 \\
& = 4r^2 + 4r + 1 + 4s^2 \\
& = 4r^2 + 4r + 4s^2 + 1 \\
& = 2(2r^2 + 2r + 2s^2) + 1 \\
\end{split}$
Since $2r^2 + 2r + 2s^2$ is an integer, then $2(2r^2 + 2r + 2s^2) + 1$ is an odd number, by the definition of odd numbers.
