## Discrete Mathematics with Applications 4th Edition

$k^2 + m^2$ 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.
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.