## Discrete Mathematics with Applications 4th Edition

$n²$ is odd because: $n$ is odd and can be write as $2k + 1$ $n² = 2(2k² + 2k) + 1$
If $n$ is an odd integer. $n$ can be write as: $n = 2 k + 1$ (with $k$ being one integer number) $k \in \mathbb{Z}$ so $\begin{split} n^2 & = (2k + 1)^2 \\ & = (2k)^2 + 2.2k.1 + 1^2 \\ & = 4k^2 + 4k + 1 \\ & = 2(2k^2 + 2k) + 1 \\ \end{split}$ Since $2k^2 + 2k$ is an integer, then $2(2k^2 + 2k) + 1$ is also odd, by the definition of odd numbers.