For n∈Z, prove n^2 is odd if and only if n is odd.

For $n\in\Z$, let us prove that $n^2$ is odd if and only if $n$ is odd.

Let $n$ be an odd number. Then $n=2k+1$ for some $k\in\Z.$ Then $n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1=2s+1,$ where $s=2k^2+2k\in\Z.$ Therefore, $n^2$ is odd.

Let $n^2$ be an odd number. Let us prove that $n$ is odd using the method by contradiction. Suppose that $n$ is not odd, and hence $n$ is even, that is $n=2k$ for some $k\in\Z.$ Then $n^2=(2k)^2=4k^2=2(2k^2)=2m,$ where $m=2k^2\in\Z.$ It follows that $n^2$ is even. This contradiction proves that $n$ is odd.