Decide which of the following are valid proofs of the following statement:

If \(a b\) is an even number, then \(a\) or \(b\) is even.

Suppose \(a\) and \(b\) are odd. That is, \(a=2k+1\) and \(b=2m+1\) for some integers \(k\) and \(m\text{.}\) Then

\begin{align*} ab \amp =(2k+1)(2m+1)\\ \amp =4km+2k+2m+1\\ \amp =2(2km+k+m)+1\text{.} \end{align*}

Therefore \(ab\) is odd.
Assume that \(a\) or \(b\) is even - say it is \(a\) (the case where \(b\) is even will be identical). That is, \(a=2k\) for some integer \(k\text{.}\) Then

\begin{align*} ab \amp =(2k)b\\ \amp =2(kb)\text{.} \end{align*}

Thus \(ab\) is even.
Suppose that \(ab\) is even but \(a\) and \(b\) are both odd. Namely, \(ab = 2n\text{,}\) \(a=2k+1\) and \(b=2j+1\) for some integers \(n\text{,}\) \(k\text{,}\) and \(j\text{.}\) Then

\begin{align*} 2n \amp =(2k+1)(2j+1)\\ 2n \amp =4kj+2k+2j+1\\ n \amp = 2kj+k+j+\frac{1}{2}\text{.} \end{align*}

But since \(2kj+k+j\) is an integer, this says that the integer \(n\) is equal to a non-integer, which is impossible.
Let \(ab\) be an even number, say \(ab=2n\text{,}\) and \(a\) be an odd number, say \(a=2k+1\text{.}\)

\begin{align*} ab \amp =(2k+1)b\\ 2n \amp =2kb+b\\ 2n-2kb\amp =b\\ 2(n-kb)\amp =b\text{.} \end{align*}

Therefore \(b\) must be even.