Let $m$ and $n$ be two integers. Prove that $mn+m$ is odd if and only if $m$ is odd and $n$ is even.

Question

—The statement:
Let $m$ and $n$ be two integers. Prove that $mn+m$ is odd if and only if $m$ is odd and $n$ is even.

—Solution:
First, we show that if $m$ is odd and $n$ is even, then $mn+m$ is odd. Assume that $m$ is odd and $n$ is even. Then $m = 2a+1$ and $n = 2b$ for integers $a$ and $b$. So $mn+m = (2a+1)(2b)+(2a+1)= 4ab +2b+2a+1=2(2ab+b+a)+1$. Since $2ab+b+a$ is an integer, $mn+m$ is odd.
Next, we verify the converse, that is, if $mn + m$ is odd, then $m$ is odd and $n$ is even. Assume that it is not the case that $m$ is odd and $n$ is even. Then either $m$ is even or $n$ is odd. Therefore, either $m$ is even or $m$ and $n$ are both odd. We consider these two cases.
Case 1. $m$ is even.
Case 2. $m$ and $n$ are both odd.

—My question:
Why didn't we consider a third case where $m$ is even and $n$ is odd

Answer 1

$mn+m=m(n+1)$ is odd iff both factors $m$ and $n+1$ are odd iff $m$ is odd and $n$ is even.