I am confused by the following exercise in Velleman's How To Prove It:
'Suppose $m$ and $n$ are integers. If $mn$ is even, then either $m$ is even or $n$ is even.
Proof: Suppose $mn$ is even. Then we can choose an integer $k$ such that $mn=2k$. If $m$ is even then there is nothing more to prove, so suppose $m$ is odd. Then $m=2j+1$ for some integer $j$. Substituting this into the equation $mn=2k$, we get $(2j+1)n=2k$, so $2jn+n=2k$, and therefore $n=2k-2jn=2(k-jn)$. To prove $n$ is even it suffices to find an integer $c$ such that $n=2c$, from the above $c=k-jn$ works. Since $k-jn$ is an integer, it follows that $n$ is even.'
There are two things I don't understand:
1) How could $n=2(k-jn)$ be legitimate? Isn't that circular? Surely if he is to solve for n all n must be on one side of the equation?
2) I am not quite sure what he did. It seems that he split them into two cases to exhaust the possibilities of m. If one can prove exhaustively that the conclusion necessarily happens, the proof is correct. But in terms of the entire equation that doesn't seem exhaustive to me, i.e. what about $n$?
Also, I understand that by existential instantiation the definition of an even number $$\exists x(m=2x) $$ leads to m=2k, and the same goes with m=2j+1.But why is it suffice to find an integer c such that n=2c? Is he going to use existential generalization to make n=2c into $$ \exists x(n=2x) $$ i.e. the definition of even number, to complete a direct proof? But this is not explicit in the proof.
I am trying to self-study this and my maths is, needless to say, dire. So I really appreciate any help on this, thank you so much!