Prove $mn$ and $m+n$ are both even if and only if $m$ and $n$ are both even.

Question

How should I begin? It is solve like a normal bi-conditional or must it be done a proof by case?

Answer 1

There is really only one case you need to demonstrate in the forward direction, this only requires that you use the definition of a number being even i.e. $n$ is even if $n=2k$ for $k\in \mathbb{Z}$.

So to begin try assuming that $mn$ and $m+n$ are both even and either $m$ is odd or $n$ is odd ( there is no need for cases here since the same argument for $n$ will apply for $m$). Try and derive a contradiction.

For the opposite way use the definition of a number being even and prove it directly. Hope that helps :)

Answer 2

A non-case simple proof:

$(m+1)(n+1)=mn+m+n+1$ is odd, hence $m+1$ and $n+1$ are both odd