Question: Question: Let $ n,m \in \mathbb{Z}$. Prove the following:

$\forall$$(m \wedge n)$, if both $m$ and $n$ are even, then so is $mn.$

Attempt: By Definition 2.3.1, we let $ n \in \mathbb{Z}$. Then, $n$ is even whenever there exists some $ k \in \mathbb{Z}$ such that $ n = 2k$ and $n$ is odd whenever there exists some $ k \in \mathbb{Z}$ such that $n = 2k+1$

Suppose $m$ and $n$ are even, then there exists some $ k \in \mathbb{Z}$ [and $l \in \mathbb{Z}$] such that $ m = 2l$ and $n = 2k$. 

For, $mn$, we have

$(2k)(2l)$

My question is whether or not I should just stop there or go further. 

So I could just continue by multiplying the $2$ 

$(4kl)$ 

and then manipulate a bit?

$(2)(2kl)$

As a result $mn$ is even.

Answer: 
It never hurts to provide more detail as long as it is pertinent to the proof.

I would have finished it off as follows:
$$mn = (2k)(2l) = 4(kl) =2(2kl)$$
and as $2kl\in \mathbb{Z}$, it is a witness to the conclusion that $mn$ is even.
You basically have it. Using modular arithmetic to condense the argument, suppose $n \equiv 0 \pmod{2}$ and $m \equiv 0 \pmod{2}$. Then we have:

$$nm \equiv 0 \pmod{2}$$

by properties of modular arithmetic. In general, if $a \equiv b \pmod{n}$ and $x \equiv y \pmod{n}$, then $ax \equiv by \pmod{n}$. It's a good exercise to prove this :)