What approach should I take to establish this logical proof? I need to design a logical math proof:
Write a detailed structured proof to prove that if m and n are integers, then either 4 divides mn or else 4 does not divide n. 
Hint: Think about the form of the statement. 
I was thinking about first proving that either 4 should divide m or n in order for the first part to be true, then use proof by cases to prove what happened when m was 4, and another case when n was 4...
But I got stuck, what do you recommend:

This is what I got so far:
Claim: (m ∈ Z ∧ n ∈ Z) => (4 | mn ˅ 4 ∤ n)
Negated claim: (m ∈ z ∧ n ∈ z) ∧ ( P(n) ∧ ¬P(mn) )
Proof by contradiction
Assume m , n ∈ z:
   Assume P(n) ∧ ¬P(mn):
      Then ∃ q ∈ R, n = 4q
      Let q0 be such that n = 4q0
      Then mn = m(4q0)
      Then mn = 4(mq0)
      Then ∃ q ∈ R, mn = 4q 
      Then 4 | mn 
   Then P(n) ∧ P(mn)
Then (m , n ∈ z) ∧ P(n) ∧ P(mn)) 

I know that is wrong, how can I fix it?
 A: You are to write your proof using only what is given: $m, n\in \mathbb Z$. You are then to prove that from this, it follows that 
$(1)\quad 4 \mid mn$ or 
$(2) \quad\lnot (4\mid n)$.
Suggestion: start by considering the two cases:


*

*$4 \mid n$, and 

*$\lnot (4 \mid n)$,


which are mutually exclusive cases, exhausting all options for $n\in \mathbb Z$. One of these two cases must necessarily be true for any given $n$: Any $n$ whatsoever must be such that $4$ divides $n$, or $4$ does not divide $n$. That's a tautology, with no need for proof. 


*

*Case (1): Suppose $4$ divides $n \in \mathbb Z.\;$ Then $\,n = 4k\,$ where $k$ is some
integer (by the definition of divisibility by $4$). Then take any $m$
whatsoever. $$mn = m(4k) = 4(mk).$$ Hence $4$ divides $mn$. Hence the
proposition is true when $4$ divides $n$, and it is true for all integers $m$.

*Case (2): Suppose $4$ does not divide $n\in \mathbb Z.\;$ Then the proposition is true
in this case, (it satisfies the disjunction "or else $4$ does not divide $n$. And it is true for any $m\in \mathbb Z$.
It either case (and we have exhausted possible cases), the proposition is true.
A: Claim: $$m,n \in \mathbb{Z} \Rightarrow (4 \mid mn \vee 4 \nmid n).$$
Proof (by contradiction):
Suppose the claim is false, that is,
\begin{align*}
\neg(m,n \in \mathbb{Z} \Rightarrow (4 \mid mn \vee 4 \nmid n)) &\equiv (m,n \in \mathbb{Z})\wedge\neg(4 \mid mn \vee 4 \nmid n) \\ 
&\equiv (m,n \in \mathbb{Z})\wedge(4 \nmid mn \wedge 4 \mid n).
\end{align*}
All that is left to show is that any part of the negation is false, ie. $(4 \nmid mn \wedge 4 \mid n)$ is impossible. Conclude that since the negation of the claim is false, the claim itself must be true.
