Proving square of nonzero integer is natural number I am learning proofs with $\mathbb N$ and have this proposition: Let $m \in\mathbb Z$. If $m \ne 0$, then $m^2 \in\mathbb N$. Previously, I have proven: For $m \in\mathbb Z$, one and only one of the following is true: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$. 
If $m \ne 0$, then there are two cases to prove:
Case 1: $m \in\mathbb N$
That's straightforward because the product of two natural numbers is a natural number
Case 2: $-m \in\mathbb N$:
\begin{align*}
-m \cdot -m \in\mathbb N\\
(-1 \cdot -1) \cdot m \cdot m \in\mathbb N\\
m \cdot m \in\mathbb N\\
\end{align*}
What do you think? 
 A: If $m\in Z$,and if $m\ne0$, you've two cases.
Case 1: $m\in Z^+$
$$ => m^2 \in Z^+ => m^2 \in N$$ (Since multiplication within positive integers is closed, and results in a natural number always.)
Case 2: $m\in Z^-$
$$ => m^2 \in Z^+ => m^2 \in N$$
(Since multiplication within negative integers always results in a positive integer as you said by typing out the below :-)\begin{align*}
-m \cdot -m \in\mathbb N\\
(-1 \cdot -1) \cdot m \cdot m \in\mathbb N\\
m \cdot m \in\mathbb N\\
\end{align*}
QED
Your proof is perfect (it's essentially the same as mine!); I'd just suggest you say that multiplication operation is closed for numbers picked from positive integers instead of 'That's straightforward because the product of two natural numbers is a natural number' (In case you want to use a bit of terminology).
Otherwise it's completely fine!
A: You need to prove that: $(-m)(-m) = m\cdot m$. 
By definition of $-m$ is that:
$m + (-m) = 0 \to (m+(-m))(m-(-m)) = 0(m+(-m)) = 0 \to m^2 - m(-m) + (-m)m - (-m)(-m) = 0 \to m^2 = (-m)(-m)$.
