Does this simple proof-by-contradiction, also require contrapositive? Simple exercise 6.2 in Hammack's Book of Proof. "Use proof by contradiction to prove"
"Suppose $n$ is an integer. If $n^2$ is odd, then $n$ is odd"
So my approach was:
Suppose instead, IF $n^2$ is odd THEN $n$ is even
Alternatively, then you have the contrapositive, IF $n$ is not even ($n$ is odd), then $n^2$ is not odd ($n^2$ is even).
$n = 2k+1$  where $k$ is an integer.  (definition of odd)
$n^2 = (2k+1)^2$
$n^2 = 4k^2 + 4k + 1$
$n^2 = 2(2k^2 + 2k) + 1$
$n^2 = 2q + 1$ where $q = 2k^2 + 2k$
therefore $n^2$ is odd by definition of odd.
Therefore we have a contradiction. Contradictory contrapositive proposition said $n^2$ is not odd, but the derivation says $n^2$ is odd.  Therefore the contradictory contrapositive is false,  therefore the original proposition is true.
Not sure if this was the efficient/correct way to prove this using Proof-By-Contradiction.

Last night I read this as a perfectly acceptable claim, but, as has been pointed out, your negation was not simply a "harder case" but instead the converse. Apologies! Your proof happened to work here because the stronger relationship (i.e. $\iff$)
A couple of things to note though. For a proof by contradiction would simply need to state that: there exists some $n^2$ which is odd, which has an even $n$ (perhaps phrased better: Suppose that you have $n^2$ which is odd for a corresponding $n$ even). With this in mind, consider the following direct proof by contradiction.
Assume that for some $n^2$ which is odd, we have $n = 2k$.
\begin{align}
 n^2 &= n\cdot n\\
&= (2k)(2k)\\
&= 2(2k^2) \\ 
2k^2 &\in \mathbb{Z} \ \ \ \ \text{call it q} \\
n^2 &= 2q \implies 2 | n^2 
\end{align}
and so we have reached a contradiction, so our assumption must be incorrect!
I do want to stress that, as with most propositions, there are multiple ways in which to prove this statement, as has been pointed out using the contrapositive is the most efficient way of proving it, but as the exercise asked to use a contradiction method, the above would work!
