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.

To prove
$$
n^2\text{ is odd}\implies n\text{ is odd}\tag{1}
$$
by contradiction, you need to prove that
$$
n^2\text{ is odd}\wedge n\text{ is even}\tag{2}
$$
is false. That is, you need to suppose that $n^2$ is odd and that $n$ is even and obtain a contradiction from those two statements.
This method of proof becomes clearer when the implication
$$
n^2\text{ is odd}\implies n\text{ is odd}
$$
is written in a logically equivalent way as
$$
\neg((n^2\text{ is odd})\wedge\neg(n\text{ is odd}))\tag{3}
$$
The proof by contradiction assumes the negation of the statement and obtains a known contradiction from it. In this case, you see that the negation of $(3)$ is $(2)$.
You propose to show
$$
n^2\text{ is odd}\implies n\text{ is even}\tag{4}
$$
is false in order to show $(1)$. That is incorrect.
For example, one could prove that
$$
x>0\implies\sin(x)\geq0
$$
is false and yet
$$
x>0\implies\sin(x)<0
$$
is also false.
In fact, what you did is show the converse of $(1)$. That is, you showed
$$
n\text{ is odd}\implies n^2\text{ is odd}
$$
In this case, in order to prove $(1)$, a proof of its contrapositive is the simplest way to go. Indeed, if $n=2k$ is even, then $n^2=(2k)^2=2(2k^2)$ is even. Here there is no real difference between the proof by contradiction and the proof by contrapositive: the hypothesis that $n^2$ is odd in $(2)$ doesn't need to be used.
