You will want to use contrapositive for proving the converse of this statement, and in most introductory proof classes the professor should make a point of this. That is to say, for the question you posed the cleanest proof is given as follows,
Claim: If $n$ is odd, then $n^2$ is odd, for all $n \in \mathbb{Z}$.
Proof: Assume that $n$ is odd, then $n=2k+1$, for some $k \in \mathbb{Z}$. Hence,
$$n^2 = (2k+1)^2= 4k^2 + 4k + 1 = 2(2k^2 + 2k) +1 $$
where $(2k^2 + 2k) \in \mathbb{Z}$. Therefore, $n^2$ is odd as desired.
Whereas, for the converse you will quickly run into trouble if you do not try a proof by contrapositive (Exercise: Try it with a direct proof and see where you get stuck!)
Claim: If $n^2$ is odd, then $n$ is odd, for all $n \in \mathbb{Z}$.
Proof: By contrapositive, the claim is logically equivalent to, "If $n$ is even then $n^2$ is even, for all $n \in \mathbb{Z}$". Assume that $n$ is even, then $n=2k$, for some $k \in \mathbb{Z}$. Hence, $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$ where $2k^2 \in \mathbb{Z}$. Therefore, $n^2$ is even as desired.
Your idea is fine. More compactly, any number is congruent to $0,1,2,3$ modulo $4$. Squaring gives $0,1,0,1$. But $5=-1$ modulo $4$, but it cannot be the case $a^2=-1\mod 4$ for any $a$, by the above.
Best Answer
you can write more generally that the product of 2 odd integers is odd
let $a = 2k+1$ and $b = 2p+1$
then $ab = (2k+1)(2p+1) = 2(2kp+k+p)+1 $ which is the general format of odd numbers