Is my proof correct: if $n$ is odd then $n^2$ is odd? 
Prove that for every integer $n,$ if $n$ is odd then $n^2$ is odd.

I wonder whether my answer to the question above is correct. Hope that someone can help me with this.

Using contrapositive, suppose $n^2$ is not odd, hence even. Then $n^2 = 2a$ for some integer $a$, and 
  $$n = 2(\frac{a}{n})$$ where $\frac{a}{n}$ is an integer. Hence $n$ is even.


Hint $\rm\ \ n\ odd\:\Rightarrow\:2\ |\ n\!-\!1\ |\ n^2\! -\! 1\:\Rightarrow\,n^2\!-\!1\:$ is even, so $\rm\,n^2\ is\ \ldots\ $ QED
More conceptually, multiplying by an odd integer preserves parity since $\rm\:(1\!+\!2k)n = n + 2kn\:$ leaves the same remainder as $\rm\:n\:$ when divided by $2$. Hence a product of odd integers is odd.
As for your proof, that odd $\rm\:n\ |\ 2a\:\Rightarrow\: n\ |\ a\ $ requires justification. You could use Euclid's Lemma, prime factorization, or Bezout, e.g. $\rm\:11\ |\ 2a\:\Rightarrow\:11\ |\ 6(2a)\!-\!11a\: =\: a,\:$ which uses the Bezout identity $\rm\: 1 = gcd(2,11) = 6\cdot 2 - 1\cdot 11.\:$ It is easy  to generalize that from $11$ to any odd integer.
However, that approach is much more work than need be. Notice that
$$\rm n\ |\ 2a\iff \exists\ k\!:\ nk\ =\ 2a\iff \dfrac{a}n\: =\: \dfrac{k}2,\ \ \ so\ \ \ n\ |\ a\iff 2\ |\ k$$
Generally proving $\rm\:2\ |\ k\:$ is easier than $\rm\:n\ |\ a\:$ since there are only $2$ residue cases to test modulo $2$, i.e. the smaller the divisor the easier the computation of the remainder. Indeed, here it's trivial by the above "more conceptual" proof, viz. since $\rm\:n\:$ is odd, $\rm\:nk\:$ and $\rm\:k\:$ have the same parity, therefore $\rm\:nk = 2a\:$ is even implies $\rm\:k\:$ is even. Thus $\rm\:2\ |\ k,\:$ so $\rm\:n\ |\ a.$
