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.


You need to show that $a/n$ is an integer. Try thinking about the prime factorizations of $a$ and $n$.
