TITLE: Is my proof correct: if $n$ is odd then $n^2$ is odd?
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: An integer can be said to be odd if $2$ does not occur in its prime factorization,
since $n$ is odd, $2$ does not not occur in the $prime$ $factorization$  of $ n$, 
if we write the prime factorization of $n^2$ ,  $2$ will not occur in this as well,
since $n^2$ will have same prime factors as that of $ n$ , but with increased powers.
hence $n^2$ will be odd for every odd  $ n$ .