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I know that this is a common question, but I want to see your take on this. The way I approached this was by contraposition (i'm still new with the proofs but any help would be appreciated, even if it's by contradiction.)
Assume $n$ is odd, thus $n^2$ is odd as well.
Since $n$ is odd then $n=2k+1$ for some integer $k$.
Then $n^2= (2k+1)^2 = 4k^2+4x+1 \rightarrow 2(2k^2+2k)+1.$ Which clearly shows that $n^2$ is odd.