# Show that if $n^2$ is even then $n$ is even, for $n$ an integer. [duplicate]

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.

## marked as duplicate by Dietrich Burde, Ethan Bolker, Community♦Mar 9 '18 at 16:34

There's a flaw in your proof. That's when you write “Assume $n$ is odd, thus $n^2$ is odd as well.” That's exactly what you are supposed to prove! Eliminate that sentence, and your proof will be just fine.