elementary number theory - Prove that if $3n^2 + 2n$ is even, then $n$ is even

Just want to check if my proof is OK

Prove that if $3n^2 + 2n$ is even, then $n$ is even

If $3n^2 + 2n$ is even, then $3n^2 + 2n = 2k$ for some $k \in \mathbb{Z}$.

$\Rightarrow n(3n+2) = 2k$

So 2 must either divide $n$ or $(3n+2)$. If 2 divides $n$, we're done. Otherwise let 2 divide $(3n+2)$. Then 2 must divide $3n \Rightarrow$ it divides $n$ (since it doesn't divide 3). Hence result.

Is this ok?

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EDIT (further question):

A proof by contradiction supposedly also works:

Assume $n$ is odd, i.e. $n=2k+1$.

Then $3(2k+1)^2 + 2(2k+1) = 3(4k^2 + 4k +1) + 4k + 2 = 12k^2 + 16k + 5$ which is not divisible 2. Hence $n$ must be even.

My issue with this: how have we proved that $n$ is even? Have we not only proven that $n$ can't be odd? We don't know it works for every even number though...?

Answer

Yes, it is just fine.

I would have done it as follows: since $3n^2+2n=n^2+2(n^2+n)$ and $2(n^2+n)$ is even, $3n^2+2n$ is even if an only if $n^2$ is even. And $n^2$ is even if and only if $n$ is even.

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Your proof by contradiction shows (correctly) that $$n\text{ odd}\implies3n^2+2n\text{ odd,}$$ which is equivalent to the statement that you wish to prove.