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

Question

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?

---

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 1

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.

---

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.

Answer 2

Here is direct proof: $$ n=(3n^2+3n)-(3n^2+2n)=6\binom{n+1}{2}-(3n^2+2n) $$ is even because it is the difference of two even numbers.

Answer 3

$3n^2+2n$ can be written as $n^2+2(n^2+n)$. As we can see $2(n^2+n)$ is even then for whole expression to be even $n^2$ should also be even(as addition of 2 even numbers is even and one being even already, $n^2$ should also be even).Now we know product of two even number is even(and $n$ at a time can be even or odd.$n$ cannot be odd because product will give an odd number).Therefore $n$ has to be even for the whole expression to be even.

Answer 4

Another way to solve this is by using modular arithmetic in the ring $\mathbb{Z}/2$.

If $3n^{2}+2n$ is even, we can write this as $$ 3n^{2}+2n \equiv 0 \mod{2}$$ which is equivalent to $$n^{2} \equiv 0 \mod{2}\\$$

When $n=0, n^{2} =0$ and when $n=1, n^{2} = 1$, so if $3n^{2}+2n$ is even, $n$ is even.