Prove: $2n^{2} + 3n$ is even if and only if $n^{3} - 4$ is even. I need to prove this statement in some way, maybe through a contradiction or contrapositive but I am stuck. I first assume that if $n^{3} - 4$ is even that $n^{3} - 4$ = $2k$ for some integer $k$ (by definition of even). Then when I try to solve for $n$ so that I can plug it into $2n^{2} + 3n$, I get $n = (2k + 4)^{1/3}$. If I try to plug this into $2n^{2} + 3n$, then I do not know how to manipulate $$2((2k + 4)^{1/3})^{2} + 3((2k + 4)^{1/3})$$ to show that it is indeed even (or equal to 2 times some integer).
Guide:

If $n^3-4$ is even, then $n^3$ has to be even, hence $n$ is even. Hopefully you can prove that $2n^2+3n$ is even from here.

Conversely, if $2n^2+3n=n(2n+3)$ is even, since $2n+3$ is odd, we have $n$ must be even. Again, hopefully you can prove that $n^3-4$ is even.
Let consider the two cases


$n=2k$ is even then


$$2n^{2} + 3n=8k^2+6k=2(4k^2+3k)$$

$$ n^{3} - 4=8k^3-4=2(4k^3-2)$$

both are even


$n=2k+1$ is odd then


$$2n^{2} + 3n=8k^2+8k+2+6k+3=2(4k^2+7k+2)+1$$

$$ n^{3} - 4=8k^3+12k^2+6k-3=2(4k^3+6k^2+3k-1)+1$$

both are odd.