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).
Prove: $2n^{2} + 3n$ is even if and only if $n^{3} – 4$ is even.
elementary-number-theoryproof-writing
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Best Answer
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.