Prove: $2n^{2} + 3n$ is even if and only if $n^{3} - 4$ is even.

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

Note that $2n^2+3n$ is even iff $n$ is even.

Similarly $n^3-4 $ is even iff $n$ is even.

Thus $2n^2+3n$ is even iff $n^3-4$ is even.

Answer 4

Another way to prove this is to show: $n^3-4$ and $2n^2+3n$ are in same equivalance class modulo $2$, i.e;$$n^3-4\equiv 2n^2+3n\pmod{2}$$ which is easy. First observe that, $n^3\equiv n\pmod{2}$, because, if $n$ is even then $n^3$ also even and if $n$ is odd then also $n^3$, also, $n^3-4\equiv n\pmod{2}$ and $2n^2+3n\equiv n\pmod{2}$. Hence done. In simple word, this means both terms leaves same remainder when divided by $2$, so, if one of them is $0\pmod{2}$ then other also will be $0\pmod{2}$.

Answer 5

A) $n^3-4$ is even.

Then $(n^3-4)+4 =n^3$ is even.

Since $2| n^3 \rightarrow 2|n$ , Euclid's lemma.

Hence $n(2n+3)$ is even (why?).

B) $n(2n+3)$ is even.

2| $n(2n+3)$, Euclid's lemma:

$2|n$ or $2|(2(n+1)+1)$;

$2|(2(n+1)+1)$ is ruled out (why?).

Hence $2|n$ and $n^3 -4$ is even (why?).

Answer 6

Lemma: If $c$ is even then $a \pm c$ if even and only if $a$ is even.

Pf: Trivial.

So $2n^2$ is even and $4$ is even. So $2n^2 +3n$ is even if and only if $3n$ is even and $n^3 -4$ is even if and only if $n^3$ is even.

So it suffices to show that $3n$ is even if and only if $n^3$ is even.

Lemma 2: If $c$ is odd then $ac$ is even if and only if $a$ is even. If $c$ is even $ac$ is even.

Pf: Trivial.

Corollary: For integers $n \ge 1$ then $a^n$ is even iff and only $a$ is even.

Pf: If $a$ is even then $a*a^{n-1} = a^n$ is even. If $a$ is odd then $a^1$ is odd. And if $a^{n-1}$ is odd then $a^n = a*a^{n-1}$ is odd and the result follows by induction.

So as $3$ is odd. $3n$ is even if and only if $n$ is even. And $n^3$ is even if and only if $n$ is even.

So it suffices to show $n$ is even if and only if $n$ is even which is a tautology.

......

In other words:

If $2n^2 + 3n$ is even then

$3n$ is even then

$n$ is even then

$n^3$ is even the

$n^3 -4$ is even.

If $n^3 - 4$ is even then

$n^3$ is even then

$n$ is even then

$3n$ is even then

$2n^2 + 3n$ is even.

Answer 7

Hint: $$ \begin{align} \left(n^3-4\right)-\left(2n^2+3n\right) &=n^3-2n^2-3n-4\\ &=n^3-n-2\left(n^2+n+2\right)\\ &=6\binom{n+1}{3}-2\left(n^2+n+2\right) \end{align} $$ is even.

Answer 8

It's a trick question. The $n^3 - 4$ is just there to confuse you and make you think there is something complicated.

$2n^2 + 3n$ is even if and only if n is even: That's because $2n^2$ is always even, 3n is even if and only if n is even, so the sum $2n^2 + 3n$ is even if n is even, and odd if n is odd.

And the same for $n^3 - 4$: $n^3$ is even if and only if n is even, and so is of course $n^3 - 4$.

So you have two expressions that are even if and only if n is even. So they are also even if and only the other is even.