Proving $n^3$ is even iff $n$ is even I am trying to prove the following statement:

Prove $n^3$ is even iff n is even.

Translated into symbols we have:

$n^3$ is even $\iff$ $n$ is even

Since it's a double implication, I started assuming n is even, then eventually concluded:
$$n \;\text{ is even }\;\implies \; n^3\;\text{ is even.}$$
However, since it's a double implication I have to conclude
$$n^3\;\text{ is even }\;\implies n \;\text{ is even.}$$
I assume $n^3$ is even. Then $n^3 = 2k$ for some integer $k$.
Then $n = (2k)^{1/3}$...
But I can't really seem to find a way to get a $2k$ equivalent expression for $n$...
Can you guide me?
 A: Although I agree that the other comment and answer given are correct and are likely the approach you are expected to use, personally I immediately note that if $n$ is written in its prime factorization form ($p_1^{e_1} \cdot p_2^{e_2} ... p_i^{e_i}$) and you observe the form that $n^3$ has as well, the stated proposition is obvious.  But, I probably only see that so easily because I am obsessed with the prime factorization patterns of numbers.
Note that if you are taking a proof-writing class, this is a perfect example of using one proof as necessary for another, since the fact that no new factors are introduced depends on the Fundamental Theorem of Arithmetic.
A: HINT: For the second implication, try proving the contrapositive of the implication. Suppose $n$ is not even (i.e., assume $n$ is odd), and prove that, then, $n^3$ is not even (i.e., $n^3$ is odd).
$$P \implies Q \equiv \lnot Q \implies \lnot P$$
A: Note that $n^3-n=n(n-1)(n+1)$. This is always even. Since the a difference  $x-y$ is even $\iff$ both $x,y$ are even or both $x,y$ are odd, you're done.
A: $
\newcommand{\even}[1]{#1\text{ is even}}
$Here is another proof, which assumes you may use the fact that an even number has at least one even factor, in other words
$$
(0) \;\;\; \even{a*b} \;\equiv\; \even{a} \lor \even{b}
$$
Using this, we need no special tricks to calculate
\begin{align}
& \even{n^3} \\
\equiv & \;\;\;\;\;\text{"using $(0)$ twice"} \\
& \even{n} \lor \even{n} \lor \even{n} \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& \even{n} \\
\end{align}
Of course you still might want to try and prove $(0)$...
A: $\leftarrow$ direct proof: If $n$ is even, then $n=2k$ where $k\in \mathbb{Z}$. So $n^3=(2k)^3=8k^3=2(4k^3)$ where $4k^3\in \mathbb{Z}$. Thus if $n$ is even, then $n^3$ is even.
$\rightarrow$ proof by contrapositive: Assume that if $n$ is odd, then $n=2l+1$ where $l\in \mathbb{Z}$. So $(2l+1)^3=8l^3+12l^2+6l+1=2(4l^3+6l^2+3l)+1$ where $4l^3+6l^2+3l\in \mathbb{Z}$. Thus if $n^3$ is even, then $n$ is even.
