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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?

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    $\begingroup$ Hint: Try to assume $n$ is odd. $\endgroup$
    – user99914
    Commented Nov 18, 2013 at 0:22

5 Answers 5

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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).

$$P \implies Q \equiv \lnot Q \implies \lnot P$$

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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.

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  • $\begingroup$ Also a very nice way to approach this proof. $\endgroup$ Commented Nov 18, 2013 at 1:01
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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.

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$ \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 $(0)$...

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$\leftarrow$ 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.

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