prove that for all $n$, $n$ is an integer, if $n^3$ is an even integer then $n$ is an even integer

Question

Prove that for all $n$ ∈ integers, If $n^3$ is an even integer then $n$ is an even integer. And prove there is no rational number $x$ such that $x^3 =2$

Answer 1

Hint

Perhaps it would be easier to show the contrapositive. Suppose $n$ is odd i.e. $n=2k+1$ for some integer $k$ and show $n^3$ is odd. To this end expand $(2k+1)^3$ using the binomial theorem.

Answer 2

Let there exist two such cases:

• $n=2k$ such that $k$ is a positive integer; and
• $n=2k+1$ such that $k$ is a positive integer.

In regard to the first case, $$\begin{align} n^3 &= (2k)^3 \\ &= 2^3\cdot k^3 \\ &= 8k^3\quad\text{even.}\end{align}$$ Thus, $n^3$ is even if $n$ is even; or, $n$ is even if $n^3$ is even.

In regard to the second case, $$\begin{align} n^3 &=(2k+1)^3 \\ &= (2k+1)(2k+1)^2 \\ &= (2k+1)\big((2k)^2+2\cdot 2k\cdot 1 + 1^2\big) \\ &= (2k+1)\big(2^2\cdot k^2 + 2\cdot 2k + 1\big) \\ &= (2k+1)(4k^2 + 4k + 1).\end{align}$$ Note that both the binomials are odd; they are not divisible by $2$ as they each have a remainder of $1$. And, since they divide into $n^3$ then it follows that $n^3$ is not divisible by $2$ either; $n^3$ is odd.

$$\therefore n^3=(2k+1)(4k^2+4k+1)\quad\text{odd.}$$ $$\Downarrow$$ $$n^3\text{ even }\Leftrightarrow n\text{ even}.\tag*{$\bigcirc$}$$

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Now, suppose that there exists a rational number $x$ such that $x^3=2$. Since $x$ is rational, there exist integers $p$ and $q$ such that $x = p/q$ which is irreducible.

$$\begin{align}\therefore 2&=\left(\frac pq\right)^3 \\ \\ &= \frac{p^3}{q^3} \\ \\ \Leftrightarrow p^3&=2q^3.\end{align}$$

Now, since $p^3$ is even, then as we proved before, $p$ must also be even. Thus, there exists an integer $r$ such that $p=2r$. $$\begin{align}\therefore 2q^3 &= (2r)^3 \\ &= 2^3\cdot r^3 \\ &= 8r^3 \\ \Leftrightarrow q^3 &= 4r^3.\end{align}$$ Now, since $q^3$ is even, there exists an integer $s$ such that $q=2s$. But, if $x$ is irreducible, then $p$ and $q$ must have a greatest common divisor of $1$; and it has been obtained that $p$ and $q$ have a greatest common divisor of $2$, thus bringing forward a contradiction.

$$\text{Consequently, there does not exist a rational number } x \text{ such that } x^3 = 2.\tag*{$\bigcirc$}$$