$Show: a\in \mathbb{Z} \Rightarrow 6\mid a^3 -a$

Question

$Show: a\in \mathbb{Z} \Rightarrow 6\mid a^3 -a$

My attempt:

NTS: $6\mid a^3 -a$, so $(2\mid a^3 -a) \land (3\mid a^3 -a)$

Assume that $a\in \mathbb{Z}$, therefore I have 2 cases:

1. a is even $\Rightarrow a=2k, k\in \mathbb{Z}$
2. a is odd $\Rightarrow a=2k+1, k\in \mathbb{Z}$

Therefore in case 1: $a^3-a=(2k)^3-2k=8k^3-2k$

and in case 2: $a^3-a=(2k+1)^3-(2k+1)=8k^3+2k$

And it led me to a blind alley. I can only factor 2 out of the expressions but there's no way to factor 3 out, so $k\in \mathbb{Z}$

Is there a way to show that using direct proof and the assumption?

I also tried to factor the $a^3-a=(a-1)(a)(a+1)$ but it isn't telling me much either.

Answer 1

$a-1,a,a+1$ are three consecutive integers, so at least one of them must be even and one of them must be a multiple of $3$.

Answer 2

At least one of the factors $a, a-1,a+1$ must be divisible by $3$ and also by $2$ and since gcd$(2,3) = 1$, the product divisible by $2\cdot 3 = 6$.

Answer 3

Let $a$ $\in\mathbb {Z}$.

Given:
$a^{3}-a=a(a-1)(a+1)$

For $a$ even, then 2 | $a$. If $a$ is odd, clearly we got that $a + 1$ or $a - 1$ is even.
So for a $\in\mathbb {Z}$, 2 | $a^{3}-a$.

Since gdc(2,3)=1, we need to show $3$| $a^{3}-a$.

For a $\in\mathbb {Z}$, $a$ can be written in one of the following forms $3k_{0}$, $3k_{1}+1$, $3k_{2}+2$, with $ k_{0}$,$k_{1}$,$k_{2}$ $\in$ $\mathbb {Z}$.

Using this forms for $a$ and the above factorization the result follows.

Answer 4

Wkt $gcd(m,n) = 1$ and $$ \begin{align*} a&\equiv b\pmod m\\ a&\equiv b\pmod n\\ \end{align*} $$ then$$ \begin{align*} a&\equiv b\pmod k \\ \end{align*} $$ where $k=mn$ -------->(1)

Wkt $gcd(2,3) = 1 $

By using Euler's theorem

$$ \begin{align*} a^2&\equiv 0/1\pmod2\\ a^3&\equiv 0/a\pmod2\\ \end{align*} $$ &

$$ \begin{align*} a^2&\equiv0/1\pmod3\\ a^3&\equiv 0/a\pmod3\\ \end{align*} $$ By using (1) we get

$$ \begin{align*} a^3&\equiv 0/a\pmod 6\\ \end{align*} $$

$a\in \mathbb{Z} \Rightarrow 6\mid a^3 -a$