I need to prove that $n^3 - n$ is divisible by $6$ by factoring it and by knowing that the product of each consecutive $3$ numbers is divisible by $2$ and $3$. I tried:

$n(n^2 - 1)$

Factoring it didn't really get me anywhere. I think I'm supposed to do something like:

$n[(n - 1)(n - 2)]$ And if I do the math in this I get back:

$n[n^2 - 3n + 2] = n^3 - 3n^2 + 2n$

which looks like a dead end, unless I can somehow prove that:

$n^3 - n = n^3 - 3n^2 + 2n$

But that's impossible, since it's not true that $2n^2 = 2n$ for every positive integer

So what else can I do? These are the instructions the book gave - factor it and remember that the product of three consecutive number is divisible by 6.

  • 1
    $\begingroup$ Hint: $n^2-1=(n-1)(n+1)$ $\endgroup$ – Guest Dec 5 '14 at 22:12
  • $\begingroup$ In general, the product of k consecutive numbers is always divisible by $k!$. $\endgroup$ – Lucian Dec 6 '14 at 2:44

You were on the right track initially.

$$n^3-n = n(n^2-1) = n(n+1)(n-1).$$

These are three consecutive integers, so at least one of them is even. Additionally, $3$ must divide one of the numbers by the same logic. Thus $n^3-n$ is divisible by $6$ since it is divisible by $2$ and $3$.


Hint:$(n^2-1)=(n+1)(n-1)$, so $$ n(n^2-1) = (n-1)n(n+1)$$

And because $6 = 2 \cdot 3 \wedge \gcd(2,3)=1$

$$6 \mid n \Longleftrightarrow 2 \mid n \wedge 3 \mid n $$

You should note too, that one of two numbers has to be even.

$$(\forall n \in \mathbb{Z})(2 \mid n~ \dot\vee ~2 \mid (n+1))$$

And similarly you can show divisible by three ($3$ has to divide one of three consecutive numbers).

If you want long explanation you can say: the remainder ($n \mod 3$) is $0$, then $3\mid n$, otherwise the remainder can be $2$ and $3 \mid (n+1)$ or $1$ and then $3 \mid (n+2)$. So $$(\forall n \in \mathbb{Z})\left(3 \mid (n-1)~ \dot\vee ~ 3 \mid n~ \dot\vee ~3 \mid (n+1)\right)$$

Pursuant to above claims.

$$\begin{split} (2 \mid n ~\dot\vee~ 2 \mid (n+1))\wedge (3 \mid (n-1) ~\dot\vee~ (3 \mid n) ~\dot\vee~ 3 \mid (n+1)) &\Longleftrightarrow\\ \left(2 \mid n(n^2-1) \wedge 3 \mid n(n^2-1)\right)&\Longleftrightarrow 6 \mid n(n^2-1) \end{split}$$



$$N^3-N = N \cdot (N^2-1) = N \cdot (N-1) \cdot (N+1)$$

$N-1, \;N$ and $N+1$ have to be three consecutive numbers as $N$ is constant.

If you have three numbers, at least one of them has to have a factor of $2$ and one of them has a factor of $3$.

Since both $2$ and $3$ are factors, $\operatorname{lcm}(2,3)=6$ is also a factor.

  • $\begingroup$ Please use latex to write the answer $\endgroup$ – Isaac YIU Math Studio Nov 16 '19 at 15:15

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