How do I prove that $6 \mid n^3- n$ for all $n ≥ 1$?

Question

How do I prove that $6 \mid n^3 - n$ for all $n \ge 1$?

I'm having difficulty understanding this problem. All help is greatly appreciated.

Answer 1

$n(n^2-1)=(n-1)n(n+1)$. Product of three consecutive integers. Product of $r$ consecutive integers always divisible by$ r!$

Answer 2

Let $n \ge 1$ be an integer. Then $$n \equiv -2,\, -1,\, 0,\, 1,\, 2,\, \text{or}\, \ 3 \pmod 6.$$ Check each case seperately and conclude that $n^3 - n \equiv 0 \pmod 6$ for all $n \ge 1$.

Answer 3

Take a look at $n^3 - n$ more closely.

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

So then we have, in general, that $n^3 - n = (n-1)\cdot (n) \cdot (n+1)$

In order for $6$ to divide into a number, we can say that both $2$ and $3$ must divide that number.

If we write $n^3 - n$ as $a*(a + 1)*(a + 2)$ it should be apparent why we must have a factor of $2$ and a factor of $3$ in this product. Let's look at the cases:

a is even: then of course we have a factor of 2.

a is odd: then of course $a + 1$ will be an even number, and we have the factor of $2$.

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(These cases are not mutually exclusive from above. Of course $a$ can be even and divisible by $3$. It is just easier to separate the two sets as opposed to including $6$ possible cases !)

a is divisible by 3: then we have the factor of $3$.

a mod 3 = 1: then we have that $(a + 2)$ mod $3 = 0$.

a mod 3 = 2: then we have that $(a + 1)$ mod $3 = 0$.