How to prove it is always divisible by 6

Question

Prove that $n(n^2 − 7)$ for is always divisible by 6. (for any natural number $n$)

I have no idea.

Answer 1

Hint: Since $6x$ is a multiple of $6$, $x(x^2 - 7)$ is divisible by $6$ iff $$x (x^2 - 7) + 6x = x^3 - x$$ is.

Additional hint: We can factor $$x^3 - x = (x - 1) x (x + 1).$$

Answer 2

You can see $n(n^2-7) = n(n^2-1-6) = (n-1)n(n+1) -6n$ , you just need to verify that $(n-1)n(n+1)$ has a factor of $6$, which is not hard.

For any 3 consecutive numbers, one of them must be a multiple of 3, and at least one of them must be multiple of 2

Answer 3

Consider the following cases:

 x mod 2 | x mod 3 | x^2-7 mod 2 | x^2-7 mod 3
---------|---------|-------------|-------------
    0    |    0    |             |
    0    |    1    |             |  1-7≡-6≡0
    0    |    2    |             |  4-7≡-3≡0
    1    |    0    |  1-7≡-6≡0   |
    1    |    1    |  1-7≡-6≡0   |  1-7≡-6≡0
    1    |    2    |  1-7≡-6≡0   |  4-7≡-3≡0

Answer 4

Using mathematical induction can prove this:

For $ n = 1, n(n^{2}-7) = -6$, so it is divisible by $6$.

Now assume the $ n=k $ case is correct, i.e. $ k(k^{2}-7) = 6p $ for some integer $p$.

For $ n = k+1, n(n^{2}-7) \\= (k+1)((k+1)^{2}-7) \\= (k + 1)(k^{2}+2k-6) \\= k^{3}+3k^{2}-4k-6 \\= 6p + 3k^{2}-3k-6 \\= 6(p-1) + 3(k)(k-1) $

Now, as $k$ is an integer, either $k$ or $k-1$ is divisible by $2$ (or, you may like to prove this using another mathematical induction). Hence the whole expression $ 6(p-1) + 3(k)(k-1) $ can be written as $ 6(p-1+q) $, where $p$, $q$ are integers and $q = k(k-1)/2$. Therefore, by the principal of mathematical induction, $n(n^{2}-7)$ must be divisible by $6$ for any natural number $n$.