Prove that the product of three consecutive natural numbers is divisible by 6

Question

My math teacher asked me this question, and I told him that in every three consecutive natural numbers we have one multiple of 3, and at least two multiples of 2. And for a number to be divisible by 6, we need it to be both divisible by 3 and 2. He said that this proof was correct, but it was not strong enough because I had used Induction. And I should use Inference. How can I prove this using Inference?

Answer 1

This part is an answer to the initial question (asking for the SUM instead of the Product)

Three consecutive natural numbers are $n-1,n,n+1$ and their sum is $3n.$ But if $n$ is odd, this sum is not divisible by $6.$

Answer related to the PRODUCT:

Product of three consecutive natural numbers is $(n-1)n(n+1)=n^3-n.$ Denote it $P.$

1. If $n$ is even ($n=2k$), then $n^3=8k^3$ is even and $P$ is even as the difference of two even numbers. Similarly, if $n$ is odd, then $n^3$ is odd and the difference of two odd numbers is even.

We conclude that $P$ is even (divisible by $2$).

2. For divisibility by $3,$ consider the $3$ cases:

   • $n\equiv0\pmod3,$ then $n^3\equiv 0\pmod{3},$ from where $P\equiv 0\pmod 3;$

   • $n\equiv 1\pmod{3},$ then $n^3\equiv 1\pmod3,$ which gives $P\equiv 0\pmod 3;$

   • $n\equiv 2 \pmod 3,$ then $n^3\equiv 2 \pmod 3$ and we have again $P\equiv 0\pmod 3.$

Conclusion: $P$ is divisible by $2$ and by $3$, therefore is divisible by $6.$

Answer 2

Out of three consecutive numbers you will always find an even number and a multiple of three. By multiplying, you get a multiple of 6.

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Here is another proof:

Let $n$ a natural. $n\times (n+1) \times (n+2)=n^3+3n^2+2n$

If $n \equiv 1 \pmod 2$, $n^3+3n \equiv 0 \pmod 2$.

If $n \equiv 0 \pmod 2$, $n^3+3n \equiv 0 \pmod 2$.

You can do the same reasoning modulo 3.

Finally, $n^3+3n^2+2n \equiv 0 \pmod{2\times 3}\Leftrightarrow \exists p \in \mathbb{N}, n(n+1)(n+2)=6p$