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I was given the following problem.

Show $n(n+1)(n+2)=6m$ for any $n \in \mathbb{Z}$.

I self study and have no professors, so I would appreciate it if someone could tell me whether my solution is right or not. Here's what I did.

Notice that

$$ n(n+1)(n+2)=(n^2+n)(n+2)=n^3+3n^2+2n \tag{1} $$

Let $ n$ be a natural number for the moment. Consider that if $n = 1$ then $n(n+1)(n+2)=6$, which is divisible by $6$. Let our inductive hypothesis be $H:6|k(k+1)(k+2)$. Now consider the case for

$$(k+1)(k+2)(k+3) = (k^2+3k+2)(k+3) = k^3+6k^2+11k+6$$

Notice that

$$ \begin{align} k^3+6k^2+11k+6 &=k^3+3k^2+2k+(3k^2+9k+6) \\ &=k(k+1)(k+2)+(3k^2+9k+6) \tag{Due to $(1)$} \\ &=6m+3k^2+9k+6 \end{align} $$

If $k = 2q, q \in \mathbb{Z}$ then $3k^2+9k+6=12q^2+18q+6 = 6(2q^2+3q+1)$. If $k = 2q + 1$ then

$$ \begin{align}3k^2+9k+6&=3(2q+1)^2+(18q+9)+6 \\ &=3(4q^2+4q+1)+15 \\ &=12q^2+12q+18 \\ &= 6(2q^2+2q+3) \end{align} $$

and therefore for any $k$ we have

$$ (k+1)(k+2)(k+3)=6m+6m'=6(m+m') \tag{$m, m' \in \mathbb{Z}$} $$

which concludes our demonstration for $n \in \mathbb{N}$. Now, for any $k(k+1)(k+2)$ we know $|k(k+1)(k+2)|=6m$, and $a|b \implies a| \mp b$. Hence, our result extends to $\mathbb{Z}$.

Is this demonstration correct? Aside from validation, I would highly appreciate alternative ways to prove this (e.g., proofs without induction).

lafinur
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    Maybe it's too complicated, among three consecutive integers there is one divisible by three and one divisible by two – Jochen Nov 11 '22 at 23:11
  • In addition to what Jochen wrote, your proof is correct. However, one simplification is to use in your induction step that $(k+1)(k+2)(\color{blue}{k}+\color{red}{3}) = (k+1)(k+2)(\color{blue}{k}) + (k+1)(k+2)(\color{red}{3}) = 6m + \color{red}{3}(k+1)(k+2)$. Since either $k+1$ or $k+2$ is even, then $2 \mid (k+1)(k+2)$, so $6 \mid \color{red}{3}(k+1)(k+2)$. – John Omielan Nov 11 '22 at 23:15
  • Your demo probably is okay but you are making it way too hard. Note that if for $n=k$ we have Our expression goes from being $a_k = \color{blue}k\color{green}{(k+1)(k+2)}$ to $a_{k+1} = \color{green}{(k+1)(k+2)}\color{red}{(k+3)}$. $3\midk+1$ or $k+2$ which are components of both. The only components that aren't component of both are $k$ and $k+3$. $3$ either divides both or none. And if $2|k$ but not $k+3$ then $2|k+2$ which is a component of both. – fleablood Nov 12 '22 at 00:17
  • ... of course, you don't have to use induction at all. Enough to not that exactly one of $n, n+1$ or $n+2$ is divisible by $3$ and at least one $n$ and $n+1$ and $n+3$ are divisible by $2$. – fleablood Nov 12 '22 at 00:19
  • If you want to be perverse $(n+1)(n+2)(n+3)= n(n+1)(n+2)+ 3(n+1)(n+2)=n(n+1)(n+2)+6\frac{(n+1)(n+2)}2$ – fleablood Nov 12 '22 at 00:21
  • @Jochen wow, that basically turns the proof into a one-liner. I didn't think about it that way. Thank you so much! – lafinur Nov 12 '22 at 00:41
  • @fleablood I highly appreciate your insight! That last comment... perverse is indeed the right word! Haha. Thank you so much. – lafinur Nov 12 '22 at 00:42

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