4

I want to show that $n^2 \mid P(n)$, where $$P(n) = \frac{n^2(n+1)^2(n+2)(n+3)}{48}$$ for every odd positive integer $n$. The approach I took involved showing that $\cfrac{P(n)}{n^2}$ is always an integer (for such $n$), but then I had to create a polynomial even more complex and then prove nine different cases. While it did provide a valid proof (as far as I know), I have a feeling it was more work than I needed.

So my question is: are there "simpler" proofs to this problem, and what are their approaches/methods? By simpler I roughly mean: prove less cases, reduce the problem to a simpler form, etc; basically a solution that takes up less "space" on paper. (I know that's not the best explanation, sorry!)

Thank you very much!

Gizmo
  • 715

3 Answers3

4

I'm not sure what you did, but showing that $$\frac{P(n)}{n^2}=\frac{(n+1)^2(n+2)(n+3)}{48}$$ is always an integer if $n$ is odd is quite straightforward. You just have to show that $(n+1)^2(n+2)(n+3)$ is always divisible by $48=2^4\cdot 3$. It's always divisible by $3$ since one of $n+1,n+2,$ and $n+3$ is a multiple of $3$.

The factors of $2$ are a little more complicated but not bad. Since $n$ is odd, $n+1$ and $n+3$ are even, so $(n+1)^2(n+3)$ gives at least $3$ factors of $2$. Moreover, one of $n+1$ and $n+3$ is a multiple of $4$, which gives one extra factor of $2$. So in total there are at least $4$ factors of $2$.

The moral here is that when thinking about divisibility questions, factor. We keep the numerator of $\frac{P(n)}{n^2}$ in its factored form, so we can identify the contributions from each individual factor. And to test divisibility by $48$, we split it into its prime factorization so we can look for each prime factor separately.

Eric Wofsey
  • 330,363
  • Oh shoot, I remember trying this approach too. But I couldn't figure out how to get the last factor of $2$ from $2^4$. Is it enough to say that: $n$ is odd, so $n+1$ or $n+3$ is divisible by $4$? That's my understanding anyway, now that I see your answer. I love the ending explanation! – Gizmo Dec 12 '18 at 00:15
  • Whether that is enough to say depends on your audience and what level of detail they expect. – Eric Wofsey Dec 12 '18 at 00:24
2

Note that $(n+1)^2(n+2)(n+3)$=$(n+1)(n+1)(n+2)(n+3)$=$(n+1)\Big(\text{Product of 3 Consecutive Integers}\Big)$

Hence $3|(n+1)^2(n+2)(n+3)$

We are left to show 16|$(n+1)^2(n+2)(n+3)$ whenever $n$ is odd.

When $n$ is odd let $n=2k+1$, for some $k \in \mathbb{N} \cup \{0\}$

Then $$(n+1)^2(n+2)(n+3)=(2k+2)^2(2k+3)(2k+4)=8(k+1)^2(2k+3)(k+2)=16m$$

Note that last statement follows because the product involves $(k+1)(k+2)$ which is product of $2$ consecutive integers and hence divisible by $2$.

Rakesh Bhatt
  • 1,436
2

Hint $\, 3\,$ divides a product of $3$ consecutive integers (proof), and even $\,n\!+\!1 = \color{#0a0}{4k}\,$ or $\,\color{#90f}{4k\!+\!2}\,$ and

$$\color{#c00}{16}\mid \underbrace{(\color{#0a0}{4k})^2}_{\Large\color{#c00}{ 4^2}}(4k\!+\!1)(4k\!+\!2)\ \ \ {\rm and}\ \ \ \color{#c00}{16}\mid \underbrace{(\color{#90f}{4k\!+\!2})^2}_{\Large \color{#c00}{2^2}}(4k\!+\!3)\underbrace{(4k\!+\!4)}_{\Large\color{#c00} 4}$$

Bill Dubuque
  • 272,048