Simpler ways to show that $n^2$ divides a polynomial? 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!
 A: 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.
A: 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 $2$.
A: Hint $\, 3\,$ divides a product of $3$ consecutive integers (proof), and even $\,n\!+\!1 = \color{#0a0}{4k}\,$ or $\,\color{#90f}{4k\!+\!2}\,$ and
$$\color{#c00}{16}\mid \underbrace{(\color{#0a0}{4k})^2}_{\Large\color{#c00}{ 4^2}}(4k\!+\!1)(4k\!+\!2)\ \ \ {\rm and}\ \ \ \color{#c00}{16}\mid \underbrace{(\color{#90f}{4k\!+\!2})^2}_{\Large \color{#c00}{2^2}}(4k\!+\!3)\underbrace{(4k\!+\!4)}_{\Large\color{#c00} 4}$$ 
