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!
Answer
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.
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