how to find all $n \in \Bbb N$ such that $n(n+1)\mid(n-1)! $

Question

A brute force search gave $8$, $9$, $14$, $15$ and then all integers greater or equal to $20$ (at least up to 1,000,000) ... You might want to investigate on that ;) — Dolma, May 04 '13 at 15:16
Well, no since the question is not $n|(n-1)!$ but $n(n+1)|(n-1)!$ — Dolma, May 04 '13 at 15:23
Yes, I was a bit quick on that one. I guess there was some rounding error on Matlab that made him think those were divisors ... my mistake :/ — Dolma, May 04 '13 at 15:58

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

2

Hint:

Here $n|(n-1)!$ for all $n$ being composite and $n>4$. And $n$ and $n+1$ are not primes.

Proof that $n|(n-1)!$ if $n>4$ is composite is given here.

$(n-1)!=(n-1)(n-2) \cdots 1$. You need to have sufficient factors in $\{n-1,n-2 ,\cdots 1\}$to cancel out both $n$ and $n+1$.

edited Apr 13 '17 at 12:20

Community

answered May 04 '13 at 15:19

Inceptio

$3$ is not a divisor of $2!=2$, $4$ is not a divisor of $3!=6$, ... – Dolma May 04 '13 at 15:21
Positive $\to$ Composite (I was in parallel universe) – Inceptio May 04 '13 at 15:22
Yeah sorry, I saw your edit after I posted my comment ;) – Dolma May 04 '13 at 15:24
Your first statement is still wrong. Hint: $4$. – TMM May 04 '13 at 15:30
If $n=r\cdot p^2,$ can $n$ divide $(n-1)!$ – lab bhattacharjee May 04 '13 at 15:33
@Inceptio: I said "Your first statement is still wrong." where your first statement is "Here $n|(n−1)!$ for all $n$ being composite." – TMM May 04 '13 at 15:35
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Note also that $n$ and $n+1$ are coprime, so do not share any common factors, so divisibility can be tested separately for $n$ and $n+1$. – Mark Bennet May 04 '13 at 15:39
@labbhattacharjee: Doesn't it? We have enough factors in $(n-1)!$ to cancel out $n$? – Inceptio May 04 '13 at 15:41
@TMM: Does it suffice the condition now? – Inceptio May 04 '13 at 15:57
@Inceptio, agreed. Sorry,for the confusion. – lab bhattacharjee May 04 '13 at 17:59

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