4
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how to find all $n \in \Bbb N$ such that $n(n+1)\mid(n-1)! $

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  • $\begingroup$ 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 ;) $\endgroup$ – Dolma May 4 '13 at 15:16
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    $\begingroup$ @Dolma: So $23 \mid 22!$? $\endgroup$ – TMM May 4 '13 at 15:19
  • $\begingroup$ Well, no since the question is not $n|(n-1)!$ but $n(n+1)|(n-1)!$ $\endgroup$ – Dolma May 4 '13 at 15:23
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    $\begingroup$ @Dolma So $(23*24) \,|\, 22!$? $\endgroup$ – Petr Pudlák May 4 '13 at 15:41
  • $\begingroup$ 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 :/ $\endgroup$ – Dolma May 4 '13 at 15:58
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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 $n+1$.

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  • $\begingroup$ $3$ is not a divisor of $2!=2$, $4$ is not a divisor of $3!=6$, ... $\endgroup$ – Dolma May 4 '13 at 15:21
  • $\begingroup$ Positive $\to$ Composite (I was in parallel universe) $\endgroup$ – Inceptio May 4 '13 at 15:22
  • $\begingroup$ Yeah sorry, I saw your edit after I posted my comment ;) $\endgroup$ – Dolma May 4 '13 at 15:24
  • $\begingroup$ Your first statement is still wrong. Hint: $4$. $\endgroup$ – TMM May 4 '13 at 15:30
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    $\begingroup$ 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 $n+1$. $\endgroup$ – Mark Bennet May 4 '13 at 15:39

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