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Is there a composite integer $n \geq 9$ such that $n \nmid (n-1)!$?

If we are not talking about composites then by Wilson's theorem we have $n \nmid (n-1)!$.

Yes
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  • There is not even one for $n>4$ – almagest Sep 07 '14 at 11:22
  • $6 \mid 5!$, I am afraid. – Yes Sep 07 '14 at 11:23
  • But you are looking for $n$ that does not divide $(n-1)!$. – almagest Sep 07 '14 at 11:25
  • Yeah, right. How to prove this? I said $6 \mid 5!$ because you said "there is not even one for $n > 4$". There does exist such $n$ if $n > 4,$ viz $n=6$. :) – Yes Sep 07 '14 at 11:26
  • consider $n = ab$, Notice that $ a \lt n$ and $b \lt n$ See http://math.stackexchange.com/questions/164852/if-n-is-composite-then-n-divides-n-1 – AgentS Sep 07 '14 at 11:27
  • @Comeseeconquer: From 9 to 10^5, there is no such that number. I examined it. So we have to prove it probably. – Mikasa Sep 07 '14 at 11:28
  • It is obvious that any prime $p$ dividing composite $n$ also divides $(n-1)!$. The question is how to prove that if $p^r$ divides $n$, then it also divides $(n-1)!$. So you need some kind of lower bound on the number dividing $(n-1)!$ – almagest Sep 07 '14 at 11:28

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If $n>4$ is composite, then either $n=p^2$ for a prime number $p>2$ (and thus $2p<p\cdot p=n$) or you can find two different numbers $n_1$ and $n_2$ with $n_1 n_2 = n$ and $n_1,n_2>1$ in which case clearly $n_1,n_2\leq n-1$.

In the first case $p$ and $2p$ are two different factors of $(n-1)!$, in the second case $n_1$ and $n_2$ are.

Clarification: With two different factors I obviously meant that $p$ and $2p$ or $n_1$ and $n_2$ are two different members of $\{1,2,\dots,n-1\}$ (and not two arbitrary divisors of $(n-1)!$ which would not suffice).

Frunobulax
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