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We need to prove that if $n$ is a composite number $>4$, then $n|(n-1)!$. I wanted to ask if my observation is correct or not. What I think is that the statement can be reduced to $n|(n-2)!$ Because $n$ and $n-1$ are always co-prime.

Ali Caglayan
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Archit Nanda
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2 Answers2

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That is true. But it's perhaps easier to note that $n = (n-a)(n-b)$ for some $a$ and $b$. Clearly, if $a \ne b$ both $(n-a)$ and $(n-b)$ are factors of $(n-1)!$. It's a little more complicated if $a=b$.

Edit: if $a=b$, since $n>4$ we have $2(n-a) < (n-a)(n-b)$, so $(n-a)$ and $2(n-a)$ are both factors of $(n-1)!$. Thus $2(n-a)^2$ is a factor of $(n-1)!$ so $n=(n-a)^2$ must also be a factor of $(n-1)!$.

user164587
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  • More precisely, $n-a$ and $2(n-a)$ are distinct integers each less than $n$, so their product divides $(n-1)!$ (the fact that they are both factors does not imply that their product is...) – rogerl Oct 26 '14 at 12:17
  • Yes. Thanks for the clarification. – user164587 Oct 27 '14 at 16:18
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Hint: $$(n-1)!=1\cdot2\cdot3\cdots(n-2)\cdot(n-1)$$