# If $n$ is composite, then $n$ divides $(n-1)!$.

I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!

Here is what I am asked to prove:

If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.

Proof:

$n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0<a,b<n$.

Case 1: If $a=b$ then $n=a^{2}$. Now $n \mid (n-1)! \implies a \mid (n-1)!$, so \begin{aligned} (n-1)! &\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &\equiv 0 \pmod n \end{aligned}

Case 2: $0<a<b<n$.

Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.

So this implies $(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$, Q.E.D.

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In case 1, there is a subcase where $a=n-a$, and the result is false. –  Jonas Meyer Jun 30 '12 at 8:32
$3!\equiv2\pmod4$ –  Mike Jun 30 '12 at 8:39
@JonasMeyer I did not consider that case. Darn, this also only happens when n=4. :( thank you. –  HowardRoark Jun 30 '12 at 8:41
@HowardRoark: Exactly (and Mike has also made it explicit). What is the source of the problem? Was the exception of $n=4$ not mentioned? Otherwise, your method looks good. –  Jonas Meyer Jun 30 '12 at 8:42
@JonasMeyer This is how the question goes: (a)calculate $(n-1)!$(mod $n$) for $n=10,12,14,$ and $15$. (b) guess a theorem and prove it. –  HowardRoark Jun 30 '12 at 8:46

Hint $\rm\,\ n\, =\, \color{#C00}a\:\!\color{#0A0}b\:|\:1\!\cdot\! 2\cdots\color{#C00} a\:\color{#0A0}{(a\!+\!1)\, (a\!+\!2) \cdots (a\!+\!b)}\cdots (ab\!-\!1)\,\$ when $\rm\,\ a\!+\!b \le ab\!-\!1$

Note $\$ That $\rm\:\color{#0A0}b\:$ divides the above $\rm\color{#0A0}{green}$ term is not deduced from the fact that it is divisible by $\rm\,b!\,$ by integrality of the binomial coefficient $\rm\:(a\!+\!b:a),\:$ as in WimC's answer. Rather, I deduce it from the much simpler fact that any sequence of $\rm\,b\,$ consecutive integers contains a multiple of $\rm\,b,\,$ which is an immediate consequence of (Euclidean) division. For completeness. here is a proof: $$\rm\ mod\ b\!:\ a\!+\!b\equiv k\in [\color{#C00}0,\color{#0A0}{b\!-\!1}]\ \Rightarrow\ b\:|\:a\!+\!b\!-\!k\in [\color{#0A0}{a\!+\!1},\color{#C00}{a\!+\!b}]$$

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+1: Very nice! Hope the OP isn't red-green colorblind. –  Cameron Buie Jun 30 '12 at 15:18
hahaha. Thank you! –  HowardRoark Jul 1 '12 at 3:22
For a moment there I forgot I am using the color removing script, I have to say that to my color-blind eyes the colorized version looks extremely bad. My mind is too busy trying to figure out what is going on with the text and ignores the content completely (despite knowing the content from a colorless read!) you might want to consider using red and blue for future reference. Those are much harder to confuse and they sit a lot better for most color blindness. (Now I'll just turn that Greasemonkey script back on, and all will be well again...) –  Asaf Karagila Jul 3 '12 at 21:17

If $n > 4$ and $n = a \cdot b$ with $a, b \geq 2$ then $a + b \leq n - 1$. Since ${a+b \choose a}$ is an integer it follows that $n = a \cdot b \mid a! \cdot b! \mid (a + b)! \mid (n - 1)!$.

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(+1) Very neat proof! –  TMM Jun 30 '12 at 10:14
(+1) for compact and precise proof. –  Aang Jun 30 '12 at 10:26
@TMM Binomial coefficients are a bit of a sledgehammer here. Instead, more simply, one needs only that every sequence of consecutive integers of length $\rm\:b\:$ contains a multiple of $\rm\:b.\:$ See my answer. –  Bill Dubuque Jun 30 '12 at 14:33
@BillDubuque Why do you consider it is a sledgehammer? Maybe it is true from a number theoretical view, but very straightforward from a combinatorial one, right? –  Pedro Tamaroff Jun 30 '12 at 15:28
@BillDubuque I ask again, Bill: In terms of combinatorics, isn't it "natural"? (I know in NT it is not trivial) –  Pedro Tamaroff Jun 30 '12 at 16:37

Since $n|(n-1)!$ and $(n-1)!\equiv0\pmod n$ are equivalent, if you can prove one, you should be done.

Take your case 2 for example. You say "then since $a|n,b|n$, and $n|(n-1)!$..." But $n|(n-1)!$ is what you set out to prove and you weren't explicit about why it's true, which is the point of a proof in the first place.

What I think you were going for is that $(n-1)!$ is the product of all positive integers $\le n-1$, which includes $a$ and $b$. Change the order of multiplication and let the product of all integers $\le n-1$ excluding $a$ and $b$ be equal to a new constant $k$. So $(n-1)!=kab=kn$. Therefore, $n|(n-1)!$ or $(n-1)!\equiv0\pmod n$.

Case 1 was a bit more explicit, but you tried using your conclusion to try to prove something else again. And you missed the loophole that Jonas's comment points out. You may want to be a bit more explicit about why the product is zero as well rather than just stating that it is.

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We will assume that $n>4$, since $4\hspace{-3pt}\not|\,3!$.

Let $p$ be the smallest factor of $n$. Since $n$ is composite, $p\le\sqrt{n}$.

If $p=\sqrt{n}$, then since $n>4$, we must have $p>2$ so that $2p<p^2=n$. Thus, $p\le n-1$ and $2p\le n-1$, and therefore, $2n=p\cdot2p\,|\,(n-1)!$

If $p<\sqrt{n}$, then $n/p>\sqrt{n}$. Thus, $p\le n-1$ and $n/p\le n-1$, and therefore, $n=p\cdot n/p\,|\,(n-1)!$

In either case, $n|(n-1)!$

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Recall the definition of the factorial: $$m! = \prod_{k=1}^m k = 1 \times 2 \times 3 \times \dotsb \times m.$$

From this is should be obvious that $ab \mid m!$ for any $1 \le a < b \le m$, since both $a$ and $b$ appear as distinct terms in the product.

In particular, let $n$ be a composite number, and let $m = n-1$. If $n$ is not the square of a prime, there exist two distinct integers $1 < a < b < n$ such that $n = ab$ (you may want to prove this — it's not difficult), and thus $n \mid (n-1)!$.

What if $n$ is the square of a prime, i.e. $n = p^2$ for some prime $p$? If $p = 2$, we have a counterexample: $4 \nmid 3! = 6$.

However, if $p > 2$, then $2p < p^2 = n$, and thus we may choose $a = p$ and $b = 2p$ to show that $2n \mid (n-1)!$, and therefore also that $n \mid (n-1)!$.

Finally, the fact that the result does not hold for any prime $n$ follows easily from the fundamental theorem of arithmetic, as the prime factorization of $(n-1)!$ will not contain $n$ if it is prime. (I'm sure there are weaker lemmas that could be used to prove this, but why bother? The FToA does it cleanly and easily.) Thus, for integers $n > 1$, $n \nmid (n-1)!$ if and only if $n$ is either prime or $4$.

(In fact, the result holds trivially for $n = 1$ too, at least under the usual definition of $0! = 1$, but that does not follow from the argument above — $1$ is not composite in the sense needed for the argument.)

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