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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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    $\begingroup$ In case 1, there is a subcase where $a=n-a$, and the result is false. $\endgroup$ Jun 30, 2012 at 8:32
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    $\begingroup$ $3!\equiv2\pmod4$ $\endgroup$
    – Mike
    Jun 30, 2012 at 8:39
  • $\begingroup$ @JonasMeyer I did not consider that case. Darn, this also only happens when n=4. :( thank you. $\endgroup$ Jun 30, 2012 at 8:41
  • $\begingroup$ @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. $\endgroup$ Jun 30, 2012 at 8:42
  • $\begingroup$ @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. $\endgroup$ Jun 30, 2012 at 8:46

5 Answers 5

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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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  • $\begingroup$ @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. $\endgroup$ Jun 30, 2012 at 14:33
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    $\begingroup$ @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? $\endgroup$
    – Pedro
    Jun 30, 2012 at 15:28
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    $\begingroup$ @Peter Integrality of binomial coefficients is a much deeper result than said divisibility result. See the note I added to my answer. $\endgroup$ Jun 30, 2012 at 15:57
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    $\begingroup$ @BillDubuque I ask again, Bill: In terms of combinatorics, isn't it "natural"? (I know in NT it is not trivial) $\endgroup$
    – Pedro
    Jun 30, 2012 at 16:37
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Hint $\rm\ n\, =\, \color{#C00}a\:\!\color{#0A0}b\mid1\!\cdot\! 2\cdots\color{#C00} a\:\color{#0A0}{(a\!+\!1)\, (a\!+\!2) \cdots (a\!+\!b)}\cdots (\color{#90f}{ab\!-\!1})=(n\!-\!1)!\,\ $ by $\rm\,\ \color{#0a0}{a\!+\!b}\le \color{#90f}{ab\!-\!1} $

Note $\rm\,\color{#0A0}b\,$ divides $\rm\color{#0A0}{green}$ product since a sequence of $\rm\,b\,$ consecutive integers has a multiple of $\rm\,b,\,$

and $\rm\,a\!+\!b \le ab\!-\!1 \!\!\iff\!\! 2\le (\underbrace{a\!-\!1}_{\large \ge\,1})(\underbrace{\color{#f70}{b\!-\!1}}_{\large\color{#f70}{\ge\, 2}}), \,$ true by $\rm\,a,b\ge 2,\,$ $\rm\underbrace{not\ both\!=\!2,}_{\large n\,=\,ab\,\ne\, 4}\,$ so $\,\color{#f70}{\rm one}$ is $\,\color{#f70}{\ge 3}$

The prior inequality implies that all of the $\rm\color{#0A0}{green}$ factors do occur in $\rm\,(ab\!-\!1)!$

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, we deduce it from the more elementary fact that a sequence of $\rm\,b\,$ consecutive integers contains a multiple of $\rm\,b,\,$ which is an immediate consequence of (Euclidean) division.

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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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    $\begingroup$ Ps. See also Wilson's theorem, which says that $(n-1)! \equiv -1 \pmod n$ if and only if $n$ is prime. $\endgroup$ Dec 13, 2015 at 19:14
  • $\begingroup$ I got the part why prime numbers dont satisfy this but How choosing $a=2p$ and $b=p$ proves it for all composite number,i think you have assumed $ab|(n-1)!$ first then you choose a,b and therefore$ab=2p^2|(n-1)!$ and using $n=ab$ you conclude $2n|(n-1)!$ how is that a proof ? Please correct me if i am getting you wrong. $\endgroup$ Mar 24, 2019 at 15:40
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    $\begingroup$ @NewBornMATH: There are four distinct cases that I consider above: a) $n$ is prime, b) $n=4=2^2$, c) $n$ is the square of some prime $p>2$, and d) $n$ is neither a prime nor the square of a prime. In cases a and b, $n$ does not divide $(n-1)!$. In case c, we have $1<p<2p<n$, and thus $p(2p)=2p^2=2n$ divides $(n-1)!$. In case d, which I consider in the third paragraph above, $n$ has a pair of integer divisors $a$ and $b = n/a$ such that $1<a<\sqrt n<b<n$ (I have omitted the proof of this, but it's not hard to show by contradiction), and thus $n=ab$ divides $(n-1)!$. $\endgroup$ Mar 25, 2019 at 5:40
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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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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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