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For all $n>k$, why is: $$ {n\choose k} \equiv 0 \pmod n$$ if $n$ is prime? Any hints anyone? I am really puzzled.

Tim Raczkowski
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2 Answers2

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You may notice that $\displaystyle \binom n k =\frac{n}{k}\binom {n-1} {k-1}$

Hence $k\binom n k = n \binom {n-1} {k-1}$

Since $n$ is prime and $n>k$, $n$ does not divide $k$, and by Euclid's lemma $n$ divides $\binom n k$.

Gabriel Romon
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    Should be $\binom{n-1}{k-1}$ – Mike May 24 '15 at 19:43
  • @Mike Of course. – Gabriel Romon May 24 '15 at 19:45
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    Remark that prime $, n\nmid k,\ n\mid k,{n\choose k},\Rightarrow, n\mid {n\choose k},$ is a consequence of Euclid's Lemma, or unique factorization / FTA (this should be explicitly mentioned since it is essential to the proof). – Bill Dubuque May 24 '15 at 21:01
  • I really appreciate all of your comments guys. I blamed myself for not requesting an actual rigorous proof but just some insights into why that was the case. Perhaps, that is why Michael Hardy's answer was so succinct. –  May 24 '15 at 21:23
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$$ \binom n k = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots 1}. $$ If $n$ is prime, then it can have no factor in common with any of the smaller numbers in the denominator; hence no part of it cancels with anything in the denominator. The $n$ in the numerator survives fully intact after all the canceling is done. Hence the product at the end is divisible by $n$.

Michael Hardy
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  • Thanks. I really appreciate your enlightening input. –  May 24 '15 at 18:15
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    @NumThcurious While the above idea is correct, much more needs to be said to obtain a rigorous proof. The proof requires unique factorization (or some equivalent such a Euclid's Lemma), so a proper proof must explicitly mention these properties. Better to use LeGrandDODOM's proof, and explicitly write: prime $, n\nmid k,\ n\mid k{n\choose k},\Rightarrow, n\mid {n\choose k}\ $ by Euclid or unique factorization. – Bill Dubuque May 24 '15 at 20:03
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    @BillDubuque : I think your claim that more is needed is mistaken. In order that $n$ be reduced by canceling, some divisor of $n$ other than $1$ must occur in the denominator. That can't happen since everything in the denominator is smaller than the smallest divisor of $n$ that is bigger than $1$. ${}\qquad{}$ – Michael Hardy May 24 '15 at 20:09
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    @Michael You are using "intuitive" consequences of unique factorization without rigorously justifying them. One of the major goals of a course in elementary number theory to to make rigorous such prior intuition. When one reads a proof like the above it is impossible to know if the (intended) proof is ether incorrect or incomplete. Proofs should not leave such doubts. – Bill Dubuque May 24 '15 at 20:20
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    @BillDubuque : The only uniqueness of factorization that we need in this case is the factorization of $n$ itself. Being prime, it can be factored only as $1\times n$. Uniqueness of factorization of all the other numbers that appear in this example doesn't matter to the result. ${}\qquad{}$ – Michael Hardy May 24 '15 at 20:23
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    @Michael Now you have confirmed that the "proof" is not incomplete but, rather, incorrect. So I will downvote this answer. If you fix it then I will be happy to remove the downvote. – Bill Dubuque May 24 '15 at 20:25
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    @BillDubuque : Can you prove it's incorrect? Why would one need uniqueness of factorization of anything other than $n$, together with the fact that the numbers in the denominator can't have any factors that divide $n$, besides $1$? ${}\qquad{}$ – Michael Hardy May 24 '15 at 20:27
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    @BillDubuque : Regardless of uniqueness, every cancelation is that of some divisor of some factor in the denominator with that same number as a divisor of some factor in the numerator. Cancelation of $1$ from the top and the bottom amounts to nothing. No divisor of any number in the numerator besides $1$ can be a divisor of $n$ so no cancelation ever touches $n$. ${}\qquad{}$ – Michael Hardy May 24 '15 at 20:30
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    @Michael There is so much handwaving in the answer that I do not know where to begin to debug it. You apparently intend to use some properties of fractions (that depend on unique factorization), but it is not clear precisely what properties you are implicitly using (there are many). – Bill Dubuque May 24 '15 at 20:31
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    $\dfrac{ab}{ac} = \dfrac b c$. Alright, we need to look at the dependence of that result on uniqueness of factorization. ${}\qquad{}$ – Michael Hardy May 24 '15 at 20:33
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    @Michael If I had to guess, I would say that perhaps the following is what was intended $\tag{}$ Lemma* $\ \ nj/k \in \Bbb Z,\ (n,k)=1,\Rightarrow, n\mid nj/k$ $\tag{}$ Proof* $\ $ By Euclid $,(n,k)=1,\ k\mid nj,\Rightarrow, k\mid j,\ $ so $,j/k = i\in\Bbb Z,,$ so $,n\mid ni=nj/k\ $ $\tag*{}$ – Bill Dubuque May 24 '15 at 20:42
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    But I should not have to guess what you intended. A mathematical proof should not leave such serious doubts. – Bill Dubuque May 24 '15 at 20:46
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    @BillDubuque : I'm not yet convinced we need to mention primality of any number involved except $n$. But I'll think this through and post more later. – Michael Hardy May 24 '15 at 20:48
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    @Michael There are many ways it can be proved because there are many properties equivalent to Euclid's Lemma / uniqueness of factorization. But I still have no clue how you intend to complete the above sketch into rigorous proof. – Bill Dubuque May 24 '15 at 20:57
  • @Michael Consider applying your argument to $, w\bar w/2 = 3,,\ w = 1!+!\sqrt{-5},$ in $,\Bbb Z[\sqrt{-5}],,$ a non-UFD. Does your argument incorrectly conclude that $, w\mid w\bar w/2,,$ i.e. $,2\mid \bar w?\ $ – Bill Dubuque May 24 '15 at 21:18
  • It seems to me that all answers posted to stackexchange that use a cancelation of the form $\dfrac{ab}{ac}=\dfrac b c$ where $a,b,c$ are integers are vulnerable to this same criticism. ${}\qquad{}$ – Michael Hardy May 24 '15 at 21:20
  • @Michael I don't know why you think the issue has something to do with cancellation. Rather, it has to do with handwaving vs. rigorous justification of crucial. fundamental properties closely related to unique factorization.If you try applying your argument on said example in $,\Bbb Z[\sqrt{-5}],$ then this may help to better understand the points I raise above. Then you can see where the argument breaks down, and determine precisely what special properties of $,\Bbb Z,$ are necessary for the argument to go through there. – Bill Dubuque May 24 '15 at 21:31