# How can we prove that a binomial coefficient $n\choose m$is divisible by the ratio of $n$ and $\gcd(n,m)$?

Prove that for every $n\geq m \geq1$ natural numbers, the following number is an integer:

$${\gcd(n,m)\over n}\cdot{n\choose m}$$

Where $\gcd$ is the greatest common divisor.

I tried to make it simpler by cancelling the $n$ from the left side, and making it $(n-1)!$ on the right: $\gcd(n, m) \cdot \frac{(n-1)!}{m!(n-m)!}$, but can't really go further.

This was problem B-2 on the 2000 Putnam exam.

• Try to use this: "Bézout's identity (also called Bézout's lemma) is a theorem in the elementary theory of numbers: let a and b be nonzero integers and let d be their greatest common divisor. Then there exist integers x and y such that ax+by=d" en.wikipedia.org/wiki/B%C3%A9zout%27s_identity Feb 25, 2015 at 18:42
• I tried considering group actions but it didn't pan out. In particular, $\Bbb Z/n\Bbb Z$ acts on $\{1,2,\cdots,n\}$ which induces an action on $\binom{n}{m}$ (the collection of $m$-subsets of $\{1,\cdots,n\}$). If the subgroup $\langle\bar{m}\rangle$ acts freely on $\binom{n}{m}$ then the claim would follow from orbit-stabilizer. Unfortunately the action needn't be free - the smallest counterexample is $m=2$, $n=4$. Feb 26, 2015 at 2:23

Write $nx+my=\gcd(n,m)$, with $x,y\in\mathbb{Z}$

Then:

$$\frac{\gcd(n,m)}{n}\binom{n}{m}=\frac{nx+my}{n}\binom{n}{m}=x\binom{n}{m}+y\,\frac{m}{n}\binom{n}{m}=x\binom{n}{m}+y\binom{n-1}{m-1}\in\mathbb{Z}$$

• Wow, I am really impressed! Very nice proof. Feb 25, 2015 at 19:03
• Next question: What's the combinatorial interpretation of the expression that is simplified here? ${}\qquad{}$ Feb 25, 2015 at 19:30

Here is a conceptual way to derive this. Below we show that it is a special case of the well-known $$\rm\color{#0a0}{Lemma}$$ below that if a fraction $$q\,$$ can be written with denominators $$\,n\,$$ and $$\,m,\,$$ then it can also be written with denominator being their gcd $$= (n,m).\,$$ This makes the proof obvious, viz.

\begin{align} nq,\ mq\in\Bbb Z\,\ &\Rightarrow\, (n,m)q \in\Bbb Z,\ \text{ so for }\ \color{#c00}q = \frac{1}{n}{n\choose m}\\ n\color{#c00}q = {n\choose m}^{\vphantom{|^{|^|}}}\in\Bbb Z,\,\ m\color{#c00}q= {n\!-\!1\choose m\!-\!1}\in\Bbb Z\,\ &\Rightarrow\ (n,m)q = \dfrac{(n,m)}n{{{n\choose m}}}\in \Bbb Z\end{align}\qquad

$$\rm\color{#0a0}{Lemma}\$$ If a fraction $$\,q = c/d\,$$ can be written with denominators $$\,n\,$$ and $$\,m,\,$$ then it can also be written with denominator being their gcd $$=(n,m)$$.

Proof $$\$$ We give a few proofs of this basic result since that proves instructive.

$$(1)\$$ Recall that a fraction can be written with denominator $$\,n\,$$ iff its least denominator $$\,d\mid n.\,$$ Therefore $$\,m,n\,$$ are denoms $$\iff d\mid m,n\iff d\mid (m,n)\iff (m,n)\:$$ is a denom.

$$(2)\ \ \dfrac{mc}d,\dfrac{nc}d\in\Bbb Z\iff d\mid mc,nc\iff d\mid (mc,nc)=(m,n)c\iff\! \dfrac{(m,n)c}d\in\Bbb Z$$

$$(3)\ \ \dfrac{mc}d, \dfrac{nc}d\in\Bbb Z\,\Rightarrow \dfrac{jmc}d,\, \dfrac{knc}d\in\Bbb Z\,\Rightarrow\,\dfrac{(jm\!+\!kn)c}d\,\overset{\large \color{#c00}{\exists\, j,k}_{\phantom{1^{1^{1}}\!\!\!\!\!}}} = \dfrac{(m,n)c}d\in\Bbb Z\$$ by $$\rm\color{#c00}{Bezout}$$

$$(4)\ \ \,q = \dfrac{C}{D} = \dfrac{c}d\Rightarrow d(C,c) = (dC,dc) = (cD,dc) = c(D,d),\,$$ so $$\ \dfrac{c}d = \dfrac{(C,c)}{(D,d)}$$

Remark  The $$\rm\color{#0a0}{Lemma}$$ is a denominator form of this ubiquitous group theory theorem:

$$\qquad$$ If $$\,q^m\! = 1 = q^n\,$$ then $$\,q^{(m,n)}=1,\$$ by $$\ {\rm ord}(q)\mid m,n\Rightarrow {\rm ord}(q)\mid (m,n)$$

The least denominator of a fraction is its order in $$\,\Bbb Q/\Bbb Z,\,$$ so the Lemma is a special case of this result. For more on this viewpoint (denominator and order ideals) see here.

• Applying this lemma to $q=\frac{1}{n}\binom{n}{m}$ is just as much pulled out of a hand as the approaches of the other answers. At the end there is not much conceptually deep/interesting going on here, it is just in one way or another playing around with fundamental properties of the gcd. Nov 5, 2021 at 13:01