The binomial coefficient $n\choose m$ is divisible by $n/\gcd(n,m)$

Question

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.

Answer 1

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

Then:

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

Answer 2

Here is a conceptual way to discover a proof. Recall that a fraction $Q$ can be written two ways, with denominators both $\,n\,$ and $\,m\!\color{#c00}\iff $ it can be written with denom being their gcd $(n,m).\, $ Since by definition, $\,0\neq d\,$ is a denom for $\,Q\!\iff\! d\:\!Q\in \Bbb Z,\,$ the above becomes in the OP

${\rm for}\,\ Q={\small {\displaystyle\dfrac{1}n{n\choose m}}}\!: \ \ \overbrace{(n,m)\:\!Q\:\!\in\:\! \Bbb Z}^{\small \textstyle (n,m)\:\!\ {\rm denom}}\color{#c00} \iff $ $\overbrace{\underbrace{n\:\!Q,\ \ m\:\!Q}_{\!\!\displaystyle\small{n\choose m}{\large ,}\:\!{n\!-\!1\choose m\!-\!1}\!\!\!\!\!}\in\, \Bbb Z}^{\textstyle\small n,m\,\ \rm denoms^{\vphantom{|^|}}},\ $ ${\rm which}\ {\mathit is}\ \bf true \ \color{#c00}\checkmark$

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Remark $\ mQ \in\Bbb Z\,$ by scaling the well-known ${\small {\displaystyle{n\choose m} = \dfrac{n}m {n\!-\!1\choose m\!-\!1}}}\,$ by $\:\!\large \frac{m}n.\,$ This (unscaled) equation is used in the same way here. This method applies to many binomial coef. equations put in form $MA = NB,\,$ viz. $\,Q = \frac{A}N\:\!$ has denoms $M,N\,$ so it also has denom their gcd $\:\!(M,N).\,$

The key linked Lemma is an additive form of the more commonly known ubiquitous group theory Lemma below (since in the above proof the least denominator of a fraction is simply its order in the rational circle group $\,\Bbb Q/\Bbb Z = \,$ rationals $\!\bmod 1),\,$ where we used $\:\!\color{#0af}{\rm U}$ = universal gcd property.

$\qquad\qquad\quad\ \overbrace{\color{#c00}{q^m} =\, 1\, =\, \color{#00}{q^n}}^{\textstyle\small m,n\,\ \rm periods}\iff\!\!\! \overbrace{q^{\color{#0a0}{(m,n)}}=1}^{\textstyle\small (m,n)\,\ \rm a\ period}\!\!\!\!,\ $ by $\,\ {\rm ord}\,q\mid m,n\!\!\overset{\color{#0af}{\rm U}\!}\iff {\rm ord}\,q\mid (m,n)$

Similarly $\:\!\ \color{#c00}{f_m(q)}=0=\color{c00}{f_n(q)}\!\iff\! \color{#0a0}{(f_m,f_n)}(q)=0\,$ for polynomials $\,f_m,f_n\in F[x],\,$ where a (monic) minimal polynomial of $\,q\,$ is its "order" (= generator of ideal of poly's with $\:\!q\:\!$ as root)

Generally all of the above are special cases of the fundamental fact that principal ideals in a PID are $\rm\color{#0a0}{closed\ under\ gcd}$: $\, m,n\in I=(d)\Rightarrow \color{#0a0}{(m,n)}\in I,\,$ i.e. $\,d\mid m,n\Rightarrow d\mid (m,n),\,$ hence nonzero ideals are generated by any "least" element (the "order" in above cases), where the size measure for "least" can be #prime factors in a PID, or least Euclidean size in a Euclidean domain, e.g. least absolute value in $\Bbb Z,\,$ and least degree in $F[x])$.

For more on this viewpoint (and denominator, order, annihilator ideals) see e.g. here and here.

Answer 3

Consider the set of subsets with $m$ elements of $\mathbb{Z}/n$. The group $\mathbb{Z}/n$ acts on it by translations:

$$k * \{a_1, \ldots, a_m\}= \{k+a_1, \ldots, k+a_m\}$$

Now if $k$ fixes $\{a_1, \ldots, a_m\}$, then $$\sum_{i=1}^m a_i = \sum_{i=1}^m (k + a_i)$$

so $m k= 0$ in $\mathbb{Z}/n$. Hence every orbit has cardinality divisible by $\frac{n}{\gcd(n,m)}$, and so has the full set.

$\bf{Added:}$

Another approach: Let $n\ge m \ge k \ge 0$ integers. We have the equality

$$\binom{n}{m} \cdot \binom{m}{k} = \binom{n}{k} \cdot \binom{n-k}{m-k}$$

since both numbers count the numbers of pairs $A\supset B$ of subsets of $\{1,\ldots, n\}$, with $|A| = m$, $|B|=k$, therefore the equality of fractions:

$$ \frac{\binom{m}{k}}{\binom{n}{k}}=\frac{\binom{n-k}{m-k}}{\binom{n}{m}} $$

Now convert the first fraction to its simplest form. The denominator will be

$$ \frac{\binom{n}{k}}{\gcd(\binom{m}{k}, \binom{n}{k})}$$

It will be the simplest form of the second fraction too. We conclude that $$ \frac{\binom{n}{k}}{\gcd(\binom{m}{k}, \binom{n}{k})} \text{ divides } \binom{n}{m}$$

For $k=1$ we obtain our result.

Note the from the above we also conclude that any two binomial coefficients $\binom{n}{m}$, $\binom{n}{k}$, with $0<k,m<n$ are not relatively prime.

$\bf{Added:}$ So what was the idea: write a fraction with denominator $\binom{n}{m}$ equal to a second fraction, and then reduce the latter one to simpler terms. One could try to juggle with this, for instance showing that $$(n+1) \mid \binom{2n}{n}$$, since

$$\frac{\binom{2n}{n+1}}{\binom{2n}{n}} = \frac{n}{n+1}$$

We want though a different generalization of the statement in the question, using $q$-binomial coefficients. Using the interpretation of them ( for $q$ a prime power) as number of subspaces, we get again the formula:

$$ \frac{\binom{m}{k}_q}{\binom{n}{k}_q}=\frac{\binom{n-k}{m-k}_q}{\binom{n}{m}_q} $$

Recall that we are dealing wit fractions from $\mathbb{Z}[q]$, a UFD. Therefore we conclude that the denominator of the reduction to lowest term of the first fraction divides $\binom{n}{m}_q$ as before. In the particular case $k=1$ we get

$$\frac{(n)_q}{(m)_q} = \frac{\binom{n-1}{m-1}_q}{\binom{n}{m}_q} $$

so $$\frac{(n)_q}{\gcd((m)_q, (n)_q))} ( = \frac{(n)_q}{(m,n)_q}) \mid \binom{n}{m}_q$$