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Let $n \geq 2$ be an integer, $p$ a prime with $p^e$ the highest power of $p$ dividing $n$. Then $\binom{n}{p^e}$ is not divisible by $p$.

I think you can do it using this formula for $\binom{n}{k}$. From that post we have that

$$ \binom{n}{r} = (\prod_{n-r\lt p\leq n} p^{k_p(n)})(\prod_{r\lt p\leq n-r} p^{k_p(n) - k_p(n-r)})(\prod_{p\leq r}p^{k_p(n)-k_p(n-r)-k_p(r)}) $$

Well $p$ falls into the class $p \leq r= p^e$. So it suffices to show that $k_p(n) - k_p(n-p^e) - k_p(p^e) = 0$ where $k_p(n) = \sum_{k\geq 1} \lfloor \frac{n}{p^k}\rfloor$. Well, $k_p(n) = k_p(mp^e) = m + mp + mp^2 + ... mp^{e-1}$, and

$$ k_p(n - p^e) = k_p((m - 1)p^e) $$

but $m-1$ could equal $\ell p^k$ for some $k$. So I'm not sure where to go from here. Alternative simpler proofs welcome.

The proof from the book goes like this: For each $n-k$ divisible by $p$ in the numerator of $\binom{n}{p^e}$, $n-k = mp^e - \ell p^i$. It says "then $i \lt e$", but I don't how that's true. Without that requirement we have either $e = i, \ e \gt i,$ or $e \lt i$. Let's see. If $e \leq i$, then $p^e - k = p^e(1 - \ell p^{i-e})$, but there is not such $p^e-k$ in the denominator so $e \gt i$. It then says $(m-k) = (p^e - k)$ and $(n-k) = (p^em - k)$ are both divisible by $p^i$ but not $p^{i+1}$. How did they get $m-k = p^e - k$?

Community
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Daniel Donnelly
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  • I am having trouble making heads or tails of your discussion on the book's proof. God forbid OPs cite the book they're referring to! (I actually checked Proofs from The Book, which is an actual book, and the closest thing to this topic is chapter 3, but it doesn't cover this theorem in particular.) – anon Sep 27 '13 at 00:44
  • Please see this http://math.stackexchange.com/questions/51469/prime-dividing-the-binomial-coefficients/51475#51475 – Bruno Joyal Sep 27 '13 at 01:11

2 Answers2

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This is a direct consequence of Lucas' Theorem.

Write $n$ and $p^e$ as base-$p$ numbers, $n$ would have exactly $e$ $0$'s at the end, and would have a digit $1\le d< p$ to the left of those $0$'s. $p^e$ also has exactly $e$ zeros, with a leading $1$.

Clearly, the binomial coefficients in the Lucas' theorem are all ones, except for the $(e+1)$-th from the right, which is $\binom{d}{1}=d<p$, so $\binom{n}{p^e}\equiv d\not\equiv 0 \pmod{p}$.

ftfish
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Personally I would use Lucas' theorem for binomial coefficient residues in $p$-digital expansions:

$$\binom{a_kp^k+\cdots+a_1p+a_0}{b_kp^k+\cdots+b_1p+b_0}\equiv\binom{a_k}{b_k}\binom{a_{k-1}}{b_{k-1}}\cdots\binom{a_1}{b_1}\binom{a_0}{b_0}\mod p.$$

(Here $0\le a_i,b_j<p$.) The theorem telescopes out inductively from the following version:

$$\binom{ap+r}{bp+s}\equiv\binom{a}{b}\binom{r}{s}\mod p,$$

where $0\le r,s<p$ and $a,b\ge0$. This follows from using the binomial theorem and freshman's dream in characteristic $p$. In "generatingfunctionological" language, we can observe

$$\sum_{b,s} \binom{ap+r}{bp+s}X^{bp+s}\equiv(1+X)^{ap+r}\equiv(1+X^p)^a(1+X)^r\equiv\sum_{b,s} \binom{a}{b}\binom{r}{s}X^{ap+r},$$

with $b$ ranging as $0\le b\le a$ and $s$ ranging as $0\le s<p$ if $b<a$ and $0\le s\le r$ if $b=a$.

Equating coefficients gives Lucas' theorem. Afterwards we have

$$\binom{mp^e}{p^e}\equiv\binom{m}{1}\equiv m\not\equiv0\mod p.$$

anon
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  • Not sure how you get $\sum_{b,s} \binom{ap+r}{bp+s}X^{bp+s}\equiv(1+X)^{ap+r}$ – Daniel Donnelly Sep 27 '13 at 00:44
  • @EnjoysMath That is the binomial theorem. (The indices have been decomposed as $k=bp+s$ with $0\le k\le ap+s$ using the division algorithm, which requires the bounds I state on $a,b,r,s$ right after the congruence.) – anon Sep 27 '13 at 00:45