Krdytkyu

This is an exercise from Apostol, which i have been struggling for a while.

Given a prime $p$, how does one show that $${n choose p} equiv biggl[frac{n}{p}biggr] (text{mod} p)$$ Note that $Bigl[frac{n}{p}Bigr]$ denotes the integral part of $frac{n}{p}$.

I would also like to know as to how does one try to solve this problem. Well, what we need is to show is whenever one divides ${n choose p}$ by a prime $p$ the remainder is the integral part of $frac{n}{p}$.

Now, $${ n choose p} = frac{n!}{p! cdot (n-p)!}$$ Now $n!$ can be written as $$n!= n cdot (n-1) cdot (n-2) cdots (n-p) cdots 2 cdot 1$$

But i am really struggling in getting the integral part.

Hint $ $ If $rm nequiv j : (mod p): $ and $rm: bigl[frac{n}{p}bigr] = k, : $ then pairing factors so that top $equiv$ bottom $rm:(mod p):$ in the binomial coefficient fractions below makes the result obvious. For example

$${17choose 7} = frac{17}3 frac{16}2 frac{15}1 color{#c00}{frac{14}7}frac{13}6frac{12}5frac{11}4,equiv, color{#c00}2, =, leftlfloorfrac{17}7rightrfloorpmod 7qquadqquad $$

since all of the fractions with bottom $rmne p,$ are $,rmequiv 1pmod p,$ by top $equiv$ bottom, and the red term with the bottom = $rm ,p = 7,$ is just $rm,color{#c00}{kp/p = k}.,$ Generally we have

$$begin{eqnarray}rm {n choose p} &=&rm frac{n:(n-1):cdots:(n-p+1)}{p:(p-1):cdots: 1} \
\
&=& rm frac{n}{j} frac{n-1}{j-1}cdotsfrac{kp+1}{1} color{#c00}{frac{kp}p} frac{kp-1}{p-1}:cdotsfrac{n-p+2}{j+2} frac{n-p+1}{j+1}end{eqnarray}$$

This is a very special case of much more general arithmetical results on binomial coefficients. For example, see Andrew Granville's very interesting survey The Arithmetic Properties of Binomial Coefficients

Remark The above proof is a special case of a very simple purely arithmetical proof that I devised to show that binomial coefficients are integral. Namely, the same idea of exploiting the innate symmetry by aligning the numerators and denominators $rm:(mod p):$ extends to yield a simple algorithm that, given a binomial coefficient and a prime $rm p$, rewrites the binomial coefficient as a product of fractions whose denominators are all coprime to $rm p$. This implies that no prime divides the lowest-terms denominator, so that we may therefore conclude that the binomial coefficient is integral. For an example and further discussion see my post here.

A useful result in such problems is Lucas's Theorem which states that

If $p$ is a prime and if $a = sum_{i=0}^{k} a_{i} p^{i}, 0 le a_i < p $ i.e. the $a_i$ are digits of $a$ in base $p$ and similarly $b = sum_{i=0}^{k} b_{i}p^{i}$ (pad with zeroes if required) then

$${a choose b} = prod_{i=0}^{k} {a_i choose b_i} mod p$$

In our case $n=a$ and $b=p$. Since $b = p$ we have $b_1 = 1$ and rest of the $b_i = 0$.

Thus

$${a choose p} = {a_1 choose 1} = a_1 mod p$$

It is easy to see that $a_1 = [frac{a}{p}] mod p$.

Here's another way to look at it. Suppose $n=kp+j,$ for $0 le j le p-1.$ Then

$$ (p-1)! {n choose p} = frac{n(n-1) cdots (n-p+1)}{p} = left( {n-j over p} right)
prod_{i=0,ine j }^{p-1} (n-i)$$
$$ equiv k(p-1)! (textrm{ mod } p). $$
Since the product runs through a complete set of non-zero residues mod p.

The binomial theorem together with the identity $(x+y)^p = x^p + y^p mod p$ is useful for this type of problem.

Here we want the the $x^p$ coefficient of $(1+x)^{Ap+B} = (1+x^p)^A (1+x)^B = (1 + Ax^p + dots)(1+X)^B$ where $0 leq B < p$. This coefficient comes only from multiplication of $(Ax^p)$ by $(1)$. So the result mod p is $A$.

Another solution is to use combinatorics. Partition the set of integers from 1 to $n$ into $A$ blocks of size $p$ and a (perhaps empty) remainder of size $B$, as above. Any subset is either a complete block or it contains a proper nonempty subset of at least one block. Rotating the elements in the first partially filled block organizes the $binom{n}{p}$ p-subsets into collections of size $p$, plus a "remainder" consisting of complete $p$-blocks, and there are $A$ of the those.

Consider ${n choose p}$ as $(n)(n-1)...(p+2)(p+1)/(n-p)!$. That can be written as $(n)/(n-p) cdot (n-1)/(n-p-1) cdot cdots cdot (p+1)/(1)$. Notice that these are all of the form $(x+p)/x$. Thus, terms where $x$ is not divisible by $p$ can be reduced modulo $p$ to $x/x=1$. The rest of the terms are going to be
$(p [n/p])/((p [n/p]-1))) cdot cdots cdot (3p)/(2p) cdot (2p)/(p)$. This product telescopes to $(p [n/p])/p = [n/p]$.