Number Theory Proof for $\binom{pA}{pB} \equiv \binom{A}{B} \pmod{p^2}$

Question

I was doing some reading on Wolstenholme's theorem, which states that for prime $p$, $\binom{pA}{pB} \equiv \binom{A}{B} \pmod{p^3}$ for $p\geq 5$. The proof of the statement is outside my scope, but I was considering why it is still true for $\pmod p$ and $\pmod {p^2}$.

In Given integers $a \ge b > 0$ and a prime number $p$, prove that ${pa \choose pb} \equiv {a \choose b} \mod p$., a fairly straightforward argument is given for the case of $\pmod p$, where they find 2 equivalent expressions and equate them.

I was wondering how I could prove this in similar spirit for the case of $\pmod {p^2}$, without bringing in many additional lemmas, etc.

Answer 1

This is identical to the proof given on Wikipedia, but I've stripped out the group theory in the hopes that will make it more accessible.

Take $a$ boxes and fill each one with the integers modulo $p$, and let $S$ be the resulting set of size $pa$.

Let $T$ be a subset of $S$ of size $pb$, noting that the set of all such $T$ has size ${pa}\choose {pb}$. Define the "box number" of $T$ to be the number of boxes such that $T$ contains some number in that box, but not all numbers in that box.

We can show without much trouble that the number of subsets with box number $k$ is divisible by $p^k$. Indeed, every set $T$ with box number $k$ has exactly $p^k$ distinct "rotations" given by adding arbitrary integers mod $p$ to the incomplete boxes. Here we really need to use the fact that $p$ is prime to see that all these rotations are distinct; this is a nice number theory exercise.

How many subsets $T$ have box number $0$? Well, for each box, we either pick the whole box or none of the box, and we need to end up with exactly $pb$ elements, so the answer is ${a}\choose{b}$.

To show the final result we want, all we have to do is show that there is no subset with box number $1$, as then we will have written ${pa}\choose{pb}$ as a sum of ${a}\choose{b}$ and something divisible by $p^2$.

But a subset with box number $1$ would be the union of a bunch of complete boxes (with size divisible by $p$), and a single incomplete box (with size not divisible by $p$), so such a subset could not possibly have size $pb$.

Answer 2

$(1+x)^{pa}=(1+x)^p(1+x)^{p(a-1)}$

Binomial expanding both sides, the coefficient of $x^{pb}$ is $$\binom{pa}{pb}=\binom{p}{0}\binom{p(a-1)}{pb}+\binom{p}{1}\binom{p(a-1)}{pb-1}+\binom{p}{2}\binom{p(a-1)}{pb-2}+\dots+\binom{p}{p}\binom{p(a-1)}{p(b-1)}$$ [Vandermonde's identity]

On the right side, all terms except the first and last are divisible by $p^2$ [because $\forall0<i<p$, the second factor $\binom{p(a-1)}{pb-i}=\frac{p(a-1)}{pb-i}\binom{p(a-1)-1}{pb-i-1}$ and $p\nmid pb-i$ and $\binom{p(a-1)-1}{pb-i-1}\in\mathbb{N}$, so $p\mid\binom{p(a-1)}{pb-i}$].

Thus, $\binom{pa}{pb}\equiv\binom{p(a-1)}{pb}+\binom{p(a-1)}{p(b-1)} \pmod{p^2}$

Next, using mathematical induction on $a$, by induction hypothesis $\binom{p(a-1)}{pb}\equiv\binom{a-1}{b}\pmod{p^2},\binom{p(a-1)}{p(b-1)}\equiv\binom{a-1}{b-1}\pmod{p^2}$, so we get $\binom{pa}{pb}\equiv\binom{a-1}{b}+\binom{a-1}{b-1}=\binom{a}{b}\pmod{p^2}$