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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.

noHopeLeft
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  • Wikipedia has a one-paragraph proof: https://en.wikipedia.org/wiki/Wolstenholme%27s_theorem#A_proof_of_the_theorem – Andrew Dudzik Jul 08 '21 at 14:34
  • Thanks Slade, but I am looking for a more elementary NT way to prove it as I am not familiar with group theory. The solution in wikipedia is similar to the second solution in the linked question, which is not what my focus is. – noHopeLeft Jul 08 '21 at 14:37
  • It's just counting orbits so there really isn't any group theory being used here. I'll elaborate in an answer when time allows. – Andrew Dudzik Jul 08 '21 at 14:50
  • Just give common denominator and bash everything – dezdichado Jul 08 '21 at 16:41

1 Answers1

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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 $pb$.

Andrew Dudzik
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