Wolstenholme's Theorem

Wolstenholme's Theorem.

Statement 1. Let $p>3$ be a prime number. Then, $\dbinom {2p-1}{p-1} \equiv 1 \pmod {p^3}$.

Statement 2. Let $p>3$ be a prime number. Then, $\dbinom {ap}{bp} \equiv \dbinom {a}{b} \pmod {p^3}$, for any naturals $a,b$.

Wolstenholme's theorem states that for a prime number p > 3, the congruence

${2p-1 \choose p-1} \equiv 1 \pmod{p^3}$ holds, where the parentheses denote a binomial coefficient. For example, with p = 7, this says that 1716 is one more than a multiple of 343. An equivalent formulation is the congruence

${ap \choose bp} \equiv {a \choose b} \pmod{p^3}.$ The theorem was first proved by Joseph Wolstenholme in 1862. In 1819, Charles Babbage showed the same congruence modulo $p^2$, which holds for all primes $p$ (for $p=2$ only in the second formulation). The second formulation of Wolstenholme's theorem is due to J. W. L. Glaisher and is inspired by Lucas' theorem.

No known composite numbers satisfy Wolstenholme's theorem and it is conjectured that there are none (see below). A prime that satisfies the congruence modulo $p^4$ is called a Wolstenholme prime (see below).

As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers:

A proof of the theorem

There is more than one way to prove Wolstenholme's theorem. Here is a proof that directly establishes Glaisher's version using both combinatorics and algebra.

For the moment let p be any prime, and let a and b be any non-negative integers. Then a set A with $ap$ elements can be divided into a rings of length p, and the rings can be rotated separately. Thus, the a-fold direct sum of the cyclic group of order p acts on the set A, and by extension it acts on the set of subsets of size $bp$. Every orbit of this group action has $p^k$ elements, where k is the number of incomplete rings, i.e., if there are k rings that only partly intersect a subset B in the orbit. There are $\textstyle {a \choose b}$ orbits of size 1 and there are no orbits of size p. Thus we first obtain Babbage's theorem

${ap \choose bp} \equiv {a \choose b} \pmod{p^2}$.

Examining the orbits of size $p^2$, we also obtain

${ap \choose bp} \equiv {a \choose b} + {a \choose 2}\left({2p \choose p} - 2\right){a -2 \choose b-1} \pmod{p^3}.$

Among other consequences, this equation tells us that the case a=2 and b=1 implies the general case of the second form of Wolstenholme's theorem.

Switching from combinatorics to algebra, both sides of this congruence are polynomials in a for each fixed value of b. The congruence therefore holds when a is any integer, positive or negative, provided that b is a fixed positive integer. In particular, if a=-1 and b=1, the congruence becomes

${-p \choose p} \equiv {-1 \choose 1} + {-1 \choose 2}\left({2p \choose p} - 2\right) \pmod{p^3}.$

This congruence becomes an equation for $\textstyle {2p \choose p}$ using the relation

${-p \choose p} = \frac{(-1)^p}2{2p \choose p}.$

When p is odd, the relation is

${2p \choose p} \equiv 6 \pmod{p^3}.$

When p≠3, we can divide both sides by 3 to complete the argument.

A similar derivation modulo $p^4$ establishes that

${ap \choose bp} \equiv {a \choose b} \pmod{p^4}$

for all positive a and b if and only if it holds when a=2 and b=1, i.e., if and only if p is a Wolstenholme prime

Prove that the converse of Wolstenholme's Theorem is true for even positive integers.