Wolstenholme's Theorem - A Proof of The Theorem

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 group C_pa 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 pk 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 orbits of size 1 and there are no orbits of size p. Thus we first obtain Babbage's theorem

Examining the orbits of size p2, we also obtain

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

This congruence becomes an equation for using the relation

When p is odd, the relation is

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

A similar derivation modulo p4 establishes that

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.