Geometric Proof of Wilson's Theorem

Wilson's Theorem states that, for a prime p,

  (p - 1)! = -1 (mod p)

or, in other words, that p divides (p - 1)! + 1. Here is a proof I came across in an old Russian magazine (1930s) authored by A. B. (Moscow) which incidentally fits perfectly my initials. The author refers to it as geometric, although nowadays it would be probably thought of a combinatorial proof. I failed to locate a more recent reference.

So let's p be a prime and number the successive vertices of a regular p-gon 1, 2, ..., p. In the regular p-gon, the vertices are joined sequentially, 1234...p. But me may also join them by skipping 1 vertex: 135..., or 2 vertices 147..., ... getting in this manner star shaped polygons. These polygons (or the corresponding permutations of the vertices) come in pairs, because skipping k vertices is the same as skipping p - k + 2 of them. It follows that the total number of star shaped polygons on the given set of p vertices is (p - 1)/2.

More general polygons can be obtained by less regular permutations, like say, 13245... Congruent polygons come from the shift of indices: 13245..., 24356..., 35467..., etc. These polygons, too, are congruent in pairs since the vertices of a polygon could be traversed in two directions. In any event, we see that the number of such irregular polygons is divisible by p.

On the other hand, all possible distinct polygons on the given set vertices can be obtained by permuting all the vertices, but one, implying (p - 1)! permutations, which in pairs define the same polygon. Thus the total number of distinct polygons on the given set of p vertices is (p - 1)!/2.

The number of irregular polygons (that is divisible by p) is then given by

  (p - 1)!/2 - (p - 1)/2 = 1/2 [(p - 1)! + 1 - p].

The latter number is divisible by p if and only if (p - 1)! + 1 is divisible by p, proving Wilson's theorem.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny