Let's use (mostly) geometrical arguments to show that


For any prime p and positive integer a, p|(a^p - a).


Arrange a^p points in a cubic lattice in \mathbb{R}^p. Specifically, each point has coordinates (x_1 ... x_p) with 1\le x_i\le a. The diagonal of the cube is the line of a points (1,\dots,1),\dots,(a,\dots,a), and deleting these from the lattice leaves a^p-a points.

Now note that the cube has a p-fold rotational symmetry around its diagonal. Concretely, this is a cyclic permutation of the coordinate labels, sending (x_1, x_2, ..., x_p) \to (x_2, x_3, ..., x_p, x_1). Points on the diagonal are fixed, but the other points are each taken through an orbit by this symmetry, and each orbit is of size exactly p.

This is because each point must return to its original location after p rotations, and if it did so in any smaller number m of rotations, then m|p, contradicting that p is prime. (Of course, one could have m=1, but this would be a fixed point, hence on the diagonal.) Therefore, the non-diagonal points of the cube are partitioned entirely into disjoint orbits of size p, and hence p|(a^p-a).


The case for negative a follows from the positive case, but of course the geometrical argument fails to have an interpretation in that case.