Abstract

It is a beautiful property of prime numbers, first proved more than three centuries ago by Fermat, that k Pk (mod p) for all prime numbers p and all integers k. Here we present a simple proof of Fermat's theorem by considering iterates of the function f(z) = z k on the complex plane. The method of proof has the advantage of generalizing the theorem to composite exponents: for every n we find a degree-n polynomial, with coefficients + 1, that is always divisible by n. This is different from Euler's generalization (k 4(n) =1 (mod n) for k and n copiime). The method of proof is potentially more general still, since it is easily adapted to other functions f. Indeed, for any set S, every function f: S -> S satisfying a certain property corresponds to a divisibility result similar to Fermat's little theorem. Let k be a positive integer and p be prime. Consider the function f(z) = zk for complex z. The pth iterate of f is evidently fP(z) = zk. Let Pp be the set of those z that are fixed under fP but not under f itself. Then IPp I = k P-k. But if z E Pp, then fJ(z) Epp for every i = 0,1,. . ., p 1; and since p is prime, the p values z,f(z),...jfP-(z) are all distinct. Hence, we can partition Pp into equivalence classes, each containing p elements, obtaining

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