Miller's Primality Test

Primality test, prime In this paper we prove the following simplification of Miller's primality criterion [2]. Theorem 1. Assume that for every integer d that is 1 mod 4 and either prime or the product of two primes, the L-function I= & (k/d) l kmS satisfies the generalized Riemann hypothesis, where {kid) denotes the Jacobi symbol, defined below. Let n be an odd integer, n > I , and write n-I-= 2t l u, with t and u integers, and u odd. Then n is a prime number if and onlv if for every prime number a < c 9 (log nj2, a # n, we have a uz 1 modn (1) Or *2i-u_=-1 mod n for some integer j, 0 < j < t. (2) Here c is some constant not depending on n, and log denotes the natural logarithm. This theorem differs in two respects from Miller's result. In the first place, we require the generalized Riemann hypothesis for a smaller set of L-functions than Miller does. This results from a simplification (of Miller's proof, which has been observed by several people and which consists in elirninating the modified Carlnichael function from the argument. In the sec-13nd Flare: we have suppressed Miller's condition that 12 is nl, pt;rfect power, i.e. n # ms for all integers m, s with s 2 2. This point could have been dealt with by applying Nfontgomerv's version of Ankeny's theorem '3 Theorem 13.11 tcl 3 character of order p that is deiined modulop2, fcjrp prime, but this would have 86 required the generalized Riemann hypothesis for the L-functions attached to such characters. Instead we give a completely elementary argument, which requires no unproved hypotheses, and which leads to the following two results. Theorem 2. Let n be a positive integer, n # 4, and assume that a "-1 3 1 mod n for every prime number a < (log nj2. l7ien n is the product of distinct prime numbers. Theorem 3. Let p be an odd prime number. 13ten we have ap-' $1 mod p2 for some prime number a < 4 l (logp)2. It ,will be clear from the proof of Theorem 3, that for every E > 0 we can take a < @em2 + e) l (log Q)~ for all p exceeding a bound depending on e; here 4K2 = 0.54134.... Theorem 3 is probably far from best possible, …