Prime Numbers

202 Chapter 4 PRIMALITY PROVING
Algorithm 4.5.1 (Agrawal–Kayal–Saxena (AKS) primality test). We are
given an integer n ≥ 2. This deterministic algorithm decides whether n is prime
or composite.
1. [Power test]
If n is a square or higher power, return “n is composite”;
2. [Setup]
Find the least integer r with the order of n in Z ∗ r exceeding lg 2 n;
If n has a proper factor in [2, ϕ(r)lgn], return “n is composite”;
// ϕ is Euler’s function.
3. [Binomial congruences]
for(1 ≤ a ≤ ϕ(r)lgn) {
if((x + a) n ≡ x n + a (mod x r − 1,n)) return “n is composite”;
}
Return “n is prime”;
Square testing in Step [Power test] may be done by Algorithm 9.2.11,
and higher-power testing may be done by a similar Newton iteration, cf.
Exercise 4.11. Note that one has only to test that n is in the form a b for
b ≤ lg n. (Note too that from Exercise 4.28, Step [Power test] may actually be
skipped entirely!) The integer r in Step [Setup] may be found by sequential
search over the integers exceeding lg 2 n. In this search, if a value of r is found
for which 1 < gcd(r, n) ϕ(r)lgn. Since Step [Setup]
involves a search for small divisors of n, it may occur that all possibilities up
to √ n are accounted for, and so n is proved prime. In this case, of course, one
need not continue to Step [Binomial congruences], but this event can occur
only for fairly small values of n. See the end of Section 4.5.4 for more notes
on AKS implementation.
We shall return to the issue of the size of r when we discuss the complexity
of the algorithm, but first we will discuss why the algorithm is correct.
Algorithm 4.5.1 is based principally on the following beautiful criterion.
Theorem 4.5.2 (Agrawal, Kayal, Saxena). Suppose n is an integer with
n ≥ 2, r is a positive integer coprime to n such that the order of n in Z ∗ r
is larger than lg 2 n,and
(x + a) n ≡ x n + a (mod x r − 1,n) (4.26)
holds for each integer a with 0 ≤ a ≤ ϕ(r)lgn. Ifn has a prime factor
p> ϕ(r)lgn, then n = p m for some positive integer m. In particular, if n
has no prime factors in [1, ϕ(r)lgn] and n is not a proper power, then n is
prime.