Prime Numbers

216 Chapter 4 PRIMALITY PROVING
so that A has order r by our hypothesis. Then, there is some integer k such
that for all nonnegative integers j, i,
j
N
(x − 1) ≡ A j x − 1(modx r i
P
− b), (x − 1) ≡ A ik x − x (mod x r − b),
where we view these as polynomials in K[x]. This follows in exactly the same
way as in the proof of Theorem 4.5.7, and further we get that
(x − 1) P i (N/P ) j
≡ A ik+j(1−k) x − 1(modx r − b).
If E is the order of x − 1inK[x]/(x r − b), then E ≥ 2 r − 1bythesame
argument as before. But again as before, there are different pairs of integers
i1,j1 and i2,j2 with Ul := P il (N/P) jl ∈ [1,N √ r]forl =1, 2andU1 ≡ U2
(mod E). This forces U1 = U2, andson is a power of p (since N is a power
of n and P is a power of p).
The reader is invited to observe the remarkable similarity of Theorem 4.3.3
to Theorem 4.5.8, where I,F,g of the former theorem correspond to d, r, b,
respectively, in the latter.
We may use Theorem 4.5.8 as the basis of a fast random algorithm that
is expected to supply primes with proofs that they are primes:
Algorithm 4.5.9 (Quartic-time variant of AKS test). We are given an integer
n>1. This random algorithm attempts to decide whether n is prime or
composite, and it decides this issue correctly whenever it terminates.
1. [Setup]
If n is a square or higher power, return “n is composite”;
Find a pair r, d of positive integers with rd2 minimal such that r|nd − 1
and d2 lg 2 n