Prime Numbers

4.5 The primality test of Agrawal, Kayal, and Saxena (AKS test) 215
the number n − 1 does not have such a divisor r. Second, even if we do have
such a number r, there is a problem of what to choose for b. Surely, if n is
prime, then there are many numbers b that will work. Indeed, just choose b
as a primitive root for n, and there are other choices as well. So, it would be
easy to find a choice for b by a random search, but we still do not know how
to solve a problem like this in deterministic polynomial time without some
extra assumption such as the ERH.
So let us throw in the towel for now on the issue of determinism. If n − 1
has a divisor r with lg 2 nlg 2 n that is not too large. In [Berrizbeitia
2002] it is shown how to quickly prove primality for n if n + 1 is divisible by
apowerof2ofsizeaboutlg 2 n. The reader may note a parallel, for in some
sense, this chapter has come full circle. We have faced the limitations of the
n−1 test,whichledustothen+1 test, and eventually to the finite field test,
where we look for a suitable divisor of n d − 1 for some relatively small integer
d. Note that it follows from Theorem 4.3.5 with x =lg 2 n that if n>16 (so
that lg 2 n>16), then there is an integer dlg 2 n and such that each prime factor of r is one
more than a divisor of d. Hence by peeling off some of these prime factors of
r if necessary, we may assume that lg 2 nd 2 lg 2 n, but essentially we have
the same thing; namely there is some d bounded by (ln ln n) O(ln ln ln ln n) such
that n d − 1 has a divisor r with d 2 lg 2 n 1, r|n d − 1, r >
d 2 lg 2 n. Suppose too that f(t) is a monic polynomial in Zn[t] of degree d,
set R as the ring Zn[t]/(f(t)), and suppose that b = b(t) ∈ R is such that
b nd −1 =1and b (n d −1)/q − 1 is a unit in R for each prime q|r. If
(x − 1) nd
≡ x nd
− 1(modx r − b)
in R[x], then n is either a prime or prime power.
The proof of Theorem 4.5.8 is very similar to that of Theorem 4.5.7, so
we will give only a sketch. Let p be a prime factor of n and let h(t) bean
irreducible factor of f(t) modulo p. SetK as the finite field Zp[t]/(h(t)), so
that K is a homomorphic image of the ring R. SetN = n d and P = p deg h ,
so that P | p d | N. Weidentifyb with its image in K and set A = b (N−1)/r ,