Prime Numbers

214 Chapter 4 PRIMALITY PROVING
for all nonnegative integers i, j. Thusforsuchi, j,
(x − 1) pi (n/p) j
≡ A j(1−k)+ik x − 1(modx r − b, p), (4.34)
sincebothsideshavethesamep j -th power (mod x r − b, p), and raising to
the p j -th power is one-to-one in Zp[x]/(x r − b). This last assertion follows
because raising to the p-th power is one-to-one in any Zp[x]/(f(x)) where
f(x) does not have any repeated irreducible factors modulo p, noting that
since gcd(x r − b, rx r−1 )=1inZp[x], the polynomial x r − b indeed does not
have any repeated factors.
Note that x − 1 is a unit in Zp[x]/(x r − b). Indeed, in Zp[x], we have
gcd(x − 1,x r − b) =gcd(x − 1, 1 − b) = 1, provided that p does not divide
b − 1. But since A = b (n−1)/r modulo p has order r>lg 2 n ≥ 1, we do indeed
have p not dividing b − 1. Let E denote the multiplicative order of x − 1in
Zp[x]/(x r − b). Note that
E ≥ 2 r − 1,
since the polynomials
(A j x − 1),
j∈S
where S runs over the proper subsets of {0, 1,...,r− 1}, are not only distinct
in Zp[x]/(x r − b), but each is a power of x − 1, by (4.32).
Consider integers i, j with 0 ≤ i, j ≤ √ r.Itmustbethattherearetwo
distinct pairs (i1,j1), (i2,j2) with
j1(1 − k)+i1k ≡ j2(1 − k)+i2k (mod r),
so that if u1 = p i1 (n/p) j1 ,u2 = p i2 (n/p) j2 ,then
(x − 1) u1 ≡ A j1(1−k)+i1k x − 1 ≡ A j2(1−k)+i2k x − 1 ≡ (x − 1) u2 (mod x r − b, p).
Hence
u1 ≡ u2 (mod E).
But u1,u2 ∈ [1,n √ r ]andE>2 r − 1 >n √ r − 1, the last inequality holding
by our hypothesis that r>lg 2 n.Thus,u1 = u2, and as we saw in the proof
of Theorem 4.5.2, this immediately leads to n beingapowerofp. ✷
This theorem may be essentially found in [Bernstein 2003] and (independently)
[Mihăilescu and Avanzi 2003]. It was originally proved in the case of
r a power of 2 by Berrizbeitia and in the case of r aprimeorprimepowerby
Cheng.
Note that using fast polynomial and integer arithmetic, the congruence
(4.30) can be checked in Õ(r ln2 n) bit operations, the notation Õ having been
introduced in Section 4.5.2. So if r can be chosen such that r = O(ln 2 n), we
thus would have the basis for a primality test of complexity Õ(ln4 n). There
are two problems with this. First, not every prime n has a divisor r of n − 1
with lg 2 n