Prime Numbers

4.5 The primality test of Agrawal, Kayal, and Saxena (AKS test) 213
4.5.4 A quartic time primality test
Since the most time-consuming step of Algorithm 4.5.1 is the checking of the
congruence (x + a) n ≡ x n + a (mod x r − 1,n) for so many values of a, this
area would be a good place to look for improvements. In Theorem 4.5.4 we
had the improvement of replacing x r − 1 with a polynomial f(x) of possibly
smaller degree. Another idea is to get binomial congruences verified “for free.”
In the following theorem, we replace x r − 1withx r − b for a suitable integer
b, and we need only verify one binomial congruence.
Theorem 4.5.7. Let n, r, b be integers with n > 1, r|n − 1, r > lg 2 n,
b n−1 ≡ 1(modn), andgcd(b (n−1)/q − 1,n)=1for each prime q|r. If
then n is a prime or prime power.
(x − 1) n ≡ x n − 1(modx r − b, n), (4.30)
Proof. Let p|n be prime and set A = b (n−1)/r mod p. ThenA has order r in
Z ∗ p, so that in particular, r|p − 1 (see Pocklington’s Theorem 4.1.3). Note that
Thus, by our hypothesis,
x n = x · x n−1 = x(x r ) (n−1)/r ≡ Ax (mod x r − b, p). (4.31)
(x − 1) n ≡ x n − 1 ≡ Ax − 1(modx r − b, p).
Also note that if f(x) ≡ g(x) (modx r − b, p), then f(A i x) ≡ g(A i x)
(mod x r − b, p) for any integer i, since(A i x) r − b ≡ x r − b (mod p). Thus,
taking f(x) =(x − 1) n and g(x) =Ax − 1, we have
(x − 1) n2
≡ (Ax − 1) n ≡ A 2 x − 1(modx r − b, p),
and more generally by induction, we get
(x − 1) nj
≡ A j x − 1(modx r − b, p) (4.32)
for every nonnegative integer j.
Note that if c is an integer and c r ≡ 1(modp), then c ≡ A k (mod p) for
some integer k; indeed, all that is used for this observation is that p is prime
and A has order r modulo p. So,wehave
x p = x · x p−1 = x(x r ) (p−1)/r ≡ b (p−1)/r x ≡ A k x (mod x r − b, p)
for some integer k. Thus, since (Ak ) p ≡ Ak (mod p), we have by induction
that
x pi
≡ A ik x (mod x r − b, p) (4.33)
for every nonnegative integer i. Wehavef(x) pi
Zp[x], so that by (4.32) and (4.33), we have
= f(xpi) for every f(x) ∈
(x − 1) pin j
≡ (A j x − 1) pi
≡ A j x pi
− 1 ≡ A j+ik x − 1(modx r − b, p)