Prime Numbers

3.6 Lucas pseudoprimes 151
Theorem 3.6.10. Suppose n is a composite number that is not a square
and not divisible by any prime up to 50000. Then n is a strong Frobenius
pseudoprime with respect to at most 1/7710 of all polynomials x2 − ax + b,
where a, b run over the integers in [1,n] with a 2
−4b
b
n = −1 and n =1.
This result should be contrasted with the Monier–Rabin theorem
(Theorem 3.5.4). If one does three random strong pseudoprime tests, that
result implies that a composite number will fail to be recognized as such at
most 1/64 of the time. Using Theorem 3.6.10, in about the same time, one has
a test that recognizes composites with failure at most 1/7710 of the time. A
recent test in [Zhang 2002] should be mentioned in this context. It combines
a strong probable prime test and a Lucas test, giving a result that is superior
to the quadratic Frobenius test in all but a thin set of cases.
3.6.5 The general Frobenius test
In the last few sections we have discussed Grantham’s Frobenius test for
quadratic polynomials. Here we briefly describe how the idea generalizes to
arbitrary monic polynomials in Z[x].
Let f(x) be a monic polynomial in Z[x] withdegreed ≥ 1. We do not
necessarily assume that f(x) is irreducible. Suppose p is an odd prime that
does not divide the discriminant, disc(f), of f(x). (The discriminant of a
monic polynomial f(x) ofdegreed may be computed as (−1) d(d−1)/2 times
the resultant of f(x) and its derivative. This resultant is the determinant of
the (2d−1)×(2d−1) matrix whose i, j entry is the coefficient of x j−i in f(x)for
i =1,...,d−1 and is the coefficient of x j−(i−d+1) in f ′ (x) fori = d,...,2d−1,
where if the power of x does not actually appear, the matrix entry is 0.) Since
disc(f) = 0 if and only if f(x) has no repeated irreducible factors of positive
degree, the hypothesis that p does not divide disc(f) automatically implies
that f has no repeated factors.
By reducing its coefficients modulo p, we may consider f(x) inFp[x].
To avoid confusion, we shall denote this polynomial by f(x). Consider the
polynomials F1(x),F2(x),...,Fd(x) inFp[x] defined by
F1(x) =gcd(x p − x, f(x)),
F2(x) =gcd(x p2
− x, f(x)/F1(x)),
.
Fd(x) =gcd(x pd
− x, f(x)/(F1(x) ···Fd−1(x))).
Then the following assertions hold:
(1) i divides deg(Fi(x)) for i =1,...,d,
(2) Fi(x) divides Fi(x p )fori =1,...,d,
(3) for
S =
i even
1
i deg(Fi(x)),