Prime Numbers

146 Chapter 3 RECOGNIZING PRIMES AND COMPOSITES
and the former, via x(a − x) ≡ b (mod (f(x),n)), leads to xn ≡ a − x
(mod (f(x),n)). Thus, n is a Frobenius pseudoprime with respect to f(x).
Now suppose n is a Frobenius pseudoprime with respect to f(x). Exercise
3.27 shows that n is a Lucas pseudoprime with respect to f(x), namely
that U ∆ n−( ≡ 0(modn). Thus, from the identity above, 2xn−(∆n)
≡
n)
V ∆ n−( n) (mod (f(x),n)). Suppose
∆
n+1
n = −1. Then x ≡ (a − x)x ≡ b
(mod (f(x),n)), so that Vn+1 ≡ 2b (mod n). Finally, suppose
∆
n =1.Then
since x is invertible modulo (f(x),n), we have xn−1 ≡ 1(mod(f(x),n)),
which gives Vn−1 ≡ 2(modn). ✷
The first Frobenius pseudoprime n with respect to x2 − x − 1 is 4181 (the
nineteenth Fibonacci number), and the first with
5
n = −1 is 5777. We thus
see that not every Lucas pseudoprime is a Frobenius pseudoprime, that is, the
Frobenius test is more stringent. In fact, the Frobenius pseudoprime test can
be very effective. For example, for x2 +5x + 5 we don’t know any examples
at all of a Frobenius pseudoprime n with
5
n = −1, though such numbers are
conjectured to exist; see Exercise 3.42.
3.6.3 Implementing the Lucas and quadratic Frobenius tests
It turns out that we can implement the Lucas test in about twice the time
of an ordinary pseudoprime test, and we can implement the Frobenius test in
about three times the time of an ordinary pseudoprime test. However, if we
approach these tests naively, the running time is somewhat more than just
claimed. To achieve the factors two and three mentioned, a little cleverness is
required.
As before, we let a, b be integers with ∆ = a 2 − 4b not a square, and
we define the sequences (Uj), (Vj) as in (3.8). We first remark that it is
easy to deal solely with the sequence (Vj). If we have Vm and Vm+1, wemay
immediately recover Um via the identity
Um =∆ −1 (2Vm+1 − aVm). (3.13)
We next remark that it is easy to compute Vm for large m from earlier values
using the following simple rule: If 0 ≤ j ≤ k, then
Vj+k = VjVk − b j Vk−j. (3.14)
Suppose now that b = 1. We record the formula (3.14) in the special cases
k = j and k = j +1:
V2j = V 2
j − 2, V2j+1 = VjVj+1 − a (in the case b =1). (3.15)
Thus, if we have the residues Vj (mod n), Vj+1 (mod n), then we may
compute, via (3.15), either the pair V2j (mod n), V2j+1 (mod n) or the pair
V2j+1 (mod n), V2j+2 (mod n), with each choice taking 2 multiplications
modulo n and an addition modulo n. Starting from V0,V1 we can recursively