Prime Numbers

182 Chapter 4 PRIMALITY PROVING
Recall that the polynomials Uk,Vk do not have positive degree; that is, they
are integers.
Definition 4.2.1. With the above notation, if n is a positive integer with
gcd(n, 2b∆) = 1, the rank of appearance of n, denoted by rf (n), is the least
positive integer r with Ur ≡ 0(modn).
This concept sometimes goes by the name “rank of apparition,” but according
to Ribenboim, this is due to a mistranslation of the French apparition. There
is nothing ghostly about the rank of appearance!
It is apparent from the definition (4.13) that (Uk) is a “divisibility
sequence,” that is, if k|j then Uk|Uj. (We allow the possibility that Uk =
Uj = 0.) It follows that if gcd(n, 2b∆) = 1, then Uj ≡ 0(modn) if and only if
j ≡ 0(modrf (n)). On the basis of Theorem 3.6.3 we thus have the following
result:
Theorem 4.2.2. With f,∆ as in (4.12) and p a prime not dividing 2b∆,
we have rf (p)|p −
∆
p .
(Recall the Legendre symbol
·
p from Definition 2.3.2.)
In analogy to Theorem 4.1.3, we have the following result:
Theorem 4.2.3 (Morrison). Let f,∆ be as in (4.12) and let n be a positive
integer with gcd(n, 2b) =1,
∆
n = −1. IfF is a divisor of n +1 and
Un+1 ≡ 0(modn), gcd(U (n+1)/q,n)=1for every prime q|F, (4.14)
then every prime p dividing n satisfies p ≡
∆
p (mod F ). In particular, if
F> √ n +1 and (4.14) holds, then n is prime.
(Recall the Jacobi symbol ·
n
from Definition 2.3.3.)
Proof. Let p be a prime factor of n. Then (4.14) implies that F divides rf (p).
So, by Theorem 4.2.2, p ≡
∆
√
p (mod F ). If, in addition, we have F> n +1,
then every prime factor p of n has p ≥ F − 1 > √ n,son is prime. ✷
If Theorem 4.2.3 is to be used in a primality test, we will need to find an
appropriate f in (4.12). As with Algorithm 4.1.7 where a is chosen at random,
we may choose a, b in (4.12) at random. When we start with a prime n, the
expected number of choices until a successful pair is found is not large, as the
following result indicates.
Theorem 4.2.4. Let p be an odd prime and let N be the number of pairs
a, b ∈{0, 1,...,p− 1} such that if f,∆ are given as in (4.12), then
∆
p = −1
and rf (p) =p +1. Then N = 1
2 (p − 1)ϕ(p +1).
We leave the proof as Exercise 4.12. A consequence of Theorem 4.2.4 is that
if n is an odd prime and if a, b are chosen randomly in {0, 1,...,n− 1} with
not both 0, then the expected number of choices until one is found where the