Prime Numbers

4.1 The n − 1 test 175
one might wonder whether any use can be made of a partial factorization of
n − 1. In particular, say
n − 1=FR, and the complete prime factorization of F is known. (4.2)
If F is fairly large as a function of n, we may fashion a primality proof for n
along the lines of (4.1), if indeed n happens to be prime. Our first result on
these lines allows us to deduce information on the prime factorization of n.
Theorem 4.1.3 (Pocklington). Suppose (4.2) holds and a is such that
a n−1 ≡ 1(modn) and gcd(a (n−1)/q − 1,n)=1for each prime q|F. (4.3)
Then every prime factor of n is congruent to 1(modF ).
Proof. Let p be a prime factor of n. From the first part of (4.3) we have that
the order of a R in Z ∗ p is a divisor of (n − 1)/R = F . From the second part
of (4.3) it is not a proper divisor of F , so is equal to F . Hence F divides the
order of Z ∗ p,whichisp − 1. ✷
Corollary 4.1.4. If (4.2) and (4.3) hold and F ≥ √ n, then n is prime.
Proof. Theorem 4.1.3 implies that each prime factor of n is congruent to 1
(mod F ), and so each prime factor of n exceeds F .ButF ≥ √ n, so each
prime factor of n exceeds √ n,son must be prime. ✷
The next result allows a still smaller value of F .
Theorem 4.1.5 (Brillhart, Lehmer, and Selfridge). Suppose (4.2) and
(4.3) both hold and suppose that n 1/3 ≤ F < n 1/2 . Consider the base F
representation of n, namely n = c2F 2 + c1F +1, where c1, c2 are integers in
[0,F − 1]. Then n is prime if and only if c 2 1 − 4c2 is not a square.
Proof. Since n ≡ 1(modF ), it follows that the base-F “units” digit of n
is 1. Thus n has its base-F representation in the form c2F 2 + c1F +1, as
claimed. Suppose n is composite. From Theorem 4.1.3, all the prime factors
of n are congruent to 1 (mod F ), so must exceed n 1/3 . We conclude that n
has exactly two prime factors:
We thus have
n = pq, p = aF +1, q = bF +1, a ≤ b.
c2F 2 + c1F +1=n =(aF + 1)(bF +1)=abF 2 +(a + b)F +1.
Our goal is to show that we must have c2 = ab and c1 = a + b, for then it will
follow that c 2 1 − 4c2 is a square.
First note that F 3 ≥ n>abF 2 ,sothatab ≤ F − 1. It follows that either
a+b ≤ F −1ora =1,b= F −1. In the latter case, n =(F +1)((F −1)F +1) =
F 3 + 1, contradicting F ≥ n 1/3 . Hence both ab and a + b are positive integers