Prime Numbers

4.2 The n + 1 test 183
f in (4.12) satisfies rf (n) =n +1 is 2(n +1)/ϕ(n + 1). If n>892271479, then
this expected number of choices is less than 4 ln ln n; see Exercise 4.16.
It is also possible to describe a primality test using the V sequence in
(4.13).
Theorem 4.2.5. Let f,∆ be as in (4.12) and let n be a positive integer with
gcd(n, 2b) =1and
∆
n = −1. IfF is an even divisor of n +1 and
V F/2 ≡ 0(modn), gcd(V F/2q,n)=1for every odd prime q|F, (4.15)
then every prime p dividing n satisfies p ≡
∆
p (mod F ). In particular, if
F> √ n +1, then n is prime.
Proof. Suppose p is an odd prime that divides both Um,Vm. Then (4.13)
implies x m ≡ (a − x) m (mod (f(x),p)) and x m ≡−(a − x) m (mod f(x),p),
so that x m ≡ 0(mod(f(x),p)). Then b m ≡ (x(a − x)) m ≡ 0(mod(f(x),p));
that is, p divides b. Sincen is coprime to 2b, and since U2m = UmVm, wehave
gcd(U2m,n)=gcd(Um,n) · gcd(Vm,n).
Thus, the first condition in (4.15) implies UF ≡ 0(modn) and gcd(U F/2,n)=
1. Now suppose q is an odd prime factor of F .WehaveU F/q = U F/2qV F/2q
coprime to n. Indeed, U F/2q divides U F/2, sothatgcd(U F/2q,n) = 1, and so
with the second condition in (4.15) we have that gcd(U F/q,n) = 1. Thus,
rf (p) =F , and as in the proof of Theorem 4.2.3, this is sufficient for the
conclusion. ✷
Just as the n − 1 is particularly well suited for Fermat numbers, the n +1
test is especially speedy for Mersenne numbers.
Theorem 4.2.6 (Lucas–Lehmer test for Mersenne primes). Consider the
sequence (vk) for k =0, 1,...,recursively defined by v0 =4and vk+1 = v 2 k −2.
Let p be an odd prime. Then Mp =2 p − 1 is prime if and only if vp−2 ≡ 0
(mod Mp).
Proof. Let f(x) =x2 − 4x + 1, so that ∆ = 12. Since Mp ≡ 3(mod4)
and Mp ≡ 1 (mod 3), we see that
∆ = −1. We apply Theorem 4.2.5 with
Mp
F =2p−1 =(Mp +1)/2. The conditions (4.15) reduce to the single condition
V2p−2 ≡ 0(modMp). But
V2m ≡ x 2m +(4−x) 2m =(x m +(4−x) m ) 2 −2x m (4−x) m ≡ V 2 m−2 (modf(x)),
since x(4 − x) ≡ 1(modf(x)); see (3.15). Also, V1 =4.Thus,V2k = vk, and
it follows from Theorem 4.2.5 that if vp−2 ≡ 0(modMp), then Mp is prime.
Suppose, conversely, that M = Mp is prime. Since
∆
M = −1,
Z[x]/(f(x),M) is isomorphic to the finite field FM 2. Thus, raising to the M
power is an automorphism and xM ≡ 4 − x (mod (f(x),M)); see the proof of
Theorem 3.6.3. We compute (x − 1) M+1 two ways. First, since (x − 1) 2 ≡ 2x