Prime Numbers

2.5 Research problems 113
discriminant. For the case 1 +
D
p > 0, give an algorithm for calculation of
all the roots.
2.38. Find a prime p such that the least primitive root of p exceeds the
number of binary bits in p. Find an example of such a prime p that is also
a Mersenne prime (i.e., some p = Mq =2 q − 1 whose least primitive root
exceeds q). These findings show that the least primitive root can exceed lg p.
For more exploration along these lines see Exercise 2.39.
2.5 Research problems
2.39. Implement a primitive root-finding algorithm, and study the statistical
occurrence of least primitive roots.
The study of least primitive roots is highly interesting. It is known on
the GRH that 2 is a primitive root of infinitely many primes, in fact for a
positive proportion α = (1 − 1/p(p − 1)) ≈ 0.3739558, the product running
over all primes (see Exercise 1.90). Again on the GRH, a positive proportion
whose least primitive root is not 2, has 3 as a primitive root and so on;
see [Hooley 1976]. It is conjectured that the least primitive root for prime
p is O((ln p)(ln ln p)); see [Bach 1997a]. It is known, on the GRH, that the
least primitive root for prime p is O ln 6 p ; see [Shoup 1992]. It is known
unconditionally that the least primitive root for prime p is O(p 1/4+ɛ ) for
every ɛ>0, and for infinitely many primes p it exceeds c ln p ln ln ln p for
some positive constant c, the latter a result of S. Graham and C. Ringrosee.
The study of the least primitive root is not unlike the study of the least
quadratic nonresidue—in this regard see Exercise 2.41.
2.40. Investigate the use of CRT in the seemingly remote domains of integer
convolution, or fast Fourier transforms, or public-key cryptography. A good
reference is [Ding et al. 1996].
2.41. Here we explore what might be called “statistical” features of the
Legendre symbol. For odd prime p, denote by N(a, b) the number of residues
whose successive quadratic characters are (a, b); that is, we wish to count
those integers x ∈ [1,p− 2] such that
x x +1
, =(a, b),
p p
with each of a, b attaining possible values ±1. Prove that
and therefore that
p−2
x
x +1
4N(a, b) = 1+a 1+b
p
p
x=1
N(a, b) = 1
4
p − 2 − b − ab − a
−1
p
.