REMARK ON A PAPER “PRIMITIVE ROOTS SATISFYING A
COPRIME CONDITION”
R. THANGADURAI
Let p be an odd prime. It has been conjectured [3] that there exists an integer
g ≤ p − 1 which is a primitive root modulo p and which is relatively prime to
p − 1. In 1976, M. Hausman [1] proved this conjecture for all primes p sufficiently
large and mentioned that the method that he adopted does not provide the value
of p from which point on the conjecture is true. In this remark, we prove the
following.
Theorem 1. Let p be a prime number such that p > e110.8 ∼ 1.318 × 1048 . Then
there exists an integer 1 < g ≤ p − 1 such that g is a primitive root modulo p and
(g, p − 1) = 1.
To prove this we need the following Lemma.
Lemma 1. (i) For any integer n > 90, we have
n
φ(n) >
,
log n
where φ(n) is the Euler Phi-function.
(ii) Let ω(n) denote the number of distinct prime divisors of n. Then we have
ω(p − 1) ≤ (1.385)
log p
log log p
for all primes p ≥ 5.
The first result in Lemma was proved by L. Moser [2] in 1951 and the second
result can be seen in [4], page 167.
To prove his result, M. Hausman [1] proved the following intermediate result.
Theorem 2. For any prime p, let Np denote the number of integers 1 < g < p−1
which are primitive roots modulo p and which are coprime to p−1. Then he proved
that
φ2 (p − 1) φ(p − 1)
Np =
+
Ep
p−1
p−1
where
√
|Ep | ≤ 4ω(p−1) p(log p).
1
2
R. THANGADURAI
Proof of Theorem 1. By Lemma 1 (ii), we see that
(1)
4ω(p−1) ≤ 4(1.385)(log p)/(log log p) < (6.83)log p/ log log p = p(log 6.83)/(log log p) .
Let > 0 be such that 0 < < 1/2. Then for all primes
2 log 6.83
p ≥ exp exp
,
1 − 2
we have
(2)
1
4ω(p−1) < p 2 − ,
which is an easy computation from (1) and (2). Therefore, Np ≥ 1 follows at
once, if we prove that
φ2 (p − 1)
φ(p − 1) 1−
>
p log p;
p−1
p−1
That is to prove that
φ(p − 1) > p1− (log p)
for all primes p satisfying
2 log 6.83
p > exp exp
.
1 − 2
Instead, by Lemma 1 (i), if we prove
p−1
> p1− log p,
log(p − 1)
then we are done. Indeed, the last inequality is same as
1
p > (log p)(log(p − 1)) + 1− .
p
Hence, if we can prove that
p > (log(p − 1) + 1)2/
holds good for a suitable choice of , then we are done. Choose = 1/11. Then
we get
2 log 6.83
exp exp
= exp exp(log(6.83)2.45 ) = exp((6.83)2.45 ) < e110.8 .
1 − 2
Choose primes p > e110.8 . Then p − 1 ≥ e110.8 . When n = e110.8 + 1,
log (log(n − 1) + 1)2/ = 22 log(111.8) ∼ 103.768.
Hence,
log n > 110.8 > 22(log 111.8) ∼ 103.768.
Therefore, for all primes p > e110.8 , we have
p−1
φ(p − 1) >
> p10/11 log p.
log(p − 1)
Therefore, Np ≥ 1 for all p > e110.8 . This completes the proof of Theorem 1.
REMARK ON A PAPER “PRIMITIVE ROOTS SATISFYING A COPRIME CONDITION” 3
A particular case of Theorem 1 says that for every prime p > e110.8 , there exists
an odd integer g ≤ p − 1 which is a primitive root modulo p. In Gauss’ Theorem,
while proving a primitive root g modulo p is also a primitive root modulo 2p, one
needs the fact that g is an odd integer. To avoid this situation, one generally do
the following. If g is even, then p + g is odd and is a primitive root modulo p.
However, by Theorem 1, there exists an odd integer g ≤ p − 1 which is a primitive
root modulo p.
Acknowledgement. I am thankful to Prof. M. Ram Murty for having very
useful discussion.
References
[1] M. Hausman, Primitive roots satisfying a coprime condition, this Monthly, 83 (1976)
720-723.
[2] L. Moser, On the equation φ(n) = π(n), Pi Mu Epsilon J., (1951), 101-110.
[3] Problems and Solutions, Problem E-2488, this Monthly, 81 (1974) 776.
[4] J. Sándor, D. S. Mitrinović and B. Crstici, Handbook on Number Theory I, Springer, The
Netherlands.
Harish-Chandra Research Institute, Chhatnag Road, Jhunsi, Allahabad 211010,
INDIA
E-mail address: [email protected]