Problems & Puzzles: Conjectures

Conjecture 70. n & n+1 primitive roots

Emmanuel Vantieghem sent the following conjecture:

If  p  is a prime number > 7, then there exist an integer  n  such that  n  and  n+1  both are primitive roots modulo p.

I have verified the truth of that statement for all  p  less than the  20000000-th prime.

Examples : (consecutive numbers that are primitive roots are underlined)
p = 7 ; the primitive roots (mod 7) are 3 ,5
p = 11 ; the primitive roots (mod 11) are  2, 6, 7, 8
p = 13 ; the primitive roots (mod 13) are  2, 6, 7, 11
p = 17 ; the primitive roots (mod 17) are  3, 5, 6, 710, 11, 12, 14
p = 19 ; the primitive roots (mod 19) are  2, 3, 10, 13, 14, 15
p = 23 ; the primitive roots (mod 23) are  5, 7, 10, 1114, 15, 17, 19, 20, 21
p = 29 ; the primitive roots (mod 29) are  2, 3, 8, 10, 1114, 1518, 19, 21, 26, 27

Q1. Can you probe this conjecture of find a counterexample?

Jan van Delden wrote:

Just a small observation:

x is a primitive root modulo p: x^a=1 mod p if and only if a=phi(p)=p-1. Or to put it differently the order of x mod p is maximal.
Furthermore the order of a general x mod p is a divisor of p-1.

Now (x+1)^a mod p = sum (x^b C(a,b) mod p, b=0..a) with C(a,b) the number of combinations (binomium of Newton).

I see no fast way to reduce such an expression to a simple form, what’s worse, the terms (x+1)^a mod p will in general be different from x^a mod p.

However a simple case would be considering x and p-x:

(p-x)^a mod p = (-1)^a *x^a mod p:
=   x^a mod p if a is even
= –x^a mod p if a is odd

Choose p such that we have p-x=x+1: p=2x+1, i.e. x=(p-1)/2 and x+1=(p+1)/2, let’s suppose x is a primitive root mod p.

We have (x+1)^a = x^a both unequal to 1 mod p if a|(p-1) and a even, both equal to 1 mod p if a=p-1, because x is a primitive root mod p.

We therefore need x^a<>-1 mod p, for all odd a|(p-1) or for all odd a|x which is the same thing.

We always have x^x=-1 mod p (if x is a primitive root mod p). So we should at least have that this a is a proper divisor of x, or in other words that x has another divisor 2. Hence p=1 mod 4 is necessary. It is also sufficient because if x^a=-1 mod p with a<=x/2, then x^(2a)=1 mod p but 2a<=x<p-1 in contradiction with x being a primitive root.

So we have:

If p=1 mod 4 and x=(p-1)/2 is a primitive root mod p  then x and x+1 are primitive roots mod p.

p=5,13,29,37,53,61,101,149,173,181,197,269,..

Maybe some similar observations can be made for other situations.

Emmanuel wrote:

Immediately after the appearance of this conjecture, W. Edwin Clark sent me a mail to tell me that, by a theorem of M. Szalay, the conjecture is true for all primes  p > 10^19 (M. Szalay, On the distribution of the primitive roots of a prime. Journal of Number Theory 7 (1975), 184-188).  So, there remains only a finite number of primes to be checked by hand.  However, I do not believe that this can be done in reasonable time.

For some primes, it is possible to prove the conjecture.  For instance, if   is of the form  2q+1, q prime > 5.  The number of primitive roots modulo p is then precisely  q-1.  If we take  q-1  numbers between 2  and  p-2  with minimal consecutive distances > 1, we should have to choose   or   among them.  But these are not primitive roots.

There is a similar but stronger result of E. Vegh, that confirms the conjecture for primes   such that  phi(p-1)/(p-1) > 1/4, phi(-) being Eulers totient function  (E. Vegh, A note on the distribution of the primitive roots of a prime, J of  Number Theory, 3 (1971), 13-18).

Here is a heuristic argument for the conjecture :

Assume   prime (not too small)  and take ad random  j = phi(p-1)  integers  r(i)  such that  1 <  r(1) < r(2) <  ... < r(j) < p-1.  Then, on the average, the frequency table for the differences  r(i+1)-r(i)  shows a remarkable maximum at the beginning, i. e. : the equality r(i+1)-r(i) = 1 occurs most often.  Now, if   is a primitive root mod p, then it is well known (and used sometimes by computer simulation of randomness !) that the sequence  g^1, g^2, ... ,g^(p-1) modulo p behaves as if it was chosen ad random.  So, if we restrict ourselves to powers of   that are relatively prime to  p-1, we get the sequence of primitive roots modulo p.  So, that sequence can be considered as a random one, and thus ...

Actually, the number of differences   for the primitive roots seems to be much greater than the differences 1 for random sequences, but that may be coincidence.  Anyhow, the fact that the conjecture is true for  p > 10^19  shows that the heuristic argument is not entirely phantasy (at least, I hope so).

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