A364867 Primes p such that the multiplicative order of 9 modulo p is (p-1)/2.

5, 7, 11, 17, 19, 23, 29, 31, 43, 47, 53, 59, 71, 79, 83, 89, 101, 107, 113, 127, 131, 137, 139, 149, 163, 167, 173, 179, 191, 197, 199, 211, 223, 227, 233, 239, 251, 257, 263, 269, 281, 283, 293, 311, 317, 331, 347, 353, 359, 379, 383, 389, 401, 419, 443, 449, 461, 463, 467, 479, 487

Offset: 1

Author

Jianing Song, Aug 11 2023

Keywords

• nonn
• easy

Comments

Primes p such that the multiplicative order of 9 modulo p is of the maximum possible value.

Primes p such that 3 or -3 (or both) is a primitive root modulo p. Proof of equivalence: let ord(a,k) be the multiplicative order of a modulo p. if ord(3,p) = p-1, then clearly ord(9,p) = (p-1)/2. If ord(-3,p) = p-1, then we also have ord(9,p) = (p-1)/2. Conversely, suppose that ord(9,p) = (p-1)/2, then ord(3,p) = p-1 or (p-1)/2, and ord(-3,p) = p-1 or (p-1)/2. If ord(3,p) = ord(-3,p) = (p-1)/2, then we have that (p-1)/2 is odd and (-1)^((p-1)/2) == 1 (mod p), a contradiction.

A prime p is a term if and only if one of the two following conditions holds: (a) 3 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of 3 modulo p is (p-1)/2 (in this case, we have p == 11 (mod 12) since 3 is a quadratic residue modulo p).

A prime p is a term if and only if one of the two following conditions holds: (a) -3 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of -3 modulo p is (p-1)/2 (in this case, we have p == 7 (mod 12) since -3 is a quadratic residue modulo p).

No terms are congruent to 1 modulo 12, since otherwise we would have 9^((p-1)/4) = (+-3)^((p-1)/2) == 1 (mod p). - Jianing Song, May 14 2024

Examples

			7 is a term since the multiplicative order of 9 modulo 7 is 3 = (7-1)/2.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000

Crossrefs

Union of A019334 and A105875.

A105881 is the subsequence of terms congruent to 3 modulo 4.

Cf. A211245 (order of 9 mod n-th prime), A216371.

Programs

Mathematica

okQ[p_] := MultiplicativeOrder[9, p] == (p - 1)/2;
Select[Prime[Range[100]], okQ] (* Jean-François Alcover, Nov 24 2024 *)

PARI

isA364867(p) = isprime(p) && (p!=3) && znorder(Mod(9, p)) == (p-1)/2

Python

from sympy import n_order, nextprime
from itertools import islice
def A364867_gen(startvalue=4): # generator of terms >= startvalue
    p = max(startvalue-1,3)
    while (p:=nextprime(p)):
        if n_order(9,p) == p-1>>1:
            yield p
A364867_list = list(islice(A364867_gen(),20)) # Chai Wah Wu, Aug 11 2023