A384948 Primes p == 3 (mod 4) such that 5 is a primitive root of integers modulo p, but 2+-i are not primitive roots of Gaussian integers modulo p.

83, 307, 347, 503, 587, 863, 947, 1103, 1223, 1523, 1567, 1667, 1787, 1907, 2063, 2087, 2267, 2663, 2683, 2687, 2903, 2963, 3167, 3343, 3347, 3623, 3803, 3863, 4283, 4463, 4523, 4643, 4967, 5147, 5303, 5387, 5507, 5563, 5807, 5843, 6047, 6203, 6607, 6863, 6983, 7187, 7247, 7523, 7583

Offset: 1

Author

Jianing Song, Jun 20 2025

Keywords

• nonn
• easy

Comments

For p = A002145(k), A385165(k) divides (p+1) * ord(5,p), since we have (2+-i)^(p+1) == 5 (mod p). Hence if 2+-i are primitive roots of Gaussian integers modulo p, then 5 is a primitive root of integers modulo p. This sequence lists p such that the converse does not hold.

Examples

			5 is a primitive root modulo 83, but the multiplicative order of 2+-i modulo 83 in Gaussian integers is not 83^2 - 1 = 6888; it is 2296 = 6888/3. In other words, 2+-i are not generators of (Z[i]/83Z[i])*.
		

Links

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

Crossrefs

By definition, subsequence of A019335, A122870, and A385168.

Programs

PARI

isprim(p) = my(f = factor(p^2-1)[,1]~); for(i=1, #f, if(Mod([2, -1; 1, 2], p)^((p^2-1)/f[i]) == 1, return(0))); return(1) \\ for a prime p == 3 (mod 4), determines if 2+-i are primitive roots modulo p
isA384948(p) = isprime(p) && (p%4==3) && znorder(Mod(5,p))==p-1 && !isprim(p)