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

1,1

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.

Links

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

Example

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])*.

Prog

(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)

Crossrefs

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

Sequence in context: A251082 A141570 A334644 * A288172 A341338 A031433

Adjacent sequences: A384945 A384946 A384947 * A384949 A384950 A384951

Keyword

nonn,easy

Author

Jianing Song, Jun 20 2025

Status

approved