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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.

%I A384948 #28 Jun 22 2025 17:36:30
%S A384948 83,307,347,503,587,863,947,1103,1223,1523,1567,1667,1787,1907,2063,
%T A384948 2087,2267,2663,2683,2687,2903,2963,3167,3343,3347,3623,3803,3863,
%U A384948 4283,4463,4523,4643,4967,5147,5303,5387,5507,5563,5807,5843,6047,6203,6607,6863,6983,7187,7247,7523,7583
%N 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.
%C A384948 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.
%H A384948 Jianing Song, <a href="/A384948/b384948.txt">Table of n, a(n) for n = 1..10000</a>
%e A384948 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])*.
%o A384948 (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
%o A384948 isA384948(p) = isprime(p) && (p%4==3) && znorder(Mod(5,p))==p-1 && !isprim(p)
%Y A384948 By definition, subsequence of A019335, A122870, and A385168.
%K A384948 nonn,easy
%O A384948 1,1
%A A384948 _Jianing Song_, Jun 20 2025