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A385179 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is congruent to 2 modulo 4.

%I A385179 #23 Jun 22 2025 21:32:32
%S A385179 11,131,211,251,491,811,919,1039,1091,1319,1399,1531,1811,1931,2011,
%T A385179 2251,2371,2531,2731,2851,3011,3079,3371,3931,4079,4451,4519,4759,
%U A385179 5011,5639,6091,6131,6211,6359,6451,6491,6571,6971,7411,7451,7559,7639,8291,8719,8971,9091,9491,9719,9839,9851,9931
%N A385179 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is congruent to 2 modulo 4.
%C A385179 Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
%C A385179 Let ord(a,m) be the multiplicative order of a modulo m. (Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i]). For a prime p == 3 (mod 4), we have that ord(2+-i,p) is divisible by ord(5,p), and that ord(2+-i,p) divides (p+1) * ord(5,p). What's more, ord(2+-i,p) divides (p^2-1)/2 if and only if 5 is a quadratic residue of integers modulo p. (See A385165).
%C A385179 As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:
%C A385179   - Of course this is true if ord(2+-i,p) is odd.
%C A385179   - If ord(2+-i,p) == 2 (mod 4) and ord(5,p) is even, then ord(2+-i,p)/ord(5,p) is odd, and so ord(2+-i,p) divides ((p+1)/4) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) is odd, a contradiction.
%C A385179   - If ord(2+-i,p) == 4 (mod 8) and ord(5,p) is even (we have ord(5,p) == 2 (mod 4) since p == 3 (mod 4)), then ord(2+-i,p)/ord(5,p) == 2 (mod 4), and so ord(2+-i,p) divides ((p+1)/2) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) == 2 (mod 4), a contradiction.
%C A385179 From the above paragraph, this sequence is also primes p == 3 (mod 4) such that ord(2+-i,p)/ord(5,p) == 2 (mod 4).
%H A385179 Jianing Song, <a href="/A385179/b385179.txt">Table of n, a(n) for n = 1..10000</a>
%e A385179 919 is a term since (2+-i)^21114 == 1 (mod 919), (2+-i)^(21114/2) !== 1 (mod 919), and we have 21114 == 2 (mod 4).
%o A385179 (PARI) ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
%o A385179 isA385179(p) = isprime(p) && p%4==3 && ord(p)%4==2
%Y A385179 Cf. A385165, A385169, A385188, A385218 (the actual multiplicative orders).
%Y A385179 Subsequence of A385167, which itself lies in the intersection of A122869 and A385168.
%K A385179 nonn,easy
%O A385179 1,1
%A A385179 _Jianing Song_, Jun 20 2025