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.

11, 131, 211, 251, 491, 811, 919, 1039, 1091, 1319, 1399, 1531, 1811, 1931, 2011, 2251, 2371, 2531, 2731, 2851, 3011, 3079, 3371, 3931, 4079, 4451, 4519, 4759, 5011, 5639, 6091, 6131, 6211, 6359, 6451, 6491, 6571, 6971, 7411, 7451, 7559, 7639, 8291, 8719, 8971, 9091, 9491, 9719, 9839, 9851, 9931

Offset: 1

Author

Jianing Song, Jun 20 2025

Keywords

• nonn
• easy

Comments

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.

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).

As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:

- Of course this is true if ord(2+-i,p) is odd.

- 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.

- 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.

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).

Examples

			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).
		

Links

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

Crossrefs

Cf. A385165, A385169, A385188, A385218 (the actual multiplicative orders).

Subsequence of A385167, which itself lies in the intersection of A122869 and A385168.

Programs

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)
isA385179(p) = isprime(p) && p%4==3 && ord(p)%4==2