A385190 Primes p == 3 (mod 4), p > 3 such that 1+-i are 24th powers modulo p.

31, 127, 191, 223, 383, 479, 863, 1151, 1439, 1471, 1823, 2111, 2143, 2207, 2399, 2591, 2687, 2879, 3167, 3359, 3391, 4127, 4703, 4799, 5087, 5279, 5471, 5503, 6047, 6079, 6143, 6271, 6719, 6911, 7103, 7487, 7583, 8191, 8287, 8447, 8543, 8831, 8863, 9311, 9439, 9631, 9791, 9887

Offset: 1

Author

Jianing Song, Jun 20 2025

Keywords

• nonn
• easy

Comments

Note that the primes congruent to 3 modulo 4 are precisely the rational primes in the ring of Gaussian integers.

Primes p == 3 (mod 4), p > 3 such that (1+-i)^((p^2-1)/24) == 1 (mod p). Note that p^2-1 is always divisible by 24 for primes p > 3.

Primes p = A002145(k) such that the multiplicative order of 1+-i modulo p (A385163(k)) divides (p^2-1)/24. Since A385165(k) = 4*ord(-4,p), this is also primes p == 3 (mod 4) such that 96*ord(-4,p) divides p^2-1, where ord(a,p) is the multiplicative order of a modulo p.

Sequence is infinite since it contains all primes congruent to 95 modulo 96.

Primes p == 3 (mod 4), p > 3 such that [1,-1;1,1]^((p^2-1)/24) or [1,1;-1,1]^((p^2-1)/24) == I_2 (mod p).

Since 96 divides p^2-1 for p being a term of this sequence, we must have p == 15 (mod 16).

Examples

			31 is a term since (1+-i)^((31^2-1)/24) = (-4)^((31^2-1)/96) = 1048576 == 1 (mod 31). Indeed, the solutions to x^24 == 1+i (mod 31) are x == {17-6*i, 16+6*i, 1+8*i, -1+13*i, 9-5*i, 3+5*i} X {+-1, +-i} (mod 31).
		

Links

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

Crossrefs

Cf. A385163, A385191 (2+-i are 24th powers), A002145.

Programs

PARI

isA385190(p) = isprime(p) && p%16==15 && Mod(-4,p)^((p^2-1)/96) == 1