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

1,1

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).

Links

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

Example

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).

Mathematica

q[p_] := Mod[p, 16] == 15 && PowerMod[-4, (p^2 - 1)/96, p] == 1; Select[Prime[Range[1220]], q] (* Amiram Eldar, Mar 28 2026 *)

Prog

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

Crossrefs

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

Sequence in context: A078656 A095322 A127578 * A333245 A158563 A079141

Adjacent sequences: A385187 A385188 A385189 * A385191 A385192 A385193

Keyword

nonn,easy

Author

Jianing Song, Jun 20 2025

Status

approved