cp's OEIS Frontend

A385191 Primes p == 3 (mod 4), p > 3 such that 2+-i are 24th powers modulo p.

%I A385191 #11 Jun 21 2025 00:45:49
%S A385191 599,691,1039,1291,1451,1759,2411,2879,3079,3491,3851,4519,4639,4919,
%T A385191 5051,5479,5519,5531,5639,5879,6011,6079,6599,6719,7079,7691,8011,
%U A385191 8039,8171,8731,9439,9839,10799,11159,11239,11411,11491,12239,12799,13291,13679,13759,13879,14011,14639
%N A385191 Primes p == 3 (mod 4), p > 3 such that 2+-i are 24th powers modulo p.
%C A385191 Note that the primes congruent to 3 modulo 4 are precisely the rational primes in the ring of Gaussian integers.
%C A385191 Primes p == 3 (mod 4), p > 3 such that (2+-i)^((p^2-1)/24) == 1 (mod p). Note that p^2-1 is always divisible by 24 for primes p > 3.
%C A385191 Primes p = A002145(k) > 3 such that the multiplicative order of 2+-i modulo p (A385165(k)) divides (p^2-1)/24.
%C A385191 Primes p == 3 (mod 4), p > 3 such that [2,-1;1,2]^((p^2-1)/24) or [2,1;-1,2]^((p^2-1)/24) == I_2 (mod p).
%C A385191 Note that if (x+-y*i)^24 == 1+-i (mod p) for some integers x, y, then (x^2+y^2)^24 == 5 (mod p), so 5 must be a quadratic residue (in rational integers) modulo p. By definition, we have p == 11, 19 (mod 20).
%H A385191 Jianing Song, <a href="/A385191/b385191.txt">Table of n, a(n) for n = 1..10000</a>
%e A385191 1759 is a term since (2+-i)^((1759^2-1)/24) = (-4)^((31^2-1)/96) = 1048576 == 1 (mod 31). Indeed, the solutions to x^24 == 2+i (mod 1759) are x == {441+580i, -43+860i, -292+683i, -251+779i, -635+872i, 736-648i} X {+-1, +-i} (mod 1759).
%o A385191 (PARI) isA385191(p) = p>3 && isprime(p) && p%4==3 && Mod([2,-1;1,2],p)^((p^2-1)/24) == 1
%Y A385191 Cf. A385165, A385190 (1+-i are 24th powers), A002145, A122869. A385188 is a subsequence.
%K A385191 nonn,easy
%O A385191 1,1
%A A385191 _Jianing Song_, Jun 20 2025