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Numbers m such that (1+i)^m + i is a Gaussian prime.
2

%I #21 Jul 29 2021 09:30:33

%S 0,1,2,3,4,6,7,8,11,14,16,19,38,47,62,79,151,163,167,214,239,254,283,

%T 367,379,1214,1367,2558,4406,8846,14699,49207,77291,160423,172486,

%U 221006,432182,1513678,2515574

%N Numbers m such that (1+i)^m + i is a Gaussian prime.

%C Equivalently, either (1+i)^m + i times its conjugate is an ordinary prime, or m == 2 (mod 4) and 2^(m/2) + (-1)^((m-2)/4) is an ordinary prime.

%C Let z = (1+i)^m + i. If z is not pure real or pure imaginary, then z is a Gaussian prime if the product of z and its conjugate is a rational prime. That product is 1 + 2^m + sin(m*Pi/4)*2^(1+m/2). z is imaginary when m=4k+2, in which case z has magnitude 2^(2k+1) + (-1)^k. These pure imaginary numbers are Gaussian primes when 2^(2k+1)-1 is a Mersenne prime and 2k+1 == 1 (mod 4); that is, when m is twice an odd number in A112633. - _T. D. Noe_, Mar 07 2011

%H <a href="/index/Ga#gaussians">Index entries for Gaussian integers and primes</a>

%t Select[Range[0,30000], PrimeQ[(1+I)^#+I, GaussianIntegers->True]&]

%Y Cf. A057429, A103329.

%K nonn

%O 1,3

%A _Ed Pegg Jr_, Aug 07 2002

%E More terms from _Mike Oakes_, Aug 07 2002

%E Edited by _Dean Hickerson_, Aug 14 2002

%E 0 prepended by _T. D. Noe_, Mar 07 2011