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Select[Prime[Range[200]], Mod[#, 8]==3&] (* Harvey P. Dale, Apr 05 2023 *)
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(MAGMAMagma) [p: p in PrimesUpTo(2000) | p mod 8 eq 3]; // Vincenzo Librandi, Aug 07 2012
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From Hilko Koning, Nov 24 2021 : (Start)
Theorem (Legendre symbol): With p an odd prime and a an integer coprime to p the Legendre symbol L(a/p) = -1 if a is a quadratic non-residue (mod p) and L(2/p) = -1 if p = = +- 3 (mod 8).
Theorem (Euler's criterion): L(a/p) = = a^((p-1)/2) (mod p) so with a = 2 and prime p = 2k + 1 then -1 = = 2^k (mod (2k+1)). So prime numbers (2k+1) = +- 3 (mod 8) are the prime numbers (2k+1) | (2^k +1).
If 2k+1 = = -3 (mod 8) then k is even and 2^k +1 is not divisible by 3 and if 2k+1 = = + 3 (mod 8) then k is odd and 2^k + 1 is divisible by 3.
Hence prime numbers 2k +1 = = 3 (mod 8 ) are prime numbers such that 3*(2k+1) | (2^k +1). Or, including the first term of the sequence, prime numbers 2k+1 with k is odd such that (2k+1) | (2^k+1).
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From Hilko Koning, Nov 24 2021 :(Start)
Theorem (Legendre symbol): With p an odd prime and a an integer coprime to p the Legendre symbol L(a/p) = -1 if a is a quadratic non-residue (mod p) and L(2/p) = -1 if p = +- 3 (mod 8).
Theorem (Euler's criterion): L(a/p) = a^((p-1)/2) (mod p) so with a =2 and prime p = 2k + 1 then -1 = 2^k (mod (2k+1)). So prime numbers (2k+1) = +- 3 (mod 8) are the prime numbers (2k+1) | (2^k +1).
If 2k+1 = -3 (mod 8) then k is even and 2^k +1 is not divisible by 3 and if 2k+1 = + 3 (mod 8) then k is odd and 2^k + 1 is divisible by 3.
Hence prime numbers 2k +1 = 3 (mod 8 ) are prime numbers such that 3(2k+1) | (2^k +1). Or, including the first term of the sequence, prime numbers 2k+1 with k is odd such that (2k+1) | (2^k+1).
(End)
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