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A328415 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has exactly one solution, where (Z/mZ)* is the multiplicative group of integers modulo m.

%I A328415 #8 Oct 18 2019 17:06:37
%S A328415 4,16,27,32,64,256,512,1024,2048,2187,4096,6561,8192,16384,59049,
%T A328415 65536,131072,177147,262144,524288,531441,1048576,1594323,2097152,
%U A328415 4194304,4782969,8388608,14348907,16777216,33554432,67108864,134217728,268435456,387420489,536870912,1073741824
%N A328415 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has exactly one solution, where (Z/mZ)* is the multiplicative group of integers modulo m.
%C A328415 Numbers k being powers of 2 or 3 such that 2*k+1 is not prime.
%C A328415 Proof. If m is a solution to (Z/mZ)* = C_2 X C_(2k) such that m is odd, then 2*m is also a solution, and vice versa. So if there is only one solution to (Z/mZ)* = C_2 X C_(2k), m must be a multiple of 4. If 8 divides m and m has odd prime factors, or if m has at least two distinct odd prime factors, then A046072(m) >= 3, a contradiction. So m = 2^e, e >= 3 or m = 4*p^e, p odd prime and e >= 1. If m = 4*p^e and p >= 5, then (Z/(3*p^e)Z)* = (Z/mZ)*. So we have m = 2^e, e >= 3 or m = 4*3^e, e >= 1, then (Z/mZ)* = C_2 X C_(2*2^(e-3)) or (Z/mZ)* = C_2 X C_(2*3^(e-1)).
%C A328415 If k = 2^(e-3) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(2^e)Z)*; if k = 3^(e-1) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(4*3^e)Z)*; on the other hand, if k is a power of 2 or a power of 3 such that 2*k+1 is not prime, then (Z/mZ)* = C_2 X C_(2k) indeed has only one solution.
%H A328415 Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
%e A328415 The only solution to (Z/mZ)* = C_2 X C_54 is m = 324, so 54/2 = 27 is a term.
%o A328415 (PARI) select(i->!isprime(2*i+1), upto(10^9)) \\ See A006899 for the function upto(n)
%Y A328415 Cf. A328412.
%K A328415 nonn
%O A328415 1,1
%A A328415 _Jianing Song_, Oct 14 2019