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.

4, 16, 27, 32, 64, 256, 512, 1024, 2048, 2187, 4096, 6561, 8192, 16384, 59049, 65536, 131072, 177147, 262144, 524288, 531441, 1048576, 1594323, 2097152, 4194304, 4782969, 8388608, 14348907, 16777216, 33554432, 67108864, 134217728, 268435456, 387420489, 536870912, 1073741824

Offset: 1

Author

Jianing Song, Oct 14 2019

Keywords

• nonn

Comments

Numbers k being powers of 2 or 3 such that 2*k+1 is not prime.

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)).

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.

Examples

			The only solution to (Z/mZ)* = C_2 X C_54 is m = 324, so 54/2 = 27 is a term.
		

Links

• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A328412.

Programs

PARI

select(i->!isprime(2*i+1), upto(10^9)) \\ See A006899 for the function upto(n)