extend to non prime power modulus

[jacksonwalters]

when N is odd and n is a power of 2, we can sometimes find a cyclic subgroup C of the multiplicative group (Z/NZ)* such that n | |C|. in this case, we are able to compute an n^th root of unity by finding a generator g of C, and computing omega = g^{|C|/n}.

note such a subgroup is not guaranteed to exist. for example, when N=45, \phi(45) = 24. we might hope for subgroups of size 1, 2, 4, 8. however, (Z/45Z)* \cong (Z/9Z)* x (Z/5Z)* \cong Z/6Z x Z/4Z, which does not have a subgroup of order 8.

by using the Chinese remainder theorem and finding generators of factors, we can compute such a cyclic subgroup.

[jacksonwalters]

it is necessary and sufficient for n to divide each totient phi(p^k) for each prime factor of the modulus N. this allows us to find an n^th root of unity for each multiplicative factor of order phi(p^k). using the Chinese remainder theorem, we can pullback each of these n^th roots to obtain an n^th mod N omega which satisfies the additional condition omega^{n/2} == -1 (mod N). this ensures that the sums sum_{j=0}^{n-1} omega^{jk} == 0 for all 1 \le k < n.

since n is a power of 2, and we need gcd(n,N) == 1 for n to be invertible, we can assume N is odd. thus each prime factor p is odd, and so the condition on the modulus becomes that n|p_i - 1 for each prime factor p_i of the modulus.