remove failing test for 2p^k

[jacksonwalters]

add a test for the case with modulus = 2*p^k for a prime p. the multiplicative group is cyclic here as well.

currently, this case is failing. the resulting polynomial is almost right, but there are a few coefficients that are off by either p^k or 2*p^k.

being off by 2*p^k is a non-issue since that's just the modulus. being off by p^k indicates something isn't quite right.

[jacksonwalters]

note that since n must be a power of two for the divide-and-conquer algorithm to work, then gcd(n,2p^k) = 2, so n is not invertible. for now, there is no obvious workaround, so even though $2p^k$ has a cyclic multiplicative group, we will not attempt to compute the NTT.

we add a gcd function and add a check when computing $n^{-1}$ to ensure that n is actually invertible.