Solving equations in Gaussian integer domain How to solve a complex equation of the form:
$z^n \equiv c~mod~p$, where $z, c$ are Gaussian integers, $n\in \mathbb{Z}^+$ and p is a prime.
1- Are there pre-conditions that need to be satisfied?
2- Is there any computer algebra software that can solve it?
3- Is there an efficient algorithm to solve it?
 A: If $p$ is prime in $\mathbb{Z}[i]$, then the quotient is a finite field $\mathbb{F}$ and then your question basically asks given $c\in \mathbb{F}$ and some natural number $n$ to find $z\in \mathbb{F}$ such that $z^n=c$. The multiplicative group of a finite field is cyclic of order $m=|\mathbb{F}|-1$, so that the element which are $n$-th power are exactly the elements in $\mathbb{F}^{\times n}$ which is the unique subgroup of order $d=\frac{m}{gcd(n,m)}$. It follows that you have such a solution iff $c^d=1$ (This is a well known trick when talking about $n$-th power residues).
If $p$ is an integer prime, then either it is a prime in $\mathbb{Z}[i]$ so that you can use the argument above, or it is a product of two distinct primes so that you can use the Chinese remainder theorem, or $p=2$ and in this case the quotient is isomorphic to $\mathbb{Z}_2[x]/x^2$ which is easy to understand.
In general, I think that finding a $z$ which satisfies the equation $z^n=c$ is not easy (i.e. can not be done quickly by a computer). Clearly, if you can find a generator for the multiplicative group, then you can find such a solution. There are some cases where this can be done easily, for example, if $n$ is coprime to the order of the multiplicative group, then taking the $n$-th power is an invertible map and you can find its inverse (standard cyclic group exercise).
EXAMPLE
For $p=8191$, we know that it is a prime in $\mathbb{Z}$ and it is 3 mod 4, so it remains prime in $\mathbb{Z}[i]$, so we get that $\mathbb{Z}[i]/p\cong \mathbb{F}_p[i]$ where $\mathbb{F}_p$ is the finite field with p elements. The field $\mathbb{F}_p[i]$ has $p^2$ elements so that its multiplicative group $<\zeta>$ is cyclic of order $p^2-1=67092480$. If we take $n=16 \mid 67092480$, and suppose that an element $c\in<\zeta>$ is an $n$-th power, so we can write $(\zeta^k)^n=c$. It then follows that $c^{\frac{p^2-1}{n}}=\zeta^{(p^2-1)k}=1^k=1$. On the other hand, if $\zeta^m$ is any element that satisfies $1=(\zeta^k)^{\frac{p^2-1}{n}}=\zeta^{k\cdot\frac{p^2-1}{n}}$, then since $\zeta$ has order $p^2-1$ we must have that $p^2-1\mid k\cdot\frac{p^2-1}{n}$, which is equivalent to $n\mid k$, namely $\zeta^k$ is an $n$-th power. For example, taking $c=i$ we get that $i^{\frac{p^2-1}{n}}=i^{4193280}=1$ because $i^4=1$, thus $i$ is a 16-th power of some element.
