Just a little help.
If $\zeta_k=e^\frac{2\pi ik}{n}$ is a primitive $n^\text{th}$ root of unity for $n\in\mathbb{Z}^+$, then $k\in\mathbb{Z}$ is relatively prime to $n$. Furthermore, $\zeta_j=\zeta_k\iff j\equiv k\pmod n$. So the set of all primitive $n^\text{th}$ roots of unity is parametrized by the set $K$ of all $1\le k<n$ relatively prime to $n$, i.e. the reduced residue system modulo $n$, which can be identified with $(\mathbb{Z}/n\mathbb{Z})^*$.
Now $s(n,p)=\zeta_k^p=e^\frac{2\pi ikp}{n}=\cos\frac{2\pi kp}{n}+i\sin\frac{2\pi kp}{n}$
and $\Re\,\sum\zeta_k^p=\sum\cos\frac{2\pi kp}{n}$, so we might hope that the full complex sum is zero or not zero under the respective conditions.
This will depend on what happens to the residues $k$ modulo $n$ when they are multiplied by $p$. For $p$ relatively prime to $n$, they get permuted ($pK=K$), so that the sum is zero (in the complex plane, both the real and imaginary parts). If, however, $d=\gcd(p,n)>1$, then $\zeta_k^p$ is a primitive root of order $\frac{n}{d}$. So how do we get a complex sum of zero from primitive roots? One way is to sum all the vertices of the regular $n$-gon, i.e. all $n^\text{th}$ roots of unity: $\sum_{j=0}^{n-1}\zeta_j=0$. However, these can be regrouped by their order: $\sum_{d|n}s(d,1)=0$, in terms of the sums $s(n,p)$ defined above. What is $\sum_{d|n}s(d,p)$?
Let $r=\href{http://en.wikipedia.org/wiki/Radical_of_an_integer}{\text{rad}}\,n$
$=\prod_{q\mid n}q$ be the product of all primes $q$ dividing $n$ (also called the radical or squarefree kernel of $n$).
What does it tell us when the positive (but not necessarily prime) integer $p$ is less than $\frac{n}r$? Or equal to $\frac{n}r$? What does that tell us about $\frac{n}p$ in relation to $r$? It must be less than $r$ and therefore relatively prime to at least one prime $q$ dividing $r$ and $n$.
You may also find the Möbius inversion formula helpful; your $\mu$ is $\mu(r)$, where my $\mu$ is the Möbius function.
Best Answer
You say you found all five roots of the equation, presumably in exponential form. Now write them in trigonometric form and simplify. I'm guessing there will only be one case in which both the $\cos \theta$ and $\sin \theta$ will be integers.