Number of terms containing primitive root of unity It is well known that the degree of the n-th cyclotomic polynomial is $\varphi(n)$, where $\varphi$ is the Euler totient function. I define the ${minimal}$ sum to be of the form 
\begin{align}
\xi_0 + \sum_{i=1 }^{k} \xi_i = 0
\end{align}
where $\xi_0$ an non-negative integer, $\xi_i$'s roots of unity of some order, and no subsum on the left sums to 0. If the sum were to have $\xi_n$, the primitive n-th root, as one of the terms, does the fact that the n-th cyclotomic polynomial have degree $\varphi(n)$ imply that this minimal sum has $\varphi(n)$ terms? 
 A: If I correctly interpreted the question, then the following example shows that it is possible to get away with less then $\phi(n)$ terms.
If $\xi$ is of order $25$, then $1+\xi^5+\xi^{10}+\xi^{15}+\xi^{20}=0$ as $\xi$ is a zero
of the cyclotomic polynomial $\Phi_{25}(x)=(x^{25}-1)/(x^5-1)$. Thus we have a sum of ten terms
$$(1+\xi)\Phi_{25}(\xi)=1+\xi^5+\xi^{10}+\xi^{15}+\xi^{20}+\xi+\xi^6+\xi^{11}+\xi^{16}+\xi^{21}=0.$$ 
Here $10<\phi(25)=20$. The rational term $1=\xi^0$ appears. A primitive root $\xi$ is one of the terms, and all the ten terms are distinct.

In general if $p\mid n$ is the smallest prime divisor of $n$, and $\xi$ is primitive of order $n$, then the sum
$$
0=(1+\xi)(1+\xi^{n/p}+\xi^{2n/p}+\cdots+\xi^{(p-1)n/p})=\sum_{k=0, k\equiv0,1\pmod{(n/p)}}^{n-1}\xi^k
$$
has $2p$ distinct powers of $\xi$. Namely those with exponents congruent to $0$ or $1$ modulo $n/p$. Both $\xi^0$ and $\xi^1$ occur in the sum. I don't know if $2p$ is the smallest possible number of terms such that all criteria are met.
