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?

  • $\begingroup$ It has degree $\varphi(n)$, but that means that the polynomial may consist of up to $\varphi(n)+1$ terms. $\endgroup$ – Hagen von Eitzen Aug 19 '14 at 12:08
  • $\begingroup$ I don't think minimal polynomial was the right term to use. I rephrased my question a bit. $\endgroup$ – user135562 Aug 19 '14 at 12:11
  • $\begingroup$ If $\xi$ is of order $8$, then $\xi^4+\xi^0=0$, and we only have two terms. Is that what you mean? The same can be achieved as $\xi^5+\xi^1=0$, and now both terms are primitive of order eight. $\endgroup$ – Jyrki Lahtonen Aug 19 '14 at 12:11
  • $\begingroup$ @JyrkiLahtonen Ah yes. I forgot 1 restriction: the sum also needs to have a constant term. I'll rephrase my question again, sorry. $\endgroup$ – user135562 Aug 19 '14 at 12:17
  • 2
    $\begingroup$ And if $\xi$ is of order $25$, then $1+\xi^5+\xi^{10}+\xi^{15}+\xi^{20}=0$. Thus we have a sum of ten terms $$1+\xi^5+\xi^{10}+\xi^{15}+\xi^{20}+\xi+\xi^6+\xi^{11}+\xi^{16}+\xi^{21}=0.$$ Here all the conditions 1-4 from my guessed list are met. And $10<\phi(25)=20$. $\endgroup$ – Jyrki Lahtonen Aug 19 '14 at 12:26

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.

  • $\begingroup$ Can we give an exact number for any n? Perhaps 2*smallest integer dividing n that's greater than 1 when n is composite and n when n is prime? $\endgroup$ – user135562 Aug 19 '14 at 12:35
  • $\begingroup$ Anyway I think I can probably generalize from here. Thanks for the insight. $\endgroup$ – user135562 Aug 19 '14 at 12:38

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