$\xi = \cos{\frac{ 2\pi}{n}}+i \sin{\frac{ 2\pi}{n}}$ , $i^2=-1, n$ is a positive integer.

if $\xi^{a_1}+\xi^{a_2}+...+\xi^{a_k}=0$ ,

$a_1,a_2,...,a_k\in \{0,1,...,n-1\}$ and $a_1,a_2,...,a_k$ are not equal to each other, what's the possible values of k? or, how many possible values are there for k?

For example, when n=24, k could be any number except 1 and 23, some people think k can not be relatively prime with n, but it's not true. For example when n=24 and k=7.$\xi^0 +\xi^8+\xi^{16}+\xi^1+\xi^7+\xi^{13}+\xi^{19}=0$.

I'm not sure how difficult this question is. I calculated some starting value of k and searched in OEIS, but find no result.

  • 2
    $\begingroup$ See Henry B. Mann, "On linear relations between roots of unity", Mathematika 12(1965), pp.107-117. More on this in my answer to math.stackexchange.com/questions/962636/… $\endgroup$ – Ewan Delanoy Nov 17 '15 at 5:02
  • $\begingroup$ If $p$ is a divisor of $n$ then $k=p$ works (with $a_k = k(n/p)$). If $p$ and $q$ are divisors of $n$ then $k=p+q$ works (by combinding the two $a_k$-sequences for $p$ and $q$ and possibly displacing them to avoid duplicates). I suspect that if $n$ is prime then $k=n$ is the only solution. $\endgroup$ – Kibble Nov 17 '15 at 5:52
  • $\begingroup$ Is it? When $p,q\mid n$ the equation $\frac{n}{p} i = \frac{n}{q}j + 1$ has no integer solutions in $i,j$ so the sequence $a_k = \{\frac{n}{p},2\frac{n}{p},\ldots,(p-1)\frac{n}{p},\frac{n}{q}+1,2\frac{n}{q}+1,\ldots,(q-1)\frac{n}{q}+1\}$ should works unless I'm missing something obvious here. $\endgroup$ – Kibble Nov 17 '15 at 6:04
  • $\begingroup$ This does not give all solutions as for $n=24$ we cannot explain $k=13,17,19,21$ this way. One can consider the sum of more divisors, but then more care is needed to avoid duplicates. $\endgroup$ – Kibble Nov 17 '15 at 6:10
  • $\begingroup$ @Kibble If $k= n_1$ works, then $k=n-n_1$ must works, so we can explain $k = 13, 17, 19, 21$. $\endgroup$ – Takanashi Nov 17 '15 at 9:18

Hint addition of 4 consecutive multiples of $i$ is $0$. So k can be any multiple of $4$.e g $i+i^2+i^3+i^4=0$ also you can use $cos\frac{2kπ}{n}+isin\frac{2kπ}{n}=0$ for $k=(1...4n-1)$. Note I have given only partial solution.

  • $\begingroup$ For n=5, k=4 doesn't work. $\endgroup$ – Takanashi Nov 17 '15 at 9:57
  • $\begingroup$ @ Jerry i have made an edit. $\endgroup$ – Archis Welankar Nov 17 '15 at 10:14

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