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December 16th, 2018, 07:05 PM  #1 
Newbie Joined: Dec 2018 From: Earth Posts: 4 Thanks: 1 
Hi everyone, I am new here. I have been trying to figure out how to attack multiple problems involving cyclotomic polynomials, roots of unity, number fields, and other number theory subjects related to those. One particular problem I am stuck on is the following: Let $n$ and $p$ be primes such that $p=1\pmod n$. Find the smallest set containing k distinct roots of unity $S=[\zeta_{n_1},\zeta_{n_2},\zeta_{n_3},\zeta_{n_4},.. .\zeta_{n_k}]$ such that $\zeta_{n} + \zeta_{n_2} + \zeta_{n_3} + \zeta_{n_4} ... + \zeta_{n_k} = 0 \pmod p$ Is there an algorithm for solving this problem for a particular n and p (especially if p is relatively large compared to n)? Estimating the behavior of this problem, one can predict that the value of $k$ is approximately $\log_n(p)$. It is also worth asking which sums are unique (because if $S$ is a solution set, then at least $n2$ other solution sets exists, and therefore all the subset sums are not necessarily unique). Another way of stating this problem is finding the smallest $u = \zeta_{n} + \zeta_{n_2} + \zeta_{n_3} + \zeta_{n_4} ... + \zeta_{n_k}$ such that the norm $N$ of the element $u$ in the field $K=\mathbb Q(\zeta_{n})$ is divisible by $p$. This also resorts to asking the following question (which will most likely answer my original question): What is the smallest integer $N$ such that $N=0\pmod p$ and $N$ is the norm of an element $u$ in the field $K=\mathbb Q(\zeta_{n})$? In particular, how can this integer $N$ be found? Sorry if I have described the problem in broad manner. Here is an example of the problem where I fix $n$ and $p$. Example: $n=31, p=1117$ We see that $p=1\pmod n$, and $\zeta_{n} + \zeta_{n_2} + \zeta_{n_3} + \zeta_{n_4} + \zeta_{n_k} = 0 \pmod p$ where $\zeta_{n} + \zeta_{n}^2 + \zeta_{n}^3 + \zeta_{n}^8 + \zeta_{n}^{13} = 0 \pmod p$ as the norm of $\zeta_{n} + \zeta_{n}^2 + \zeta_{n}^3 + \zeta_{n}^8 + \zeta_{n}^{13}$ in $K=\mathbb Q(\zeta_{n})$ is $1455451 = 1117*1303 = 0 \pmod p$. However, the "smallest" $N$ that is divisible by $p$ is $N=139625=5^3*1117=0\pmod p$ which is generated by norm the element $u = \zeta_{n} + \zeta_{n}^2 + \zeta_{n}^4$ (the norm of $u$ in $K$ is $N$). I am currently trying to attack this problem in general for any prime $n$ and a prime $p=1\pmod n$ under the conditions $p<2^{(n1)/2}$ and $N<2^n$ (Indeed if $p<2^{(n1)/2}$ holds, then $N<2^n$ should also hold). Any ideas? Will give thanks for any helpful responses. Again, sorry if the description of the problem I am trying to solve is too broad. Last edited by skipjack; December 17th, 2018 at 01:59 AM. 

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roots, sum, unity, vanishing 
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