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Let $p$ be prime, and $z_0, z_1, ..., z_{p-1}$ be all the $p$-th roots of unity, i.e. solutions of the equation $z^p = 1.$

Is it true or false that a combination of two (or more, in general) of the roots can give us another root of the same order?

In mathematical terms, does there exist indices $i_1,i_2,...,i_s, j,$ such that $z_j = \sum_{k=1}^s z_{i_k}?$

It seems to be that this is not possible, but I also don't have proof of that.

Thank you!

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Does your sum need to be a root of unity of same order $p$? – Loïc Teyssier Apr 9 '13 at 11:56
Cyclotomic polynomials show some ways that sums of roots of unity can be $0$. Slight modifications show ways that roots of unity can add up to another root of unity. I think not all of these come directly from cyclotomic polynomials. – Douglas Zare Apr 9 '13 at 12:18
yes, the order must stay the same – Liss Apr 9 '13 at 12:42
and p is prime, true! will it still hold? ... – Liss Apr 9 '13 at 12:43
up vote 17 down vote accepted

If $p$ is intended to be prime, you should say so explicitly. I'll assume $p$ is prime. Then the subset sums are distinct except that the sum of all $p$th roots of unity is $0$, the sum over the empty set. Any coincidence of subset sums $\sum_{i \in I} \zeta_p^i = \sum_{j\in J} \zeta_p^j $produces a polynomial of degree at most $p-1$ with coefficients in $\lbrace-1,0,1\rbrace$ so that $\zeta_p$ is a root. This polynomial must be a multiple of the minimum polynomial for $\zeta_p$, the $p$th cyclotomic polynomial $\Phi_p(x) =1 + x + ... + x^{p-1}$. The only possibilities are scalar multiples corresponding to $1 + \zeta_p + ... + \zeta_p^{p-1} = 0$.

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Yes, $p$ was intended to be prime, I included it now in the question. Thanks for the answer, and considering two sums from the left and right was the thing that I needed, as well as obtaining that only a linear combination produced from the fact that the sum of the roots is zero is a possible equality. And its great that it turns out the answer is $no$, if the number of entries $|I|+|J|< p...$ – Liss Apr 9 '13 at 13:41
@Liss: in case you should also need information what is and is not possible in more general situations the following paper and references there could be useful – user9072 Apr 9 '13 at 13:46
Thank you, @quid! – Liss Apr 9 '13 at 16:40

True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…

Edit: Now that you edited the question and $p$ becomes prime, there is a more general answer. As Douglas pointed out the key word here is cyclotomic polynomials $P_p(z)=\sum_{n=0}^{p-1}z^n$, whose roots are precisely the $p^{th}$-root of unity except $1$. In that case


gives you such a relation as long as $z_j\neq 1$.

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Since the question uses $p$ I assume that it was intended to be prime. It thus seems a bit unfortunate to me to give examples for $4$ and $6$. Even more so, since it seems genuinely easier to me to come up with examples for even order. – user9072 Apr 9 '13 at 12:23
Well, in some papers $p$ stands for e.g. a point in a euclidian space. I don't know where we are supposed to stop double-guessing the OP's intent. Personnally I draw the line at the minimal type-casting needed to make sense of the statement. As for the rest of your comment, why, good for you. I don't think the question is so involved that we shall start to compete about who's got the shortest (answer). – Loïc Teyssier Apr 9 '13 at 12:32
I did not mean to say I have an easier way to give an example for even order, but that in my opinion the case allowing even order is easier than the case where one would insist on odd order. It seems quite possible that this answer does not answer the (intended) question at all. If you could add an example for prime $p$, by contrast it surely would. – user9072 Apr 9 '13 at 12:43
Sure. I do that now. – Loïc Teyssier Apr 9 '13 at 12:47

It is known that for every positive integer $n$, the primitive $n$-th roots of unity are linearly independent over $\mathbb{Q}$ if and only if $n$ is square free.

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