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Let $\theta_1,\dots,\theta_N$ be complex numbers with absolute value 1. Assume that for any natural number $r\geq 1$ the sum $$\theta_1^r+\dots+\theta_N^r$$ is a constant independent of $r$.

Does it follow that $\theta_1=\dots=\theta_N=1$?

MKO
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    This could prove to be useful: https://en.wikipedia.org/wiki/Newton%27s_identities Moreover, the case for when the constant would be $0$ is dealt with here (it might inspire you potentially?): https://math.stackexchange.com/a/96219/1104384 – Bruno B Apr 26 '23 at 16:16
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    @BrunoB That's not quite the case for the constant of 0. Sva has infinitely many equations, while the link has it hold only for the first $N-1$ integers (and doesn't restrict to absolute value 1). In particular, we end up with $ \sum \theta_i ^ N = N$. – Calvin Lin Apr 26 '23 at 16:26
  • @CalvinLin Oh I see, thanks. Well, I guess that proves that the constant cannot be $0$ then. – Bruno B Apr 26 '23 at 16:37
  • Could you tell us a little more about where the problem comes from and what have you tried yet? – cs89 Apr 26 '23 at 19:33

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I think a more general statement is true. Let $x_1,\dots,x_N$ be complex numbers such that $$x_1^r+\dots+x_N^r=c,\,\,\,\, (1)$$ is a constant independent of $r\geq 1$.

Claim. Each $x_i$ is either equal to 0 or 1.

Proof. Let us assume that there is $x_j\ne 0,1$. Let us omit all $x_i=0$ or $1$ in (1) and change the constant $c$ correspondingly. Thus $x_i\ne 0,1$ for all $i$. Multiplying (1) by a parameter $t$ and summing up over $r\geq 1$ we get $$\sum_{i=1}^N\frac{x_i t}{1-x_it}=c\frac{t}{1-t}.\,\,\,\,(2)$$ Case 1. Assume $c\ne 0$. Then in the right hand side there is a pole at $t=1$. Hence it should be in the left hand side. Hence there exists $i$ such that $x_i=1$ which is a contradiction.

Case 2. Assume that $c=0$. Then (2) implies that $\sum_{i=1}^N\frac{x_i }{1-x_it}=0$. In other words $$\sum_{i=1}^N\frac{1}{t-x_i^{-1}}=0.$$ This is impossible. QED

MKO
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