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Let $k$ be an odd rational integer, $p$ a rational prime and $\zeta_p$ a primitive $p$th root of unity. Let $\sigma$ a generator of $Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})$, i.e., $\sigma(\zeta_p)=\zeta_p^g$ for some primitive root $g$ modolo $p$. What are the elements $z$ in $\mathbb{Q}(\zeta_p)$ with $|z|=|\sigma(z)|=\cdots=|\sigma^{p-2}(z)|$? In particular, I want to find those $z$ such that $|\sigma^r(z)|^2=kp+1$ for $r=0,1,\ldots,p-2$.

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Note that the condition $|\sigma(z)|=|z|$ can be written $\sigma(z) \sigma(\bar{z}) = z \bar{z}$, which implies $\sigma^r(z) \sigma^r(\bar{z}) = z \bar{z}$ for all $r$. So you're asking what are the elements $z \in \mathbf{Q}(\zeta_p)^\times$ such that $\sigma(z)/z$ has modulus 1. Note that the map $z \mapsto \sigma(z)/z$ is surjective onto the elements of norm 1 (Hilbert 90), thus it attains all elements of modulus 1. – François Brunault Sep 23 '12 at 10:49
Also, note that $|z|^2=z\bar{z} = kp+1$ implies $|\sigma^r(z)|^2= kp+1$ for all $r$. If there exists $z_0$ such that $|z_0|^2 = kp+1$, then all other $z$ are of the form $z=z_0 \cdot u/\bar{u}$ with $u \neq 0$. I don't see any reason why such $z_0$ should exist in general. For example if $p=3$ and $k=1$ then $z_0=2$ works, but if $p=3$ and $k=3$ then the equation $|z|^2=10$ has no solution in $\mathbf{Q}(\zeta_3)$. – François Brunault Sep 23 '12 at 11:22
Oh,I think my second question is to find the so called $\sqrt{kp+1}$-Weil numbers in cyclotomic fieds. In general, $m$-Weil numbers are not easy to classify. – Binzhou Xia Sep 25 '12 at 4:00

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