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Let $p \neq 2$ be an odd prime and $\zeta_p$ be a root of unity. It is well known that $(1-\zeta_p)^{p-1}$ is divisible by $p$ inside $\mathbb{Z}[\zeta_p]$, i.e., there exists $a \in \mathbb{Z}[\zeta_p]$ such that $(1-\zeta_p)^{p-1}= p \cdot a$ inside $\mathbb{Z}[\zeta_p]$.
(from now on I drop the lower $p$ index, so $\zeta:= \zeta_p$)

One can say even more: Namely one can show that $c$ is a unit and can be determined explicitly as

$$ (\zeta - 1)^{p-1}/p =:c= \prod_{i=1}^{p-1} \frac{\zeta - 1}{\zeta^i - 1} $$

[See eg. this answer]. Especially, $(\zeta - 1)^{p-1}$ and $p$ are associated elements in $\mathbb{Z}[\zeta]$.
Clearly $c$ has norm $\pm 1$ (as norm map maps units to units) and I was wondering if it is possible to express $c$ also as a power of $\zeta_p$.

Trying with $p=3,5$ leads me to conjecture that $c$ can be also expressed as $c= (\zeta)^{\frac{p-1}{2}}$, but up to now I failed to find a formal proof. Therefore:

Question: Is it true that $c:= (\zeta - 1)^{p-1}/p= (\zeta)^{\frac{p-1}{2}}$ for any odd prime $p$ and if yes how to show it?

user267839
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  • @Daniel Czmmx: The discussion you're refering to I already mentioned above. This would reduce the problem to if (and if yes, how to see) that $(\zeta)^{\frac{p-1}{2}}$ coinsides with $\prod_{i=1}^{p-1} \frac{\zeta - 1}{\zeta^i - 1}$. Do you have an idea how to prove it? – user267839 Mar 12 '25 at 18:56
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    For $p = 3$ I got $c = -\zeta$. When $p = 5$, $c$ is not even a root of unity. One of us made a mistake in computation... – Daniel Arreola Mar 12 '25 at 20:01
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    If $f(\zeta)=0$, where $f(x)=(x-1)^{p-1}-px^k$ of degree $p-1$ for some $0\le k\le p-1$, then $f=\lambda m$, where $m(x)=x^{p-1}+\ldots+x+1$ is the minimal polynomial of $\zeta$. But $1<p-1<(p-1)(p-2)/2$ are the first 3 coefficients of $(x-1)^{p-1}$ for $p-1\ge4$, so one can't perturb at most one of them to make 'em all equal – te4 Mar 12 '25 at 20:44
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    In fact, it's not a root of unity for $p>3$. Minimal polynomial of $(1-\zeta_p)^{p-1}/p$ has degree $d$ at most $p-1$. If it is $\zeta_n^m$ for $(m,n)=1$, then $\varphi(n)=d$. But then $\zeta_n^m\in\mathbb Q(\zeta_n)\cap\mathbb Q(\zeta_p)$ is irrational (see https://math.stackexchange.com/q/291029/), so $p\mid n$, but then $p-1\le\varphi(n)=d\le p-1$, so $n=p$ (otherwise, one of inequalities is strict), which is the case I've already considered – te4 Mar 13 '25 at 00:47

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