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Is there a way to get the number of solutions of an equation like $x^n=1$ in the p-adic field $\mathbb Q_p$, where $p$ is a prime and $n$ a positive integer?

I know that for $n=p-1$ there are $p-1$ roots of unity. Are there any theorems for other cases?

Best Regards

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up vote 3 down vote accepted

If $(n,p)=1$, then there are exactly as many roots as there are roots to $x^{n}-1=0$ in $\mathbb F_p=\mathbb Z/p\mathbb Z$. Any solution in $F_p$ can be extended uniquely to a solution in $\mathbb Z_p$ using Hensel's lemma.

The number of solutions to $x^{n}-1=0$ in $\mathbb F_p$ is $(n,p-1)$.

The case where $p|n$ is different. Blanking on what to do here. In think $x^p=1$ has only one solution, $x=1$, and that we can then show that if $n=p^kn'$ where $p\not\mid n'$ then $x^n-1=0$ has as many solutions as $x^{n'}-1=0$. So again that $(n,p-1)$ (since $(p^k,p-1)=1$ and thus $(n,p-1)=(n',p-1)$.)

The case $p=2$ is a special case - for example, $x^2=1$ has two roots in $\mathbb Q_2$.

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Thank you. I have just one little question for the case (n,p)=1: Why are there not more solutions in $\mathbb{Q}_p$ than in $\mathbb{Z}_p$? – user61052 Feb 5 '13 at 19:42
Because any non-zero element of $z\in\mathbb Q_p$ can be written uniquely as $p^{a}w$ where a\in\mathbb Z, $w\in \mathbb Z_p$ and $w\not\equiv 0\pmod p$. But then $z^n = p^{an}w^n$. So if $z^n=1$, then $p^{an}=1$ and thus $a=0$, so $z\in\mathbb Z_p$. – Thomas Andrews Feb 5 '13 at 19:48

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