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Show that $\Bbb Z_p$ contains all the $(p-1)$th roots of unity. For which primes $p$ does $\Bbb Z_p$ contains primitive fourth roots of unity? Here $\Bbb Z_p$ is the ring of $p$-adic integers. Proving that it has a $(p-1)$th root of unity is easy, but all roots is another matter. Please help me with these questions.

I think for the second question, $p$ has to be $5$, but maybe there are other answers that I didn't think of.

Edit: I solved the first question, only need the second one. From what I can see, to be the primitive $4$th root, $4\mid(p-1)$, is that right?

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Hint: use Hensel's lemma. $\text{ }$


Added because you made an effort:

Second hint: what are the roots of $x^{p-1}-1$ in $\mathbf F_p$?

Third hint: for which $p$ does $x^2+1$ have a root in $\mathbf F_p$?

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  • $\begingroup$ I tried to apply hensel's lemma but failed miserably, can you explain it to me a bit more?. Thanks $\endgroup$ – user108680 Nov 22 '13 at 3:47
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    $\begingroup$ @user108680 Please improve your answer to include what you have, so that I can see where you are stuck. Thanks. $\endgroup$ – Bruno Joyal Nov 22 '13 at 3:48
  • $\begingroup$ This is what i tried.Let f(x)=x^(n-1)-1,there exist x such that gcd(x,p)=1, just let x=1 and x^(n-1)=1 mod p. f'(1)=/=0 since it's still 1. so there exist a solution in Zp that f(x)=0 $\endgroup$ – user108680 Nov 22 '13 at 4:17
  • $\begingroup$ ,And part 2, since with part 1, Zp has all (p-1)th root, it's obvious that if p=5, Zp had all 4th roots, one of them is a primitive root. But that answer is too stupid $\endgroup$ – user108680 Nov 22 '13 at 4:19
  • $\begingroup$ In the previous comment,i meant f(x)=x^(p-1)-1 $\endgroup$ – user108680 Nov 22 '13 at 4:28

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