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in $\mathbb{Z}/p^n$ where $ p \equiv 1 $ mod $ 4 $

I have been told as a 'hint' to use the isomorphism $\mathbb{Z}/p^n \cong \mathbb{F}^\times_{p} \times \mathbb{Z}/p^{n-1}$ but I don't understand how this will help?

I also need to show that for any $ a\in (\mathbb{Z}/p^n)^\times $ the number of solutions to the following congruence is 0 or 4.

$ x^{4} \equiv a$ mod $p^{n} $

Any pointers would be greatly appreciated.

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  • $\begingroup$ Have you already studied Hensel's Lemma? $\endgroup$ – DonAntonio Mar 12 '14 at 12:29
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    $\begingroup$ Yes I have, I just don't know how to apply it here. $\endgroup$ – user134925 Mar 12 '14 at 12:38
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For $\;n=1\;$ we get

$$x^4=1\iff (x-1)(x+1)(x^2+1)=0$$

and since $\;p=1\pmod 4\;$ we can write $\;x^2+1\;$ as a product of two linear polynomials.

Thus, we get four different solutions of $\;x^4-1=0\pmod p\;$ , and since $\;4\omega^3\neq0\;\;,\;\;w\;$ a solution of $\;x^4-1=0\pmod p\;$ , we can apply Hensel's Lemma and lift each solution to a unique solution $\;\pmod {p^n}\;,\;\;n\in\Bbb N\;$

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  • $\begingroup$ thank you. why have you checked if $4w^3 = 0 $? $\endgroup$ – user134925 Mar 12 '14 at 13:10
  • $\begingroup$ That is one of the conditions to apply H.L., of course... $\endgroup$ – DonAntonio Mar 12 '14 at 13:19
  • $\begingroup$ okay, we just learnt it a different way. could you point me in the right direction for the second part please? $\endgroup$ – user134925 Mar 12 '14 at 14:03
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    $\begingroup$ Think of the following (arithmetic modulo $\;p\;$ all along): $$x^4=a\iff \left(x^2-\sqrt a\right)\left(x^2+\sqrt a\right)$$ so we must have that both $\;\sqrt a\;,\;\;-\sqrt a\;$ are quadratic residues, and in this case we have the same as with $\;x^4=1\;$...otherwise there's no solution at all. $\endgroup$ – DonAntonio Mar 12 '14 at 14:27
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It can be shown that $-1$ is a square $\pmod p$ when $p\equiv 1 \pmod 4$. This proves the statement for $n=1$. You can use induction. If $\alpha$ is a root for $n$, look for the root for $n+1$ among $\alpha+kp^n$, $0\leq k \leq p-1$.

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