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I came to the problem saying that there exists a number $x\in \Bbb Z_7$ such that $x^2=2$ but there is no such $x$ in $\Bbb Z_5$. Could anyone give an explanation of this? How to actually find the expansion of square root 2 in $\Bbb Z_7$? Could we generalize this to cubic root?

Thanks!

Scorpio19891119
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    The general procedure of these things is that, first you want to solve your equation in $\mathbb{F}_p$. In your case, it's clear that $x^2 = 2$ has a solution mod 7 (say 3), but not mod 5. Once you solved it in mod $p$, Hensel's lemma allows you to lift the solution uniquely to mod $p^2$, $p^3$ and so forth, which allows you to put them all together and solve it in $\mathbb{Z}_p$. This can be generalized to cubic root, or more general polynomial equations, as long as the condition of Hensel's lemma is satisfied. –  Dec 03 '13 at 21:50
  • Have you looked at Hensel's? – LASV Dec 03 '13 at 21:51
  • @Luis Unfortunately, we did NOT talk about that theorem. – Scorpio19891119 Dec 03 '13 at 22:12
  • Ah I see. In general, this is the best method (That I know) that we guarantee a solution over the p-adics. It is vey much like Newton's method and not too difficult to digest, so I would consider taking a look at it! You do understand why there is no solution in $\mathbb{Z}_5$, yes? – LASV Dec 03 '13 at 22:16
  • @Luis Yes, after referring to the Hensel's Theorem. But how to show this from definition? – Scorpio19891119 Dec 03 '13 at 22:24
  • Well from definition, you know you have a solution $\mod 7$. Namely $3^2=2$. Show that you can lift this solution $\mod p^2$. i.e. $(3+7x)^2=2$ has a solution $\mod p^2$. Then keep on lifting the solution $\mod p^n$ for all $n$. Thus you need to show that you can do this for all $n$. @Sanchez Can you think of a better way? – LASV Dec 03 '13 at 22:27
  • If you follow the really nice proof in Gouvea of Hensel's lemma, you will pretty much prove the result. (Hence, regardless, you are using Hensel's lemma without knowing it =) ) – LASV Dec 03 '13 at 22:36
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    You don't need Hensel's lemma to know that there is no solution in $\mathbb{Z}_5$: The natural homomorphism $\mathbb{Z}_5 \to \mathbb{F}_5$ shows that if there is a squareroot of 2 in $\mathbb{Z}_5$, then there is one in $\mathbb{F}_5$, and you know that the latter is false. It's the converse where Hensel's came in, and as Luis said, you will probably just prove Hensel's on the way without realizing it. I do think this is the only way of doing it. –  Dec 03 '13 at 23:13
  • A mildly related example showing the beginning of the calculation of a 5-adic $\sqrt{-1}$. I agree with the others that this really is all about Hensel's lemma. – Jyrki Lahtonen Dec 04 '13 at 13:48
  • We have all been assuming that by ${\bf Z}_7$ you mean the 7-adic integers, and not the integers modulo 7. Is this a correct assumption? – Gerry Myerson Dec 06 '13 at 09:02

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