If p is a prime, show that the only solutions of $x^2\equiv 1 \pmod p $ are $x=1$ and $x\equiv -1 \pmod p$. (from herstein's abstract algebra chapter2 section4 lagrange's theorem problem 15, this section is unfamiliar to me, and the questions are all damned hard!)
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This is the umpteenth time this question is being asked. I would like to know the reason why a veteran of the site does not recognize this as a duplicate without further ado. – Jyrki Lahtonen Apr 24 '14 at 12:39
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@Jyrk It is not easy to search for equations on SE, so for questions like this it is often much easier to ask/answer than to locate a prior duplicate. – Bill Dubuque Apr 24 '14 at 15:45
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Hint: $x^2 = 1$ if and only if $(x-1)(x+1)=0$.
Dune
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I did notice that, but then I didn't know what to do next and stucked. – pxc3110 Apr 24 '14 at 11:00
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@pxc3110 $x$ is a zero divisor if there is a $y \neq 0$ with $xy = 0$. Note that you deal with an equation of the form $xy = 0$ so in order to answer your question you have to determine all zero divisors of the corresponding ring (i.e. $\mathbb{Z}/p$). – Dune Apr 24 '14 at 11:13
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thanks. But if you're using ring theory, I think there might be some other ways of solving this, which is more easy for me to understand right now. – pxc3110 Apr 24 '14 at 11:19
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@pxc3110: This is group theory. What Dune tries to explain you is that because $p$ is prime, then if $xy = 0 \mod p$, then $x = 0 \mod p$ or $y = 0\mod p$. If $p$ is not prime, then this is not true anymore. Try with $4$ if you want. – user88595 Apr 24 '14 at 11:42
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@user88595 Thank you for your explanation. But that "or" in "x=0 mod or y=0 mod" doesn't sound good. I'm asked to prove x=0mod and y=0mod and they should be the only solutions – pxc3110 Apr 24 '14 at 11:50
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@pxc3110: You can't have $x = 1$ and $x = -1$ can you? You need to understand carefully: the reasoning I end with is $x = 1$ or $x = -1$. This implies that both are solutions, but they are not solutions together. $1$ and $-1$ are solutions but $x = 1$ and $x = -1$ isn't possible. – user88595 Apr 24 '14 at 12:04
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If a prime divides a product of integers then it divides at least one of the factors.
If $x$ is a solution of $x^2\equiv 1\pmod p$, apply this factlet to $(x-1)(x+1)$.
Hagen von Eitzen
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