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Math Help - Modules

  1. #1
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    Modules

    Let p be any odd prime number and set A = {0, 1, 2, . . . , p − 1}. Show there are exactly
    two solutions x belongs to 2 A which satisfy (x^2) ≡ 1 (mod p).
    Last edited by smithhall; April 29th 2009 at 02:40 PM.
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  2. #2
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    Arithmetic mod p

    Hello smithhall
    Quote Originally Posted by smithhall View Post
    Let p be any odd prime number and set A = {0, 1, 2, . . . , p − 1}. Show there are exactly
    two solutions x belongs to 2 A which satisfy (x^2) ≡ 1 (mod p).
    I assume that the 2 shouldn't be there!

    x^2 \equiv 1 \mod p

    \Rightarrow x^2 = np+1,\, n \in \mathbb{N}

    \Rightarrow (x+1)(x-1) = np

    \Rightarrow x = 1, n = 0, \text{ or }x+1 = \frac{np}{x-1}

    \Rightarrow (x-1)|n, since x-1<p and p is prime

    \Rightarrow n = m(x-1), m\in\mathbb{N}

    \Rightarrow x+1 = mp \equiv 0 \mod p

    \Rightarrow x \equiv -1 \mod p

    \Rightarrow x = p-1, since 0<x\le (p-1)

    So if p>2, there are just two distinct solutions: x = 1, p-1

    Grandad
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