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Math Help - if a^2 \equiv 1, then a \equiv \pm 1 (mod p)

  1. #1
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    if a^2 \equiv 1, then a \equiv \pm 1 (mod p)

    if a^2 \equiv 1, then a \equiv \pm 1 \ \mbox{(mod p)}

    Not sure how to prove this.
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  2. #2
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    Well a^2 \equiv 1 mod(p) if and only if p | (a-1) \cdot (a+1) so.....
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  3. #3
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    I wasn't sure if I could assume the mod p in if part because it wasn't written in the book.
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    MHF Contributor Bruno J.'s Avatar
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    Also : in general, a polynomial of degree n has at most n roots \mod p. In this case, we know \pm 1 are two roots, so there can't be any others.
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  5. #5
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    Quote Originally Posted by Bruno J. View Post
    Also : in general, a polynomial of degree n has at most n roots \mod p. In this case, we know \pm 1 are two roots, so there can't be any others.
    Why is this true?
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  6. #6
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    Quote Originally Posted by mathman88 View Post
    Why is this true?
    x^3=8 has at most 3 real solutions; however, we know it has x=2 and a conjugate pair of complex numbers.
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