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Math Help - Self-Invertible

  1. #1
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    Self-Invertible

    A positive integer k is self invertible modulo p iff. k=\pm1 \ \mbox{(mod p)}

    How to prove this?
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    A positive integer k is self invertible modulo p iff. k=\pm1 \ \mbox{(mod p)}

    How to prove this?
    I assume you meant k\equiv\pm1(mod\ p).

    By definition an integer b is called an inverse of k modulo a prime p if b satisfies kx\equiv1(mod\ p).

    Here k is an inverse of itself, therefore k^2\equiv1(mod\ p) ...
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    A positive integer k is self invertible modulo p iff. k=\pm1 \ \mbox{(mod p)}

    How to prove this?

    I pressume p is a prime, and then k is self-invertible modulo p iff

    k^2=1\!\!\pmod p\iff (k-1)(k+1)=0\!\!\pmod p\iff k-1=0\!\!\pmod p\,\,or\,\,\,k+1=0\!\1\pmod p

    The second biconditional following from the fact that p is prime.

    Tonio
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