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Math Help - a^p=b^p mod p ==> a^p=b^p mod p^2

  1. #1
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    a^p=b^p mod p ==> a^p=b^p mod p^2

    Show that a^p=b^p mod p ==> a^p=b^p mod p^2 (where p is prime).

    Any thoughts?
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  2. #2
    Super Member PaulRS's Avatar
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    Oct 2007
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    First note that a^p\equiv{b^p}(\bmod.p) implies a\equiv{b}(\bmod.p) by Fermat's little Theorem.
    Now, try using the binomial Theorem and note that \binom{p}{k}\equiv{0}(\bmod.p) for k=1,...,p-1
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