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P1^m1 mod 2^n = p2^m2 mod 2^n

  1. Sep 5, 2010 #1
    Let p1 and p2 be primes and m1 and m2 be integers when is:

    When is p1^m1 mod 2^n = p2^m2 mod 2^n true?

    I think this problem has applications to hash-tables.
  2. jcsd
  3. Sep 5, 2010 #2


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    They're equal when they're equal, there really isn't much to say. I don't see what being prime has to do with it.

    As an abstract Abelian group, the odd numbers with multiplication are isomorphic to the product of the group with 2 elements and the cyclic group with 2n-2 elements.
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