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Calculation of Large Exponents Modulo a Prime

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data

    In my Number Theory class, we learned how to calculate the value of large exponents modulo primes using Euler's Theorem. I understand how to do this with exponents larger than the value of the totient function of the prime, which is p-1, but what about when the exponent is actually smaller than that value? For example, if I have [itex]95^{65}\ mod\ 131[/itex], I don't see how I can reduce the exponent the way I would for something like [itex]95^{261} \equiv 95^{2(130)+1} \equiv 95\ (mod\ 131)[/itex].



    2. Relevant equations

    [itex]x^{p-1} \equiv 1\ (mod\ p)[/itex]

    and the more general form for a composite number. Also, any basic properties of exponents and moduli that I'm missing.



    3. The attempt at a solution

    I'm just not sure where to begin. Obviously, 65 is half of 130, and we know that [itex]95^{130} \equiv 1\ (mod\ 131)[/itex], but how can I use that?
     
  2. jcsd
  3. Oct 29, 2011 #2

    SammyS

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    You have pretty nearly answered your own question.

    You have that [itex]95^{65\cdot2}\equiv1 (\text{mod} 131)\,.[/itex] Then [itex]95^{65\cdot2} - 1\equiv0 (\text{mod} 131)\,.[/itex]

    Use the fact that [itex]95^{65\cdot2}=(95^ {65})^2\,.[/itex]

    Factor the difference of squares and see where that leads you.
     
  4. Oct 30, 2011 #3

    I like Serena

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    Hi Vespero! :smile:

    The generic method is to do this:
    [tex]95^{65} \equiv 95^{2\cdot 32+1} \equiv (95^2)^{32} \cdot 95 \equiv (95^2 \textrm{ mod } 131)^{32} \cdot 95 (\textrm{mod } 131)[/tex]
    Repeat, until you are done.

    If you want, you can factor 95 first into 5 x 19 to get easier calculations.

    This is quite a bit of work, although you can recycle some of your work.

    @SammyS: I believe your method leads to 2 solutions...
     
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