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Math Help - A Couple Problems with Primes

  1. #1
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    A Couple Problems with Primes

    1) Show that 1^p + 2^p + 3^p + .... + (p-1)^p = 0 (mod p) when p is an odd prime

    2) Show that if p is a prime and p > 3, then 2^{p-2} + 3^{p-2} + 6^{p-2} = 1 (mod p)

    For 1), I really don't understand which theorem or identity I'm supposed to put into use to prove this. Obviously it works every time I just don't understand how this works in terms of technicality.

    2) is the same story. I don't see how it works and therefore I can't prove it formally.
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  2. #2
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    Quote Originally Posted by Janu42 View Post
    1) Show that 1^p + 2^p + 3^p + .... + (p-1)^p = 0 (mod p) when p is an odd prime

    2) Show that if p is a prime and p > 3, then 2^{p-2} + 3^{p-2} + 6^{p-2} = 1 (mod p)

    For 1), I really don't understand which theorem or identity I'm supposed to put into use to prove this. Obviously it works every time I just don't understand how this works in terms of technicality.

    2) is the same story. I don't see how it works and therefore I can't prove it formally.
    For #1, use Fermat's Little Theorem.

    For #2, I'm not sure if the following is valid, but if not then I think it's close. Using Fermat's Little Theorem,

    2^{p-2} + 3^{p-2} + 6^{p-2} \equiv 2^{-1} + 3^{-1} + 6^{-1} \equiv \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \equiv 1\pmod{p}
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  3. #3
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    The form of the terms in both sums suggests using Fermat's Theorem.
    Let S=2^{p-2}+3^{p-2}+6^{p-2}. Then 6S=3\cdot2^{p-1}+2\cdot3^{p-1}+6^{p-1}.
    Now Fermat's Theorem implies that 6S\equiv3+2+1=6\bmod p - show this!. Use the fact that (p,6)=1- again, why?
    Therefore, you can cancell to get S\equiv1\bmod p.
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