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Math Help - need help solving this problem

  1. #1
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    need help solving this problem

    show that 1^p+2^p+3^p+...+(p-1)^p is congruent to 0 (mod p) when p is an odd prime
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  2. #2
    Moo
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    Hello,

    It can be proved for any odd integer...

    Separate the sum in two parts :

    S_1=1^p+2^p+\dots+\left(\frac{p-1}{2}\right)^p

    S_2=\left(\frac{p+1}{2}\right)^p+\left(\frac{p+1}{  2}+1\right)^p+\dots+(p-1)^p

    In each sum, there are \frac{p-1}{2} terms.

    And in the second one, just notice that :

    \begin{aligned}S_2 &=\left(p-\tfrac{p-1}{2}\right)^p+\left(p-\left(\tfrac{p-1}{2}-1\right)\right)^p+\dots+(p-2)^p+(p-1)^p \\<br />
&\equiv \left(-\tfrac{p-1}{2}\right)^p+\left(-\left(\tfrac{p-1}{2}-1\right)\right)^p+\dots+(-2)^p+(-1)^p \quad(\bmod p) \\<br />
&\equiv -\left(\tfrac{p-1}{2}\right)^p-\dots-2^p-1^p \quad(\bmod p) \quad (\text{p is odd})\\<br />
&\equiv -S_1 \quad(\bmod p) \quad \quad \blacksquare \end{aligned}
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Or a bit more simply (no offense Moo ) note that

    a^p \equiv a \mod p

    by Fermat's little theorem. Hence

    1^p+...+(p-1)^p \equiv 1+...+(p-1) = p(p-1)/2 \equiv 0 \mod p
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  4. #4
    Moo
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    Quote Originally Posted by Bruno J. View Post
    Or a bit more simply (no offense Moo ) note that

    a^p \equiv a \mod p

    by Fermat's little theorem. Hence

    1^p+...+(p-1)^p \equiv 1+...+(p-1) = p(p-1)/2 \equiv 0 \mod p
    No offence

    It's just that I've taken the proof from the case p is an odd integer, not necessarily a prime number

    Well, surely yours is easier and better for this case
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