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Math Help - powers modulo p and primitive root question

  1. #1
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    powers modulo p and primitive root question

    Let p be a prime number.
    What is the value of 1 + 2 + 3 +......+(p-1) (mod p)

    How would I begin to work this problem?
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  2. #2
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    Quote Originally Posted by duggaboy View Post
    Let p be a prime number.
    What is the value of 1 + 2 + 3 +......+(p-1) (mod p)

    How would I begin to work this problem?
    Note that (p-k) = -k mod p, k = 1,2,3,4,...

    1 + 2 + 3 +......+(p-1) (mod p) = 1 + (p-1)+ 2 + (p-2) +......+(p-1)/2 + (p+1)/2 (mod p) = p + p +......+p (mod p) = 0
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    ok is the idea here that p + p +...+p (modp) =0 because the modulo and the prime number are the same its always eqaul to zero? I'm a little confused why we want it to equal 0....

    Thank you so very much : )
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  4. #4
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    Quote Originally Posted by duggaboy View Post
    ok is the idea here that p + p +...+p (modp) =0 because the modulo and the prime number are the same its always eqaul to zero? I'm a little confused why we want it to equal 0....

    Thank you so very much : )
    Ya you are right!

    p mod p = 0 is the reason
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  5. #5
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    Quote Originally Posted by duggaboy View Post
    Let p be a prime number.
    What is the value of 1 + 2 + 3 +......+(p-1) (mod p)

    How would I begin to work this problem?
    There is another way: 1+2+...+(p-1) = p\cdot \tfrac{p-1}{2} , if p is odd.
    And so this is a multiple of p thus it reduces to 0 mod p.
    If p is even then it reduces to 1.

    Try a harder problem: find 1^k+2^k+...+(p-1)^k (\bmod p) where k is positive integer.
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    ok....so if k =2 then we it would be....

    1^2 + 2^2 + 3^2 +....+(p-1)^2

    (p-1)(p+1)/2p ??

    Or would you have to use another approach?? Ultimately you still want 0 mod p right?
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  7. #7
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    Quote Originally Posted by duggaboy View Post
    ok....so if k =2 then we it would be....

    1^2 + 2^2 + 3^2 +....+(p-1)^2

    (p-1)(p+1)/2p ??

    Or would you have to use another approach?? Ultimately you still want 0 mod p right?
    I am not exactly sure what you are doing.
    Here is a hint, p as a primitive root a and so \{ 1,2,...,p-1\} = \{ a,a^2,...,a^{p-1} \}.
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  8. #8
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    n n n
    phi(p) = (p - 1) * p 1 * (p - 1) * p 2 ... (p - 1) * p k
    1 1 2 2 k k



    phi(p) = p-1.

    Somthing like this? to get 0 mod p
    how would you handle this one? I am very interested in findining out
    Last edited by duggaboy; June 17th 2008 at 09:32 AM.
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