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Math Help - order question, with congruence

  1. #1
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    order question, with congruence

    Show that if a has order t (mod p) then
    a^(t-1) + a^(t-2)+....+1 is congruent to 0 (mod p)
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  2. #2
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    Quote Originally Posted by scubasteve123 View Post
    Show that if a has order t (mod p) then
    a^(t-1) + a^(t-2)+....+1 is congruent to 0 (mod p)

    a^t=1\!\!\pmod p\Longrightarrow a^t-1=0\!\!\pmod p  \Longrightarrow(a-1)(a^{t-1}+a^{t-2}+...+a+1)=0\!\!\pmod p...

    Tonio
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  3. #3
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    I dont really understand how you did that.
    a^t = 1 mod p makes sense clearly by definition of order
    but then in the third step im a bit lost. Can you expand?
    Thanks so much for ur help
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  4. #4
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    Quote Originally Posted by scubasteve123 View Post
    I dont really understand how you did that.
    a^t = 1 mod p makes sense clearly by definition of order
    but then in the third step im a bit lost. Can you expand?
    Thanks so much for ur help

    The third step is high school algebra:

    a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})\,,\,\,\forall\,n\in \mathbb{N}

    This follows from the formula for the sum of geometric sequences.

    Tonio
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