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Math Help - order t (mod p)

  1. #1
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    Post order t (mod p)

    If a, a not equal to 1 has order t (mod p),
    show that
    a^(t-1) + a^(t-2) + .......+1=0 (mod p).
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  2. #2
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    a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}

    Can you factorise a^t - 1?
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  3. #3
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    Quote Originally Posted by SimonM View Post
    a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}

    Can you factorise a^t - 1?

    I don't know.
    Do you think that's how it works?
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  4. #4
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    Quote Originally Posted by SimonM View Post
    a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}

    Can you factorise a^t - 1?
    Quote Originally Posted by Sally_Math View Post
    I don't know.
    Do you think that's how it works?
    SimonM gave you basically the solution, if you factorize a^t - 1 you get (a-1)(a^{t-1}+a^{t-2}+...+a+1)\equiv 0(\bmod p). Since a-1\not \equiv 0(\bmod p) (unless t=1, but ignore that case) it means a^{t-1} + ... + a + 1\equiv 0(\bmod p).
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