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Thread: order t (mod p)

  1. #1
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    Post order t (mod p)

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

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

    Can you factorise $\displaystyle a^t - 1$?

    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
    $\displaystyle a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}$

    Can you factorise $\displaystyle a^t - 1$?
    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 $\displaystyle a^t - 1$ you get $\displaystyle (a-1)(a^{t-1}+a^{t-2}+...+a+1)\equiv 0(\bmod p)$. Since $\displaystyle a-1\not \equiv 0(\bmod p)$ (unless $\displaystyle t=1$, but ignore that case) it means $\displaystyle a^{t-1} + ... + a + 1\equiv 0(\bmod p)$.
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