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Thread: order

  1. #1
    Sep 2008


    Prove that if 3 is the order of x (mod p) then 6 is the order of x+1 (mod p). The hint says factor $\displaystyle x^3 - 1$, which is $\displaystyle (x-1)(x^2+x+1)$, but I don't see any relation between $\displaystyle x^3-1$ and $\displaystyle (x-1)^6$. Some help please.
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  2. #2
    Super Member PaulRS's Avatar
    Oct 2007
    $\displaystyle x^3-1=(x-1)(x^2+x+1)=0$ -everything modulo p-

    $\displaystyle x-1=0$ is not possible since the order of x is 3, then $\displaystyle x^2+x+1=0$

    BUt then $\displaystyle -x^2 = x+1$ try to go on from here .
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