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Given a prime $p$ and an integer $a$. If the order of $a$ modulo $p$ is 6, is it true that the order of $(1-a)$ modulo $p$ is also 6? If so prove it, if not give a counterexample.

It seems right. So I aimed to show that

$(1-a)^6 \equiv 1 \pmod p$, and also

$(1-a)^i \ne 1 \pmod p$ for all $i = 1, 2, 3, 4, 5$

Am I on the right track? Or is there any other way of thinking?

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    You're on the right track. Try proving that if $b$ is a primitive root (mod $p$) (that is, every nonzero residue is a power of $b$) then $(1 - b)$ is a primitive root (mod $p$).2012-10-29
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    http://math.stackexchange.com/questions/220493/order-of-elements-modulo-p2012-10-29

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