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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 $p$). – Jonah Sinick Oct 29 '12 at 4:41

Note that $a$ has order $6$ iff $a$ is a solution of the congruence $x^6-1\equiv 0\pmod{p}$, but does not have order $\lt 6$. We can factor $x^6-1$ as $$x^6-1=(x^3-1)(x^3+1)=(x^2-1)(x^2+x+1)(x^2-x+1).$$ If $a^3-1\equiv 0\pmod p$ or $a+1\equiv 0\pmod{p}$, then $a$ has order $\lt 6$. So any element of order $6$ must be a solution of the congruence $x^2-x+1\equiv 0\pmod{p}$.

Conversely, let $p$ be a prime $\ge 7$. Suppose that $a^2-a+1\equiv 0\pmod p$. It is clear that $a\not\equiv 1\pmod{p}$. And we cannot have $a\equiv -1\pmod{p}$ unless $(-1)^2-(-1)+1\equiv 0\pmod{p}$, which forces $p=3$. And we cannot have $a^2+a+1\equiv 0\pmod{p}$, for that would imply that $(a^2+a+1)-(a^2-a+1)\equiv 0\pmod{p}$. This is impossible, since $2a\equiv 0\pmod{p}$ implies that $a\equiv 0\pmod{p}$, and thus that $a^2-a+1\not\equiv 0\pmod{p}$.

So for $p \ge 7$, the elements of order $6$ are precisely the solutions of the congruence $x^2-x+1\equiv 0\pmod p$. Suppose that $a$ is such a solution. Then $(1-a)^2-(1-a)+1=(1-2a+a^2)-(1-a)+1=a^2-a+1\equiv 0\pmod{p}$, and therefore $1-a$ is also a solution of $x^2-x+1\equiv 0\pmod{p}$. (The congruence has no more than $2$ solutions.)

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Hint: Think about the fact that $(-1)^2=1^2=1$. Something similar happens with cubes. $p$ cannot be $2$ or $3$ (why?).

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So does that imply that since $a^6 \equiv 1$, we have $(a^3)^2 \equiv 1 \pmod p$ and so $(-a^3)^2 \equiv 1 \pmod p$? – guest Oct 29 '12 at 7:27

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