<s>[INST] Order of elements modulo p Let $p$ be prime. Suppose that $x\in Z$ has order 6 mod p. Prove that $(1-x)$ has order 6 mod p as well.
I know that I need to show that the order can't be 2 or 3 (4 and 5 are trivial cases), but I'm even having difficulty showing that $(1-x)^6$ is 1. I've expanded $x^{6}-1$ to get a nice congruence that might be useful...but I can't seem to apply it.
 [/INST] If $ord_px=6\implies  p\mid(x^6-1)$ ,i.e,  $p\mid(x^3-1)(x^3+1)$
If $p\mid(x^3-1), x^3\equiv 1\pmod p\implies ord_px=3\ne 6$
So, $p\mid(x^3+1)\implies p\mid (x+1)(x^2-x+1)$ and $x^3\equiv -1\pmod p$
If $p\mid(x+1), x\equiv -1\pmod p\implies x^2\equiv 1\pmod p\implies ord_px=2\ne 6$
So, $p\mid(x^2-x+1)$
(i)  $ x^2\equiv x-1\pmod p$ 
So, $1-x=(-1)(x-1)\equiv  x^3\cdot x^2\pmod p$ as  $x^3\equiv -1\pmod p$
So, $1-x\equiv x^5\implies ord_p(1-x)=ord_p(x^5)$
We know, $ord_ma=d, ord_m(a^k)=\frac{d}{(d,k)}$ (Proof @Page#95)
So, $ord_m(x^5)=\frac{6}{(6,5)}=6$
(ii) $x-x^2\equiv1\pmod p$
$\implies 1-x\equiv x^{-1}$ dividing either side by $x$ as $(x,p)=1$
So, $ord_p(1-x)=ord_p(x^{-1})$
So, $ord_m(x^{-1})=\frac{6}{(6,-1)}=6$
Observation :
If $x^5\equiv x\pmod p, x^4\equiv 1\implies ord_px\mid 4$, but  $ord_px=6$
If $x\equiv x^{-1}\pmod p,x^2\equiv 1\implies ord_px\mid 2$
In fact, $x^{-1}\equiv x^5\pmod p$ as $x^6\equiv1$
We also know, that are exactly $\phi(6)=2$ in-congruent elements that belong to order $6\pmod p$ if $6\mid \phi(p)\implies p\equiv 1\pmod 6$.
In that case, we can conclude $x,x^{-1}\equiv x^5\equiv(1-x)$ are those two.
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