# order

• Oct 5th 2009, 03:23 PM
dori1123
order
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 $x^3 - 1$, which is $(x-1)(x^2+x+1)$, but I don't see any relation between $x^3-1$ and $(x-1)^6$. Some help please.
• Oct 5th 2009, 04:48 PM
PaulRS
$x^3-1=(x-1)(x^2+x+1)=0$ -everything modulo p-

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

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