order

**dori1123** · October 5th 2009, 03:23 PM

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.

**PaulRS** · October 5th 2009, 04:48 PM

$x^3-1=(x-1)(x^2+x+1)=0$ -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

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