Thanks for the hint but I probably should have asked about equations of the form $\displaystyle x^n \equiv 1 mod p$ with p prime as the above hint for this one hasn't helped me.
Factorizing that I can only see two solutions, 1 and 18.
How come?? $\displaystyle x^6-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$$\displaystyle
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. The roots $\displaystyle 1,18=-1\!\!\!\pmod {19}$ are immediate, and we're left now with the quadratics, so let's check the discriminant which, by the way, is identical in both!: $\displaystyle \Delta=1-4=-3$ , and the questions of questions now is whether $\displaystyle -3$ is a quadratic residue modulo 19...but this is trivial and we don't even need quadatic reciprocity, right? The complete set of roots is $\displaystyle -1,1,7,12,8,11$ Tonio
I can't find anything in our notes about this and it's stuck in the middle of a tutorial consisting of primitive root questions so I don't know how to approach it.