Originally Posted by

**Ant** I think I've found a proof for the following but would really appreciate if someone could either point out where I've gone wrong (there are a couple of steps which I'm not 100% happy with) or perhaps suggest a more simple approach. Anyway, here is the problem:

Let p be prime. Let p == 2 mod 3 and p-1 = 4q (for some prime q)>

1) a) Prove $\displaystyle 3^4 \not\equiv 1 \, \, \: (mod p)$

My proof:

$\displaystyle 3^4 \equiv 1 \, (mod p) \,\:\:\Leftrightarrow\:\:3^2 \equiv -1\,(mod p)\Leftrightarrow\:\:3^2\equiv\(p-1)\,(mod p) \:\:\Leftrightarrow\:\:3\equiv \frac{p-1}{3}\,(mod\, p)\:\:\Leftrightarrow\:\: 3+kp \equiv \frac{p-1}{3} \:\:\Leftrightarrow\:\: 9 +3kp = p-1\:\:\Leftrightarrow\:\:\(p-1)\equiv0\,(mod3)\:\:\Leftrightarrow\:\:\ p\equiv1(mod3)$

which contradictions our assumption that $\displaystyle p\equiv 2\,(modp) $

Could anyone tell me if this proof is correct? Thanks!