<s>[INST] Solve $x^8 \equiv 3 \pmod {13}$ I need to find all solutions to $x^8 \equiv 3 \pmod {13}$.
What I've tried: I know $2$ is a primitive root modulo $13$.
So it is equivalent to solve
$2^{8t} \equiv 2^4 \pmod {13}$
Then I get $t = 2 + 3k$.
I think I'm wrong... and if not, what are the final solutions?? 
 [/INST] In $\mathbb{F}_{13}$,
$$ x^8-3 = x^8-16 = (x^4-4)(x^4+4)=(x^2-2)(x^2+2)(x^2+2x+2)(x^2-2x+2) $$
hence we just have to check which of the numbers $-2,2,-1$ are quadratic residues. Since $13$ is a prime of the form $8k+5$, the only quadratic residue among the previous ones is $-1$, so $x^8-3$ splits as:
$$ x^8-3 = (x-4)(x+4)(x-6)(x+6)(x^2-2)(x^2+2) $$
and the solutions of $x^8\equiv 3\pmod{13}$ are given by $x\in\{\pm 4,\pm 6\}\pmod{13}$.
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