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> $\textbf{Problem:}$ Let $a$ be a positive integer and $p$ a prime divisor of $a^3-3a+1$, with $p \neq 3$. Prove that $p$ is of the form $9k+1$ or $9k-1$, where $k$ is an integer.
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I tried to complete the cube which didn't turn out to be anything good. If the condition asked for $p$ to be only of the form $9k+1$, I would try to show that $a$ has order $9 \pmod{p}$. But the given condition seems somewhat odd to me.
So, overall I could hardly make any real progress.

Any kind of hint or solutions are appreciated.