Solution
= Solution
Write $p=3k+2$. A solution has $p\nmid x$, so <Fermat's little theorem> gives
$$
1\equiv x^{p-1}=x^{3k+1}=(x^3)^kx\equiv(-1)^kx\pmod p.
$$
Thus $x\equiv(-1)^k\pmod p$. If $p$ is odd, then $k$ is odd and $x\equiv-1$. The remaining <prime> case $p=2$ has $k=0$ and $x\equiv1\equiv-1\pmod2$ as well. Since $-1$ itself always satisfies the equation, \b[the unique solution is $x\equiv-1\pmod p$].