 Solution

Write $p=3k+2$. A solution has $p\nmid x$, so Fermat's little theorem givesThus $x\equiv (-1{)}^k(\mathrm{mod}p)$. If $p$ is odd, then $k$ is odd and $x\equiv -1$. The remaining prime case $p=2$ has $k=0$ and $x\equiv 1\equiv -1(\mathrm{mod}2)$ as well. Since $-1$ itself always satisfies the equation, the unique solution is $x\equiv -1(\mathrm{mod}p)$.