When is $x^3 - 2$ congruent to $ 0 \pmod p$ not solvable? I want to find for what primes $p$ is the congruence $x^3 \equiv 2$ (mod $p$) is not solvable. I already know that $p = 7$ is one result from my own work but I'm trying to generalize this to find for what primes $p$ modulo an integer where the congruence is not solvable. I'm not sure, but I think that I can use cubic reciprocity but I've hit a wall and I haven't been able to move on from there. Any help would be appreciated.
 A: It is soluble if $p\equiv0$ or $2\pmod3$. If $p\equiv1\pmod3$
then we can write $p=|a+b\omega|^2$ where $a$, $b\in \Bbb Z$ and
$\omega=\frac12(-1+i\sqrt3)$, that is $p=a^2-ab+b^2$.
In addition, we can assume that $\pi$ is primary, that is $3\mid b$.
Write $\pi=a+b\omega$.
By cubic reciprocity, $x^3\equiv2\pmod p$ is soluble if
$\left(\frac2{\pi}\right)_3=1$. (This is a cubic Legendre symbol).
But $\left(\frac2{\pi}\right)_3=\left(\frac{\pi}2\right)_3$
(cubic reciprocity). Now $\left(\frac{\pi}2\right)_3=1$ iff $b$
is even. So $x^3\equiv2\pmod p$ is insoluble if $p\equiv1\pmod 3$
and $p=a^2-ab+b^2$ with $b$ an odd multiple of $p$.
A: In order for $x^3-2=0\pmod p$ not to be solvable (no integer solutions $x$), $p=1\pmod 3$ and $2$ is a cubic non-residue $\pmod p$. 
The harder question to ask is, when is $2$ a cubic non-residue? 
Many articles, and research on cubic reciprocity are known, and $2$ is a cubic residue $\pmod p$ if and only if $p=x^2+27y^2$. If $p=1\pmod 3$ is not of this form, then $x^3-2=0\pmod p$ is not solvable.
