I'm stuck on a problem and I'm wondering if I'm missing something obvious.
Let $p \equiv 2 \mod 3$ be an odd prime. The problem is to show that every value is a cubic residue mod $p$.
So far I have reduced the problem to showing that for $a \not\equiv b \mod p$, it is never true that $a^2 + ab + b^2 = 0 \mod p$. I originally felt like there was some way to apply CRT, but I couldn't find a way. I see that $a^2 + ab + b^2$ is always $0$ or $1 \mod 3$, never $2$. And I'd like to apply that fact. I rewrote the original modular equation as $a^2 + ab + b^2 = pm = (6k - 1)m$ for some integers $k$ and $m$, but then I get an "$m$" in there that I don't want.
I also tried to look for a pattern in the values of $a$ and $b$ that make $a^2 + ab + b^2 \equiv 0 \mod p$ when $p \not\equiv 2 \mod 3$, but I couldn't see any obvious pattern in the small primes, and it didn't look promising enough for me to go to the trouble of generating a large table of values. Although, if I don't have any more ideas, that will be my next step.
A couple of ways to see this.
You can, indeed, show that $a^2+ab+b^2\equiv0$ cannot happen non-trivially. If you set $x=a/b$ here, you get $$ x^2+x+1=0 $$ with $x\in\Bbb{F}_p$. But the discriminant of that quadratic is $-3$ so it has no solutions unless $-3$ is a quadratic residue modulo $p$. That implies $p\equiv1\pmod3$.
A possibly easier way is to use the fact that $\Bbb{F}_p^*$ is cyclic (equivalently, there is a primitive root $g$ modulo $p$). If $3\nmid p-1$ then cubing in that group is injective, hence surjective.