TITLE: If $p$ is congruent to $2 \pmod 3$, how can I prove that all $a$, $1 \le a \le p-1$ are cubic residues $\mod p$?
QUESTION [8 upvotes]: If $p$ is congruent to $2 \pmod 3$, how can I prove that all $a$, $1 \le a \le p-1$ are cubic residues $\mod p$?
Here's what I've done:
$1^3$ congruent to $1 \pmod p$ thus, 1 is a cubic residue,
Also
$(p-1)^3=(p-1) \pmod p$ implies 
$$(p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) $$ implies $$p|(p-1)^3 -(p-1)$$
Thus  $p-1$ is a cubic residue $\mod p$.
Now I don't know how to show for $1\lt a \lt p-1$ are all cubic residues when $p$ is congruent to $2\pmod 3$.

REPLY [11 votes]: We use more machinery than necessary, Let $g$ be a primitive root of $p$. Then for some $k$ we have $a\equiv g^k\pmod{p}$. 
Let $p=3n+2$. Then by Fermat's Theorem we have $g^{3n+1}\equiv 1\pmod{p}$, and $g^{6n+2}\equiv 1\pmod{p}$. 
Note that 
$$a\equiv g^k\equiv g^{k+3n+1}\equiv g^{k+6n+2}\pmod{p}.$$
Exactly one of $k$, $k+3n+1$, and $k+6n+2$ is divisible by $3$. Call it $3d$. Then 
$$a \equiv (g^d)^3 \pmod{p},$$ 
so we have obtained a  semi-explicit expression for the cube root of $a$ modulo $p$.