$x$ is a cubic residue mod p if it is of the form $a^3\pmod{p}$ for some residue $a$. Show if $p\equiv 1\pmod{3}$, then $x\pmod{p}$ is a cubic residue iff $x^{(p-1)/3} \equiv 1\pmod{p}$. Also, show if $p\equiv 2\pmod{3}$, then all $x\pmod{p}$ are cubic residues.
For the forward implication, I can do this easily by assuming $x\equiv a^3\pmod{p}$ for some $1\leq a\leq p-1$. Then $x^{(p-1)/3}\equiv (a^3)^{(p-1)/3} \equiv a^{p-1} \equiv 1\pmod{p}$ by Fermat's Theorem since p is prime so $\textrm{gcd}\,(a,p)=1$.
For the reverse implication, this is what I have so far. If $p\equiv 1\pmod{3}$, then $p=3k+1$ for some integer $k$, so if $x^{(p-1)/3} \equiv1\pmod{p}$, then $x^{(3k+1-1)/3}\equiv1\pmod{p}$, so $x^k \equiv1\pmod{p}$. But this is where I'm stuck. I don't know if what I've done is useful or not.
Finally, I am stuck on the last part as well where $p\equiv2\pmod{3}$. I think I may want to use the fact that for some r, I can say that the set $\{r,r^2,r^3,…,r^{p-1}\equiv1\}$ is the same set $\pmod{p}$ as $\{1,2,3,…,p-1\}$, because then this set represents all possible values of $x$. But I'm not sure how to use this.
Best Answer
For $p \equiv 1 ($mod $3)$. Use the fact that $-3$ is a quadratic residue for $p$ to show that there are exactly three solutions for $x^3-1 \equiv 0 ($mod $p)$. The key step is to make use of the identity $4(x^2+x+1)=(2x+1)^2+3$. Then we can conclude there are exactly $(p-1)/3$ non-zero cubic residues. All of them have to satisfy the equation $x^{(p-1)/3}-1 \equiv 0 ($mod $p)$. Since there are at most $(p-1)/3$ solutions for $x^{(p-1)/3}-1 \equiv 0 ($mod $p)$, the set of non-zero cubic residues coincides with the set of the solutions.
For $p \equiv 2 ($mod $3)$, Bezout's lemma gives us $a,b \in \mathbb{Z}$ such that $1=3a+(p-1)b$. For any $r$, $r \equiv r^{(3a+(p-1)b)} \equiv r^{3a} \equiv (r^a)^3 $mod $p$.