Exercise 3.1.18* (Cubic residues)

Proof. Let $p$ be a prime number.

• Suppose that $p\equiv 2(\mathit{mod}3)$. If $a\wedge p=1$, $a^{p-1}\equiv 1(\mathit{mod}3)$, and $3\mid 2p-1$, thus

  $$a\equiv a\cdot a^{p-1}\equiv a^{2p-1}\equiv {\left(a^{(2p-1)∕3}\right)}^3(\mathit{mod}p).$$

  If $x=a^{(2p-1)∕3}$, then $a\equiv x^3(\mathit{mod}p)$, thus every $a$ such that $a\wedge p=1$ is a cubic residue modulo $p$.

• Suppose now that $p\equiv 1(\mathit{mod}3)$. Let $a$ be an integer such that $a\wedge p=1$. By Theorem 2.37,

  $$\exists <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z},a\equiv x^3(\mathit{mod}p)⟺a^{(p-1)∕3}\equiv 1(modp).$$

  Let $g$ be a primitive root modulo $p$. Then $a\equiv g^t(\mathit{mod}p)$ for some integer $t$ such that $0\le t<p-1$. Since the order of $g$ is $p-1$,

  $$\begin{array}{llll}\hfill a^{\frac{p-1}{3}}\equiv 1(\mathit{mod}p)& ⟺g^{t\frac{p-1}{3}}\equiv 1(\mathit{mod}p)& \hfill & \\ \hfill & ⟺p-1\mid t\frac{p-1}{3}& \hfill & \\ \hfill & ⟺3\mid t.& \hfill & \end{array}$$

  Therefore, if $\gamma =g^3$, the set of cubic residues modulo $p$ is

  $$C=\{1,g^3,g^6,\dots ,g^{p-4}\}=\{\gamma ,{\gamma }^2,\dots ,{\gamma }^{(p-1)∕3}\},$$

  (where ${\gamma }^{(p-1)∕3}\equiv 1(\mathit{mod}3)$), thus $\mathrm{Card}(C)=(p-1)∕3$.

  Exactly one-third of the members of a reduced residue system are cubic residues.