Exercise 2.9.7 (Solution of $x^3 \equiv a \pmod p$ if $p \equiv 2 \pmod 3$)

Proof.

• Existence.

  If $p\equiv 2(\mathit{mod}3)$, then $(2p-1)∕3$ is an integer. Put $x\equiv a^{(2p-1)∕3}(\mathit{mod}p)$. Since $a\wedge p=1$, by Fermat’s theorem,

  $$\begin{array}{llll}\hfill x^3& \equiv {\left(a^{(2p-1)∕3}\right)}^3& \hfill & \\ \hfill & \equiv a^{2p-1}& \hfill & \\ \hfill & \equiv a\cdot a^{p-1}& \hfill & \\ \hfill & \equiv a(\mathit{mod}p),& \hfill & \end{array}$$

  so $x^3\equiv a(\mathit{mod}p)$.

• Unicity.

  Assume for contradiction that the congruence $x^3\equiv a(\mathit{mod}p)$ has two solutions $u,v$ such that $u\not\equiv v(\mathit{mod}p)$. Then $u\not\equiv 0(\mathit{mod}p)$, otherwise $p\mid u^3$, thus $p\mid a$, in contradiction with the hypothesis $a\wedge p=1$.

  Let $\bar{u}$ be an inverse of $u$ modulo $p$, and $\omega =\bar{u}v$. Then ${\omega }^3={\bar{u}}^3{v}^3\equiv 1(\mathit{mod}p)$, but $\omega \not\equiv 1(\mathit{mod}p)$, otherwise $u\equiv v(\mathit{mod}p)$. Therefore the order of $\omega$ is $3$. Since ${\omega }^{p-1}\equiv 1(\mathit{mod}p)$, $3\mid p-1$, thus $p\equiv 1(\mathit{mod}3)$. This is impossible, since $p\equiv 2(\mathit{mod}3)$. This contradiction shows that the solution of the congruence $x^3\equiv a(\mathit{mod}p)$ is unique.