Let $p \equiv 2 \mod{3}$. For any $a \in Z$ such that $ p \nmid a$ , show that there exists $x \in \mathbb{Z}_p$ with $x^3 = a$. Let $p \equiv 2 \mod{3}$. For any $a \in Z$ such that $ p \nmid a$ , show that there exists $x \in \mathbb{Z}_p$ with $x^3 = a$.
I've tried using Hensel's lemma and the fact that if $p \equiv 2 \mod{3}$ then the map $x \mapsto x^3$ is a bijection in $\mathbb{F}_p$ but have got stuck.
 A: This is the easy case for modular cube roots (assuming you mean $a \in \mathbb{Z}_p)$. The explicit solution for
$a \in \mathbb{Z}_p, a\ne 0$ is $x=a^{\frac{2p-1}{3}}.\,$ First note that
$3|2p-1$ and $a^p=a.$ Now the easy calculation
$$x^3 = \left(a^{\frac{2p-1}{3}}\right)^3 = a^{2p-1}=(a^p)^2a^{-1}=a^2a^{-1}=a$$
A: Once you’ve shown that $x\mapsto x^3$ is onto in $\mathbb F_p$, you really have it. For now, to find the (and there is only one) cube root of $a\in\mathbb Z_p$ when $a\notin p\mathbb Z_p$, it’s enough to find the cube root of any $b\equiv1\pmod p$, in other words $b=1+pz$ for suitable $z\in\mathbb Z_p$. But you can also write this as $b=1+9pz'$, with again $z'\in\mathbb Z_p$. But you can easily see that the binomial expansion for $(1+9t)^{1/3}$ has no $3$’s in the denominators, in fact this series has all coefficients in $\mathbb Z$, but that’s another story. But once you know that, substituting $pz'$ for $t$ in the series gives you a $p$-adically convergent solution to the (modified) problem. Note that this bypasses Hensel (or it would if he hadn’t invented $p$-adic numbers).
