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Put $f(X)=X^3+5X-12$. Any root in ${\mathbb{Q}}_p$ is a p-adic integer: if its valuation were negative, $X^3$ would be the unique term of least valuation.

For $p=2$, reduction modulo two has roots zero and one. The root zero is simple because $f^{\prime }(0)=5$ is odd, so it lifts uniquely. An odd integer $x$ satisfies $x^2\equiv 1(\mathrm{mod}8)$, and henceThus there is no odd $2$-adic root and the number of roots is one.

For $p=3$,All three roots are simple because $f^{\prime }(X)=3X^2+5\equiv 2(\mathrm{mod}3)$. Each lifts uniquely, giving three roots in ${\mathbb{Q}}_3$.

For $p=5$, reduction gives $X^3-2$, whose unique root is $X=3$; it is simple because $3X^2\not\equiv 0(\mathrm{mod}5)$. It lifts uniquely, so there is one root in ${\mathbb{Q}}_5$.

An element of the p-adic integers is a unit in a ring exactly when its reduction modulo $p$ is nonzero. More explicitly, if $\alpha \not\equiv 0(\mathrm{mod}p)$, its inverses modulo $p^n$ are unique and compatible, so they define $\beta \in {\mathbb{Z}}_p$ with $\alpha \beta =1$. The converse follows by reducing $\alpha \beta =1$ modulo $p$. Part ii therefore proves that $\alpha$ topologically generates the additive group if and only if $\alpha$ is a p-adic unit.

The ring of p-adic integers is the inverse limitEquivalently, each element has a unique convergent expansion ${\sum }_{j\ge 0}{b}_j{p}^j$ with digits $0\le b_j<p$. It is a complete discrete valuation ring with maximal ideal $(p)$ and residue field ${\mathbb{F}}_p$.