= Solution
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 $f'(0)=5$ is odd, so it lifts uniquely. An odd integer $x$ satisfies $x^2\equiv1\pmod8$, and hence
$$
f(x)=x(x^2+5)-12\equiv6x-4\not\equiv0\pmod8.
$$
Thus there is no odd $2$-adic root and the number of roots is one.
For $p=3$,
$$
f(X)\equiv X^3-X=X(X-1)(X+1)\pmod3.
$$
All three roots are simple because $f'(X)=3X^2+5\equiv2\pmod3$. 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 $3X^2\not\equiv0\pmod5$. It lifts uniquely, so there is one root in $\mathbb Q_5$.
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