Exercise 2.6.3 (Solve $x^3 + x + 57 \equiv 0 \pmod{5^3}$)

Proof. Here $x_1=4\equiv -1(\mathit{mod}5)$ is the only solution of $x^3+x+57\equiv 0(\mathit{mod}5)$.

Here $f^{\prime }(x)=3x^2+1$, and $f^{\prime }(-1)=4\not\equiv 0(\mathit{mod}5)$, so $-1$ is a nonsingular root. So $4$ lifts to a unique root modulo $p^j$ for all $j$. Since $f(4)=125=5^3\equiv 0(\mathit{mod}125)$, $4$ is the unique solution modulo $125$.

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