Hensel's Lemma for Composite Numbers

Theorem
Let $b \in \Z \setminus \set {-1, 0, 1}$ be an integer.

Let $k > 0$ be a positive integer.

Let $\map f X \in \Z \sqbrk X$ be a polynomial.

Let $x_k \in \Z$ such that:
 * $\map f {x_k} \equiv 0 \pmod {b^k}$
 * $\gcd \set {\map {f'} {x_k}, b} = 1$

Then for every integer $l \ge 0$, there exists an integer $x_{k + l}$ such that:
 * $\map f {x_{k + l} } \equiv 0 \pmod {b^{k + l} }$
 * $x_{k + l} \equiv x_k \pmod {b^k}$

and any two integers satisfying these congruences are congruent modulo $b^{k + l}$.

Moreover, for all $l \ge 0$ and any solutions $x_{k + l}$ and $x_{k + l + 1}$:
 * $x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$
 * $x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$

Proof
We use induction on $l$.

The base case $l = 0$ is trivial.

Let $l \ge 0$ be such that a solution $x_{k + l}$ exists and is unique up to a multiple of $b^{k + l}$.

Choose a solution $x_{k + l}$.

Each solution $x_{k + l + 1}$ is also a solution of the previous congruence.

By uniqueness, it has to satisfy $x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$, hence is of the form $x_{k + l} + t b^{k + l}$ with $t \in \Z$.

Let $d = \deg f$.

We have:

Because $\map {f'} {x_{k + l} } \equiv \map {f'} {x_k} \pmod b$, $\map {f'} {x_{k + l} }$ is invertible modulo $b$.

Thus $x_{k + l} + t b^{k + l}$ is a solution modulo $b^{k + l + 1}$ :
 * $t \equiv - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod b$

Thus, necessarily:
 * $x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$

which proves the existence and uniqueness.

By induction, we have shown uniqueness and existence for all $l \ge 0$, as well as the relations:
 * $x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$
 * $x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$

Also see

 * Hensel's Lemma