Hensel's Lemma/First Form

Theorem
Let $p$ be a prime number.

Let $k>0$ be a positive integer.

Let $f(X) \in \Z[X]$ be a polynomial.

Let $x_k\in\Z$ such that:
 * $f(x_k)\equiv 0 \pmod{p^k}$.
 * $f'(x_k)\not\equiv 0 \pmod{p}$.

Then for every integer $l\geq 0$, there exists an integer $x_{k+l}$ such that:
 * $f(x_{k+l})\equiv 0 \pmod{p^{k+l}}$
 * $x_{k+l}\equiv x_k\pmod{p^k}$

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

Moreover, for all $l\geq0$ and any solutions $x_{k+l}$ and $x_{k+l+1}$:
 * $x_{k+l+1}\equiv x_{k+l}-\frac{f(x_{k+l})}{f'(x_{k+l})}\pmod{p^{k+l+1}}$
 * $x_{k+l+1}\equiv x_{k+l}\pmod{p^{k+l}}$

Proof
We use induction on $l$.

The base case $l=0$ is trivial.

Let $l\geq0$ be such that a solution $x_{k+l}$ exists and is unique up to a multiple of $p^{k+l}$.

Choose a solution $x_{k+l}$ satisfying:
 * $f(x_{k+l})\equiv 0 \pmod{p^{k+l}}$
 * $x_{k+l}\equiv x_k\pmod{p^k}$

By Congruence by Divisor of Modulus, each solution $x_{k+l+1}$ is also a solution of the previous congruence.

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

Let $d=\deg f$.

We have, for all $t\in\Z$:

Because $f'(x_{k+l})\equiv f'(x_k)\not\equiv 0 \pmod p$, $f'(x_{k+l})$ is invertible modulo $p$.

Thus $x_{k+l} + tp^{k+l}$ is a solution modulo $p^{k+l+1}$ :
 * $t\equiv -\frac{f(x_{k+l})}{f'(x_{k+l})}\pmod{p}$

Thus, necessarily:
 * $x_{k+l+1}\equiv x_{k+l} - \frac{f(x_{k+l})}{f'(x_{k+l})} \pmod {p^{k+l+1}}$.

which proves the existence and uniqueness.

By induction, we have shown uniqueness and existence for all $l\geq0$, as well as the relations:
 * $x_{k+l+1}\equiv x_{k+l}-\frac{f(x_{k+l})}{f'(x_{k+l})}\pmod{p^{k+l+1}}$
 * $x_{k+l+1}\equiv x_{k+l} \pmod{p^{k+l}}$

Also see

 * Hensel's Lemma for Composite Numbers