Hensel’s lemma for integers

Let $f(x)$ be a polynomial with integer coefficients, $p$ a prime number, and $n$ a positive integer.  Assume that an integer $a$ (and naturally its whole residue class modulo $p^n$) satisfies the congruence

$f(x)\equiv \mathrm{\hspace{0.33em}0}(mod{p}^n).$(1)

The solution  $x=a$  of (1) may be refined in its residue class modulo $p^n$ to a solution  $x=a+r{p}^n$  of the congruence

$f(x)\equiv \mathrm{\hspace{0.33em}0}(mod{p}^{n+1}).$(2)

This refinement is unique modulo $p^{n+1}$ iff  $f^{\prime }(a)\not\equiv 0(modp)$.

Proof.  Now we have  $f(a)=s{p}^n$.  We have to find an $r$ such that

$$f(a+r{p}^n)\equiv \mathrm{\hspace{0.33em}0}(mod{p}^{n+1}).$$

The short Taylor theorem requires that

$$f(a+r{p}^n)\equiv f(a)+r{f}^{\prime }(a)p^n(mod{r}^2{p}^{2n})$$

where  $2n\ge n+1$, whence this congruence can be simplified to

$$s{p}^n+r{f}^{\prime }(a)p^n\equiv \mathrm{\hspace{0.33em}0}(mod{p}^{n+1}).$$

Thus the integer $r$ must satisfy the linear congruence

$$s+r{f}^{\prime }(a)\equiv \mathrm{\hspace{0.33em}0}(modp).$$

When  $f^{\prime }(a)\not\equiv 0$,  this congruence has a unique solution modulo $p$ (see linear congruence); thus we have the refinement $a^{\prime }=a+r{p}^n$  which is unique modulo $p^{n+1}$.

When  $f^{\prime }(a)\equiv 0$  and  $s\not\equiv 0(modp)$,  the congruence evidently is impossible.

In the case  $f^{\prime }(a)\equiv s\equiv 0(modp)$  the congruence (2) is identically true in the residue class of $a$ modulo $p^n$.   □