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Starting from $a_0$, Newton iteration sets $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Under the strong form of Hensel lemma inequality, the correction valuations grow at least geometrically, so $(a_n)$ is a Cauchy sequence converging to a root.

The strong form of Hensel lemma says the following. Let $v$ be a discrete valuation on a complete field $K$. If $f\in {\mathcal{O}}_K[X]$ andthen $f$ has a root $x\in {\mathcal{O}}_K$ satisfying $v(x-x_0)>v(f^{\prime }(x_0))$.

To prove it, apply Newton iteration over a valued field:The initial inequality says that the first correction has valuation greater than $v(f^{\prime }(x_0))$. Taylor expansion then shows inductively that $v(f^{\prime }(x_n))=v(f^{\prime }(x_0))$, while the valuations of the corrections tend to infinity. Hence $(x_n)$ is a Cauchy sequence. Completeness gives a limit $x$, and continuity gives $f(x)=0$.

Now decompose the multiplicative group asThe P-adic valuation givesand cubing is the identity on $\{\pm 1\}$. Put $U_2=1+9{\mathbb{Z}}_3$. Expansion gives $U_1^3\subseteq U_2$. Conversely, for $u=1+9a\in U_2$, choose $b\equiv a(\mathrm{mod}3)$ and put $x_0=1+3b$. For $f(X)=X^3-u$,so the strong form of Hensel lemma produces a cube root in $U_1$. Thus $U_1^3=U_2$. Finally,is an isomorphism. Combining the valuation and principal-unit factors proves the cube-class group of the 3-adic numbers identity

Assume $|a{|}_2>1$, equivalently $v_2(a)<0$, and writeFor $f(Y)=Y^2-(1+4/a^4)$ at $Y=1$,The strong form of Hensel lemma therefore gives $y\in {\mathbb{Q}}_2$ with $y^2=1+4/a^4$. Taking $X=a^{2}y$ yields $X^2=a^4+4$. Hence every $a$ with $|a{|}_2>1$, and in particular every sufficiently large $a$, works.