Digit lifting for square roots of 2-adic units ($x_{n+1}=x_n+{\epsilon }_n{2}^{n-1}$)

An odd square is congruent to one modulo eight. Conversely, for $u\equiv 1(\mathrm{mod}8)$, begin with $x_3=1$ and $x_n^2\equiv u(\mathrm{mod}{2}^n)$. For $n\ge 3$, adding ${\epsilon }_n{2}^{n-1}$ changes the square by ${\epsilon }_n{2}^n{x}_n$ modulo $2^{n+1}$. Since $x_n$ is odd, one choice of ${\epsilon }_n\in \{0,1\}$ gives the next congruence. The corrections tend to zero 2-adically, so completeness produces an odd square root. This directly proves the unit square criterion without needing a unit derivative in Hensel lemma.