= Digit lifting for square roots of 2-adic units
{title2=$x_{n+1}=x_n+\varepsilon_n2^{n-1}$}
An odd square is congruent to one modulo eight. Conversely, for $u\equiv1\pmod8$, begin with $x_3=1$ and $x_n^2\equiv u\pmod{2^n}$. For $n\ge3$, adding $\varepsilon_n2^{n-1}$ changes the square by $\varepsilon_n2^nx_n$ modulo $2^{n+1}$. Since $x_n$ is odd, one choice of $\varepsilon_n\in\{0,1\}$ 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>.
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