Ancient ritual recovered, part 2

How does that method work? Say you have a number $x$ which can be written as $\sum_{i=0}^n 100^ix_i$, where the $x_i$ are positive integers less than 100, that is, they are pairs of digits in $x$. Start with $x_n$ and find the largest digit $a_n$ whose square is no greater than $x_n$, that is, $x_n=a_n^2+r_n$.

Now consider $100x_n+x_{n-1}$, that is, the number formed from the first two pairs of digits in $x$. Its square root is a little more than (or equal to) $10a_n+a_{n-1}$ where $a_{n-1}$ is the largest digit which, appended to $a_n$, gives you a number whose square is no greater than $100x_n+x_{n-1}$. That is, $100x_n+x_{n-1} = (10a_n+a_{n-1})^2+r_{n-1}$. But $(10a_n+a_{n-1})^2 = 100a_n^2+(20a_n+a_{n-1})a_{n-1}$ and so $100r_n+x_{n-1} = (20a_n+a_{n-1})a_{n-1}+r_{n-1}$. The number on the left is just the remainder from the previous step with the next pair of digits from $x$ appended, and the first term on the right is twice the result from the previous step with a new digit appended, multiplied by that new digit. Their difference gives our new remainder, $r_{n-1}$.

Iterate that process and you get each digit $a_n$ of the square root.