My impression is that you have developed $(x_0-\sqrt{S})^{2^n}$ with Newton binomial formula, separated the terms with even index from the terms with odd index to get an expression of the form $A_n-\sqrt{S}B_n$ to approach $\sqrt{S}$ by $A_n/B_n$.

More precisely, you have $$A_n = \frac{(x_0+\sqrt{S})^{2^n}+ (x_0-\sqrt{S})^{2^n}}{2}$$ and $$B_n = \frac{(x_0+\sqrt{S})^{2^n}-(x_0-\sqrt{S})^{2^n}}{2\sqrt{S}}.$$

The successive approximations you get are the same as approximations provided by Newton method applied to the function $f : x \mapsto x^2-s$. Indeed, $A_0/B_0=x_0$ and for every $n \ge 0$, $$\frac{A_n}{B_n} - \frac{f(A_n/B_n)}{f'(A_n/B_n)} = \frac{A_n}{B_n} - \frac{A_n^2/B_n^2 - s}{2A_n/B_n} = \frac{A_n^2+sB_n^2}{2A_nB_n} = \frac{A_{n+1}/2}{B_{n+1}/2}.$$

My impression is that you have developed $(x_0-\sqrt{S})^{2^n}$ with Newton binomial formula, separated the terms with even index from the terms with odd index to get an expression of the form $A_n-\sqrt{S}B_n$ to approach $\sqrt{S}$ by $A_n/B_n$.

More precisely, you have $$A_n = \frac{(x_0+\sqrt{S})^{2^n}+ (x_0-\sqrt{S})^{2^n}}{2}$$ and $$B_n = \frac{(x_0+\sqrt{S})^{2^n}-(x_0-\sqrt{S})^{2^n}}{2\sqrt{S}}.$$

The successive approximations you get are the same as approximations provided by Newton method applied to the function $f : x \mapsto x^2-s$. Indeed, $A_0/B_0=x_0$ and for every $n \ge 0$, $$\frac{A_n}{B_n} - \frac{f(A_n/B_n)}{f'(A_n/B_n)} = \frac{A_n}{B_n} - \frac{A_n^2/B_n^2 - s}{2A_n/B_n} = \frac{A_n^2+sB_n^2}{2A_nB_n} = \frac{A_{n+1}}{B_{n+1}},$$ since $4(A_n^2+sB_n^2) = 2A_{n+1}$ and $4A_nB_n = 2B_{n+1}$.