Consider the polynomial $f(x)$ such that $f(x) = x^2 - a$ for some constant $a$. Using Newton-Raphson, we can approximate the roots of this polynomial, which will be $\pm \sqrt{a}$, as follows: $$x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right)$$
Is there a solution to this recurrence? Substituting $x_n = {\lambda}^n$ and $x_n = A^{2^n}$ give the following equations: $$2{\lambda}^{2n+1} = {\lambda}^{2n} + a$$ and $$2A^{3 \cdot 2^n} = A^{2\cdot2^n} + a$$ respectively. I have tried solving both by substituting $A^{2^n} = y$ as well as ${\lambda}^{2^n} = z$, which gives a quadratic as well as a cubic. I am unable to solve the cubic, but solving the quadratic gives: $$\lambda = \left({\frac{1\pm\sqrt{1+8a}}{4}}\right)^{\frac{1}{2^n}}$$ Ofcourse, the actual equation $\lambda^{2n} = \frac{1\pm\sqrt{1+8a}}{4}$ has $2n$ roots, and the above is just one of the roots, and there is no general formula for solving afaik for $n>2$. Does this mean that there are multiple solutions?
I have also tried using the inverse difference operator, $\Delta^{-1}$. This would give:$$x_n = \Delta^{-1}\left({\frac{a}{x_n} - \frac{x_n}{2}}\right)$$ which is just the sum from $0$ to $n$ of ${\frac{a}{x_n} - \frac{x_n}{2}}$, which again I have been unable to solve. All of this makes me feel that there isn't a solution to the recurrence given by Newton-Raphson for the case $f(x) = x^2 - a$. If this is the case, could someone please provide me a hint on how to prove this argument?