# Inverting the solutions of Pell's equations

Suppose I have a solution of a Pell's equation $$y_1,x_1$$, then I can find the subsequent solutions by recursively using the relation $$(y_{i},x_{i})=x_{i-1}y_1+y_{i-1}x_1,x_{i-1}x_1+Dy_{i-1}y_1$$ i.e. $$(x_1+y_1\sqrt{D})^i$$. But, if I have ONLY the $$i^{th}$$ solution, is there a way to get the first solution easily? It is straightforward if $$i=2$$, but for any number above that, if I solve by Wolfram, the expression for the solution seems to explode. So, is there an easy recursive way to get back the first solution, given the $$i^{th}$$ solution?

• WA gives a clean solution. Note that the denominator is $\pm 1$. – lhf Jul 13 '19 at 11:21
• Well, that is for $i=2$ as I mentioned in the original question. For i>2, knowing only the $i^{th}$ solution makes it a painful process to revert. – silvestre_dubois Jul 13 '19 at 11:25

From $$y_{i}=x_{i-1}y_1+y_{i-1}x_1, \qquad x_{i}=x_{i-1}x_1+Dy_{i-1}y_1$$ we get $$x_1 = \frac{x_{i-1} x_{i} - y_{i-1} y_{i} D}{x_{i-1}^2 - y_{i-1}^2 D}, \qquad y_1 = \frac{x_{i-1} y_{i} - x_{i} y_{i-1}}{x_{i-1}^2 - y_{i-1}^2 D}$$ Note that the denominator is $$\pm1$$.
If you're only given $$z=x_i+y_i\sqrt{D}=(x_1+y_1\sqrt{D})^i$$, then you could take the $$i$$-th root of $$z$$ and compare its fractional part with the fractional part of the integer multiples of $$\sqrt{D}$$. This will give $$y_1$$ and then $$x_1$$. But you may need high-precision approximation of $$\sqrt{D}$$.
• The point is I know only the $i^{th}$ solution and nothing else. Therefore, my original question. – silvestre_dubois Jul 13 '19 at 11:24