we get the continued fraction of $\sf$ as $[2; 1, 1, 4, 1, 1, 4, \ldots]$.

The Chakravala method is an ancient Indian algorithm to solve Pell's equation. It is based on the Brahmagupta's identity of quadratic decomposition.

$(x_{1}^{2} - n \cdot y_{1}^{2}) \cdot (x_{2}^{2} - n \cdot y_{2}^{2}) = (x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2})^{2} - n \cdot (x_{1} \cdot y_{2} + x_{2} \cdot y_{1})^{2}$

$(x_{1}^{2} - n \cdot y_{1}^{2}) \cdot (x_{2}^{2} - n \cdot y_{2}^{2}) = (x_{1} \cdot x_{2} - n \cdot y_{1} \cdot y_{2})^{2} - n \cdot (x_{1} \cdot y_{2} - x_{2} \cdot y_{1})^{2}$

Using above Brahmagupta's identity, If $(x_{1}, y_{1}, k_{1})$ and $(x_{2}, y_{2}, k_{2})$ satisfy $(x_{1}^{2} - y_1^{2}) \cdot (x_{2}^{2} - y_2^{2}) = k_{1} \cdot k_{2} $, then $(x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2}, x_{1} \cdot y_{2} + y_{1} \cdot x_{2}, k_{1} \cdot k_{2})$ is also a solution of $(x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2})^{2} - n \cdot (x_{1} \cdot y_{2} + x_{2} \cdot y_{1})^{2} = k_{1} \cdot k_{2}$

[//]:#(First, we choose $y_{1} = 1$ and choose $x_{1} = \lfloor \sqrt n \rfloor$ such that $k_{1}$ is a small integer.)

[//]:#()

[//]:#(Then, Iteratively we adjust $x_{1}$ and $y_{1}$ so that $k_{1} = 1$. The solution is given by $(x_{1}, y_{1})$ to original Pell's equation.)

[//]:#()

[//]:#( Choosing m. At e)

-[Pell's equation - Wikipedia](https://en.wikipedia.org/wiki/Pell%27s_equation)

-[Periodic Continued Fractions](https://en.wikipedia.org/wiki/Periodic_continued_fraction)

-[Chakralava Method - Wikipedia](https://en.wikipedia.org/wiki/Chakravala_method)

-[Chakralava Method](https://www.isibang.ac.in/~sury/chakravala.pdf)

-[Codeforces Pell's Equation](https://codeforces.com/topic/116937/en20)

-[Project Euler 66](https://projecteuler.net/problem=66)

-[Hackerrank ProjectEuler-066](https://www.hackerrank.com/contests/projecteuler/challenges/euler066/problem)