Finding two solutions to $x^2 - 6y^2 = 1$ using continued fractions

Question

Can anyone show me how to find the solutions to $x^2-6y^2=1$ by using continued fractions? I know how to find the fractions for $\sqrt6$ but do not know how to proceed. THANK YOU!!!

Answer 1

The easiest way is to guess one positive solution, namely in this case, $(x,y) = (5,2)$. This happens to be the smallest positive solution as well. Once this is done, use Brahmagupta's identity to construct more solutions.

Answer 2

Suppose you are able to find a solution to $x^2-6y^2=1$ like $u_1=5$ and $v_1=2$ as indicated in the answer given by user17762. Then ALL other positive integer solutions $(u_n,v_n)$ can be found by $$u_n+\sqrt{6}v_n=(u_1+\sqrt{6}v_1)^n.$$ Thus another solution will be $$u_2+\sqrt{6}v_2=(u_1+\sqrt{6}v_1)^2=(5+2\sqrt{6})^2=49+20\sqrt{6}.$$

In general, you can get a recurrence relation as follows: $$u_n+\sqrt{6}v_n=(u_{n-1}+\sqrt{6}v_{n-1})(u_1+\sqrt{6}v_1)=(u_{n-1}+\sqrt{6}v_{n-1})(5+2\sqrt{6}).$$ Upon comparison you get: \begin{align*} u_n & = 5u_{n-1}+12v_{n-1}\\ v_n & = 2u_{n-1}+5v_{n-1} \end{align*} This is exactly the recurence relation satisfied by the convergents of the continued fraction of $\sqrt{6}$.