Pell's equation (or a special case of a second order diophantine equation) Question Find integers $x,y$ such that $$x^2-119y^2=1.$$
So far I've tried computing the continued fraction of $\sqrt{119}$ to find the minimal solution, but either I messed up or I don't know where to stop computing a rough approximation of said square root.
Please help.

COMMENT.-To solve the Pell-Fermat equation $a^2-db^2=1$ (or $-1$) the true problem is the calculation of the fundamental unit of $\mathbb Q(\sqrt d)$. Another way is the following algorithm (Pierre Samuel):
Since $a_n+b_n\sqrt d=(a_1+b_1\sqrt d)^n$ we have easily the relation $a_1b_n+b_1b_n=b_{n+1}$ where the sequence of $b_n$ is increasing so one can calculate the numbers $db^2$ for $b\ge1$ and stop when $db^2+1$ is an square. In your case $d=119$ and we have the numbers
$$119\cdot1\\119\cdot4=476\\119\cdot9\\\cdots\\\cdots\\119\cdot100=11900\\119\cdot121=14399$$
it is apparent that $14399+1=14400$ which is an obvious square so we get $a_1+b_1\sqrt{11}=\sqrt{14400}+11\sqrt{119}=120+11\sqrt{119}$.
Whatever the used algorithm, it is clear that the procedure could be tedious for many values of $d$. In this example discarding of the ten first numbers was immediate at a glance.
