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.

The algorithm in the first answer works for the general case, but in this specific case, noticing that $119$ is very close to the perfect square $121$ suggests that $y=11$ is worth a look. Then, $119\cdot 121=(120-1)(120+1)=120^2-1$ leads directly to $x=120, y=11$ as a solution.
