<s>[INST] 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.
 [/INST] We will compute the continued fraction by the following algorithm (may or may not be the most efficient way in doing this, but it's the way I was taught!):
We set the following $$x:=\sqrt{119},\quad x_0=x,\quad a_i=\lfloor x_i\rfloor,\quad x_{i+1}=(x_i-a_i)^{-1},$$
and then we stop once we reach a loop in $x_i$ terms and list those terms as the continued fraction.
So, carrying out this procedure gives $$x_0=\sqrt{119},\quad a_0=\lfloor\sqrt{119}\rfloor=10,$$ $$x_1=\frac{1}{\sqrt{119}-10}=\frac{\sqrt{119}+10}{19},\quad a_1=\lfloor\frac{\sqrt{119}+10}{19}\rfloor=1,$$ $$x_2=\frac{1}{\frac{\sqrt{119}+10}{19}-1}=\frac{1}{\frac{\sqrt{119}-9}{19}}=\frac{19}{\sqrt{119}-9}=\frac{19(\sqrt{119}+9)}{38}=\frac{\sqrt{119}+9}{2},\quad a_2=\lfloor \frac{\sqrt{119}+9}{2}\rfloor=9,$$ $$x_3=\frac{1}{\frac{\sqrt{119}+9}{2}-9}=\frac{\sqrt{119}+9}{19},\quad a_3=\lfloor\frac{\sqrt{119}+9}{19}\rfloor=1,$$ $$x_4=\frac{1}{\frac{\sqrt{119}+9}{19}-1}=10+\sqrt{119},\quad a_4=\lfloor10+\sqrt{119}\rfloor=20,$$ $$x_5=\frac{1}{10+\sqrt{119}-20}=\frac{1}{\sqrt{119}-10}=x_1.$$ Since we have reached a loop in terms, we stop. Thus $$\sqrt{119}=[10;\overline{1,9,1,20}].$$ From here, we note that the period of this continued fraction is $4$ and so, the fundamental solution arises in the third partial convergent. We compute it as follows $$[10;1,9,1]=10+\frac{1}{1+\frac{1}{9+\frac{1}{1+1}}}=\frac{120}{11}.$$ So the pair $(120,11)$ is the smallest solution. Indeed $$120^2-119\times 11^2=1.$$ Finally, to generate all the solutions to this equation, we take the form $$(120+11\sqrt{119})^n,$$ where $n\in\Bbb N$.
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