= Solution
The continued-fraction algorithm gives
$$
\sqrt{11}=3+\frac{\sqrt{11}-3}{1},\qquad
\frac1{\sqrt{11}-3}=\frac{\sqrt{11}+3}{2},\qquad
\frac1{(\sqrt{11}+3)/2-3}=\sqrt{11}+3,
$$
after which the process repeats. Thus
$$
\boxed{\sqrt{11}=[3;\overline{3,6}]}.
$$
The first two convergents are $3/1$ and $10/3$, of <field norms> $-2$ and $1$ respectively. The even period gives the fundamental solution
$$
\boxed{x=10,\qquad y=3}.
$$
It is also directly the smallest positive solution: for $y=1,2$, the values $1+11y^2$ are $12,45$, neither square; $y=3$ gives $100$. Cubing by the preceding <Pell equation> group law gives
$$
\boxed{(10,3)^{\circ3}=(3970,1197)}.
$$
Indeed $3970^2-11\cdot1197^2=1$.
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