Lagrange's theorem for regular continued fractions

If the continued fraction of $\alpha$ is eventually periodic, then [F1] shows that $\alpha$ is a quadratic irrational.

Suppose now that $\alpha$ is a quadratic irrational. By [F2], only finitely. [F2, given] many complete quotients ${\alpha }_n$ occur, so there exist indices $m<n$ with ${\alpha }_m={\alpha }_n\mathrm{.}$

From ${\alpha }_m={\alpha }_n$ and the determinism in [F3], the next digits agree: $a_m=a_n$, and then the next complete quotients agree: ${\alpha }_{m+1}={\alpha }_{n+1}$. Repeating this argument inductively gives $a_{m+j}=a_{n+j}\text{for every }j\ge 0,$ so the continued-fraction digits repeat with period $n-m$ from the index $m$ onward. Thus the continued fraction is eventually periodic.

Steps 1.1 and 2.1 prove both directions of the equivalence.