 Solution

Directly,Thus take $\alpha =aC+A$, $\beta =aD+B$, $\gamma =C$, and $\delta =D$. They are nonnegative and

Each finite block of a continued fraction acts by a fractional linear transformation with an integral matrix of determinant $\pm 1$. If the block $a_1,\dots ,a_N$ repeats, removing one period leaves the same tail $x$, sofor integers $A,B,C,D$. Hence $C{x}^2+(D-A)x-B=0$, with the linear case allowed when $C=0$.

For period $(a,b)$,soThe continued fraction is positive, so it is the positive rootwhile the other root is negative.