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For a quadratic irrational $\sqrt{D}$, write its complete quotient as $(\sqrt{D}+m_j)/d_j$. The recurrence $m_{j+1}=d_j{a}_j-m_j$, $d_{j+1}=(D-m_{j+1}^2)/d_j$, $a_{j+1}=\lfloor (\sqrt{D}+m_{j+1})/d_{j+1}\rfloor$ computes its continued fraction. For $D=19$, the successive pairs after the initial quotient are $(4,3),(2,5),(3,2),(3,5),(2,3),(4,1)$, and then the cycle repeats. They give the displayed period of length six.

For a real number $\alpha$, set ${\alpha }_0=\alpha$ and $a_j=\lfloor {\alpha }_j\rfloor$. If ${\alpha }_j-a_j\ne 0$, define ${\alpha }_{j+1}=({\alpha }_j-a_j{)}^{-1}$. The resulting simple continued fraction is $[a_0;a_1,a_2,\dots ]$, where $a_0$ is an integer and subsequent $a_j$ are positive integers. It terminates when a complete quotient is integral; otherwise its convergents tend to $\alpha$.

For the continued fraction of the square root of nineteen, the complete quotients after $a_0=4$ areTheir integer parts are $2,1,3,1,2,8$. Taking the reciprocal after subtracting $8$ from the last quotient returns the first one, so

Let $D_n=p_{n-1}{q}_{n-2}-p_{n-2}{q}_{n-1}=(-1{)}^{n-2}$. Solving the given Möbius transformation for the complete quotient and rationalizing givesThereforeThe denominator $D_n$ is $\pm 1$, so both are integers. The algebraic norm denominator is nonzero because $\sqrt{d}$ is an irrational number.