 Solution

From part (b), $a_{2m+1}=3$ and $a_{2m+2}=6$. Eliminating the intervening odd-indexed convergent from the standard recurrence shows that either sequence $x_m=p_{2m}$ or $x_m=q_{2m}$ obeysThe initial even convergents arePut $u=10+3\sqrt{11}$. Since $N(u)=100-99=1$ and $u+u^{-1}=20$, the sequencesatisfies the same recurrence. Its first two values are $3+\sqrt{11}$ and $63+19\sqrt{11}$, soTaking norms in $\mathbb{Q}(\sqrt{11})$ yieldsThis applies in particular whenever the original index $n=2m\ge 2$ is even.