= 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 $x_m=q_{2m}$ obeys
$$
x_{m+2}=20x_{m+1}-x_m.
$$
The initial even convergents are
$$
\frac{p_0}{q_0}=\frac31,
\qquad
\frac{p_2}{q_2}=\frac{63}{19}.
$$
Put $u=10+3\sqrt{11}$. Since $N(u)=100-99=1$ and $u+u^{-1}=20$, the sequence
$$
w_m=(3+\sqrt{11})u^m
$$
satisfies the same recurrence. Its first two values are $3+\sqrt{11}$ and $63+19\sqrt{11}$, so
$$
p_{2m}+q_{2m}\sqrt{11}=(3+\sqrt{11})(10+3\sqrt{11})^m.
$$
Taking norms in $\mathbb Q(\sqrt{11})$ yields
$$
p_{2m}^2-11q_{2m}^2
=(9-11)(100-99)^m
=\boxed{-2}.
$$
This applies in particular whenever the original index $n=2m\geq2$ is even.
Back to article page
