Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 11G (Number Theory)

Setand for $n\ge 0$ defineThen $p_n/q_n=[a_0,\dots ,a_n]$ is the $n$th convergent.

For a variable final tail $x$, induction on $n$, or multiplication of the continued-fraction matrices, givesIndeed, the identity is immediate for $n=0$, and replacing the tail by $a_{n+1}+1/x$ gives the recurrence above. Taking $x=\gamma >0$ proves the formula.

The continued-fraction algorithm starts with $a_0=3$. Rationalizing the successive remainders givesandThe same remainder has returned, so the digits $3,6$ repeat. Therefore

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.