Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 11F (Number Theory)

Setand recursivelyThen $p_n/q_n=[a_0,ldots,a_n]$. The determinant identityfollows by induction. Writing the remaining complete quotient as ${\theta }_{n+1}>1$ givesso $\theta$ lies strictly between consecutive convergents and their order alternates. For odd $n$,Moreover $|\theta -p_n/q_n|<1/(q_n{q}_{n+1})$, and $q_n\to \infty$, proving convergence.

The continued-fraction algorithm gives

The convergents beginSincea strictly positive solution is $(x,y)=(8,3)$.

Reducing $x^2-7y^2=-1$ modulo seven gives $x^2\equiv -1\equiv 6(\mathrm{mod}7)$. The quadratic residues modulo seven are $0,1,2,4$, so no solution exists.

Reducing $x^2-7y^2=5$ modulo seven gives $x^2\equiv 5(\mathrm{mod}7)$, again impossible because five is not a quadratic residue modulo seven.

The denominators of the convergents areThey increase strictly from $q_2$ onward, and none equals $13$. Equivalently, $34/13=[2;1,1,1,1,2]$, which ceases to agree with $[2;\overline{1,1,1,4}]$ after the third partial quotient. Hence there is no positive index $x$ with $p_x/q_x=34/13$.