Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 1F (Number Theory)

Apply the reciprocal-and-integer-part algorithm for the continued fraction. The successive complete quotients after the integer part two are $(\sqrt{7}+2)/3$, $(\sqrt{7}+1)/2$, $(\sqrt{7}+1)/3$, and $\sqrt{7}+2$. Their integer parts are $1,1,1,4$, and taking the next reciprocal repeats the first complete quotient. ThereforeThe first four continued fraction convergents are $2,3,5/2,8/3$. The last gives $8^2-7\cdot 3^2=1$. Squaring its unit $8+3\sqrt{7}$ gives $127+48\sqrt{7}$, whose norm is again one. Thus two positive solutions of the Pell equation areDirect verification gives $127^2-7\cdot 48^2=16129-16128=1$.