Bounded representatives for $x^2-7y^2=57$

The least positive norm-one solution for $D=7$ is $8+3\sqrt{7}$, since $8^2-7\cdot 3^2=1$ and the smaller positive values $y=1,2$ give $x^2=8,29$, neither a square. Substituting ${\epsilon }_7=8+3\sqrt{7}$ and $N=57$ into [F1] gives $\mathrm{\mid }{x}_0\mathrm{\mid }\le \frac{\sqrt{57}}{2}(\sqrt{8+3\sqrt{7}}+(8+3\sqrt{7}{)}^{-1\mathrm{/}2})<17,$ $\mathrm{\mid }{y}_0\mathrm{\mid }\le \frac{\sqrt{57}}{2\sqrt{7}}(\sqrt{8+3\sqrt{7}}+(8+3\sqrt{7}{)}^{-1\mathrm{/}2})<7.$

By [F2], it is enough to enumerate solutions in that finite box. The integer solutions there are exactly $(\pm 8,\pm 1)$ and $(\pm 13,\pm 4)$, because $8^2-7\cdot 1^2=57,{13}^2-7\cdot 4^2=57,$ and no other pair with $\mathrm{\mid }x\mathrm{\mid }\le 16$, $\mathrm{\mid }y\mathrm{\mid }\le 6$ satisfies the equation. To compare their Pell orbits, note that every nonzero power of ${\epsilon }_7=8+3\sqrt{7}$ has absolute value either at least ${\epsilon }_7$ or at most ${\epsilon }_7^{-1}$. But among the eight displayed solutions, every quotient of two distinct positive representatives has absolute value between $\frac{8-\sqrt{7}}{13+4\sqrt{7}}\text{and}\frac{13+4\sqrt{7}}{8-\sqrt{7}},$ and this interval lies strictly inside $({\epsilon }_7^{-1},{\epsilon }_7)$. So no ratio of two distinct displayed solutions is a power of ${\epsilon }_7$, and negative ratios are impossible because every power of ${\epsilon }_7$ is positive. Therefore these eight boxed solutions represent eight distinct Pell orbits.