Check that $r = 55 + 12\sqrt{21}$ produces the least positive solution of
$$x^2− 21y^2 = 1$$
Determine two more positive solutions of $x^2 − 21y^2 = 1$.
The following table confirms that values of $y$ in the range $[1,11]$ do not produce integer solutions for $x$.
| y | 1+21*y^2 | x=sqrt(1+21*y^2) |
| 1 | 22 | 4.69041575982343 |
| 2 | 85 | 9.21954445729289 |
| 3 | 190 | 13.7840487520902 |
| 4 | 337 | 18.3575597506858 |
| 5 | 526 | 22.9346898823594 |
| 6 | 757 | 27.5136329843952 |
| 7 | 1030 | 32.0936130717624 |
| 8 | 1345 | 36.6742416417845 |
| 9 | 1702 | 41.2553026894725 |
| 10 | 2101 | 45.8366665454634 |
| 11 | 2542 | 50.4182506638221 |
| 12 | 3025 | 55 |
And so $(x,y)=(55,12)$ is the least positive solution.
Two further solutions are $r^2$ and $r^3$.
$$ r^2 = 6049 + 1320 \sqrt(21) $$
$$r^3 = 665335 + 145188 \sqrt{21}$$
And so two further positive solutions are $(x,y)=(6049,1320)$ and $(x,y)=(665335, 145188)$.