Here, they write the solution with both
(4d,d) and (12d,3d) and similarly (d,d) and (5d,5d).
Aren't these two sets redundant as the second can be expressed as the first as d varies over the integers?
The solution I obtained was (a,a) ,(4a,a) and (19a,a); $$a\in\mathbb{Z}$$. Is this correct?