Question

Find sum of all natural number n such that n
square + 19n + 65 is perfect square.

Answer 1

Let n2−19n+99=k2n2−19n+99=k2 where k∈Zk∈Z

4n2−76n+396=4k24n2−76n+396=4k2 or (2n−19)2−35=(2k)2(2n−19)2−35=(2k)2

(2n−19)2−(2k)2=35(2n−19)2−(2k)2=35

now we have to take two perfect square whose difference is =35=35

so one pair is (62,12)(62,12)

now my question is how can i calculate for other ordered pairs