28. If x and y are natural numbers, find the number pairs (x, y) for which x^2 - y^2 = 31.

Answer 1

well, (n+1)^2 - n^2 = 2n+1, so one pair will be when n=15, so
16^2 - 15^2 = 256-225 = 31
think of (n+3)^2 - n^2
and so on.
(why not possible for (n+2)^2 - n^2 ?)

Answer 2

pata nahi