Q3/015) n^2+19n+130=f(n)

Find the sum of all the value of n for which f(n) is whole square.?

LHS = (n + 19/2)^2 - 361/4 + 130
= (n + 19/2)^2 + 159/4 = m^2

mulitply by 4 to get

(2n + 19)^2 + 159 = 4m^2 or 159 = (2m)^2 - (2n+19)^2 = (2m + 2n + 19)(2m-2n - 19)

now factors of 159 = 159 * 1, 53 * 3

taking 159 *1 we have 2m + 2n + 19 = 159 and 2m-2n - 19 = 1

or m+n = 70 and m-n = 10 => m = 40, n= 30

taking 53 * 3 we have 2m + 2n + 19 = 53 and 2m - 2n - 19 = 3
=> m+ n = 17 and m-n = 11 => m = 14, n= 3
  so n = 3 or 30

so sum= 33