A number theory problem by Sayantan Saha

Let $x^2 = n^2-18n-4 = (n-9)^2-85$ , or $85 = (n-9)^2-x^2$ , or in alternate form of $85 = (n-9-x)(n-9+x)$ . As $85 = 5\times17$ and 5 and 17 are prime, we can get that

$(n-9-x, n-9+x) = (-85, -1), (-17, -5), (1, 85), (5, 17)$

or simply $n-9 = -43, -11, 43, 11$ . The sum of $n-9$ equals to zero, so the sum of $n$ equals to $0+9(4) = 36$ . To be more specific, all value of $n = -34, -2, 52, 20$