**[RMO-1992]:** *Determine the set of integers n for which \(n^2 + 19n + 92\) is a perfect square*

**Solution**

Since the 7 is a prime number, there are only two possibilities for the above case

**Case 1:** \(2m - 2n - 19 = 1 \text{ and } 2m + 2n + 19 = 7\)

**Case 2:** \(2m - 2n - 19 = 7 \text{ and } 2m + 2n + 19 = 1\)

The first case gives the solution \((m = 2, n = -8)\). The second case gives the solution \((m = 2, n = -11)\). Thus, the only values for \(n\) for which \(n^2 + 19n + 92\) is a perfect square are \(-8\) and \(-11\)