# Problem #2

[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 $$-11$$