Problem #2

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

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Solution

$$n^2 + 19n + 92 = m^2$$ $$=> \left(n + \frac{19}{2}\right)^2 + \frac{7}{4} = m^2$$ $$=> m^2 - \left(n + \frac{19}{2}\right)^2 = \frac{7}{4}$$ $$=> \left(m - n - \frac{19}{2}\right)\left(m + n + \frac{19}{2}\right) = \frac{7}{4}$$ $$=> (2m - 2n -19)(2m + 2n + 19) = 7$$

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 \(-11\)